PHYS 231 - UNIVERSITY PHYSICS I
- Ohm’s Law and its applications
Question Bank - Set 4
Liberty University
Question 1
Question
A 12 V battery is connected to a resistor, resulting in a current of 3 A flowing
through the circuit. If the resistance of the resistor is 4 Ω, what is the power
dissipated by the resistor?
Solution
Step 1: Recall Ohm’s Law, which states that V=IR, where Vis the voltage
across a resistor, Iis the current flowing through the resistor, and Ris the
resistance of the resistor.
Step 2: Given that the voltage Vis 12 V, the current Iis 3 A, and the
resistance Ris 4 Ω, we can use Ohm’s Law to find the power Pdissipated by
the resistor.
Step 3: First, calculate the voltage across the resistor using Ohm’s Law:
V=IR = 3 A×4 Ω = 12 V
Step 4: Now that we have the voltage across the resistor, we can calculate
the power dissipated by the resistor using the formula P=V I:
P=V×I= 12 V×3A= 36 W
Step 5: Therefore, the power dissipated by the resistor is 36 W.
Question 2
Question
A copper wire has a resistance of 8 Ω at 20◦C. If the wire is heated to 80◦C,
what will be its resistance? The temperature coefficient of resistivity for copper
is 0.00428 K−1.
Solution
Step 1: Given that the initial resistance of the copper wire is 8 Ω at 20◦C, we
want to find the resistance at 80◦C. The formula relating resistance, initial
resistance, and temperature change is:
RT=R0[1 + α(T−T0)]
where RTis the resistance at temperature T,R0is the resistance at reference
temperature T0,αis the temperature coefficient of resistivity, Tis the final
temperature, and T0is the reference temperature.
Step 2: Plugging in the given values into the equation, we get:
R80 = 8 Ω[1 + 0.00428 ×(80 −20)]
Step 3: Calculating the expression inside the brackets first:
1+0.00428 ×60 = 1 + 0.2568 = 1.2568
Step 4: Now, calculate the resistance at 80◦C:
R80 = 8 Ω ×1.2568 = 10.0544 Ω
Therefore, the resistance of the copper wire at 80◦C is 10.0544 Ω.
Question 3
Question
A circuit consists of a 10 V battery connected in series to a resistor. If the
current in the circuit is 2 A, what is the resistance of the resistor?
Solution
Step 1: Recall Ohm’s Law, which states that the voltage across a resistor is equal
to the current flowing through it multiplied by the resistance. Mathematically,
this can be written as V=IR, where Vis the voltage, Iis the current, and R
is the resistance.
Step 2: In this case, the voltage across the resistor is 10 V and the cur-
rent flowing through it is 2 A. We can rearrange Ohm’s Law to solve for the
resistance: R=V
I.
Step 3: Substitute the given values into the formula: R=10 V
2A.
Step 4: Perform the division to find the resistance: R= 5 Ω.
Step 5: Therefore, the resistance of the resistor in the circuit is 5 Ohms.
2
Question 4
Question
A circuit consists of a battery with voltage V= 12 V and three resistors con-
nected in series. The resistors have resistances R1= 5 ohms, R2= 10 ohms,
and R3= 15 ohms. Calculate the total current flowing through the circuit and
the voltage drop across each resistor.
Solution
Step 1: Calculate the total resistance of the circuit. The total resistance Rtotal
of resistors in series is the sum of the individual resistances:
Rtotal =R1+R2+R3= 5 Ω + 10 Ω + 15 Ω = 30 Ω
Step 2: Calculate the total current flowing through the circuit using Ohm’s
Law V=IR. The total current Iis given by I=V
Rtotal .
I=12 V
30 Ω = 0.4A
Step 3: Calculate the voltage drop across each resistor. The voltage drop Vi
across a resistor Riis given by Ohm’s Law Vi=IRi. For R1:
V1=I·R1= 0.4A·5 Ω = 2 V
For R2:
V2=I·R2= 0.4A·10 Ω = 4 V
For R3:
V3=I·R3= 0.4A·15 Ω = 6 V
Therefore, the total current flowing through the circuit is 0.4 A, and the
voltage drop across R1,R2, and R3are 2 V, 4 V, and 6 V, respectively.
Question 5
Question
A circuit consists of a resistor with resistance R= 15 Ω connected to a battery
with emf ε= 12 V. Calculate the current flowing through the resistor.
Solution
Step 1: Determine the relationship between voltage, current, and resistance in a
circuit. According to Ohm’s Law, the relationship between voltage (V), current
(I), and resistance (R) in a circuit is given by:
V=IR
3
where V= voltage (in volts), I= current (in amperes), R= resistance (in
ohms).
Step 2: Identify the given values. The resistance of the resistor, R= 15 Ω,
and the emf of the battery, ε= 12 V.
Step 3: Apply Ohm’s Law to find the current. Substitute the given values
into Ohm’s Law:
ε=IR
12 = I×15
I=12
15
I= 0.8A
Step 4: State the final answer. The current flowing through the resistor is
0.8amperes.
Question 6
Question
A resistor is connected to a 12 V battery, and a current of 3 A flows through it.
If the resistance of the resistor is R, calculate the resistance R.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor
is equal to the current (I) flowing through it multiplied by the resistance (R)
of the resistor. Mathematically, this can be written as:
V=I×R
Step 2: In this case, the voltage across the resistor is 12V, the current flowing
through it is 3A, and we are trying to find the resistance, R. We can substitute
these values into Ohm’s Law:
12 = 3 ×R
Step 3: Now, we solve for the resistance R:
R=12
3= 4 Ω
Step 4: Therefore, the resistance of the resistor is 4 Ω.
4
Question 7
Question
A circuit consists of a resistor with resistance R, a capacitor with capacitance C,
and an inductor with inductance Lconnected in series to an alternating current
(AC) voltage source. The voltage across the resistor is VR= 5 sin(100t)V, the
voltage across the capacitor is VC= 3 sin(100t+π
4)V, and the voltage across
the inductor is VL= 4 sin(100t−π
3)V. Given that the current in the circuit is
I= 2 sin(100t+π
6)A, determine the values of R,L, and Cin the circuit.
Solution
Step 1: Write down the phasor representation for the voltages and current in
the circuit. The phasor representation of a sinusoidal function Asin(ωt +ϕ)is
ˆ
A=A
√2∠ϕ. For VR= 5 sin(100t)V: ˆ
VR=5
√2∠0◦V. For VC= 3 sin(100t+π
4)
V: ˆ
VC=3
√2∠45◦V. For VL= 4 sin(100t−π
3)V: ˆ
VL=4
√2∠−60◦V. For
I= 2 sin(100t+π
6)A: ˆ
I=2
√2∠30◦A.
Step 2: Apply Kirchhoff’s voltage law (KVL) to the circuit:
ˆ
VR+ˆ
VL+ˆ
VC=ˆ
Vsource
Substitute in the phasor representations:
5
√2
∠0◦+4
√2
∠−60◦+3
√2
∠45◦=ˆ
Vsource
5∠0◦+ 4∠−60◦+ 3∠45◦=ˆ
Vsource
Step 3: Calculate the phasor representation of the source voltage. Since the
source voltage is not explicitly given, we can represent it as ˆ
Vsource =Vsource∠0◦.
Now solve for Vsource:
5∠0◦+ 4∠−60◦+ 3∠45◦=Vsource∠0◦
Vsource∠0◦= 5 + 4 cos(60◦) + 3 cos(−45◦) + j(4 sin(−60◦) + 3 sin(45◦))
Vsource = 5 + 2 + 3√2
2+j(−2√3 + 3√2
2)
Vsource = 7 + 3√2
2−2√3 + j3√2
2−2√3
Step 4: Compare the phasor representation of the source voltage with the
expression for the impedance of the circuit Z=R+j(ωL−1
ωC ). The impedance
of the circuit must be equal to the phasor representation of the source voltage.
Matching real and imaginary parts, we have: Real parts: R= 7 Imaginary
parts: ωL −1
ωC =3√2
2−2
5
Question 8
Question
A circuit consists of a resistor with resistance R, an inductor with inductance
L, and a capacitor with capacitance Cconnected in series. The voltage across
the circuit is given by V(t) = V0cos(ωt), where V0= 12 V and ω= 50 s−1.
The resistor has a resistance of R= 5 Ω, the inductor has an inductance of
L= 0.1H, and the capacitor has a capacitance of C= 0.01 F. Determine the
current as a function of time for this circuit.
Solution
Step 1: First, we need to find the expression for the total impedance of the
circuit. The impedance ZRof the resistor is simply its resistance R. The
impedance ZLof the inductor is given by ZL=jωL. The impedance ZCof the
capacitor is given by ZC=1
jωC . The total impedance Ztotal of the series circuit
is the sum of the impedances of the resistor, inductor, and capacitor:
Ztotal =R+ZL+ZC=R+jωL +1
jωC
Step2: Next, we can find the current I(t)as a function of time. Since the
frequency of the voltage source is ω= 50 s−1, we can express the current as:
I(t) = V0
Ztotal
cos(ωt)
Step 3: Now, substitute the given values into the expression for total impedance
to get:
Ztotal = 5 + j×50 ×0.1 + 1
j×50 ×0.01
Step 4: Simplify the expression for the total impedance:
Ztotal = 5 + j5 + 1
j0.5= 5 + j5 + j2 = 5 + j7
Step 5: Substitute the total impedance into the expression for current:
I(t) = 12
5 + j7cos(50t)
Therefore, the current as a function of time for the given series circuit is:
I(t) = 12
5 + j7cos(50t)
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Question 9
Question
A circuit consists of a resistor with resistance R= 10 Ω, a capacitor with capaci-
tance C= 2 µF , and an inductor with inductance L= 0.5Hconnected in series
to a sinusoidal voltage source with frequency ω= 100 rad/s. If the maximum
voltage supplied by the source is Vmax = 20 V, what is the maximum current in
the circuit?
Solution
Step 1: Calculate the impedance of each component. The impedance of a
resistor is given by ZR=R. The impedance of a capacitor is given by ZC=1
jωC .
The impedance of an inductor is given by ZL=jωL.
Step 2: Calculate the total impedance of the circuit. The total impedance
Ztotal of components in series is the sum of their individual impedances.
Ztotal =ZR+ZC+ZL=R+1
jωC +jωL
Step 3: Convert the total impedance to polar form. To convert Ztotal to
polar form, we need to find the magnitude and phase angle. The magnitude
|Ztotal|=√Re(Ztotal)2+Im(Ztotal)2. The phase angle θ= arctan (Im(Ztotal )
Re(Ztotal )).
Step 4: Calculate the maximum current in the circuit. The maximum current
Imax is given by Imax =Vmax
|Ztotal |.
Substitute the given values into the expressions above to find Imax.
Question 10
Question
A circuit consists of a 12V battery connected to three resistors in series. The
first resistor has a resistance of 4Ω, the second resistor has a resistance of 8Ω,
and the third resistor has a resistance of R3Ω. If the current through the circuit
is 1.5A, find the value of R3.
Solution
Step 1: Calculate the total resistance of the circuit. The total resistance Rtotal
of resistors in series is the sum of individual resistances:
Rtotal = 4Ω + 8Ω + R3Ω = 12Ω + R3Ω
Step 2: Apply Ohm’s Law to find the total resistance. Ohm’s Law states
that V=IR, where Vis the voltage, Iis the current, and Ris the resistance.
7
In this case, V= 12V and I= 1.5A. Therefore, the total resistance can be
calculated as:
12V= 1.5A×(12Ω + R3Ω)
12V = 18Ω+1.5R3Ω
Step 3: Solve for R3. Subtract 18Ωfrom both sides of the equation:
12V−18Ω = 1.5R3Ω
−6V= 1.5R3Ω
R3=−6V
1.5Ω =−4Ω
Therefore, the value of R3is 4Ω.
Question 11
Question
A resistor with resistance R1= 10 Ω is connected in series with another resistor
with resistance R2= 20 Ω. A potential difference of V= 120 Vis applied across
the combination. Calculate the current passing through each resistor.
Solution
Step 1: Calculate the total resistance of the combination. The total resistance
Rtotal of resistors in series is given by:
Rtotal =R1+R2= 10 Ω + 20 Ω = 30 Ω
Step 2: Calculate the total current passing through the combination using
Ohm’s Law V=IR. The total current Itotal passing through the combination
is:
Itotal =V
Rtotal
=120 V
30 Ω = 4 A
Step 3: Calculate the current passing through each resistor using the current
divider rule. The current passing through R1, denoted as I1, is:
I1=Itotal ×Rtotal
R1
= 4 A×30 Ω
10 Ω = 12 A
The current passing through R2, denoted as I2, is:
I2=Itotal ×Rtotal
R2
= 4 A×30 Ω
20 Ω = 6 A
Thus, the current passing through R1is 12 Aand the current passing through
R2is 6A.
8
Question 12
Question
A resistor with a resistance of 4 Ω is connected to a battery that delivers a
current of 2A. Find the power dissipated in the resistor.
Solution
Step 1: Recall that the power dissipated in a resistor can be calculated using
the formula P=I2R, where Pis the power, Iis the current, and Ris the
resistance.
Step 2: Given that the resistance R= 4 Ω and the current I= 2 A, we can
plug these values into the formula:
P= (2 A)2×4 Ω
Step 3: Simplify the expression:
P= 4 A2×4 Ω = 16 W
Step 4: Therefore, the power dissipated in the resistor is 16 W .
Question 13
Question
A copper wire has a resistance of 5 ohms. If a current of 2 amperes flows through
the wire, find the voltage across the wire.
Solution
Step 1: Recall Ohm’s Law, which states that voltage (V) is equal to current (I)
multiplied by resistance (R). The formula can be written as:
V=I×R
Step 2: Given that the resistance Ris 5 ohms and the current Iis 2 amperes,
we can substitute these values into Ohm’s Law to find the voltage V:
V= 2 A×5 Ω
Step 3: Calculate the voltage across the wire:
V= 10 V
Therefore, the voltage across the wire is 10 volts.
9
Question 14
Question
A circuit consists of a resistor with resistance R, an inductor with inductance L,
and a capacitor with capacitance Cconnected in series to an AC voltage source
with voltage V=V0sin(ωt). The circuit reaches a steady state, and the voltage
across the resistor is VR=VR0sin(ωt −ϕ). Calculate the phase difference ϕ
between the voltage across the resistor and the applied voltage.
Solution
Step 1: Write down the expression for the total impedance Zof the circuit in
terms of R,L, and C. The total impedance of the circuit is given by:
Z=R+jωL −j
ωC
Step 2: Express the current Ithrough the circuit in terms of the total
impedance Zand the applied voltage V. Using Ohm’s Law in the form V=IZ,
we have:
I=V
Z=V0sin(ωt)
Z
Step 3: Write down the expression for the voltage across the resistor VRin
terms of the current Iand the resistor R. The voltage across the resistor is
given by:
VR=IR =I·R
Step 4: Express the phase difference ϕbetween VRand Vin terms of the
impedance Z. Comparing the expressions for VRand V:
VR0sin(ωt −ϕ) = I·R=V0sin(ωt)
Z·R
⇒VR0=V0R
|Z|and ϕ=Arg(Z)
Step 5: Calculate the magnitude of Zand its argument, then find the phase
difference ϕ. Using the expression for Zfrom Step 1:
|Z|=√R2+(ωL −1
ωC )2
Arg(Z) = tan−1(ωL −1
ωC
R)
Therefore, the phase difference ϕis:
ϕ= tan−1(ωL −1
ωC
R)
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Question 15
Question
A circuit consists of a resistor, an inductor, and a capacitor connected in series to
an AC voltage source. The values of the components are R= 20 Ω,L= 0.05 H,
and C= 1 µF. The voltage source has an amplitude of V0= 10 V and operates
at a frequency of f= 100 Hz. Calculate the impedance of the circuit and the
phase angle between the current and the voltage.
Solution
Step 1: Calculate the angular frequency ω. Given that f= 100 Hz, we can find
the angular frequency using the formula ω= 2πf . Thus,
ω= 2π×100 = 200πrad/s.
Step 2: Calculate the reactance of the inductor and the capacitor. The
reactance of an inductor XLis given by XL=ωL, and the reactance of a
capacitor XCis given by XC=1
ωC . Substituting the given values,
XL= 200π×0.05 = 10πΩand XC=1
200π×1×10−6= 5 ×103πΩ.
Step 3: Calculate the impedance of the circuit. The impedance of the circuit
Zis the vector sum of the resistive, inductive, and capacitive impedance. Since
the components are in series, we have
Z=R+XL−XC= 20+10π−5×103π= 10π−5×103π+20 = (10−5×103)π+20 Ω = −4990π+20 Ω.
Step 4: Calculate the phase angle ϕbetween the current and voltage. The
phase angle ϕis given by
ϕ= arctan (XL−XC
R).
Plugging in the values,
ϕ= arctan (10π−5×103π
20 )= arctan(−2490π/10) = arctan(−249π).
Question 16
Question
A circuit contains a resistor with resistance 10 Ω, an inductor with inductance
0.02 H, and a capacitor with capacitance 8 µF connected in series to a 12 V
battery. Calculate the impedance of the circuit and the current passing through
the circuit.
11
Solution
Step 1: Calculate the total impedance of the circuit. The impedance (Z) of an
LRC series circuit is given by:
Z=√R2+ (XL−XC)2
where: R= resistance = 10 Ω,XL= inductive reactance = ωL,XC= capacitive
reactance = 1
ωC ,ω= angular frequency = 2π
T.
First, calculate the angular frequency:
ω=2π
T
Given that the frequency f=1
T= 60 Hz. Therefore, ω= 2π×60 = 120π
rad/s.
Next, calculate XLand XC:
XL=ωL = 120π×0.02 = 2.4πΩ
XC=1
ωC =1
120π×8×10−6=106
960Ω = 625
6πΩ
Now substitute these values into the impedance formula:
Z=√102+ (2.4π−625
6π)2
Z=√100 + (2.4π−625
6π)2
Step 2: Calculate the current passing through the circuit. The current (I)
flowing through the circuit is given by Ohm’s law:
I=V
Z
where V= voltage from the battery = 12 V.
Now, substitute the calculated impedance value:
I=12
Z
Now, you can calculate the impedance (Z) and then find the current passing
through the circuit.
Question 17
Question
A student sets up a simple circuit with a resistor, battery, and an ammeter.
The ammeter reads a current of 0.5 A when the voltage across the resistor is 6
V. The student then decides to double the resistance in the circuit and observes
that the current reading on the ammeter drops to 0.25 A. Calculate the original
resistance in the circuit and the new resistance after doubling it.
12
Solution
Step 1: Let’s denote the original resistance as Rand the new resistance after
doubling it as 2R.
Step 2: Using Ohm’s Law V=IR, where Vis the voltage, Iis the current,
and Ris the resistance, we can set up two equations:
For the original circuit:
6 = 0.5×R
Simplifying, we get:
R=6
0.5= 12 Ω
Step 3: For the circuit with doubled resistance:
6 = 0.25 ×2R
Simplifying, we get:
2R=6
0.25 = 24 Ω
Therefore, the new resistance 2Ris 24 Ωafter doubling it.
Step 4: The original resistance was found to be 12 Ω. Thus, the original
resistance in the circuit is 12 Ω, and the new resistance after doubling it is 24
Ω.
Question 18
Question
A circuit consists of a 12 V battery connected in series with a resistor and an
unknown device. When a current of 2 A flows through the circuit, the potential
drop across the resistor is 4 V. Determine the resistance of the unknown device.
Solution
Step 1: Recall Ohm’s Law, which states that the potential difference (V) across a
resistor is equal to the current (I) flowing through it multiplied by the resistance
(R) of the resistor. Mathematically, this can be expressed as V=I·R.
Step 2: From the problem, we are given that the potential drop across the
resistor is 4 V when a current of 2 A flows through the circuit. Therefore, we
have 4V= 2 A×R.
Step 3: To find the resistance Rof the unknown device, we can rearrange
the formula V=I·Rto solve for R. So, R=V
I.
Step 4: Substituting the given values, we find R=4V
2A= 2 Ω.
Step 5: Hence, the resistance of the unknown device in the circuit is 2 ohms.
13
Question 19
Question
A circuit consists of a 12 V battery, a resistor with resistance R, and a capacitor
with capacitance C. Initially, the capacitor is uncharged and acts like a short
circuit. When the circuit is closed, the current through the circuit is found to be
0.5A. After some time, the capacitor is fully charged and the current through
the circuit is found to be 0.1A. Calculate the resistance Rof the resistor in the
circuit.
Solution
1. Let’s denote the resistance of the resistor as R, the capacitance of the capac-
itor as C, and the potential difference across the resistor as VR. The potential
difference across the capacitor is VC= 12 V.
2. When the circuit is first closed and the capacitor acts like a short circuit,
the current in the circuit is equal to Iinitial = 0.5A. This current can be split
between the resistor and capacitor according to the equivalent resistance Req =
Rin series with the capacitor. Since the capacitor acts like a short circuit, the
full potential difference of 12 V is across the resistor:
VR= 12 V=R×Iinitial
R=VR
Iinitial
=12
0.5= 24Ω
3. When the capacitor is fully charged and the current through the circuit
is Ifinal = 0.1A, the capacitor acts like an open circuit. Now, the potential
difference VCis fully across the resistor. We can write an expression for the
potential difference VRin terms of Rand Cusing Ohm’s Law and the formula
for the charging of a capacitor:
VR=R×Ifinal =VC(1 −e−t
RC )
Since the capacitor is fully charged, VR=VC= 12 V and Ifinal = 0.1A.
Substituting these values into the equation gives:
R×0.1 = 12(1 −e−t
RC )
R=12(1 −e−t
RC )
0.1
4. Since we are looking for the resistance R, we need to find the time constant
τ=RC. We can find τby noting that in a fully charged circuit, e−t
RC ≈0.
Therefore:
R=12(1 −0)
0.1= 120Ω
5. Therefore, the resistance Rof the resistor in the circuit is 120 Ω .
14
Question 20
Question
A 10 V battery is connected to a resistor of unknown resistance. When a current
of 2 A flows through the resistor, the power dissipated is 20 W. Calculate the
resistance of the resistor.
Solution
Step 1: We can start by using the formula for power in a resistor: P=I2·R,
where Pis the power dissipated, Iis the current flowing through the resistor,
and Ris the resistance. Given that P= 20 W and I= 2 A, we can rearrange
the formula to solve for R:
R=P
I2
R=20
(2)2
R=20
4
R= 5 Ω
Therefore, the resistance of the resistor is 5 Ω.
Question 21
Question
A circuit consists of a 12 V battery, a 4 Ωresistor, and an unknown resistor
R. When the total resistance in the circuit is 6 Ω, the power dissipated in the
unknown resistor is 24 W. Find the value of R.
Solution
Step 1: Let’s first calculate the current, I, flowing through the circuit using
Ohm’s Law, V=IR, where Vis the voltage of the battery.
Given: V= 12 V, R1= 4 Ω, Rtotal = 6 Ω
Since Rtotal =R1+R, we can find Ras R=Rtotal −R1
R= 6 −4 = 2 Ω
Step 2: Next, we find the current, I, as I=V
Rtotal .
I=12
6= 2 A
15
Step 3: Now, we can find the power, P, dissipated by the unknown resistor
using the formula P=I2R.
Given: P= 24 W, R= 2 Ω
24 = (2)2×2
24 = 4 ×2
24 = 8
This equation is not met, so there must be an error in the problem statement
or calculation.
Question 22
Question
A circuit consists of a resistor with resistance R= 100 Ω connected to a battery
with emf E= 12 V. The circuit also has an ammeter connected in series and
a voltmeter connected in parallel to the resistor. If the ammeter reads 0.1A,
what reading will the voltmeter display?
Solution
Step 1: Recall Ohm’s Law which states that the current passing through a
resistor is directly proportional to the potential difference across the resistor.
The formula for Ohm’s Law is V=IR, where Vis the voltage (potential
difference) across the resistor, Iis the current passing through the resistor, and
Ris the resistance of the resistor.
Step 2: Given in the question, R= 100 Ω and I= 0.1A. We can use Ohm’s
Law to find the voltage across the resistor: V=I·R.
Step 3: Substituting the given values into the formula, we get V= 0.1A·
100 Ω = 10 V.
Step 4: Since the voltmeter is connected in parallel to the resistor, it will
read the same voltage as the resistor. Therefore, the voltmeter will display
10 V .
Question 23
Question
A circuit consists of a resistor with resistance R= 15 Ω and an inductor with
inductance L= 0.04 H. The circuit is connected to a voltage source with an
emf of V= 20 V and a frequency of f= 50 Hz. Calculate the current flowing
through the circuit.
16
Solution
Step 1: Calculate the reactance of the inductor using the formula XL= 2πf L.
XL= 2π×50 ×0.04 = 4πΩ
Step 2: Calculate the total impedance of the circuit using the formula Z=
√R2+X2
L.
Z=√152+ (4π)2≈√225 + 39.48 ≈√264.48 ≈16.27 Ω
Step 3: Calculate the current flowing through the circuit using Ohm’s Law:
I=V
Z.
I=20
16.27 ≈1.23 A
Therefore, the current flowing through the circuit is approximately 1.23 A.
Question 24
Question
A circuit consists of a resistor with resistance R, an inductor with inductance
L, a capacitor with capacitance C, and an AC voltage source V=V0sin(ωt).
The instantaneous current i(t)through the circuit is given by the equation:
i(t) = i0sin(ωt +ϕ)
where i0= 3 A, ω= 50 rad/s, and ϕ=π
6rad.
Determine the voltage amplitude V0applied by the source.
Solution
Step 1: Ohm’s Law states that for a circuit, the voltage Vapplied across a com-
ponent is equal to the product of the current Iflowing through the component
and the resistance Rof the component.
V=IR
For the given circuit, the voltage across the resistor is VR=i(t)·R. Since the
current through the resistor is i(t)and the resistance of the resistor is R, we
have:
VR=i(t)·R
Step 2: In an AC circuit with a resistor, inductor, and capacitor in series,
the total voltage across the components will be the same as the voltage of the
AC source. Therefore, the voltages across the resistor, inductor, and capacitor
must sum up to the applied voltage V=V0sin(ωt).
V=VR+VL+VC
17
Now, let’s find the expressions for VLand VC.
Step 3: The voltage across an inductor in an AC circuit is given by:
VL=Ldi
dt
where Lis the inductance of the inductor. Since i(t) = i0sin(ωt +ϕ), we can
find di
dt by differentiating i(t)with respect to t.
di
dt =d
dt(i0sin(ωt +ϕ)) = i0ωcos(ωt +ϕ)
Step 4: Substituting into the expression for VL, we get:
VL=Ldi
dt =Li0ωcos(ωt +ϕ)
Step 5: The voltage across a capacitor in an AC circuit is given by:
VC=1
C∫i dt
where Cis the capacitance of the capacitor. Since i(t) = i0sin(ωt +ϕ), we can
find ∫i dt by integrating i(t)with respect to t.
∫i dt =∫i0sin(ωt +ϕ)dt =−i0
ωcos(ωt +ϕ)
Step 6: Substituting into the expression for VC, we have:
VC=1
C∫i dt =−i0
ωC cos(ωt +ϕ)
Step 7: Now, substituting VR,VL, and VCinto the equation V=VR+VL+
VC, we get:
V0sin(ωt) = i0Rsin(ωt +ϕ) + Li0ωcos(ωt +ϕ)−i0
ωC cos(ωt +ϕ)
Step 8: We know that the voltage amplitudes of sin(ωt),cos(ωt), and cos(ωt)
terms on both sides of the equation must be equal. So, the coefficient of sin(ωt)
term on the right side must be V0. Thus, we can determine V0by equating
coefficients:
V0=i0R
V0= 3 A×R
Question 25
Question
A circuit consists of a resistor with resistance R, an inductor with induc-
tance L, and a capacitor with capacitance Cconnected in series to an AC
voltage source with frequency f. The impedance of the circuit is given by
Z=√R2+(ωL −1
ωC )2, where ω= 2πf. Determine the angular frequency ω
at which the impedance of the circuit is minimized.
18
Solution
Step 1: To find the angular frequency ωat which the impedance of the circuit is
minimized, we need to find the minimum value of the expression for impedance
Z. This minimum occurs when the derivative of Zwith respect to ωis zero.
Step 2: Calculate the derivative of Zwith respect to ω:
dZ
dω =1
2√R2+(ωL −1
ωC )2·2·(ωL −1
ωC )(L−1
ω2C)
Step 3: Set the derivative equal to zero and solve for ω:
ωL −1
ωC
√R2+(ωL −1
ωC )2= 0
Step 4: Simplify the equation and solve for ω:
ωL −1
ωC = 0
ω2=1
LC
ω=√1
LC
Therefore, the angular frequency ωat which the impedance of the circuit is
minimized is ω=√1
LC .
19
Solution
Step 1: Given that the initial resistance of the copper wire is 8 Ω at 20◦C, we
want to find the resistance at 80◦C. The formula relating resistance, initial
resistance, and temperature change is:
RT=R0[1 + α(T−T0)]
where RTis the resistance at temperature T,R0is the resistance at reference
temperature T0,αis the temperature coefficient of resistivity, Tis the final
temperature, and T0is the reference temperature.
Step 2: Plugging in the given values into the equation, we get:
R80 = 8 Ω[1 + 0.00428 ×(80 −20)]
Step 3: Calculating the expression inside the brackets first:
1+0.00428 ×60 = 1 + 0.2568 = 1.2568
Step 4: Now, calculate the resistance at 80◦C:
R80 = 8 Ω ×1.2568 = 10.0544 Ω
Therefore, the resistance of the copper wire at 80◦C is 10.0544 Ω.
Question 3
Question
A circuit consists of a 10 V battery connected in series to a resistor. If the
current in the circuit is 2 A, what is the resistance of the resistor?
Solution
Step 1: Recall Ohm’s Law, which states that the voltage across a resistor is equal
to the current flowing through it multiplied by the resistance. Mathematically,
this can be written as V=IR, where Vis the voltage, Iis the current, and R
is the resistance.
Step 2: In this case, the voltage across the resistor is 10 V and the cur-
rent flowing through it is 2 A. We can rearrange Ohm’s Law to solve for the
resistance: R=V
I.
Step 3: Substitute the given values into the formula: R=10 V
2A.
Step 4: Perform the division to find the resistance: R= 5 Ω.
Step 5: Therefore, the resistance of the resistor in the circuit is 5 Ohms.
2
Question 4
Question
A circuit consists of a battery with voltage V= 12 V and three resistors con-
nected in series. The resistors have resistances R1= 5 ohms, R2= 10 ohms,
and R3= 15 ohms. Calculate the total current flowing through the circuit and
the voltage drop across each resistor.
Solution
Step 1: Calculate the total resistance of the circuit. The total resistance Rtotal
of resistors in series is the sum of the individual resistances:
Rtotal =R1+R2+R3= 5 Ω + 10 Ω + 15 Ω = 30 Ω
Step 2: Calculate the total current flowing through the circuit using Ohm’s
Law V=IR. The total current Iis given by I=V
Rtotal .
I=12 V
30 Ω = 0.4A
Step 3: Calculate the voltage drop across each resistor. The voltage drop Vi
across a resistor Riis given by Ohm’s Law Vi=IRi. For R1:
V1=I·R1= 0.4A·5 Ω = 2 V
For R2:
V2=I·R2= 0.4A·10 Ω = 4 V
For R3:
V3=I·R3= 0.4A·15 Ω = 6 V
Therefore, the total current flowing through the circuit is 0.4 A, and the
voltage drop across R1,R2, and R3are 2 V, 4 V, and 6 V, respectively.
Question 5
Question
A circuit consists of a resistor with resistance R= 15 Ω connected to a battery
with emf ε= 12 V. Calculate the current flowing through the resistor.
Solution
Step 1: Determine the relationship between voltage, current, and resistance in a
circuit. According to Ohm’s Law, the relationship between voltage (V), current
(I), and resistance (R) in a circuit is given by:
V=IR
3
where V= voltage (in volts), I= current (in amperes), R= resistance (in
ohms).
Step 2: Identify the given values. The resistance of the resistor, R= 15 Ω,
and the emf of the battery, ε= 12 V.
Step 3: Apply Ohm’s Law to find the current. Substitute the given values
into Ohm’s Law:
ε=IR
12 = I×15
I=12
15
I= 0.8A
Step 4: State the final answer. The current flowing through the resistor is
0.8amperes.
Question 6
Question
A resistor is connected to a 12 V battery, and a current of 3 A flows through it.
If the resistance of the resistor is R, calculate the resistance R.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor
is equal to the current (I) flowing through it multiplied by the resistance (R)
of the resistor. Mathematically, this can be written as:
V=I×R
Step 2: In this case, the voltage across the resistor is 12V, the current flowing
through it is 3A, and we are trying to find the resistance, R. We can substitute
these values into Ohm’s Law:
12 = 3 ×R
Step 3: Now, we solve for the resistance R:
R=12
3= 4 Ω
Step 4: Therefore, the resistance of the resistor is 4 Ω.
4
Question 7
Question
A circuit consists of a resistor with resistance R, a capacitor with capacitance C,
and an inductor with inductance Lconnected in series to an alternating current
(AC) voltage source. The voltage across the resistor is VR= 5 sin(100t)V, the
voltage across the capacitor is VC= 3 sin(100t+π
4)V, and the voltage across
the inductor is VL= 4 sin(100t−π
3)V. Given that the current in the circuit is
I= 2 sin(100t+π
6)A, determine the values of R,L, and Cin the circuit.
Solution
Step 1: Write down the phasor representation for the voltages and current in
the circuit. The phasor representation of a sinusoidal function Asin(ωt +ϕ)is
ˆ
A=A
√2∠ϕ. For VR= 5 sin(100t)V: ˆ
VR=5
√2∠0◦V. For VC= 3 sin(100t+π
4)
V: ˆ
VC=3
√2∠45◦V. For VL= 4 sin(100t−π
3)V: ˆ
VL=4
√2∠−60◦V. For
I= 2 sin(100t+π
6)A: ˆ
I=2
√2∠30◦A.
Step 2: Apply Kirchhoff’s voltage law (KVL) to the circuit:
ˆ
VR+ˆ
VL+ˆ
VC=ˆ
Vsource
Substitute in the phasor representations:
5
√2
∠0◦+4
√2
∠−60◦+3
√2
∠45◦=ˆ
Vsource
5∠0◦+ 4∠−60◦+ 3∠45◦=ˆ
Vsource
Step 3: Calculate the phasor representation of the source voltage. Since the
source voltage is not explicitly given, we can represent it as ˆ
Vsource =Vsource∠0◦.
Now solve for Vsource:
5∠0◦+ 4∠−60◦+ 3∠45◦=Vsource∠0◦
Vsource∠0◦= 5 + 4 cos(60◦) + 3 cos(−45◦) + j(4 sin(−60◦) + 3 sin(45◦))
Vsource = 5 + 2 + 3√2
2+j(−2√3 + 3√2
2)
Vsource = 7 + 3√2
2−2√3 + j3√2
2−2√3
Step 4: Compare the phasor representation of the source voltage with the
expression for the impedance of the circuit Z=R+j(ωL−1
ωC ). The impedance
of the circuit must be equal to the phasor representation of the source voltage.
Matching real and imaginary parts, we have: Real parts: R= 7 Imaginary
parts: ωL −1
ωC =3√2
2−2
5
Question 8
Question
A circuit consists of a resistor with resistance R, an inductor with inductance
L, and a capacitor with capacitance Cconnected in series. The voltage across
the circuit is given by V(t) = V0cos(ωt), where V0= 12 V and ω= 50 s−1.
The resistor has a resistance of R= 5 Ω, the inductor has an inductance of
L= 0.1H, and the capacitor has a capacitance of C= 0.01 F. Determine the
current as a function of time for this circuit.
Solution
Step 1: First, we need to find the expression for the total impedance of the
circuit. The impedance ZRof the resistor is simply its resistance R. The
impedance ZLof the inductor is given by ZL=jωL. The impedance ZCof the
capacitor is given by ZC=1
jωC . The total impedance Ztotal of the series circuit
is the sum of the impedances of the resistor, inductor, and capacitor:
Ztotal =R+ZL+ZC=R+jωL +1
jωC
Step2: Next, we can find the current I(t)as a function of time. Since the
frequency of the voltage source is ω= 50 s−1, we can express the current as:
I(t) = V0
Ztotal
cos(ωt)
Step 3: Now, substitute the given values into the expression for total impedance
to get:
Ztotal = 5 + j×50 ×0.1 + 1
j×50 ×0.01
Step 4: Simplify the expression for the total impedance:
Ztotal = 5 + j5 + 1
j0.5= 5 + j5 + j2 = 5 + j7
Step 5: Substitute the total impedance into the expression for current:
I(t) = 12
5 + j7cos(50t)
Therefore, the current as a function of time for the given series circuit is:
I(t) = 12
5 + j7cos(50t)
6
Question 9
Question
A circuit consists of a resistor with resistance R= 10 Ω, a capacitor with capaci-
tance C= 2 µF , and an inductor with inductance L= 0.5Hconnected in series
to a sinusoidal voltage source with frequency ω= 100 rad/s. If the maximum
voltage supplied by the source is Vmax = 20 V, what is the maximum current in
the circuit?
Solution
Step 1: Calculate the impedance of each component. The impedance of a
resistor is given by ZR=R. The impedance of a capacitor is given by ZC=1
jωC .
The impedance of an inductor is given by ZL=jωL.
Step 2: Calculate the total impedance of the circuit. The total impedance
Ztotal of components in series is the sum of their individual impedances.
Ztotal =ZR+ZC+ZL=R+1
jωC +jωL
Step 3: Convert the total impedance to polar form. To convert Ztotal to
polar form, we need to find the magnitude and phase angle. The magnitude
|Ztotal|=√Re(Ztotal)2+Im(Ztotal)2. The phase angle θ= arctan (Im(Ztotal )
Re(Ztotal )).
Step 4: Calculate the maximum current in the circuit. The maximum current
Imax is given by Imax =Vmax
|Ztotal |.
Substitute the given values into the expressions above to find Imax.
Question 10
Question
A circuit consists of a 12V battery connected to three resistors in series. The
first resistor has a resistance of 4Ω, the second resistor has a resistance of 8Ω,
and the third resistor has a resistance of R3Ω. If the current through the circuit
is 1.5A, find the value of R3.
Solution
Step 1: Calculate the total resistance of the circuit. The total resistance Rtotal
of resistors in series is the sum of individual resistances:
Rtotal = 4Ω + 8Ω + R3Ω = 12Ω + R3Ω
Step 2: Apply Ohm’s Law to find the total resistance. Ohm’s Law states
that V=IR, where Vis the voltage, Iis the current, and Ris the resistance.
7
In this case, V= 12V and I= 1.5A. Therefore, the total resistance can be
calculated as:
12V= 1.5A×(12Ω + R3Ω)
12V = 18Ω+1.5R3Ω
Step 3: Solve for R3. Subtract 18Ωfrom both sides of the equation:
12V−18Ω = 1.5R3Ω
−6V= 1.5R3Ω
R3=−6V
1.5Ω =−4Ω
Therefore, the value of R3is 4Ω.
Question 11
Question
A resistor with resistance R1= 10 Ω is connected in series with another resistor
with resistance R2= 20 Ω. A potential difference of V= 120 Vis applied across
the combination. Calculate the current passing through each resistor.
Solution
Step 1: Calculate the total resistance of the combination. The total resistance
Rtotal of resistors in series is given by:
Rtotal =R1+R2= 10 Ω + 20 Ω = 30 Ω
Step 2: Calculate the total current passing through the combination using
Ohm’s Law V=IR. The total current Itotal passing through the combination
is:
Itotal =V
Rtotal
=120 V
30 Ω = 4 A
Step 3: Calculate the current passing through each resistor using the current
divider rule. The current passing through R1, denoted as I1, is:
I1=Itotal ×Rtotal
R1
= 4 A×30 Ω
10 Ω = 12 A
The current passing through R2, denoted as I2, is:
I2=Itotal ×Rtotal
R2
= 4 A×30 Ω
20 Ω = 6 A
Thus, the current passing through R1is 12 Aand the current passing through
R2is 6A.
8
Question 12
Question
A resistor with a resistance of 4 Ω is connected to a battery that delivers a
current of 2A. Find the power dissipated in the resistor.
Solution
Step 1: Recall that the power dissipated in a resistor can be calculated using
the formula P=I2R, where Pis the power, Iis the current, and Ris the
resistance.
Step 2: Given that the resistance R= 4 Ω and the current I= 2 A, we can
plug these values into the formula:
P= (2 A)2×4 Ω
Step 3: Simplify the expression:
P= 4 A2×4 Ω = 16 W
Step 4: Therefore, the power dissipated in the resistor is 16 W .
Question 13
Question
A copper wire has a resistance of 5 ohms. If a current of 2 amperes flows through
the wire, find the voltage across the wire.
Solution
Step 1: Recall Ohm’s Law, which states that voltage (V) is equal to current (I)
multiplied by resistance (R). The formula can be written as:
V=I×R
Step 2: Given that the resistance Ris 5 ohms and the current Iis 2 amperes,
we can substitute these values into Ohm’s Law to find the voltage V:
V= 2 A×5 Ω
Step 3: Calculate the voltage across the wire:
V= 10 V
Therefore, the voltage across the wire is 10 volts.
9
Question 14
Question
A circuit consists of a resistor with resistance R, an inductor with inductance L,
and a capacitor with capacitance Cconnected in series to an AC voltage source
with voltage V=V0sin(ωt). The circuit reaches a steady state, and the voltage
across the resistor is VR=VR0sin(ωt −ϕ). Calculate the phase difference ϕ
between the voltage across the resistor and the applied voltage.
Solution
Step 1: Write down the expression for the total impedance Zof the circuit in
terms of R,L, and C. The total impedance of the circuit is given by:
Z=R+jωL −j
ωC
Step 2: Express the current Ithrough the circuit in terms of the total
impedance Zand the applied voltage V. Using Ohm’s Law in the form V=IZ,
we have:
I=V
Z=V0sin(ωt)
Z
Step 3: Write down the expression for the voltage across the resistor VRin
terms of the current Iand the resistor R. The voltage across the resistor is
given by:
VR=IR =I·R
Step 4: Express the phase difference ϕbetween VRand Vin terms of the
impedance Z. Comparing the expressions for VRand V:
VR0sin(ωt −ϕ) = I·R=V0sin(ωt)
Z·R
⇒VR0=V0R
|Z|and ϕ=Arg(Z)
Step 5: Calculate the magnitude of Zand its argument, then find the phase
difference ϕ. Using the expression for Zfrom Step 1:
|Z|=√R2+(ωL −1
ωC )2
Arg(Z) = tan−1(ωL −1
ωC
R)
Therefore, the phase difference ϕis:
ϕ= tan−1(ωL −1
ωC
R)
10
Question 15
Question
A circuit consists of a resistor, an inductor, and a capacitor connected in series to
an AC voltage source. The values of the components are R= 20 Ω,L= 0.05 H,
and C= 1 µF. The voltage source has an amplitude of V0= 10 V and operates
at a frequency of f= 100 Hz. Calculate the impedance of the circuit and the
phase angle between the current and the voltage.
Solution
Step 1: Calculate the angular frequency ω. Given that f= 100 Hz, we can find
the angular frequency using the formula ω= 2πf . Thus,
ω= 2π×100 = 200πrad/s.
Step 2: Calculate the reactance of the inductor and the capacitor. The
reactance of an inductor XLis given by XL=ωL, and the reactance of a
capacitor XCis given by XC=1
ωC . Substituting the given values,
XL= 200π×0.05 = 10πΩand XC=1
200π×1×10−6= 5 ×103πΩ.
Step 3: Calculate the impedance of the circuit. The impedance of the circuit
Zis the vector sum of the resistive, inductive, and capacitive impedance. Since
the components are in series, we have
Z=R+XL−XC= 20+10π−5×103π= 10π−5×103π+20 = (10−5×103)π+20 Ω = −4990π+20 Ω.
Step 4: Calculate the phase angle ϕbetween the current and voltage. The
phase angle ϕis given by
ϕ= arctan (XL−XC
R).
Plugging in the values,
ϕ= arctan (10π−5×103π
20 )= arctan(−2490π/10) = arctan(−249π).
Question 16
Question
A circuit contains a resistor with resistance 10 Ω, an inductor with inductance
0.02 H, and a capacitor with capacitance 8 µF connected in series to a 12 V
battery. Calculate the impedance of the circuit and the current passing through
the circuit.
11
Solution
Step 1: Calculate the total impedance of the circuit. The impedance (Z) of an
LRC series circuit is given by:
Z=√R2+ (XL−XC)2
where: R= resistance = 10 Ω,XL= inductive reactance = ωL,XC= capacitive
reactance = 1
ωC ,ω= angular frequency = 2π
T.
First, calculate the angular frequency:
ω=2π
T
Given that the frequency f=1
T= 60 Hz. Therefore, ω= 2π×60 = 120π
rad/s.
Next, calculate XLand XC:
XL=ωL = 120π×0.02 = 2.4πΩ
XC=1
ωC =1
120π×8×10−6=106
960Ω = 625
6πΩ
Now substitute these values into the impedance formula:
Z=√102+ (2.4π−625
6π)2
Z=√100 + (2.4π−625
6π)2
Step 2: Calculate the current passing through the circuit. The current (I)
flowing through the circuit is given by Ohm’s law:
I=V
Z
where V= voltage from the battery = 12 V.
Now, substitute the calculated impedance value:
I=12
Z
Now, you can calculate the impedance (Z) and then find the current passing
through the circuit.
Question 17
Question
A student sets up a simple circuit with a resistor, battery, and an ammeter.
The ammeter reads a current of 0.5 A when the voltage across the resistor is 6
V. The student then decides to double the resistance in the circuit and observes
that the current reading on the ammeter drops to 0.25 A. Calculate the original
resistance in the circuit and the new resistance after doubling it.
12
Solution
Step 1: Let’s denote the original resistance as Rand the new resistance after
doubling it as 2R.
Step 2: Using Ohm’s Law V=IR, where Vis the voltage, Iis the current,
and Ris the resistance, we can set up two equations:
For the original circuit:
6 = 0.5×R
Simplifying, we get:
R=6
0.5= 12 Ω
Step 3: For the circuit with doubled resistance:
6 = 0.25 ×2R
Simplifying, we get:
2R=6
0.25 = 24 Ω
Therefore, the new resistance 2Ris 24 Ωafter doubling it.
Step 4: The original resistance was found to be 12 Ω. Thus, the original
resistance in the circuit is 12 Ω, and the new resistance after doubling it is 24
Ω.
Question 18
Question
A circuit consists of a 12 V battery connected in series with a resistor and an
unknown device. When a current of 2 A flows through the circuit, the potential
drop across the resistor is 4 V. Determine the resistance of the unknown device.
Solution
Step 1: Recall Ohm’s Law, which states that the potential difference (V) across a
resistor is equal to the current (I) flowing through it multiplied by the resistance
(R) of the resistor. Mathematically, this can be expressed as V=I·R.
Step 2: From the problem, we are given that the potential drop across the
resistor is 4 V when a current of 2 A flows through the circuit. Therefore, we
have 4V= 2 A×R.
Step 3: To find the resistance Rof the unknown device, we can rearrange
the formula V=I·Rto solve for R. So, R=V
I.
Step 4: Substituting the given values, we find R=4V
2A= 2 Ω.
Step 5: Hence, the resistance of the unknown device in the circuit is 2 ohms.
13
Question 19
Question
A circuit consists of a 12 V battery, a resistor with resistance R, and a capacitor
with capacitance C. Initially, the capacitor is uncharged and acts like a short
circuit. When the circuit is closed, the current through the circuit is found to be
0.5A. After some time, the capacitor is fully charged and the current through
the circuit is found to be 0.1A. Calculate the resistance Rof the resistor in the
circuit.
Solution
1. Let’s denote the resistance of the resistor as R, the capacitance of the capac-
itor as C, and the potential difference across the resistor as VR. The potential
difference across the capacitor is VC= 12 V.
2. When the circuit is first closed and the capacitor acts like a short circuit,
the current in the circuit is equal to Iinitial = 0.5A. This current can be split
between the resistor and capacitor according to the equivalent resistance Req =
Rin series with the capacitor. Since the capacitor acts like a short circuit, the
full potential difference of 12 V is across the resistor:
VR= 12 V=R×Iinitial
R=VR
Iinitial
=12
0.5= 24Ω
3. When the capacitor is fully charged and the current through the circuit
is Ifinal = 0.1A, the capacitor acts like an open circuit. Now, the potential
difference VCis fully across the resistor. We can write an expression for the
potential difference VRin terms of Rand Cusing Ohm’s Law and the formula
for the charging of a capacitor:
VR=R×Ifinal =VC(1 −e−t
RC )
Since the capacitor is fully charged, VR=VC= 12 V and Ifinal = 0.1A.
Substituting these values into the equation gives:
R×0.1 = 12(1 −e−t
RC )
R=12(1 −e−t
RC )
0.1
4. Since we are looking for the resistance R, we need to find the time constant
τ=RC. We can find τby noting that in a fully charged circuit, e−t
RC ≈0.
Therefore:
R=12(1 −0)
0.1= 120Ω
5. Therefore, the resistance Rof the resistor in the circuit is 120 Ω .
14
Question 20
Question
A 10 V battery is connected to a resistor of unknown resistance. When a current
of 2 A flows through the resistor, the power dissipated is 20 W. Calculate the
resistance of the resistor.
Solution
Step 1: We can start by using the formula for power in a resistor: P=I2·R,
where Pis the power dissipated, Iis the current flowing through the resistor,
and Ris the resistance. Given that P= 20 W and I= 2 A, we can rearrange
the formula to solve for R:
R=P
I2
R=20
(2)2
R=20
4
R= 5 Ω
Therefore, the resistance of the resistor is 5 Ω.
Question 21
Question
A circuit consists of a 12 V battery, a 4 Ωresistor, and an unknown resistor
R. When the total resistance in the circuit is 6 Ω, the power dissipated in the
unknown resistor is 24 W. Find the value of R.
Solution
Step 1: Let’s first calculate the current, I, flowing through the circuit using
Ohm’s Law, V=IR, where Vis the voltage of the battery.
Given: V= 12 V, R1= 4 Ω, Rtotal = 6 Ω
Since Rtotal =R1+R, we can find Ras R=Rtotal −R1
R= 6 −4 = 2 Ω
Step 2: Next, we find the current, I, as I=V
Rtotal .
I=12
6= 2 A
15
Step 3: Now, we can find the power, P, dissipated by the unknown resistor
using the formula P=I2R.
Given: P= 24 W, R= 2 Ω
24 = (2)2×2
24 = 4 ×2
24 = 8
This equation is not met, so there must be an error in the problem statement
or calculation.
Question 22
Question
A circuit consists of a resistor with resistance R= 100 Ω connected to a battery
with emf E= 12 V. The circuit also has an ammeter connected in series and
a voltmeter connected in parallel to the resistor. If the ammeter reads 0.1A,
what reading will the voltmeter display?
Solution
Step 1: Recall Ohm’s Law which states that the current passing through a
resistor is directly proportional to the potential difference across the resistor.
The formula for Ohm’s Law is V=IR, where Vis the voltage (potential
difference) across the resistor, Iis the current passing through the resistor, and
Ris the resistance of the resistor.
Step 2: Given in the question, R= 100 Ω and I= 0.1A. We can use Ohm’s
Law to find the voltage across the resistor: V=I·R.
Step 3: Substituting the given values into the formula, we get V= 0.1A·
100 Ω = 10 V.
Step 4: Since the voltmeter is connected in parallel to the resistor, it will
read the same voltage as the resistor. Therefore, the voltmeter will display
10 V .
Question 23
Question
A circuit consists of a resistor with resistance R= 15 Ω and an inductor with
inductance L= 0.04 H. The circuit is connected to a voltage source with an
emf of V= 20 V and a frequency of f= 50 Hz. Calculate the current flowing
through the circuit.
16
Solution
Step 1: Calculate the reactance of the inductor using the formula XL= 2πf L.
XL= 2π×50 ×0.04 = 4πΩ
Step 2: Calculate the total impedance of the circuit using the formula Z=
√R2+X2
L.
Z=√152+ (4π)2≈√225 + 39.48 ≈√264.48 ≈16.27 Ω
Step 3: Calculate the current flowing through the circuit using Ohm’s Law:
I=V
Z.
I=20
16.27 ≈1.23 A
Therefore, the current flowing through the circuit is approximately 1.23 A.
Question 24
Question
A circuit consists of a resistor with resistance R, an inductor with inductance
L, a capacitor with capacitance C, and an AC voltage source V=V0sin(ωt).
The instantaneous current i(t)through the circuit is given by the equation:
i(t) = i0sin(ωt +ϕ)
where i0= 3 A, ω= 50 rad/s, and ϕ=π
6rad.
Determine the voltage amplitude V0applied by the source.
Solution
Step 1: Ohm’s Law states that for a circuit, the voltage Vapplied across a com-
ponent is equal to the product of the current Iflowing through the component
and the resistance Rof the component.
V=IR
For the given circuit, the voltage across the resistor is VR=i(t)·R. Since the
current through the resistor is i(t)and the resistance of the resistor is R, we
have:
VR=i(t)·R
Step 2: In an AC circuit with a resistor, inductor, and capacitor in series,
the total voltage across the components will be the same as the voltage of the
AC source. Therefore, the voltages across the resistor, inductor, and capacitor
must sum up to the applied voltage V=V0sin(ωt).
V=VR+VL+VC
17
Now, let’s find the expressions for VLand VC.
Step 3: The voltage across an inductor in an AC circuit is given by:
VL=Ldi
dt
where Lis the inductance of the inductor. Since i(t) = i0sin(ωt +ϕ), we can
find di
dt by differentiating i(t)with respect to t.
di
dt =d
dt(i0sin(ωt +ϕ)) = i0ωcos(ωt +ϕ)
Step 4: Substituting into the expression for VL, we get:
VL=Ldi
dt =Li0ωcos(ωt +ϕ)
Step 5: The voltage across a capacitor in an AC circuit is given by:
VC=1
C∫i dt
where Cis the capacitance of the capacitor. Since i(t) = i0sin(ωt +ϕ), we can
find ∫i dt by integrating i(t)with respect to t.
∫i dt =∫i0sin(ωt +ϕ)dt =−i0
ωcos(ωt +ϕ)
Step 6: Substituting into the expression for VC, we have:
VC=1
C∫i dt =−i0
ωC cos(ωt +ϕ)
Step 7: Now, substituting VR,VL, and VCinto the equation V=VR+VL+
VC, we get:
V0sin(ωt) = i0Rsin(ωt +ϕ) + Li0ωcos(ωt +ϕ)−i0
ωC cos(ωt +ϕ)
Step 8: We know that the voltage amplitudes of sin(ωt),cos(ωt), and cos(ωt)
terms on both sides of the equation must be equal. So, the coefficient of sin(ωt)
term on the right side must be V0. Thus, we can determine V0by equating
coefficients:
V0=i0R
V0= 3 A×R
Question 25
Question
A circuit consists of a resistor with resistance R, an inductor with induc-
tance L, and a capacitor with capacitance Cconnected in series to an AC
voltage source with frequency f. The impedance of the circuit is given by
Z=√R2+(ωL −1
ωC )2, where ω= 2πf. Determine the angular frequency ω
at which the impedance of the circuit is minimized.
18
Solution
Step 1: To find the angular frequency ωat which the impedance of the circuit is
minimized, we need to find the minimum value of the expression for impedance
Z. This minimum occurs when the derivative of Zwith respect to ωis zero.
Step 2: Calculate the derivative of Zwith respect to ω:
dZ
dω =1
2√R2+(ωL −1
ωC )2·2·(ωL −1
ωC )(L−1
ω2C)
Step 3: Set the derivative equal to zero and solve for ω:
ωL −1
ωC
√R2+(ωL −1
ωC )2= 0
Step 4: Simplify the equation and solve for ω:
ωL −1
ωC = 0
ω2=1
LC
ω=√1
LC
Therefore, the angular frequency ωat which the impedance of the circuit is
minimized is ω=√1
LC .
19
Solution
Step 1: Given that the initial resistance of the copper wire is 8 Ω at 20◦C, we
want to find the resistance at 80◦C. The formula relating resistance, initial
resistance, and temperature change is:
RT=R0[1 + α(T−T0)]
where RTis the resistance at temperature T,R0is the resistance at reference
temperature T0,αis the temperature coefficient of resistivity, Tis the final
temperature, and T0is the reference temperature.
Step 2: Plugging in the given values into the equation, we get:
R80 = 8 Ω[1 + 0.00428 ×(80 −20)]
Step 3: Calculating the expression inside the brackets first:
1+0.00428 ×60 = 1 + 0.2568 = 1.2568
Step 4: Now, calculate the resistance at 80◦C:
R80 = 8 Ω ×1.2568 = 10.0544 Ω
Therefore, the resistance of the copper wire at 80◦C is 10.0544 Ω.
Question 3
Question
A circuit consists of a 10 V battery connected in series to a resistor. If the
current in the circuit is 2 A, what is the resistance of the resistor?
Solution
Step 1: Recall Ohm’s Law, which states that the voltage across a resistor is equal
to the current flowing through it multiplied by the resistance. Mathematically,
this can be written as V=IR, where Vis the voltage, Iis the current, and R
is the resistance.
Step 2: In this case, the voltage across the resistor is 10 V and the cur-
rent flowing through it is 2 A. We can rearrange Ohm’s Law to solve for the
resistance: R=V
I.
Step 3: Substitute the given values into the formula: R=10 V
2A.
Step 4: Perform the division to find the resistance: R= 5 Ω.
Step 5: Therefore, the resistance of the resistor in the circuit is 5 Ohms.
2
Question 4
Question
A circuit consists of a battery with voltage V= 12 V and three resistors con-
nected in series. The resistors have resistances R1= 5 ohms, R2= 10 ohms,
and R3= 15 ohms. Calculate the total current flowing through the circuit and
the voltage drop across each resistor.
Solution
Step 1: Calculate the total resistance of the circuit. The total resistance Rtotal
of resistors in series is the sum of the individual resistances:
Rtotal =R1+R2+R3= 5 Ω + 10 Ω + 15 Ω = 30 Ω
Step 2: Calculate the total current flowing through the circuit using Ohm’s
Law V=IR. The total current Iis given by I=V
Rtotal .
I=12 V
30 Ω = 0.4A
Step 3: Calculate the voltage drop across each resistor. The voltage drop Vi
across a resistor Riis given by Ohm’s Law Vi=IRi. For R1:
V1=I·R1= 0.4A·5 Ω = 2 V
For R2:
V2=I·R2= 0.4A·10 Ω = 4 V
For R3:
V3=I·R3= 0.4A·15 Ω = 6 V
Therefore, the total current flowing through the circuit is 0.4 A, and the
voltage drop across R1,R2, and R3are 2 V, 4 V, and 6 V, respectively.
Question 5
Question
A circuit consists of a resistor with resistance R= 15 Ω connected to a battery
with emf ε= 12 V. Calculate the current flowing through the resistor.
Solution
Step 1: Determine the relationship between voltage, current, and resistance in a
circuit. According to Ohm’s Law, the relationship between voltage (V), current
(I), and resistance (R) in a circuit is given by:
V=IR
3
where V= voltage (in volts), I= current (in amperes), R= resistance (in
ohms).
Step 2: Identify the given values. The resistance of the resistor, R= 15 Ω,
and the emf of the battery, ε= 12 V.
Step 3: Apply Ohm’s Law to find the current. Substitute the given values
into Ohm’s Law:
ε=IR
12 = I×15
I=12
15
I= 0.8A
Step 4: State the final answer. The current flowing through the resistor is
0.8amperes.
Question 6
Question
A resistor is connected to a 12 V battery, and a current of 3 A flows through it.
If the resistance of the resistor is R, calculate the resistance R.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor
is equal to the current (I) flowing through it multiplied by the resistance (R)
of the resistor. Mathematically, this can be written as:
V=I×R
Step 2: In this case, the voltage across the resistor is 12V, the current flowing
through it is 3A, and we are trying to find the resistance, R. We can substitute
these values into Ohm’s Law:
12 = 3 ×R
Step 3: Now, we solve for the resistance R:
R=12
3= 4 Ω
Step 4: Therefore, the resistance of the resistor is 4 Ω.
4
Question 7
Question
A circuit consists of a resistor with resistance R, a capacitor with capacitance C,
and an inductor with inductance Lconnected in series to an alternating current
(AC) voltage source. The voltage across the resistor is VR= 5 sin(100t)V, the
voltage across the capacitor is VC= 3 sin(100t+π
4)V, and the voltage across
the inductor is VL= 4 sin(100t−π
3)V. Given that the current in the circuit is
I= 2 sin(100t+π
6)A, determine the values of R,L, and Cin the circuit.
Solution
Step 1: Write down the phasor representation for the voltages and current in
the circuit. The phasor representation of a sinusoidal function Asin(ωt +ϕ)is
ˆ
A=A
√2∠ϕ. For VR= 5 sin(100t)V: ˆ
VR=5
√2∠0◦V. For VC= 3 sin(100t+π
4)
V: ˆ
VC=3
√2∠45◦V. For VL= 4 sin(100t−π
3)V: ˆ
VL=4
√2∠−60◦V. For
I= 2 sin(100t+π
6)A: ˆ
I=2
√2∠30◦A.
Step 2: Apply Kirchhoff’s voltage law (KVL) to the circuit:
ˆ
VR+ˆ
VL+ˆ
VC=ˆ
Vsource
Substitute in the phasor representations:
5
√2
∠0◦+4
√2
∠−60◦+3
√2
∠45◦=ˆ
Vsource
5∠0◦+ 4∠−60◦+ 3∠45◦=ˆ
Vsource
Step 3: Calculate the phasor representation of the source voltage. Since the
source voltage is not explicitly given, we can represent it as ˆ
Vsource =Vsource∠0◦.
Now solve for Vsource:
5∠0◦+ 4∠−60◦+ 3∠45◦=Vsource∠0◦
Vsource∠0◦= 5 + 4 cos(60◦) + 3 cos(−45◦) + j(4 sin(−60◦) + 3 sin(45◦))
Vsource = 5 + 2 + 3√2
2+j(−2√3 + 3√2
2)
Vsource = 7 + 3√2
2−2√3 + j3√2
2−2√3
Step 4: Compare the phasor representation of the source voltage with the
expression for the impedance of the circuit Z=R+j(ωL−1
ωC ). The impedance
of the circuit must be equal to the phasor representation of the source voltage.
Matching real and imaginary parts, we have: Real parts: R= 7 Imaginary
parts: ωL −1
ωC =3√2
2−2
5
Question 8
Question
A circuit consists of a resistor with resistance R, an inductor with inductance
L, and a capacitor with capacitance Cconnected in series. The voltage across
the circuit is given by V(t) = V0cos(ωt), where V0= 12 V and ω= 50 s−1.
The resistor has a resistance of R= 5 Ω, the inductor has an inductance of
L= 0.1H, and the capacitor has a capacitance of C= 0.01 F. Determine the
current as a function of time for this circuit.
Solution
Step 1: First, we need to find the expression for the total impedance of the
circuit. The impedance ZRof the resistor is simply its resistance R. The
impedance ZLof the inductor is given by ZL=jωL. The impedance ZCof the
capacitor is given by ZC=1
jωC . The total impedance Ztotal of the series circuit
is the sum of the impedances of the resistor, inductor, and capacitor:
Ztotal =R+ZL+ZC=R+jωL +1
jωC
Step2: Next, we can find the current I(t)as a function of time. Since the
frequency of the voltage source is ω= 50 s−1, we can express the current as:
I(t) = V0
Ztotal
cos(ωt)
Step 3: Now, substitute the given values into the expression for total impedance
to get:
Ztotal = 5 + j×50 ×0.1 + 1
j×50 ×0.01
Step 4: Simplify the expression for the total impedance:
Ztotal = 5 + j5 + 1
j0.5= 5 + j5 + j2 = 5 + j7
Step 5: Substitute the total impedance into the expression for current:
I(t) = 12
5 + j7cos(50t)
Therefore, the current as a function of time for the given series circuit is:
I(t) = 12
5 + j7cos(50t)
6
Question 9
Question
A circuit consists of a resistor with resistance R= 10 Ω, a capacitor with capaci-
tance C= 2 µF , and an inductor with inductance L= 0.5Hconnected in series
to a sinusoidal voltage source with frequency ω= 100 rad/s. If the maximum
voltage supplied by the source is Vmax = 20 V, what is the maximum current in
the circuit?
Solution
Step 1: Calculate the impedance of each component. The impedance of a
resistor is given by ZR=R. The impedance of a capacitor is given by ZC=1
jωC .
The impedance of an inductor is given by ZL=jωL.
Step 2: Calculate the total impedance of the circuit. The total impedance
Ztotal of components in series is the sum of their individual impedances.
Ztotal =ZR+ZC+ZL=R+1
jωC +jωL
Step 3: Convert the total impedance to polar form. To convert Ztotal to
polar form, we need to find the magnitude and phase angle. The magnitude
|Ztotal|=√Re(Ztotal)2+Im(Ztotal)2. The phase angle θ= arctan (Im(Ztotal )
Re(Ztotal )).
Step 4: Calculate the maximum current in the circuit. The maximum current
Imax is given by Imax =Vmax
|Ztotal |.
Substitute the given values into the expressions above to find Imax.
Question 10
Question
A circuit consists of a 12V battery connected to three resistors in series. The
first resistor has a resistance of 4Ω, the second resistor has a resistance of 8Ω,
and the third resistor has a resistance of R3Ω. If the current through the circuit
is 1.5A, find the value of R3.
Solution
Step 1: Calculate the total resistance of the circuit. The total resistance Rtotal
of resistors in series is the sum of individual resistances:
Rtotal = 4Ω + 8Ω + R3Ω = 12Ω + R3Ω
Step 2: Apply Ohm’s Law to find the total resistance. Ohm’s Law states
that V=IR, where Vis the voltage, Iis the current, and Ris the resistance.
7
In this case, V= 12V and I= 1.5A. Therefore, the total resistance can be
calculated as:
12V= 1.5A×(12Ω + R3Ω)
12V = 18Ω+1.5R3Ω
Step 3: Solve for R3. Subtract 18Ωfrom both sides of the equation:
12V−18Ω = 1.5R3Ω
−6V= 1.5R3Ω
R3=−6V
1.5Ω =−4Ω
Therefore, the value of R3is 4Ω.
Question 11
Question
A resistor with resistance R1= 10 Ω is connected in series with another resistor
with resistance R2= 20 Ω. A potential difference of V= 120 Vis applied across
the combination. Calculate the current passing through each resistor.
Solution
Step 1: Calculate the total resistance of the combination. The total resistance
Rtotal of resistors in series is given by:
Rtotal =R1+R2= 10 Ω + 20 Ω = 30 Ω
Step 2: Calculate the total current passing through the combination using
Ohm’s Law V=IR. The total current Itotal passing through the combination
is:
Itotal =V
Rtotal
=120 V
30 Ω = 4 A
Step 3: Calculate the current passing through each resistor using the current
divider rule. The current passing through R1, denoted as I1, is:
I1=Itotal ×Rtotal
R1
= 4 A×30 Ω
10 Ω = 12 A
The current passing through R2, denoted as I2, is:
I2=Itotal ×Rtotal
R2
= 4 A×30 Ω
20 Ω = 6 A
Thus, the current passing through R1is 12 Aand the current passing through
R2is 6A.
8
Question 12
Question
A resistor with a resistance of 4 Ω is connected to a battery that delivers a
current of 2A. Find the power dissipated in the resistor.
Solution
Step 1: Recall that the power dissipated in a resistor can be calculated using
the formula P=I2R, where Pis the power, Iis the current, and Ris the
resistance.
Step 2: Given that the resistance R= 4 Ω and the current I= 2 A, we can
plug these values into the formula:
P= (2 A)2×4 Ω
Step 3: Simplify the expression:
P= 4 A2×4 Ω = 16 W
Step 4: Therefore, the power dissipated in the resistor is 16 W .
Question 13
Question
A copper wire has a resistance of 5 ohms. If a current of 2 amperes flows through
the wire, find the voltage across the wire.
Solution
Step 1: Recall Ohm’s Law, which states that voltage (V) is equal to current (I)
multiplied by resistance (R). The formula can be written as:
V=I×R
Step 2: Given that the resistance Ris 5 ohms and the current Iis 2 amperes,
we can substitute these values into Ohm’s Law to find the voltage V:
V= 2 A×5 Ω
Step 3: Calculate the voltage across the wire:
V= 10 V
Therefore, the voltage across the wire is 10 volts.
9
Question 14
Question
A circuit consists of a resistor with resistance R, an inductor with inductance L,
and a capacitor with capacitance Cconnected in series to an AC voltage source
with voltage V=V0sin(ωt). The circuit reaches a steady state, and the voltage
across the resistor is VR=VR0sin(ωt −ϕ). Calculate the phase difference ϕ
between the voltage across the resistor and the applied voltage.
Solution
Step 1: Write down the expression for the total impedance Zof the circuit in
terms of R,L, and C. The total impedance of the circuit is given by:
Z=R+jωL −j
ωC
Step 2: Express the current Ithrough the circuit in terms of the total
impedance Zand the applied voltage V. Using Ohm’s Law in the form V=IZ,
we have:
I=V
Z=V0sin(ωt)
Z
Step 3: Write down the expression for the voltage across the resistor VRin
terms of the current Iand the resistor R. The voltage across the resistor is
given by:
VR=IR =I·R
Step 4: Express the phase difference ϕbetween VRand Vin terms of the
impedance Z. Comparing the expressions for VRand V:
VR0sin(ωt −ϕ) = I·R=V0sin(ωt)
Z·R
⇒VR0=V0R
|Z|and ϕ=Arg(Z)
Step 5: Calculate the magnitude of Zand its argument, then find the phase
difference ϕ. Using the expression for Zfrom Step 1:
|Z|=√R2+(ωL −1
ωC )2
Arg(Z) = tan−1(ωL −1
ωC
R)
Therefore, the phase difference ϕis:
ϕ= tan−1(ωL −1
ωC
R)
10
Question 15
Question
A circuit consists of a resistor, an inductor, and a capacitor connected in series to
an AC voltage source. The values of the components are R= 20 Ω,L= 0.05 H,
and C= 1 µF. The voltage source has an amplitude of V0= 10 V and operates
at a frequency of f= 100 Hz. Calculate the impedance of the circuit and the
phase angle between the current and the voltage.
Solution
Step 1: Calculate the angular frequency ω. Given that f= 100 Hz, we can find
the angular frequency using the formula ω= 2πf . Thus,
ω= 2π×100 = 200πrad/s.
Step 2: Calculate the reactance of the inductor and the capacitor. The
reactance of an inductor XLis given by XL=ωL, and the reactance of a
capacitor XCis given by XC=1
ωC . Substituting the given values,
XL= 200π×0.05 = 10πΩand XC=1
200π×1×10−6= 5 ×103πΩ.
Step 3: Calculate the impedance of the circuit. The impedance of the circuit
Zis the vector sum of the resistive, inductive, and capacitive impedance. Since
the components are in series, we have
Z=R+XL−XC= 20+10π−5×103π= 10π−5×103π+20 = (10−5×103)π+20 Ω = −4990π+20 Ω.
Step 4: Calculate the phase angle ϕbetween the current and voltage. The
phase angle ϕis given by
ϕ= arctan (XL−XC
R).
Plugging in the values,
ϕ= arctan (10π−5×103π
20 )= arctan(−2490π/10) = arctan(−249π).
Question 16
Question
A circuit contains a resistor with resistance 10 Ω, an inductor with inductance
0.02 H, and a capacitor with capacitance 8 µF connected in series to a 12 V
battery. Calculate the impedance of the circuit and the current passing through
the circuit.
11
Solution
Step 1: Calculate the total impedance of the circuit. The impedance (Z) of an
LRC series circuit is given by:
Z=√R2+ (XL−XC)2
where: R= resistance = 10 Ω,XL= inductive reactance = ωL,XC= capacitive
reactance = 1
ωC ,ω= angular frequency = 2π
T.
First, calculate the angular frequency:
ω=2π
T
Given that the frequency f=1
T= 60 Hz. Therefore, ω= 2π×60 = 120π
rad/s.
Next, calculate XLand XC:
XL=ωL = 120π×0.02 = 2.4πΩ
XC=1
ωC =1
120π×8×10−6=106
960Ω = 625
6πΩ
Now substitute these values into the impedance formula:
Z=√102+ (2.4π−625
6π)2
Z=√100 + (2.4π−625
6π)2
Step 2: Calculate the current passing through the circuit. The current (I)
flowing through the circuit is given by Ohm’s law:
I=V
Z
where V= voltage from the battery = 12 V.
Now, substitute the calculated impedance value:
I=12
Z
Now, you can calculate the impedance (Z) and then find the current passing
through the circuit.
Question 17
Question
A student sets up a simple circuit with a resistor, battery, and an ammeter.
The ammeter reads a current of 0.5 A when the voltage across the resistor is 6
V. The student then decides to double the resistance in the circuit and observes
that the current reading on the ammeter drops to 0.25 A. Calculate the original
resistance in the circuit and the new resistance after doubling it.
12
Solution
Step 1: Let’s denote the original resistance as Rand the new resistance after
doubling it as 2R.
Step 2: Using Ohm’s Law V=IR, where Vis the voltage, Iis the current,
and Ris the resistance, we can set up two equations:
For the original circuit:
6 = 0.5×R
Simplifying, we get:
R=6
0.5= 12 Ω
Step 3: For the circuit with doubled resistance:
6 = 0.25 ×2R
Simplifying, we get:
2R=6
0.25 = 24 Ω
Therefore, the new resistance 2Ris 24 Ωafter doubling it.
Step 4: The original resistance was found to be 12 Ω. Thus, the original
resistance in the circuit is 12 Ω, and the new resistance after doubling it is 24
Ω.
Question 18
Question
A circuit consists of a 12 V battery connected in series with a resistor and an
unknown device. When a current of 2 A flows through the circuit, the potential
drop across the resistor is 4 V. Determine the resistance of the unknown device.
Solution
Step 1: Recall Ohm’s Law, which states that the potential difference (V) across a
resistor is equal to the current (I) flowing through it multiplied by the resistance
(R) of the resistor. Mathematically, this can be expressed as V=I·R.
Step 2: From the problem, we are given that the potential drop across the
resistor is 4 V when a current of 2 A flows through the circuit. Therefore, we
have 4V= 2 A×R.
Step 3: To find the resistance Rof the unknown device, we can rearrange
the formula V=I·Rto solve for R. So, R=V
I.
Step 4: Substituting the given values, we find R=4V
2A= 2 Ω.
Step 5: Hence, the resistance of the unknown device in the circuit is 2 ohms.
13
Question 19
Question
A circuit consists of a 12 V battery, a resistor with resistance R, and a capacitor
with capacitance C. Initially, the capacitor is uncharged and acts like a short
circuit. When the circuit is closed, the current through the circuit is found to be
0.5A. After some time, the capacitor is fully charged and the current through
the circuit is found to be 0.1A. Calculate the resistance Rof the resistor in the
circuit.
Solution
1. Let’s denote the resistance of the resistor as R, the capacitance of the capac-
itor as C, and the potential difference across the resistor as VR. The potential
difference across the capacitor is VC= 12 V.
2. When the circuit is first closed and the capacitor acts like a short circuit,
the current in the circuit is equal to Iinitial = 0.5A. This current can be split
between the resistor and capacitor according to the equivalent resistance Req =
Rin series with the capacitor. Since the capacitor acts like a short circuit, the
full potential difference of 12 V is across the resistor:
VR= 12 V=R×Iinitial
R=VR
Iinitial
=12
0.5= 24Ω
3. When the capacitor is fully charged and the current through the circuit
is Ifinal = 0.1A, the capacitor acts like an open circuit. Now, the potential
difference VCis fully across the resistor. We can write an expression for the
potential difference VRin terms of Rand Cusing Ohm’s Law and the formula
for the charging of a capacitor:
VR=R×Ifinal =VC(1 −e−t
RC )
Since the capacitor is fully charged, VR=VC= 12 V and Ifinal = 0.1A.
Substituting these values into the equation gives:
R×0.1 = 12(1 −e−t
RC )
R=12(1 −e−t
RC )
0.1
4. Since we are looking for the resistance R, we need to find the time constant
τ=RC. We can find τby noting that in a fully charged circuit, e−t
RC ≈0.
Therefore:
R=12(1 −0)
0.1= 120Ω
5. Therefore, the resistance Rof the resistor in the circuit is 120 Ω .
14
Question 20
Question
A 10 V battery is connected to a resistor of unknown resistance. When a current
of 2 A flows through the resistor, the power dissipated is 20 W. Calculate the
resistance of the resistor.
Solution
Step 1: We can start by using the formula for power in a resistor: P=I2·R,
where Pis the power dissipated, Iis the current flowing through the resistor,
and Ris the resistance. Given that P= 20 W and I= 2 A, we can rearrange
the formula to solve for R:
R=P
I2
R=20
(2)2
R=20
4
R= 5 Ω
Therefore, the resistance of the resistor is 5 Ω.
Question 21
Question
A circuit consists of a 12 V battery, a 4 Ωresistor, and an unknown resistor
R. When the total resistance in the circuit is 6 Ω, the power dissipated in the
unknown resistor is 24 W. Find the value of R.
Solution
Step 1: Let’s first calculate the current, I, flowing through the circuit using
Ohm’s Law, V=IR, where Vis the voltage of the battery.
Given: V= 12 V, R1= 4 Ω, Rtotal = 6 Ω
Since Rtotal =R1+R, we can find Ras R=Rtotal −R1
R= 6 −4 = 2 Ω
Step 2: Next, we find the current, I, as I=V
Rtotal .
I=12
6= 2 A
15
Step 3: Now, we can find the power, P, dissipated by the unknown resistor
using the formula P=I2R.
Given: P= 24 W, R= 2 Ω
24 = (2)2×2
24 = 4 ×2
24 = 8
This equation is not met, so there must be an error in the problem statement
or calculation.
Question 22
Question
A circuit consists of a resistor with resistance R= 100 Ω connected to a battery
with emf E= 12 V. The circuit also has an ammeter connected in series and
a voltmeter connected in parallel to the resistor. If the ammeter reads 0.1A,
what reading will the voltmeter display?
Solution
Step 1: Recall Ohm’s Law which states that the current passing through a
resistor is directly proportional to the potential difference across the resistor.
The formula for Ohm’s Law is V=IR, where Vis the voltage (potential
difference) across the resistor, Iis the current passing through the resistor, and
Ris the resistance of the resistor.
Step 2: Given in the question, R= 100 Ω and I= 0.1A. We can use Ohm’s
Law to find the voltage across the resistor: V=I·R.
Step 3: Substituting the given values into the formula, we get V= 0.1A·
100 Ω = 10 V.
Step 4: Since the voltmeter is connected in parallel to the resistor, it will
read the same voltage as the resistor. Therefore, the voltmeter will display
10 V .
Question 23
Question
A circuit consists of a resistor with resistance R= 15 Ω and an inductor with
inductance L= 0.04 H. The circuit is connected to a voltage source with an
emf of V= 20 V and a frequency of f= 50 Hz. Calculate the current flowing
through the circuit.
16
Solution
Step 1: Calculate the reactance of the inductor using the formula XL= 2πf L.
XL= 2π×50 ×0.04 = 4πΩ
Step 2: Calculate the total impedance of the circuit using the formula Z=
√R2+X2
L.
Z=√152+ (4π)2≈√225 + 39.48 ≈√264.48 ≈16.27 Ω
Step 3: Calculate the current flowing through the circuit using Ohm’s Law:
I=V
Z.
I=20
16.27 ≈1.23 A
Therefore, the current flowing through the circuit is approximately 1.23 A.
Question 24
Question
A circuit consists of a resistor with resistance R, an inductor with inductance
L, a capacitor with capacitance C, and an AC voltage source V=V0sin(ωt).
The instantaneous current i(t)through the circuit is given by the equation:
i(t) = i0sin(ωt +ϕ)
where i0= 3 A, ω= 50 rad/s, and ϕ=π
6rad.
Determine the voltage amplitude V0applied by the source.
Solution
Step 1: Ohm’s Law states that for a circuit, the voltage Vapplied across a com-
ponent is equal to the product of the current Iflowing through the component
and the resistance Rof the component.
V=IR
For the given circuit, the voltage across the resistor is VR=i(t)·R. Since the
current through the resistor is i(t)and the resistance of the resistor is R, we
have:
VR=i(t)·R
Step 2: In an AC circuit with a resistor, inductor, and capacitor in series,
the total voltage across the components will be the same as the voltage of the
AC source. Therefore, the voltages across the resistor, inductor, and capacitor
must sum up to the applied voltage V=V0sin(ωt).
V=VR+VL+VC
17
Now, let’s find the expressions for VLand VC.
Step 3: The voltage across an inductor in an AC circuit is given by:
VL=Ldi
dt
where Lis the inductance of the inductor. Since i(t) = i0sin(ωt +ϕ), we can
find di
dt by differentiating i(t)with respect to t.
di
dt =d
dt(i0sin(ωt +ϕ)) = i0ωcos(ωt +ϕ)
Step 4: Substituting into the expression for VL, we get:
VL=Ldi
dt =Li0ωcos(ωt +ϕ)
Step 5: The voltage across a capacitor in an AC circuit is given by:
VC=1
C∫i dt
where Cis the capacitance of the capacitor. Since i(t) = i0sin(ωt +ϕ), we can
find ∫i dt by integrating i(t)with respect to t.
∫i dt =∫i0sin(ωt +ϕ)dt =−i0
ωcos(ωt +ϕ)
Step 6: Substituting into the expression for VC, we have:
VC=1
C∫i dt =−i0
ωC cos(ωt +ϕ)
Step 7: Now, substituting VR,VL, and VCinto the equation V=VR+VL+
VC, we get:
V0sin(ωt) = i0Rsin(ωt +ϕ) + Li0ωcos(ωt +ϕ)−i0
ωC cos(ωt +ϕ)
Step 8: We know that the voltage amplitudes of sin(ωt),cos(ωt), and cos(ωt)
terms on both sides of the equation must be equal. So, the coefficient of sin(ωt)
term on the right side must be V0. Thus, we can determine V0by equating
coefficients:
V0=i0R
V0= 3 A×R
Question 25
Question
A circuit consists of a resistor with resistance R, an inductor with induc-
tance L, and a capacitor with capacitance Cconnected in series to an AC
voltage source with frequency f. The impedance of the circuit is given by
Z=√R2+(ωL −1
ωC )2, where ω= 2πf. Determine the angular frequency ω
at which the impedance of the circuit is minimized.
18
Solution
Step 1: To find the angular frequency ωat which the impedance of the circuit is
minimized, we need to find the minimum value of the expression for impedance
Z. This minimum occurs when the derivative of Zwith respect to ωis zero.
Step 2: Calculate the derivative of Zwith respect to ω:
dZ
dω =1
2√R2+(ωL −1
ωC )2·2·(ωL −1
ωC )(L−1
ω2C)
Step 3: Set the derivative equal to zero and solve for ω:
ωL −1
ωC
√R2+(ωL −1
ωC )2= 0
Step 4: Simplify the equation and solve for ω:
ωL −1
ωC = 0
ω2=1
LC
ω=√1
LC
Therefore, the angular frequency ωat which the impedance of the circuit is
minimized is ω=√1
LC .
19
Solution
Step 1: Given that the initial resistance of the copper wire is 8 Ω at 20◦C, we
want to find the resistance at 80◦C. The formula relating resistance, initial
resistance, and temperature change is:
RT=R0[1 + α(T−T0)]
where RTis the resistance at temperature T,R0is the resistance at reference
temperature T0,αis the temperature coefficient of resistivity, Tis the final
temperature, and T0is the reference temperature.
Step 2: Plugging in the given values into the equation, we get:
R80 = 8 Ω[1 + 0.00428 ×(80 −20)]
Step 3: Calculating the expression inside the brackets first:
1+0.00428 ×60 = 1 + 0.2568 = 1.2568
Step 4: Now, calculate the resistance at 80◦C:
R80 = 8 Ω ×1.2568 = 10.0544 Ω
Therefore, the resistance of the copper wire at 80◦C is 10.0544 Ω.
Question 3
Question
A circuit consists of a 10 V battery connected in series to a resistor. If the
current in the circuit is 2 A, what is the resistance of the resistor?
Solution
Step 1: Recall Ohm’s Law, which states that the voltage across a resistor is equal
to the current flowing through it multiplied by the resistance. Mathematically,
this can be written as V=IR, where Vis the voltage, Iis the current, and R
is the resistance.
Step 2: In this case, the voltage across the resistor is 10 V and the cur-
rent flowing through it is 2 A. We can rearrange Ohm’s Law to solve for the
resistance: R=V
I.
Step 3: Substitute the given values into the formula: R=10 V
2A.
Step 4: Perform the division to find the resistance: R= 5 Ω.
Step 5: Therefore, the resistance of the resistor in the circuit is 5 Ohms.
2
Question 4
Question
A circuit consists of a battery with voltage V= 12 V and three resistors con-
nected in series. The resistors have resistances R1= 5 ohms, R2= 10 ohms,
and R3= 15 ohms. Calculate the total current flowing through the circuit and
the voltage drop across each resistor.
Solution
Step 1: Calculate the total resistance of the circuit. The total resistance Rtotal
of resistors in series is the sum of the individual resistances:
Rtotal =R1+R2+R3= 5 Ω + 10 Ω + 15 Ω = 30 Ω
Step 2: Calculate the total current flowing through the circuit using Ohm’s
Law V=IR. The total current Iis given by I=V
Rtotal .
I=12 V
30 Ω = 0.4A
Step 3: Calculate the voltage drop across each resistor. The voltage drop Vi
across a resistor Riis given by Ohm’s Law Vi=IRi. For R1:
V1=I·R1= 0.4A·5 Ω = 2 V
For R2:
V2=I·R2= 0.4A·10 Ω = 4 V
For R3:
V3=I·R3= 0.4A·15 Ω = 6 V
Therefore, the total current flowing through the circuit is 0.4 A, and the
voltage drop across R1,R2, and R3are 2 V, 4 V, and 6 V, respectively.
Question 5
Question
A circuit consists of a resistor with resistance R= 15 Ω connected to a battery
with emf ε= 12 V. Calculate the current flowing through the resistor.
Solution
Step 1: Determine the relationship between voltage, current, and resistance in a
circuit. According to Ohm’s Law, the relationship between voltage (V), current
(I), and resistance (R) in a circuit is given by:
V=IR
3
where V= voltage (in volts), I= current (in amperes), R= resistance (in
ohms).
Step 2: Identify the given values. The resistance of the resistor, R= 15 Ω,
and the emf of the battery, ε= 12 V.
Step 3: Apply Ohm’s Law to find the current. Substitute the given values
into Ohm’s Law:
ε=IR
12 = I×15
I=12
15
I= 0.8A
Step 4: State the final answer. The current flowing through the resistor is
0.8amperes.
Question 6
Question
A resistor is connected to a 12 V battery, and a current of 3 A flows through it.
If the resistance of the resistor is R, calculate the resistance R.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor
is equal to the current (I) flowing through it multiplied by the resistance (R)
of the resistor. Mathematically, this can be written as:
V=I×R
Step 2: In this case, the voltage across the resistor is 12V, the current flowing
through it is 3A, and we are trying to find the resistance, R. We can substitute
these values into Ohm’s Law:
12 = 3 ×R
Step 3: Now, we solve for the resistance R:
R=12
3= 4 Ω
Step 4: Therefore, the resistance of the resistor is 4 Ω.
4
Question 7
Question
A circuit consists of a resistor with resistance R, a capacitor with capacitance C,
and an inductor with inductance Lconnected in series to an alternating current
(AC) voltage source. The voltage across the resistor is VR= 5 sin(100t)V, the
voltage across the capacitor is VC= 3 sin(100t+π
4)V, and the voltage across
the inductor is VL= 4 sin(100t−π
3)V. Given that the current in the circuit is
I= 2 sin(100t+π
6)A, determine the values of R,L, and Cin the circuit.
Solution
Step 1: Write down the phasor representation for the voltages and current in
the circuit. The phasor representation of a sinusoidal function Asin(ωt +ϕ)is
ˆ
A=A
√2∠ϕ. For VR= 5 sin(100t)V: ˆ
VR=5
√2∠0◦V. For VC= 3 sin(100t+π
4)
V: ˆ
VC=3
√2∠45◦V. For VL= 4 sin(100t−π
3)V: ˆ
VL=4
√2∠−60◦V. For
I= 2 sin(100t+π
6)A: ˆ
I=2
√2∠30◦A.
Step 2: Apply Kirchhoff’s voltage law (KVL) to the circuit:
ˆ
VR+ˆ
VL+ˆ
VC=ˆ
Vsource
Substitute in the phasor representations:
5
√2
∠0◦+4
√2
∠−60◦+3
√2
∠45◦=ˆ
Vsource
5∠0◦+ 4∠−60◦+ 3∠45◦=ˆ
Vsource
Step 3: Calculate the phasor representation of the source voltage. Since the
source voltage is not explicitly given, we can represent it as ˆ
Vsource =Vsource∠0◦.
Now solve for Vsource:
5∠0◦+ 4∠−60◦+ 3∠45◦=Vsource∠0◦
Vsource∠0◦= 5 + 4 cos(60◦) + 3 cos(−45◦) + j(4 sin(−60◦) + 3 sin(45◦))
Vsource = 5 + 2 + 3√2
2+j(−2√3 + 3√2
2)
Vsource = 7 + 3√2
2−2√3 + j3√2
2−2√3
Step 4: Compare the phasor representation of the source voltage with the
expression for the impedance of the circuit Z=R+j(ωL−1
ωC ). The impedance
of the circuit must be equal to the phasor representation of the source voltage.
Matching real and imaginary parts, we have: Real parts: R= 7 Imaginary
parts: ωL −1
ωC =3√2
2−2
5
Question 8
Question
A circuit consists of a resistor with resistance R, an inductor with inductance
L, and a capacitor with capacitance Cconnected in series. The voltage across
the circuit is given by V(t) = V0cos(ωt), where V0= 12 V and ω= 50 s−1.
The resistor has a resistance of R= 5 Ω, the inductor has an inductance of
L= 0.1H, and the capacitor has a capacitance of C= 0.01 F. Determine the
current as a function of time for this circuit.
Solution
Step 1: First, we need to find the expression for the total impedance of the
circuit. The impedance ZRof the resistor is simply its resistance R. The
impedance ZLof the inductor is given by ZL=jωL. The impedance ZCof the
capacitor is given by ZC=1
jωC . The total impedance Ztotal of the series circuit
is the sum of the impedances of the resistor, inductor, and capacitor:
Ztotal =R+ZL+ZC=R+jωL +1
jωC
Step2: Next, we can find the current I(t)as a function of time. Since the
frequency of the voltage source is ω= 50 s−1, we can express the current as:
I(t) = V0
Ztotal
cos(ωt)
Step 3: Now, substitute the given values into the expression for total impedance
to get:
Ztotal = 5 + j×50 ×0.1 + 1
j×50 ×0.01
Step 4: Simplify the expression for the total impedance:
Ztotal = 5 + j5 + 1
j0.5= 5 + j5 + j2 = 5 + j7
Step 5: Substitute the total impedance into the expression for current:
I(t) = 12
5 + j7cos(50t)
Therefore, the current as a function of time for the given series circuit is:
I(t) = 12
5 + j7cos(50t)
6
Question 9
Question
A circuit consists of a resistor with resistance R= 10 Ω, a capacitor with capaci-
tance C= 2 µF , and an inductor with inductance L= 0.5Hconnected in series
to a sinusoidal voltage source with frequency ω= 100 rad/s. If the maximum
voltage supplied by the source is Vmax = 20 V, what is the maximum current in
the circuit?
Solution
Step 1: Calculate the impedance of each component. The impedance of a
resistor is given by ZR=R. The impedance of a capacitor is given by ZC=1
jωC .
The impedance of an inductor is given by ZL=jωL.
Step 2: Calculate the total impedance of the circuit. The total impedance
Ztotal of components in series is the sum of their individual impedances.
Ztotal =ZR+ZC+ZL=R+1
jωC +jωL
Step 3: Convert the total impedance to polar form. To convert Ztotal to
polar form, we need to find the magnitude and phase angle. The magnitude
|Ztotal|=√Re(Ztotal)2+Im(Ztotal)2. The phase angle θ= arctan (Im(Ztotal )
Re(Ztotal )).
Step 4: Calculate the maximum current in the circuit. The maximum current
Imax is given by Imax =Vmax
|Ztotal |.
Substitute the given values into the expressions above to find Imax.
Question 10
Question
A circuit consists of a 12V battery connected to three resistors in series. The
first resistor has a resistance of 4Ω, the second resistor has a resistance of 8Ω,
and the third resistor has a resistance of R3Ω. If the current through the circuit
is 1.5A, find the value of R3.
Solution
Step 1: Calculate the total resistance of the circuit. The total resistance Rtotal
of resistors in series is the sum of individual resistances:
Rtotal = 4Ω + 8Ω + R3Ω = 12Ω + R3Ω
Step 2: Apply Ohm’s Law to find the total resistance. Ohm’s Law states
that V=IR, where Vis the voltage, Iis the current, and Ris the resistance.
7
In this case, V= 12V and I= 1.5A. Therefore, the total resistance can be
calculated as:
12V= 1.5A×(12Ω + R3Ω)
12V = 18Ω+1.5R3Ω
Step 3: Solve for R3. Subtract 18Ωfrom both sides of the equation:
12V−18Ω = 1.5R3Ω
−6V= 1.5R3Ω
R3=−6V
1.5Ω =−4Ω
Therefore, the value of R3is 4Ω.
Question 11
Question
A resistor with resistance R1= 10 Ω is connected in series with another resistor
with resistance R2= 20 Ω. A potential difference of V= 120 Vis applied across
the combination. Calculate the current passing through each resistor.
Solution
Step 1: Calculate the total resistance of the combination. The total resistance
Rtotal of resistors in series is given by:
Rtotal =R1+R2= 10 Ω + 20 Ω = 30 Ω
Step 2: Calculate the total current passing through the combination using
Ohm’s Law V=IR. The total current Itotal passing through the combination
is:
Itotal =V
Rtotal
=120 V
30 Ω = 4 A
Step 3: Calculate the current passing through each resistor using the current
divider rule. The current passing through R1, denoted as I1, is:
I1=Itotal ×Rtotal
R1
= 4 A×30 Ω
10 Ω = 12 A
The current passing through R2, denoted as I2, is:
I2=Itotal ×Rtotal
R2
= 4 A×30 Ω
20 Ω = 6 A
Thus, the current passing through R1is 12 Aand the current passing through
R2is 6A.
8
Question 12
Question
A resistor with a resistance of 4 Ω is connected to a battery that delivers a
current of 2A. Find the power dissipated in the resistor.
Solution
Step 1: Recall that the power dissipated in a resistor can be calculated using
the formula P=I2R, where Pis the power, Iis the current, and Ris the
resistance.
Step 2: Given that the resistance R= 4 Ω and the current I= 2 A, we can
plug these values into the formula:
P= (2 A)2×4 Ω
Step 3: Simplify the expression:
P= 4 A2×4 Ω = 16 W
Step 4: Therefore, the power dissipated in the resistor is 16 W .
Question 13
Question
A copper wire has a resistance of 5 ohms. If a current of 2 amperes flows through
the wire, find the voltage across the wire.
Solution
Step 1: Recall Ohm’s Law, which states that voltage (V) is equal to current (I)
multiplied by resistance (R). The formula can be written as:
V=I×R
Step 2: Given that the resistance Ris 5 ohms and the current Iis 2 amperes,
we can substitute these values into Ohm’s Law to find the voltage V:
V= 2 A×5 Ω
Step 3: Calculate the voltage across the wire:
V= 10 V
Therefore, the voltage across the wire is 10 volts.
9
Question 14
Question
A circuit consists of a resistor with resistance R, an inductor with inductance L,
and a capacitor with capacitance Cconnected in series to an AC voltage source
with voltage V=V0sin(ωt). The circuit reaches a steady state, and the voltage
across the resistor is VR=VR0sin(ωt −ϕ). Calculate the phase difference ϕ
between the voltage across the resistor and the applied voltage.
Solution
Step 1: Write down the expression for the total impedance Zof the circuit in
terms of R,L, and C. The total impedance of the circuit is given by:
Z=R+jωL −j
ωC
Step 2: Express the current Ithrough the circuit in terms of the total
impedance Zand the applied voltage V. Using Ohm’s Law in the form V=IZ,
we have:
I=V
Z=V0sin(ωt)
Z
Step 3: Write down the expression for the voltage across the resistor VRin
terms of the current Iand the resistor R. The voltage across the resistor is
given by:
VR=IR =I·R
Step 4: Express the phase difference ϕbetween VRand Vin terms of the
impedance Z. Comparing the expressions for VRand V:
VR0sin(ωt −ϕ) = I·R=V0sin(ωt)
Z·R
⇒VR0=V0R
|Z|and ϕ=Arg(Z)
Step 5: Calculate the magnitude of Zand its argument, then find the phase
difference ϕ. Using the expression for Zfrom Step 1:
|Z|=√R2+(ωL −1
ωC )2
Arg(Z) = tan−1(ωL −1
ωC
R)
Therefore, the phase difference ϕis:
ϕ= tan−1(ωL −1
ωC
R)
10
Question 15
Question
A circuit consists of a resistor, an inductor, and a capacitor connected in series to
an AC voltage source. The values of the components are R= 20 Ω,L= 0.05 H,
and C= 1 µF. The voltage source has an amplitude of V0= 10 V and operates
at a frequency of f= 100 Hz. Calculate the impedance of the circuit and the
phase angle between the current and the voltage.
Solution
Step 1: Calculate the angular frequency ω. Given that f= 100 Hz, we can find
the angular frequency using the formula ω= 2πf . Thus,
ω= 2π×100 = 200πrad/s.
Step 2: Calculate the reactance of the inductor and the capacitor. The
reactance of an inductor XLis given by XL=ωL, and the reactance of a
capacitor XCis given by XC=1
ωC . Substituting the given values,
XL= 200π×0.05 = 10πΩand XC=1
200π×1×10−6= 5 ×103πΩ.
Step 3: Calculate the impedance of the circuit. The impedance of the circuit
Zis the vector sum of the resistive, inductive, and capacitive impedance. Since
the components are in series, we have
Z=R+XL−XC= 20+10π−5×103π= 10π−5×103π+20 = (10−5×103)π+20 Ω = −4990π+20 Ω.
Step 4: Calculate the phase angle ϕbetween the current and voltage. The
phase angle ϕis given by
ϕ= arctan (XL−XC
R).
Plugging in the values,
ϕ= arctan (10π−5×103π
20 )= arctan(−2490π/10) = arctan(−249π).
Question 16
Question
A circuit contains a resistor with resistance 10 Ω, an inductor with inductance
0.02 H, and a capacitor with capacitance 8 µF connected in series to a 12 V
battery. Calculate the impedance of the circuit and the current passing through
the circuit.
11
Solution
Step 1: Calculate the total impedance of the circuit. The impedance (Z) of an
LRC series circuit is given by:
Z=√R2+ (XL−XC)2
where: R= resistance = 10 Ω,XL= inductive reactance = ωL,XC= capacitive
reactance = 1
ωC ,ω= angular frequency = 2π
T.
First, calculate the angular frequency:
ω=2π
T
Given that the frequency f=1
T= 60 Hz. Therefore, ω= 2π×60 = 120π
rad/s.
Next, calculate XLand XC:
XL=ωL = 120π×0.02 = 2.4πΩ
XC=1
ωC =1
120π×8×10−6=106
960Ω = 625
6πΩ
Now substitute these values into the impedance formula:
Z=√102+ (2.4π−625
6π)2
Z=√100 + (2.4π−625
6π)2
Step 2: Calculate the current passing through the circuit. The current (I)
flowing through the circuit is given by Ohm’s law:
I=V
Z
where V= voltage from the battery = 12 V.
Now, substitute the calculated impedance value:
I=12
Z
Now, you can calculate the impedance (Z) and then find the current passing
through the circuit.
Question 17
Question
A student sets up a simple circuit with a resistor, battery, and an ammeter.
The ammeter reads a current of 0.5 A when the voltage across the resistor is 6
V. The student then decides to double the resistance in the circuit and observes
that the current reading on the ammeter drops to 0.25 A. Calculate the original
resistance in the circuit and the new resistance after doubling it.
12
Solution
Step 1: Let’s denote the original resistance as Rand the new resistance after
doubling it as 2R.
Step 2: Using Ohm’s Law V=IR, where Vis the voltage, Iis the current,
and Ris the resistance, we can set up two equations:
For the original circuit:
6 = 0.5×R
Simplifying, we get:
R=6
0.5= 12 Ω
Step 3: For the circuit with doubled resistance:
6 = 0.25 ×2R
Simplifying, we get:
2R=6
0.25 = 24 Ω
Therefore, the new resistance 2Ris 24 Ωafter doubling it.
Step 4: The original resistance was found to be 12 Ω. Thus, the original
resistance in the circuit is 12 Ω, and the new resistance after doubling it is 24
Ω.
Question 18
Question
A circuit consists of a 12 V battery connected in series with a resistor and an
unknown device. When a current of 2 A flows through the circuit, the potential
drop across the resistor is 4 V. Determine the resistance of the unknown device.
Solution
Step 1: Recall Ohm’s Law, which states that the potential difference (V) across a
resistor is equal to the current (I) flowing through it multiplied by the resistance
(R) of the resistor. Mathematically, this can be expressed as V=I·R.
Step 2: From the problem, we are given that the potential drop across the
resistor is 4 V when a current of 2 A flows through the circuit. Therefore, we
have 4V= 2 A×R.
Step 3: To find the resistance Rof the unknown device, we can rearrange
the formula V=I·Rto solve for R. So, R=V
I.
Step 4: Substituting the given values, we find R=4V
2A= 2 Ω.
Step 5: Hence, the resistance of the unknown device in the circuit is 2 ohms.
13
Question 19
Question
A circuit consists of a 12 V battery, a resistor with resistance R, and a capacitor
with capacitance C. Initially, the capacitor is uncharged and acts like a short
circuit. When the circuit is closed, the current through the circuit is found to be
0.5A. After some time, the capacitor is fully charged and the current through
the circuit is found to be 0.1A. Calculate the resistance Rof the resistor in the
circuit.
Solution
1. Let’s denote the resistance of the resistor as R, the capacitance of the capac-
itor as C, and the potential difference across the resistor as VR. The potential
difference across the capacitor is VC= 12 V.
2. When the circuit is first closed and the capacitor acts like a short circuit,
the current in the circuit is equal to Iinitial = 0.5A. This current can be split
between the resistor and capacitor according to the equivalent resistance Req =
Rin series with the capacitor. Since the capacitor acts like a short circuit, the
full potential difference of 12 V is across the resistor:
VR= 12 V=R×Iinitial
R=VR
Iinitial
=12
0.5= 24Ω
3. When the capacitor is fully charged and the current through the circuit
is Ifinal = 0.1A, the capacitor acts like an open circuit. Now, the potential
difference VCis fully across the resistor. We can write an expression for the
potential difference VRin terms of Rand Cusing Ohm’s Law and the formula
for the charging of a capacitor:
VR=R×Ifinal =VC(1 −e−t
RC )
Since the capacitor is fully charged, VR=VC= 12 V and Ifinal = 0.1A.
Substituting these values into the equation gives:
R×0.1 = 12(1 −e−t
RC )
R=12(1 −e−t
RC )
0.1
4. Since we are looking for the resistance R, we need to find the time constant
τ=RC. We can find τby noting that in a fully charged circuit, e−t
RC ≈0.
Therefore:
R=12(1 −0)
0.1= 120Ω
5. Therefore, the resistance Rof the resistor in the circuit is 120 Ω .
14
Question 20
Question
A 10 V battery is connected to a resistor of unknown resistance. When a current
of 2 A flows through the resistor, the power dissipated is 20 W. Calculate the
resistance of the resistor.
Solution
Step 1: We can start by using the formula for power in a resistor: P=I2·R,
where Pis the power dissipated, Iis the current flowing through the resistor,
and Ris the resistance. Given that P= 20 W and I= 2 A, we can rearrange
the formula to solve for R:
R=P
I2
R=20
(2)2
R=20
4
R= 5 Ω
Therefore, the resistance of the resistor is 5 Ω.
Question 21
Question
A circuit consists of a 12 V battery, a 4 Ωresistor, and an unknown resistor
R. When the total resistance in the circuit is 6 Ω, the power dissipated in the
unknown resistor is 24 W. Find the value of R.
Solution
Step 1: Let’s first calculate the current, I, flowing through the circuit using
Ohm’s Law, V=IR, where Vis the voltage of the battery.
Given: V= 12 V, R1= 4 Ω, Rtotal = 6 Ω
Since Rtotal =R1+R, we can find Ras R=Rtotal −R1
R= 6 −4 = 2 Ω
Step 2: Next, we find the current, I, as I=V
Rtotal .
I=12
6= 2 A
15
Step 3: Now, we can find the power, P, dissipated by the unknown resistor
using the formula P=I2R.
Given: P= 24 W, R= 2 Ω
24 = (2)2×2
24 = 4 ×2
24 = 8
This equation is not met, so there must be an error in the problem statement
or calculation.
Question 22
Question
A circuit consists of a resistor with resistance R= 100 Ω connected to a battery
with emf E= 12 V. The circuit also has an ammeter connected in series and
a voltmeter connected in parallel to the resistor. If the ammeter reads 0.1A,
what reading will the voltmeter display?
Solution
Step 1: Recall Ohm’s Law which states that the current passing through a
resistor is directly proportional to the potential difference across the resistor.
The formula for Ohm’s Law is V=IR, where Vis the voltage (potential
difference) across the resistor, Iis the current passing through the resistor, and
Ris the resistance of the resistor.
Step 2: Given in the question, R= 100 Ω and I= 0.1A. We can use Ohm’s
Law to find the voltage across the resistor: V=I·R.
Step 3: Substituting the given values into the formula, we get V= 0.1A·
100 Ω = 10 V.
Step 4: Since the voltmeter is connected in parallel to the resistor, it will
read the same voltage as the resistor. Therefore, the voltmeter will display
10 V .
Question 23
Question
A circuit consists of a resistor with resistance R= 15 Ω and an inductor with
inductance L= 0.04 H. The circuit is connected to a voltage source with an
emf of V= 20 V and a frequency of f= 50 Hz. Calculate the current flowing
through the circuit.
16
Solution
Step 1: Calculate the reactance of the inductor using the formula XL= 2πf L.
XL= 2π×50 ×0.04 = 4πΩ
Step 2: Calculate the total impedance of the circuit using the formula Z=
√R2+X2
L.
Z=√152+ (4π)2≈√225 + 39.48 ≈√264.48 ≈16.27 Ω
Step 3: Calculate the current flowing through the circuit using Ohm’s Law:
I=V
Z.
I=20
16.27 ≈1.23 A
Therefore, the current flowing through the circuit is approximately 1.23 A.
Question 24
Question
A circuit consists of a resistor with resistance R, an inductor with inductance
L, a capacitor with capacitance C, and an AC voltage source V=V0sin(ωt).
The instantaneous current i(t)through the circuit is given by the equation:
i(t) = i0sin(ωt +ϕ)
where i0= 3 A, ω= 50 rad/s, and ϕ=π
6rad.
Determine the voltage amplitude V0applied by the source.
Solution
Step 1: Ohm’s Law states that for a circuit, the voltage Vapplied across a com-
ponent is equal to the product of the current Iflowing through the component
and the resistance Rof the component.
V=IR
For the given circuit, the voltage across the resistor is VR=i(t)·R. Since the
current through the resistor is i(t)and the resistance of the resistor is R, we
have:
VR=i(t)·R
Step 2: In an AC circuit with a resistor, inductor, and capacitor in series,
the total voltage across the components will be the same as the voltage of the
AC source. Therefore, the voltages across the resistor, inductor, and capacitor
must sum up to the applied voltage V=V0sin(ωt).
V=VR+VL+VC
17
Now, let’s find the expressions for VLand VC.
Step 3: The voltage across an inductor in an AC circuit is given by:
VL=Ldi
dt
where Lis the inductance of the inductor. Since i(t) = i0sin(ωt +ϕ), we can
find di
dt by differentiating i(t)with respect to t.
di
dt =d
dt(i0sin(ωt +ϕ)) = i0ωcos(ωt +ϕ)
Step 4: Substituting into the expression for VL, we get:
VL=Ldi
dt =Li0ωcos(ωt +ϕ)
Step 5: The voltage across a capacitor in an AC circuit is given by:
VC=1
C∫i dt
where Cis the capacitance of the capacitor. Since i(t) = i0sin(ωt +ϕ), we can
find ∫i dt by integrating i(t)with respect to t.
∫i dt =∫i0sin(ωt +ϕ)dt =−i0
ωcos(ωt +ϕ)
Step 6: Substituting into the expression for VC, we have:
VC=1
C∫i dt =−i0
ωC cos(ωt +ϕ)
Step 7: Now, substituting VR,VL, and VCinto the equation V=VR+VL+
VC, we get:
V0sin(ωt) = i0Rsin(ωt +ϕ) + Li0ωcos(ωt +ϕ)−i0
ωC cos(ωt +ϕ)
Step 8: We know that the voltage amplitudes of sin(ωt),cos(ωt), and cos(ωt)
terms on both sides of the equation must be equal. So, the coefficient of sin(ωt)
term on the right side must be V0. Thus, we can determine V0by equating
coefficients:
V0=i0R
V0= 3 A×R
Question 25
Question
A circuit consists of a resistor with resistance R, an inductor with induc-
tance L, and a capacitor with capacitance Cconnected in series to an AC
voltage source with frequency f. The impedance of the circuit is given by
Z=√R2+(ωL −1
ωC )2, where ω= 2πf. Determine the angular frequency ω
at which the impedance of the circuit is minimized.
18
Solution
Step 1: To find the angular frequency ωat which the impedance of the circuit is
minimized, we need to find the minimum value of the expression for impedance
Z. This minimum occurs when the derivative of Zwith respect to ωis zero.
Step 2: Calculate the derivative of Zwith respect to ω:
dZ
dω =1
2√R2+(ωL −1
ωC )2·2·(ωL −1
ωC )(L−1
ω2C)
Step 3: Set the derivative equal to zero and solve for ω:
ωL −1
ωC
√R2+(ωL −1
ωC )2= 0
Step 4: Simplify the equation and solve for ω:
ωL −1
ωC = 0
ω2=1
LC
ω=√1
LC
Therefore, the angular frequency ωat which the impedance of the circuit is
minimized is ω=√1
LC .
19
Solution
Step 1: Given that the initial resistance of the copper wire is 8 Ω at 20◦C, we
want to find the resistance at 80◦C. The formula relating resistance, initial
resistance, and temperature change is:
RT=R0[1 + α(T−T0)]
where RTis the resistance at temperature T,R0is the resistance at reference
temperature T0,αis the temperature coefficient of resistivity, Tis the final
temperature, and T0is the reference temperature.
Step 2: Plugging in the given values into the equation, we get:
R80 = 8 Ω[1 + 0.00428 ×(80 −20)]
Step 3: Calculating the expression inside the brackets first:
1+0.00428 ×60 = 1 + 0.2568 = 1.2568
Step 4: Now, calculate the resistance at 80◦C:
R80 = 8 Ω ×1.2568 = 10.0544 Ω
Therefore, the resistance of the copper wire at 80◦C is 10.0544 Ω.
Question 3
Question
A circuit consists of a 10 V battery connected in series to a resistor. If the
current in the circuit is 2 A, what is the resistance of the resistor?
Solution
Step 1: Recall Ohm’s Law, which states that the voltage across a resistor is equal
to the current flowing through it multiplied by the resistance. Mathematically,
this can be written as V=IR, where Vis the voltage, Iis the current, and R
is the resistance.
Step 2: In this case, the voltage across the resistor is 10 V and the cur-
rent flowing through it is 2 A. We can rearrange Ohm’s Law to solve for the
resistance: R=V
I.
Step 3: Substitute the given values into the formula: R=10 V
2A.
Step 4: Perform the division to find the resistance: R= 5 Ω.
Step 5: Therefore, the resistance of the resistor in the circuit is 5 Ohms.
2
Question 4
Question
A circuit consists of a battery with voltage V= 12 V and three resistors con-
nected in series. The resistors have resistances R1= 5 ohms, R2= 10 ohms,
and R3= 15 ohms. Calculate the total current flowing through the circuit and
the voltage drop across each resistor.
Solution
Step 1: Calculate the total resistance of the circuit. The total resistance Rtotal
of resistors in series is the sum of the individual resistances:
Rtotal =R1+R2+R3= 5 Ω + 10 Ω + 15 Ω = 30 Ω
Step 2: Calculate the total current flowing through the circuit using Ohm’s
Law V=IR. The total current Iis given by I=V
Rtotal .
I=12 V
30 Ω = 0.4A
Step 3: Calculate the voltage drop across each resistor. The voltage drop Vi
across a resistor Riis given by Ohm’s Law Vi=IRi. For R1:
V1=I·R1= 0.4A·5 Ω = 2 V
For R2:
V2=I·R2= 0.4A·10 Ω = 4 V
For R3:
V3=I·R3= 0.4A·15 Ω = 6 V
Therefore, the total current flowing through the circuit is 0.4 A, and the
voltage drop across R1,R2, and R3are 2 V, 4 V, and 6 V, respectively.
Question 5
Question
A circuit consists of a resistor with resistance R= 15 Ω connected to a battery
with emf ε= 12 V. Calculate the current flowing through the resistor.
Solution
Step 1: Determine the relationship between voltage, current, and resistance in a
circuit. According to Ohm’s Law, the relationship between voltage (V), current
(I), and resistance (R) in a circuit is given by:
V=IR
3
where V= voltage (in volts), I= current (in amperes), R= resistance (in
ohms).
Step 2: Identify the given values. The resistance of the resistor, R= 15 Ω,
and the emf of the battery, ε= 12 V.
Step 3: Apply Ohm’s Law to find the current. Substitute the given values
into Ohm’s Law:
ε=IR
12 = I×15
I=12
15
I= 0.8A
Step 4: State the final answer. The current flowing through the resistor is
0.8amperes.
Question 6
Question
A resistor is connected to a 12 V battery, and a current of 3 A flows through it.
If the resistance of the resistor is R, calculate the resistance R.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor
is equal to the current (I) flowing through it multiplied by the resistance (R)
of the resistor. Mathematically, this can be written as:
V=I×R
Step 2: In this case, the voltage across the resistor is 12V, the current flowing
through it is 3A, and we are trying to find the resistance, R. We can substitute
these values into Ohm’s Law:
12 = 3 ×R
Step 3: Now, we solve for the resistance R:
R=12
3= 4 Ω
Step 4: Therefore, the resistance of the resistor is 4 Ω.
4
Question 7
Question
A circuit consists of a resistor with resistance R, a capacitor with capacitance C,
and an inductor with inductance Lconnected in series to an alternating current
(AC) voltage source. The voltage across the resistor is VR= 5 sin(100t)V, the
voltage across the capacitor is VC= 3 sin(100t+π
4)V, and the voltage across
the inductor is VL= 4 sin(100t−π
3)V. Given that the current in the circuit is
I= 2 sin(100t+π
6)A, determine the values of R,L, and Cin the circuit.
Solution
Step 1: Write down the phasor representation for the voltages and current in
the circuit. The phasor representation of a sinusoidal function Asin(ωt +ϕ)is
ˆ
A=A
√2∠ϕ. For VR= 5 sin(100t)V: ˆ
VR=5
√2∠0◦V. For VC= 3 sin(100t+π
4)
V: ˆ
VC=3
√2∠45◦V. For VL= 4 sin(100t−π
3)V: ˆ
VL=4
√2∠−60◦V. For
I= 2 sin(100t+π
6)A: ˆ
I=2
√2∠30◦A.
Step 2: Apply Kirchhoff’s voltage law (KVL) to the circuit:
ˆ
VR+ˆ
VL+ˆ
VC=ˆ
Vsource
Substitute in the phasor representations:
5
√2
∠0◦+4
√2
∠−60◦+3
√2
∠45◦=ˆ
Vsource
5∠0◦+ 4∠−60◦+ 3∠45◦=ˆ
Vsource
Step 3: Calculate the phasor representation of the source voltage. Since the
source voltage is not explicitly given, we can represent it as ˆ
Vsource =Vsource∠0◦.
Now solve for Vsource:
5∠0◦+ 4∠−60◦+ 3∠45◦=Vsource∠0◦
Vsource∠0◦= 5 + 4 cos(60◦) + 3 cos(−45◦) + j(4 sin(−60◦) + 3 sin(45◦))
Vsource = 5 + 2 + 3√2
2+j(−2√3 + 3√2
2)
Vsource = 7 + 3√2
2−2√3 + j3√2
2−2√3
Step 4: Compare the phasor representation of the source voltage with the
expression for the impedance of the circuit Z=R+j(ωL−1
ωC ). The impedance
of the circuit must be equal to the phasor representation of the source voltage.
Matching real and imaginary parts, we have: Real parts: R= 7 Imaginary
parts: ωL −1
ωC =3√2
2−2
5
Question 8
Question
A circuit consists of a resistor with resistance R, an inductor with inductance
L, and a capacitor with capacitance Cconnected in series. The voltage across
the circuit is given by V(t) = V0cos(ωt), where V0= 12 V and ω= 50 s−1.
The resistor has a resistance of R= 5 Ω, the inductor has an inductance of
L= 0.1H, and the capacitor has a capacitance of C= 0.01 F. Determine the
current as a function of time for this circuit.
Solution
Step 1: First, we need to find the expression for the total impedance of the
circuit. The impedance ZRof the resistor is simply its resistance R. The
impedance ZLof the inductor is given by ZL=jωL. The impedance ZCof the
capacitor is given by ZC=1
jωC . The total impedance Ztotal of the series circuit
is the sum of the impedances of the resistor, inductor, and capacitor:
Ztotal =R+ZL+ZC=R+jωL +1
jωC
Step2: Next, we can find the current I(t)as a function of time. Since the
frequency of the voltage source is ω= 50 s−1, we can express the current as:
I(t) = V0
Ztotal
cos(ωt)
Step 3: Now, substitute the given values into the expression for total impedance
to get:
Ztotal = 5 + j×50 ×0.1 + 1
j×50 ×0.01
Step 4: Simplify the expression for the total impedance:
Ztotal = 5 + j5 + 1
j0.5= 5 + j5 + j2 = 5 + j7
Step 5: Substitute the total impedance into the expression for current:
I(t) = 12
5 + j7cos(50t)
Therefore, the current as a function of time for the given series circuit is:
I(t) = 12
5 + j7cos(50t)
6
Question 9
Question
A circuit consists of a resistor with resistance R= 10 Ω, a capacitor with capaci-
tance C= 2 µF , and an inductor with inductance L= 0.5Hconnected in series
to a sinusoidal voltage source with frequency ω= 100 rad/s. If the maximum
voltage supplied by the source is Vmax = 20 V, what is the maximum current in
the circuit?
Solution
Step 1: Calculate the impedance of each component. The impedance of a
resistor is given by ZR=R. The impedance of a capacitor is given by ZC=1
jωC .
The impedance of an inductor is given by ZL=jωL.
Step 2: Calculate the total impedance of the circuit. The total impedance
Ztotal of components in series is the sum of their individual impedances.
Ztotal =ZR+ZC+ZL=R+1
jωC +jωL
Step 3: Convert the total impedance to polar form. To convert Ztotal to
polar form, we need to find the magnitude and phase angle. The magnitude
|Ztotal|=√Re(Ztotal)2+Im(Ztotal)2. The phase angle θ= arctan (Im(Ztotal )
Re(Ztotal )).
Step 4: Calculate the maximum current in the circuit. The maximum current
Imax is given by Imax =Vmax
|Ztotal |.
Substitute the given values into the expressions above to find Imax.
Question 10
Question
A circuit consists of a 12V battery connected to three resistors in series. The
first resistor has a resistance of 4Ω, the second resistor has a resistance of 8Ω,
and the third resistor has a resistance of R3Ω. If the current through the circuit
is 1.5A, find the value of R3.
Solution
Step 1: Calculate the total resistance of the circuit. The total resistance Rtotal
of resistors in series is the sum of individual resistances:
Rtotal = 4Ω + 8Ω + R3Ω = 12Ω + R3Ω
Step 2: Apply Ohm’s Law to find the total resistance. Ohm’s Law states
that V=IR, where Vis the voltage, Iis the current, and Ris the resistance.
7
In this case, V= 12V and I= 1.5A. Therefore, the total resistance can be
calculated as:
12V= 1.5A×(12Ω + R3Ω)
12V = 18Ω+1.5R3Ω
Step 3: Solve for R3. Subtract 18Ωfrom both sides of the equation:
12V−18Ω = 1.5R3Ω
−6V= 1.5R3Ω
R3=−6V
1.5Ω =−4Ω
Therefore, the value of R3is 4Ω.
Question 11
Question
A resistor with resistance R1= 10 Ω is connected in series with another resistor
with resistance R2= 20 Ω. A potential difference of V= 120 Vis applied across
the combination. Calculate the current passing through each resistor.
Solution
Step 1: Calculate the total resistance of the combination. The total resistance
Rtotal of resistors in series is given by:
Rtotal =R1+R2= 10 Ω + 20 Ω = 30 Ω
Step 2: Calculate the total current passing through the combination using
Ohm’s Law V=IR. The total current Itotal passing through the combination
is:
Itotal =V
Rtotal
=120 V
30 Ω = 4 A
Step 3: Calculate the current passing through each resistor using the current
divider rule. The current passing through R1, denoted as I1, is:
I1=Itotal ×Rtotal
R1
= 4 A×30 Ω
10 Ω = 12 A
The current passing through R2, denoted as I2, is:
I2=Itotal ×Rtotal
R2
= 4 A×30 Ω
20 Ω = 6 A
Thus, the current passing through R1is 12 Aand the current passing through
R2is 6A.
8
Question 12
Question
A resistor with a resistance of 4 Ω is connected to a battery that delivers a
current of 2A. Find the power dissipated in the resistor.
Solution
Step 1: Recall that the power dissipated in a resistor can be calculated using
the formula P=I2R, where Pis the power, Iis the current, and Ris the
resistance.
Step 2: Given that the resistance R= 4 Ω and the current I= 2 A, we can
plug these values into the formula:
P= (2 A)2×4 Ω
Step 3: Simplify the expression:
P= 4 A2×4 Ω = 16 W
Step 4: Therefore, the power dissipated in the resistor is 16 W .
Question 13
Question
A copper wire has a resistance of 5 ohms. If a current of 2 amperes flows through
the wire, find the voltage across the wire.
Solution
Step 1: Recall Ohm’s Law, which states that voltage (V) is equal to current (I)
multiplied by resistance (R). The formula can be written as:
V=I×R
Step 2: Given that the resistance Ris 5 ohms and the current Iis 2 amperes,
we can substitute these values into Ohm’s Law to find the voltage V:
V= 2 A×5 Ω
Step 3: Calculate the voltage across the wire:
V= 10 V
Therefore, the voltage across the wire is 10 volts.
9
Question 14
Question
A circuit consists of a resistor with resistance R, an inductor with inductance L,
and a capacitor with capacitance Cconnected in series to an AC voltage source
with voltage V=V0sin(ωt). The circuit reaches a steady state, and the voltage
across the resistor is VR=VR0sin(ωt −ϕ). Calculate the phase difference ϕ
between the voltage across the resistor and the applied voltage.
Solution
Step 1: Write down the expression for the total impedance Zof the circuit in
terms of R,L, and C. The total impedance of the circuit is given by:
Z=R+jωL −j
ωC
Step 2: Express the current Ithrough the circuit in terms of the total
impedance Zand the applied voltage V. Using Ohm’s Law in the form V=IZ,
we have:
I=V
Z=V0sin(ωt)
Z
Step 3: Write down the expression for the voltage across the resistor VRin
terms of the current Iand the resistor R. The voltage across the resistor is
given by:
VR=IR =I·R
Step 4: Express the phase difference ϕbetween VRand Vin terms of the
impedance Z. Comparing the expressions for VRand V:
VR0sin(ωt −ϕ) = I·R=V0sin(ωt)
Z·R
⇒VR0=V0R
|Z|and ϕ=Arg(Z)
Step 5: Calculate the magnitude of Zand its argument, then find the phase
difference ϕ. Using the expression for Zfrom Step 1:
|Z|=√R2+(ωL −1
ωC )2
Arg(Z) = tan−1(ωL −1
ωC
R)
Therefore, the phase difference ϕis:
ϕ= tan−1(ωL −1
ωC
R)
10
Question 15
Question
A circuit consists of a resistor, an inductor, and a capacitor connected in series to
an AC voltage source. The values of the components are R= 20 Ω,L= 0.05 H,
and C= 1 µF. The voltage source has an amplitude of V0= 10 V and operates
at a frequency of f= 100 Hz. Calculate the impedance of the circuit and the
phase angle between the current and the voltage.
Solution
Step 1: Calculate the angular frequency ω. Given that f= 100 Hz, we can find
the angular frequency using the formula ω= 2πf . Thus,
ω= 2π×100 = 200πrad/s.
Step 2: Calculate the reactance of the inductor and the capacitor. The
reactance of an inductor XLis given by XL=ωL, and the reactance of a
capacitor XCis given by XC=1
ωC . Substituting the given values,
XL= 200π×0.05 = 10πΩand XC=1
200π×1×10−6= 5 ×103πΩ.
Step 3: Calculate the impedance of the circuit. The impedance of the circuit
Zis the vector sum of the resistive, inductive, and capacitive impedance. Since
the components are in series, we have
Z=R+XL−XC= 20+10π−5×103π= 10π−5×103π+20 = (10−5×103)π+20 Ω = −4990π+20 Ω.
Step 4: Calculate the phase angle ϕbetween the current and voltage. The
phase angle ϕis given by
ϕ= arctan (XL−XC
R).
Plugging in the values,
ϕ= arctan (10π−5×103π
20 )= arctan(−2490π/10) = arctan(−249π).
Question 16
Question
A circuit contains a resistor with resistance 10 Ω, an inductor with inductance
0.02 H, and a capacitor with capacitance 8 µF connected in series to a 12 V
battery. Calculate the impedance of the circuit and the current passing through
the circuit.
11
Solution
Step 1: Calculate the total impedance of the circuit. The impedance (Z) of an
LRC series circuit is given by:
Z=√R2+ (XL−XC)2
where: R= resistance = 10 Ω,XL= inductive reactance = ωL,XC= capacitive
reactance = 1
ωC ,ω= angular frequency = 2π
T.
First, calculate the angular frequency:
ω=2π
T
Given that the frequency f=1
T= 60 Hz. Therefore, ω= 2π×60 = 120π
rad/s.
Next, calculate XLand XC:
XL=ωL = 120π×0.02 = 2.4πΩ
XC=1
ωC =1
120π×8×10−6=106
960Ω = 625
6πΩ
Now substitute these values into the impedance formula:
Z=√102+ (2.4π−625
6π)2
Z=√100 + (2.4π−625
6π)2
Step 2: Calculate the current passing through the circuit. The current (I)
flowing through the circuit is given by Ohm’s law:
I=V
Z
where V= voltage from the battery = 12 V.
Now, substitute the calculated impedance value:
I=12
Z
Now, you can calculate the impedance (Z) and then find the current passing
through the circuit.
Question 17
Question
A student sets up a simple circuit with a resistor, battery, and an ammeter.
The ammeter reads a current of 0.5 A when the voltage across the resistor is 6
V. The student then decides to double the resistance in the circuit and observes
that the current reading on the ammeter drops to 0.25 A. Calculate the original
resistance in the circuit and the new resistance after doubling it.
12
Solution
Step 1: Let’s denote the original resistance as Rand the new resistance after
doubling it as 2R.
Step 2: Using Ohm’s Law V=IR, where Vis the voltage, Iis the current,
and Ris the resistance, we can set up two equations:
For the original circuit:
6 = 0.5×R
Simplifying, we get:
R=6
0.5= 12 Ω
Step 3: For the circuit with doubled resistance:
6 = 0.25 ×2R
Simplifying, we get:
2R=6
0.25 = 24 Ω
Therefore, the new resistance 2Ris 24 Ωafter doubling it.
Step 4: The original resistance was found to be 12 Ω. Thus, the original
resistance in the circuit is 12 Ω, and the new resistance after doubling it is 24
Ω.
Question 18
Question
A circuit consists of a 12 V battery connected in series with a resistor and an
unknown device. When a current of 2 A flows through the circuit, the potential
drop across the resistor is 4 V. Determine the resistance of the unknown device.
Solution
Step 1: Recall Ohm’s Law, which states that the potential difference (V) across a
resistor is equal to the current (I) flowing through it multiplied by the resistance
(R) of the resistor. Mathematically, this can be expressed as V=I·R.
Step 2: From the problem, we are given that the potential drop across the
resistor is 4 V when a current of 2 A flows through the circuit. Therefore, we
have 4V= 2 A×R.
Step 3: To find the resistance Rof the unknown device, we can rearrange
the formula V=I·Rto solve for R. So, R=V
I.
Step 4: Substituting the given values, we find R=4V
2A= 2 Ω.
Step 5: Hence, the resistance of the unknown device in the circuit is 2 ohms.
13
Question 19
Question
A circuit consists of a 12 V battery, a resistor with resistance R, and a capacitor
with capacitance C. Initially, the capacitor is uncharged and acts like a short
circuit. When the circuit is closed, the current through the circuit is found to be
0.5A. After some time, the capacitor is fully charged and the current through
the circuit is found to be 0.1A. Calculate the resistance Rof the resistor in the
circuit.
Solution
1. Let’s denote the resistance of the resistor as R, the capacitance of the capac-
itor as C, and the potential difference across the resistor as VR. The potential
difference across the capacitor is VC= 12 V.
2. When the circuit is first closed and the capacitor acts like a short circuit,
the current in the circuit is equal to Iinitial = 0.5A. This current can be split
between the resistor and capacitor according to the equivalent resistance Req =
Rin series with the capacitor. Since the capacitor acts like a short circuit, the
full potential difference of 12 V is across the resistor:
VR= 12 V=R×Iinitial
R=VR
Iinitial
=12
0.5= 24Ω
3. When the capacitor is fully charged and the current through the circuit
is Ifinal = 0.1A, the capacitor acts like an open circuit. Now, the potential
difference VCis fully across the resistor. We can write an expression for the
potential difference VRin terms of Rand Cusing Ohm’s Law and the formula
for the charging of a capacitor:
VR=R×Ifinal =VC(1 −e−t
RC )
Since the capacitor is fully charged, VR=VC= 12 V and Ifinal = 0.1A.
Substituting these values into the equation gives:
R×0.1 = 12(1 −e−t
RC )
R=12(1 −e−t
RC )
0.1
4. Since we are looking for the resistance R, we need to find the time constant
τ=RC. We can find τby noting that in a fully charged circuit, e−t
RC ≈0.
Therefore:
R=12(1 −0)
0.1= 120Ω
5. Therefore, the resistance Rof the resistor in the circuit is 120 Ω .
14
Question 20
Question
A 10 V battery is connected to a resistor of unknown resistance. When a current
of 2 A flows through the resistor, the power dissipated is 20 W. Calculate the
resistance of the resistor.
Solution
Step 1: We can start by using the formula for power in a resistor: P=I2·R,
where Pis the power dissipated, Iis the current flowing through the resistor,
and Ris the resistance. Given that P= 20 W and I= 2 A, we can rearrange
the formula to solve for R:
R=P
I2
R=20
(2)2
R=20
4
R= 5 Ω
Therefore, the resistance of the resistor is 5 Ω.
Question 21
Question
A circuit consists of a 12 V battery, a 4 Ωresistor, and an unknown resistor
R. When the total resistance in the circuit is 6 Ω, the power dissipated in the
unknown resistor is 24 W. Find the value of R.
Solution
Step 1: Let’s first calculate the current, I, flowing through the circuit using
Ohm’s Law, V=IR, where Vis the voltage of the battery.
Given: V= 12 V, R1= 4 Ω, Rtotal = 6 Ω
Since Rtotal =R1+R, we can find Ras R=Rtotal −R1
R= 6 −4 = 2 Ω
Step 2: Next, we find the current, I, as I=V
Rtotal .
I=12
6= 2 A
15
Step 3: Now, we can find the power, P, dissipated by the unknown resistor
using the formula P=I2R.
Given: P= 24 W, R= 2 Ω
24 = (2)2×2
24 = 4 ×2
24 = 8
This equation is not met, so there must be an error in the problem statement
or calculation.
Question 22
Question
A circuit consists of a resistor with resistance R= 100 Ω connected to a battery
with emf E= 12 V. The circuit also has an ammeter connected in series and
a voltmeter connected in parallel to the resistor. If the ammeter reads 0.1A,
what reading will the voltmeter display?
Solution
Step 1: Recall Ohm’s Law which states that the current passing through a
resistor is directly proportional to the potential difference across the resistor.
The formula for Ohm’s Law is V=IR, where Vis the voltage (potential
difference) across the resistor, Iis the current passing through the resistor, and
Ris the resistance of the resistor.
Step 2: Given in the question, R= 100 Ω and I= 0.1A. We can use Ohm’s
Law to find the voltage across the resistor: V=I·R.
Step 3: Substituting the given values into the formula, we get V= 0.1A·
100 Ω = 10 V.
Step 4: Since the voltmeter is connected in parallel to the resistor, it will
read the same voltage as the resistor. Therefore, the voltmeter will display
10 V .
Question 23
Question
A circuit consists of a resistor with resistance R= 15 Ω and an inductor with
inductance L= 0.04 H. The circuit is connected to a voltage source with an
emf of V= 20 V and a frequency of f= 50 Hz. Calculate the current flowing
through the circuit.
16
Solution
Step 1: Calculate the reactance of the inductor using the formula XL= 2πf L.
XL= 2π×50 ×0.04 = 4πΩ
Step 2: Calculate the total impedance of the circuit using the formula Z=
√R2+X2
L.
Z=√152+ (4π)2≈√225 + 39.48 ≈√264.48 ≈16.27 Ω
Step 3: Calculate the current flowing through the circuit using Ohm’s Law:
I=V
Z.
I=20
16.27 ≈1.23 A
Therefore, the current flowing through the circuit is approximately 1.23 A.
Question 24
Question
A circuit consists of a resistor with resistance R, an inductor with inductance
L, a capacitor with capacitance C, and an AC voltage source V=V0sin(ωt).
The instantaneous current i(t)through the circuit is given by the equation:
i(t) = i0sin(ωt +ϕ)
where i0= 3 A, ω= 50 rad/s, and ϕ=π
6rad.
Determine the voltage amplitude V0applied by the source.
Solution
Step 1: Ohm’s Law states that for a circuit, the voltage Vapplied across a com-
ponent is equal to the product of the current Iflowing through the component
and the resistance Rof the component.
V=IR
For the given circuit, the voltage across the resistor is VR=i(t)·R. Since the
current through the resistor is i(t)and the resistance of the resistor is R, we
have:
VR=i(t)·R
Step 2: In an AC circuit with a resistor, inductor, and capacitor in series,
the total voltage across the components will be the same as the voltage of the
AC source. Therefore, the voltages across the resistor, inductor, and capacitor
must sum up to the applied voltage V=V0sin(ωt).
V=VR+VL+VC
17
Now, let’s find the expressions for VLand VC.
Step 3: The voltage across an inductor in an AC circuit is given by:
VL=Ldi
dt
where Lis the inductance of the inductor. Since i(t) = i0sin(ωt +ϕ), we can
find di
dt by differentiating i(t)with respect to t.
di
dt =d
dt(i0sin(ωt +ϕ)) = i0ωcos(ωt +ϕ)
Step 4: Substituting into the expression for VL, we get:
VL=Ldi
dt =Li0ωcos(ωt +ϕ)
Step 5: The voltage across a capacitor in an AC circuit is given by:
VC=1
C∫i dt
where Cis the capacitance of the capacitor. Since i(t) = i0sin(ωt +ϕ), we can
find ∫i dt by integrating i(t)with respect to t.
∫i dt =∫i0sin(ωt +ϕ)dt =−i0
ωcos(ωt +ϕ)
Step 6: Substituting into the expression for VC, we have:
VC=1
C∫i dt =−i0
ωC cos(ωt +ϕ)
Step 7: Now, substituting VR,VL, and VCinto the equation V=VR+VL+
VC, we get:
V0sin(ωt) = i0Rsin(ωt +ϕ) + Li0ωcos(ωt +ϕ)−i0
ωC cos(ωt +ϕ)
Step 8: We know that the voltage amplitudes of sin(ωt),cos(ωt), and cos(ωt)
terms on both sides of the equation must be equal. So, the coefficient of sin(ωt)
term on the right side must be V0. Thus, we can determine V0by equating
coefficients:
V0=i0R
V0= 3 A×R
Question 25
Question
A circuit consists of a resistor with resistance R, an inductor with induc-
tance L, and a capacitor with capacitance Cconnected in series to an AC
voltage source with frequency f. The impedance of the circuit is given by
Z=√R2+(ωL −1
ωC )2, where ω= 2πf. Determine the angular frequency ω
at which the impedance of the circuit is minimized.
18
Solution
Step 1: To find the angular frequency ωat which the impedance of the circuit is
minimized, we need to find the minimum value of the expression for impedance
Z. This minimum occurs when the derivative of Zwith respect to ωis zero.
Step 2: Calculate the derivative of Zwith respect to ω:
dZ
dω =1
2√R2+(ωL −1
ωC )2·2·(ωL −1
ωC )(L−1
ω2C)
Step 3: Set the derivative equal to zero and solve for ω:
ωL −1
ωC
√R2+(ωL −1
ωC )2= 0
Step 4: Simplify the equation and solve for ω:
ωL −1
ωC = 0
ω2=1
LC
ω=√1
LC
Therefore, the angular frequency ωat which the impedance of the circuit is
minimized is ω=√1
LC .
19
Solution
Step 1: Given that the initial resistance of the copper wire is 8 Ω at 20◦C, we
want to find the resistance at 80◦C. The formula relating resistance, initial
resistance, and temperature change is:
RT=R0[1 + α(T−T0)]
where RTis the resistance at temperature T,R0is the resistance at reference
temperature T0,αis the temperature coefficient of resistivity, Tis the final
temperature, and T0is the reference temperature.
Step 2: Plugging in the given values into the equation, we get:
R80 = 8 Ω[1 + 0.00428 ×(80 −20)]
Step 3: Calculating the expression inside the brackets first:
1+0.00428 ×60 = 1 + 0.2568 = 1.2568
Step 4: Now, calculate the resistance at 80◦C:
R80 = 8 Ω ×1.2568 = 10.0544 Ω
Therefore, the resistance of the copper wire at 80◦C is 10.0544 Ω.
Question 3
Question
A circuit consists of a 10 V battery connected in series to a resistor. If the
current in the circuit is 2 A, what is the resistance of the resistor?
Solution
Step 1: Recall Ohm’s Law, which states that the voltage across a resistor is equal
to the current flowing through it multiplied by the resistance. Mathematically,
this can be written as V=IR, where Vis the voltage, Iis the current, and R
is the resistance.
Step 2: In this case, the voltage across the resistor is 10 V and the cur-
rent flowing through it is 2 A. We can rearrange Ohm’s Law to solve for the
resistance: R=V
I.
Step 3: Substitute the given values into the formula: R=10 V
2A.
Step 4: Perform the division to find the resistance: R= 5 Ω.
Step 5: Therefore, the resistance of the resistor in the circuit is 5 Ohms.
2
Question 4
Question
A circuit consists of a battery with voltage V= 12 V and three resistors con-
nected in series. The resistors have resistances R1= 5 ohms, R2= 10 ohms,
and R3= 15 ohms. Calculate the total current flowing through the circuit and
the voltage drop across each resistor.
Solution
Step 1: Calculate the total resistance of the circuit. The total resistance Rtotal
of resistors in series is the sum of the individual resistances:
Rtotal =R1+R2+R3= 5 Ω + 10 Ω + 15 Ω = 30 Ω
Step 2: Calculate the total current flowing through the circuit using Ohm’s
Law V=IR. The total current Iis given by I=V
Rtotal .
I=12 V
30 Ω = 0.4A
Step 3: Calculate the voltage drop across each resistor. The voltage drop Vi
across a resistor Riis given by Ohm’s Law Vi=IRi. For R1:
V1=I·R1= 0.4A·5 Ω = 2 V
For R2:
V2=I·R2= 0.4A·10 Ω = 4 V
For R3:
V3=I·R3= 0.4A·15 Ω = 6 V
Therefore, the total current flowing through the circuit is 0.4 A, and the
voltage drop across R1,R2, and R3are 2 V, 4 V, and 6 V, respectively.
Question 5
Question
A circuit consists of a resistor with resistance R= 15 Ω connected to a battery
with emf ε= 12 V. Calculate the current flowing through the resistor.
Solution
Step 1: Determine the relationship between voltage, current, and resistance in a
circuit. According to Ohm’s Law, the relationship between voltage (V), current
(I), and resistance (R) in a circuit is given by:
V=IR
3
where V= voltage (in volts), I= current (in amperes), R= resistance (in
ohms).
Step 2: Identify the given values. The resistance of the resistor, R= 15 Ω,
and the emf of the battery, ε= 12 V.
Step 3: Apply Ohm’s Law to find the current. Substitute the given values
into Ohm’s Law:
ε=IR
12 = I×15
I=12
15
I= 0.8A
Step 4: State the final answer. The current flowing through the resistor is
0.8amperes.
Question 6
Question
A resistor is connected to a 12 V battery, and a current of 3 A flows through it.
If the resistance of the resistor is R, calculate the resistance R.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor
is equal to the current (I) flowing through it multiplied by the resistance (R)
of the resistor. Mathematically, this can be written as:
V=I×R
Step 2: In this case, the voltage across the resistor is 12V, the current flowing
through it is 3A, and we are trying to find the resistance, R. We can substitute
these values into Ohm’s Law:
12 = 3 ×R
Step 3: Now, we solve for the resistance R:
R=12
3= 4 Ω
Step 4: Therefore, the resistance of the resistor is 4 Ω.
4
Question 7
Question
A circuit consists of a resistor with resistance R, a capacitor with capacitance C,
and an inductor with inductance Lconnected in series to an alternating current
(AC) voltage source. The voltage across the resistor is VR= 5 sin(100t)V, the
voltage across the capacitor is VC= 3 sin(100t+π
4)V, and the voltage across
the inductor is VL= 4 sin(100t−π
3)V. Given that the current in the circuit is
I= 2 sin(100t+π
6)A, determine the values of R,L, and Cin the circuit.
Solution
Step 1: Write down the phasor representation for the voltages and current in
the circuit. The phasor representation of a sinusoidal function Asin(ωt +ϕ)is
ˆ
A=A
√2∠ϕ. For VR= 5 sin(100t)V: ˆ
VR=5
√2∠0◦V. For VC= 3 sin(100t+π
4)
V: ˆ
VC=3
√2∠45◦V. For VL= 4 sin(100t−π
3)V: ˆ
VL=4
√2∠−60◦V. For
I= 2 sin(100t+π
6)A: ˆ
I=2
√2∠30◦A.
Step 2: Apply Kirchhoff’s voltage law (KVL) to the circuit:
ˆ
VR+ˆ
VL+ˆ
VC=ˆ
Vsource
Substitute in the phasor representations:
5
√2
∠0◦+4
√2
∠−60◦+3
√2
∠45◦=ˆ
Vsource
5∠0◦+ 4∠−60◦+ 3∠45◦=ˆ
Vsource
Step 3: Calculate the phasor representation of the source voltage. Since the
source voltage is not explicitly given, we can represent it as ˆ
Vsource =Vsource∠0◦.
Now solve for Vsource:
5∠0◦+ 4∠−60◦+ 3∠45◦=Vsource∠0◦
Vsource∠0◦= 5 + 4 cos(60◦) + 3 cos(−45◦) + j(4 sin(−60◦) + 3 sin(45◦))
Vsource = 5 + 2 + 3√2
2+j(−2√3 + 3√2
2)
Vsource = 7 + 3√2
2−2√3 + j3√2
2−2√3
Step 4: Compare the phasor representation of the source voltage with the
expression for the impedance of the circuit Z=R+j(ωL−1
ωC ). The impedance
of the circuit must be equal to the phasor representation of the source voltage.
Matching real and imaginary parts, we have: Real parts: R= 7 Imaginary
parts: ωL −1
ωC =3√2
2−2
5
Question 8
Question
A circuit consists of a resistor with resistance R, an inductor with inductance
L, and a capacitor with capacitance Cconnected in series. The voltage across
the circuit is given by V(t) = V0cos(ωt), where V0= 12 V and ω= 50 s−1.
The resistor has a resistance of R= 5 Ω, the inductor has an inductance of
L= 0.1H, and the capacitor has a capacitance of C= 0.01 F. Determine the
current as a function of time for this circuit.
Solution
Step 1: First, we need to find the expression for the total impedance of the
circuit. The impedance ZRof the resistor is simply its resistance R. The
impedance ZLof the inductor is given by ZL=jωL. The impedance ZCof the
capacitor is given by ZC=1
jωC . The total impedance Ztotal of the series circuit
is the sum of the impedances of the resistor, inductor, and capacitor:
Ztotal =R+ZL+ZC=R+jωL +1
jωC
Step2: Next, we can find the current I(t)as a function of time. Since the
frequency of the voltage source is ω= 50 s−1, we can express the current as:
I(t) = V0
Ztotal
cos(ωt)
Step 3: Now, substitute the given values into the expression for total impedance
to get:
Ztotal = 5 + j×50 ×0.1 + 1
j×50 ×0.01
Step 4: Simplify the expression for the total impedance:
Ztotal = 5 + j5 + 1
j0.5= 5 + j5 + j2 = 5 + j7
Step 5: Substitute the total impedance into the expression for current:
I(t) = 12
5 + j7cos(50t)
Therefore, the current as a function of time for the given series circuit is:
I(t) = 12
5 + j7cos(50t)
6
Question 9
Question
A circuit consists of a resistor with resistance R= 10 Ω, a capacitor with capaci-
tance C= 2 µF , and an inductor with inductance L= 0.5Hconnected in series
to a sinusoidal voltage source with frequency ω= 100 rad/s. If the maximum
voltage supplied by the source is Vmax = 20 V, what is the maximum current in
the circuit?
Solution
Step 1: Calculate the impedance of each component. The impedance of a
resistor is given by ZR=R. The impedance of a capacitor is given by ZC=1
jωC .
The impedance of an inductor is given by ZL=jωL.
Step 2: Calculate the total impedance of the circuit. The total impedance
Ztotal of components in series is the sum of their individual impedances.
Ztotal =ZR+ZC+ZL=R+1
jωC +jωL
Step 3: Convert the total impedance to polar form. To convert Ztotal to
polar form, we need to find the magnitude and phase angle. The magnitude
|Ztotal|=√Re(Ztotal)2+Im(Ztotal)2. The phase angle θ= arctan (Im(Ztotal )
Re(Ztotal )).
Step 4: Calculate the maximum current in the circuit. The maximum current
Imax is given by Imax =Vmax
|Ztotal |.
Substitute the given values into the expressions above to find Imax.
Question 10
Question
A circuit consists of a 12V battery connected to three resistors in series. The
first resistor has a resistance of 4Ω, the second resistor has a resistance of 8Ω,
and the third resistor has a resistance of R3Ω. If the current through the circuit
is 1.5A, find the value of R3.
Solution
Step 1: Calculate the total resistance of the circuit. The total resistance Rtotal
of resistors in series is the sum of individual resistances:
Rtotal = 4Ω + 8Ω + R3Ω = 12Ω + R3Ω
Step 2: Apply Ohm’s Law to find the total resistance. Ohm’s Law states
that V=IR, where Vis the voltage, Iis the current, and Ris the resistance.
7
In this case, V= 12V and I= 1.5A. Therefore, the total resistance can be
calculated as:
12V= 1.5A×(12Ω + R3Ω)
12V = 18Ω+1.5R3Ω
Step 3: Solve for R3. Subtract 18Ωfrom both sides of the equation:
12V−18Ω = 1.5R3Ω
−6V= 1.5R3Ω
R3=−6V
1.5Ω =−4Ω
Therefore, the value of R3is 4Ω.
Question 11
Question
A resistor with resistance R1= 10 Ω is connected in series with another resistor
with resistance R2= 20 Ω. A potential difference of V= 120 Vis applied across
the combination. Calculate the current passing through each resistor.
Solution
Step 1: Calculate the total resistance of the combination. The total resistance
Rtotal of resistors in series is given by:
Rtotal =R1+R2= 10 Ω + 20 Ω = 30 Ω
Step 2: Calculate the total current passing through the combination using
Ohm’s Law V=IR. The total current Itotal passing through the combination
is:
Itotal =V
Rtotal
=120 V
30 Ω = 4 A
Step 3: Calculate the current passing through each resistor using the current
divider rule. The current passing through R1, denoted as I1, is:
I1=Itotal ×Rtotal
R1
= 4 A×30 Ω
10 Ω = 12 A
The current passing through R2, denoted as I2, is:
I2=Itotal ×Rtotal
R2
= 4 A×30 Ω
20 Ω = 6 A
Thus, the current passing through R1is 12 Aand the current passing through
R2is 6A.
8
Question 12
Question
A resistor with a resistance of 4 Ω is connected to a battery that delivers a
current of 2A. Find the power dissipated in the resistor.
Solution
Step 1: Recall that the power dissipated in a resistor can be calculated using
the formula P=I2R, where Pis the power, Iis the current, and Ris the
resistance.
Step 2: Given that the resistance R= 4 Ω and the current I= 2 A, we can
plug these values into the formula:
P= (2 A)2×4 Ω
Step 3: Simplify the expression:
P= 4 A2×4 Ω = 16 W
Step 4: Therefore, the power dissipated in the resistor is 16 W .
Question 13
Question
A copper wire has a resistance of 5 ohms. If a current of 2 amperes flows through
the wire, find the voltage across the wire.
Solution
Step 1: Recall Ohm’s Law, which states that voltage (V) is equal to current (I)
multiplied by resistance (R). The formula can be written as:
V=I×R
Step 2: Given that the resistance Ris 5 ohms and the current Iis 2 amperes,
we can substitute these values into Ohm’s Law to find the voltage V:
V= 2 A×5 Ω
Step 3: Calculate the voltage across the wire:
V= 10 V
Therefore, the voltage across the wire is 10 volts.
9
Question 14
Question
A circuit consists of a resistor with resistance R, an inductor with inductance L,
and a capacitor with capacitance Cconnected in series to an AC voltage source
with voltage V=V0sin(ωt). The circuit reaches a steady state, and the voltage
across the resistor is VR=VR0sin(ωt −ϕ). Calculate the phase difference ϕ
between the voltage across the resistor and the applied voltage.
Solution
Step 1: Write down the expression for the total impedance Zof the circuit in
terms of R,L, and C. The total impedance of the circuit is given by:
Z=R+jωL −j
ωC
Step 2: Express the current Ithrough the circuit in terms of the total
impedance Zand the applied voltage V. Using Ohm’s Law in the form V=IZ,
we have:
I=V
Z=V0sin(ωt)
Z
Step 3: Write down the expression for the voltage across the resistor VRin
terms of the current Iand the resistor R. The voltage across the resistor is
given by:
VR=IR =I·R
Step 4: Express the phase difference ϕbetween VRand Vin terms of the
impedance Z. Comparing the expressions for VRand V:
VR0sin(ωt −ϕ) = I·R=V0sin(ωt)
Z·R
⇒VR0=V0R
|Z|and ϕ=Arg(Z)
Step 5: Calculate the magnitude of Zand its argument, then find the phase
difference ϕ. Using the expression for Zfrom Step 1:
|Z|=√R2+(ωL −1
ωC )2
Arg(Z) = tan−1(ωL −1
ωC
R)
Therefore, the phase difference ϕis:
ϕ= tan−1(ωL −1
ωC
R)
10
Question 15
Question
A circuit consists of a resistor, an inductor, and a capacitor connected in series to
an AC voltage source. The values of the components are R= 20 Ω,L= 0.05 H,
and C= 1 µF. The voltage source has an amplitude of V0= 10 V and operates
at a frequency of f= 100 Hz. Calculate the impedance of the circuit and the
phase angle between the current and the voltage.
Solution
Step 1: Calculate the angular frequency ω. Given that f= 100 Hz, we can find
the angular frequency using the formula ω= 2πf . Thus,
ω= 2π×100 = 200πrad/s.
Step 2: Calculate the reactance of the inductor and the capacitor. The
reactance of an inductor XLis given by XL=ωL, and the reactance of a
capacitor XCis given by XC=1
ωC . Substituting the given values,
XL= 200π×0.05 = 10πΩand XC=1
200π×1×10−6= 5 ×103πΩ.
Step 3: Calculate the impedance of the circuit. The impedance of the circuit
Zis the vector sum of the resistive, inductive, and capacitive impedance. Since
the components are in series, we have
Z=R+XL−XC= 20+10π−5×103π= 10π−5×103π+20 = (10−5×103)π+20 Ω = −4990π+20 Ω.
Step 4: Calculate the phase angle ϕbetween the current and voltage. The
phase angle ϕis given by
ϕ= arctan (XL−XC
R).
Plugging in the values,
ϕ= arctan (10π−5×103π
20 )= arctan(−2490π/10) = arctan(−249π).
Question 16
Question
A circuit contains a resistor with resistance 10 Ω, an inductor with inductance
0.02 H, and a capacitor with capacitance 8 µF connected in series to a 12 V
battery. Calculate the impedance of the circuit and the current passing through
the circuit.
11
Solution
Step 1: Calculate the total impedance of the circuit. The impedance (Z) of an
LRC series circuit is given by:
Z=√R2+ (XL−XC)2
where: R= resistance = 10 Ω,XL= inductive reactance = ωL,XC= capacitive
reactance = 1
ωC ,ω= angular frequency = 2π
T.
First, calculate the angular frequency:
ω=2π
T
Given that the frequency f=1
T= 60 Hz. Therefore, ω= 2π×60 = 120π
rad/s.
Next, calculate XLand XC:
XL=ωL = 120π×0.02 = 2.4πΩ
XC=1
ωC =1
120π×8×10−6=106
960Ω = 625
6πΩ
Now substitute these values into the impedance formula:
Z=√102+ (2.4π−625
6π)2
Z=√100 + (2.4π−625
6π)2
Step 2: Calculate the current passing through the circuit. The current (I)
flowing through the circuit is given by Ohm’s law:
I=V
Z
where V= voltage from the battery = 12 V.
Now, substitute the calculated impedance value:
I=12
Z
Now, you can calculate the impedance (Z) and then find the current passing
through the circuit.
Question 17
Question
A student sets up a simple circuit with a resistor, battery, and an ammeter.
The ammeter reads a current of 0.5 A when the voltage across the resistor is 6
V. The student then decides to double the resistance in the circuit and observes
that the current reading on the ammeter drops to 0.25 A. Calculate the original
resistance in the circuit and the new resistance after doubling it.
12
Solution
Step 1: Let’s denote the original resistance as Rand the new resistance after
doubling it as 2R.
Step 2: Using Ohm’s Law V=IR, where Vis the voltage, Iis the current,
and Ris the resistance, we can set up two equations:
For the original circuit:
6 = 0.5×R
Simplifying, we get:
R=6
0.5= 12 Ω
Step 3: For the circuit with doubled resistance:
6 = 0.25 ×2R
Simplifying, we get:
2R=6
0.25 = 24 Ω
Therefore, the new resistance 2Ris 24 Ωafter doubling it.
Step 4: The original resistance was found to be 12 Ω. Thus, the original
resistance in the circuit is 12 Ω, and the new resistance after doubling it is 24
Ω.
Question 18
Question
A circuit consists of a 12 V battery connected in series with a resistor and an
unknown device. When a current of 2 A flows through the circuit, the potential
drop across the resistor is 4 V. Determine the resistance of the unknown device.
Solution
Step 1: Recall Ohm’s Law, which states that the potential difference (V) across a
resistor is equal to the current (I) flowing through it multiplied by the resistance
(R) of the resistor. Mathematically, this can be expressed as V=I·R.
Step 2: From the problem, we are given that the potential drop across the
resistor is 4 V when a current of 2 A flows through the circuit. Therefore, we
have 4V= 2 A×R.
Step 3: To find the resistance Rof the unknown device, we can rearrange
the formula V=I·Rto solve for R. So, R=V
I.
Step 4: Substituting the given values, we find R=4V
2A= 2 Ω.
Step 5: Hence, the resistance of the unknown device in the circuit is 2 ohms.
13
Question 19
Question
A circuit consists of a 12 V battery, a resistor with resistance R, and a capacitor
with capacitance C. Initially, the capacitor is uncharged and acts like a short
circuit. When the circuit is closed, the current through the circuit is found to be
0.5A. After some time, the capacitor is fully charged and the current through
the circuit is found to be 0.1A. Calculate the resistance Rof the resistor in the
circuit.
Solution
1. Let’s denote the resistance of the resistor as R, the capacitance of the capac-
itor as C, and the potential difference across the resistor as VR. The potential
difference across the capacitor is VC= 12 V.
2. When the circuit is first closed and the capacitor acts like a short circuit,
the current in the circuit is equal to Iinitial = 0.5A. This current can be split
between the resistor and capacitor according to the equivalent resistance Req =
Rin series with the capacitor. Since the capacitor acts like a short circuit, the
full potential difference of 12 V is across the resistor:
VR= 12 V=R×Iinitial
R=VR
Iinitial
=12
0.5= 24Ω
3. When the capacitor is fully charged and the current through the circuit
is Ifinal = 0.1A, the capacitor acts like an open circuit. Now, the potential
difference VCis fully across the resistor. We can write an expression for the
potential difference VRin terms of Rand Cusing Ohm’s Law and the formula
for the charging of a capacitor:
VR=R×Ifinal =VC(1 −e−t
RC )
Since the capacitor is fully charged, VR=VC= 12 V and Ifinal = 0.1A.
Substituting these values into the equation gives:
R×0.1 = 12(1 −e−t
RC )
R=12(1 −e−t
RC )
0.1
4. Since we are looking for the resistance R, we need to find the time constant
τ=RC. We can find τby noting that in a fully charged circuit, e−t
RC ≈0.
Therefore:
R=12(1 −0)
0.1= 120Ω
5. Therefore, the resistance Rof the resistor in the circuit is 120 Ω .
14
Question 20
Question
A 10 V battery is connected to a resistor of unknown resistance. When a current
of 2 A flows through the resistor, the power dissipated is 20 W. Calculate the
resistance of the resistor.
Solution
Step 1: We can start by using the formula for power in a resistor: P=I2·R,
where Pis the power dissipated, Iis the current flowing through the resistor,
and Ris the resistance. Given that P= 20 W and I= 2 A, we can rearrange
the formula to solve for R:
R=P
I2
R=20
(2)2
R=20
4
R= 5 Ω
Therefore, the resistance of the resistor is 5 Ω.
Question 21
Question
A circuit consists of a 12 V battery, a 4 Ωresistor, and an unknown resistor
R. When the total resistance in the circuit is 6 Ω, the power dissipated in the
unknown resistor is 24 W. Find the value of R.
Solution
Step 1: Let’s first calculate the current, I, flowing through the circuit using
Ohm’s Law, V=IR, where Vis the voltage of the battery.
Given: V= 12 V, R1= 4 Ω, Rtotal = 6 Ω
Since Rtotal =R1+R, we can find Ras R=Rtotal −R1
R= 6 −4 = 2 Ω
Step 2: Next, we find the current, I, as I=V
Rtotal .
I=12
6= 2 A
15
Step 3: Now, we can find the power, P, dissipated by the unknown resistor
using the formula P=I2R.
Given: P= 24 W, R= 2 Ω
24 = (2)2×2
24 = 4 ×2
24 = 8
This equation is not met, so there must be an error in the problem statement
or calculation.
Question 22
Question
A circuit consists of a resistor with resistance R= 100 Ω connected to a battery
with emf E= 12 V. The circuit also has an ammeter connected in series and
a voltmeter connected in parallel to the resistor. If the ammeter reads 0.1A,
what reading will the voltmeter display?
Solution
Step 1: Recall Ohm’s Law which states that the current passing through a
resistor is directly proportional to the potential difference across the resistor.
The formula for Ohm’s Law is V=IR, where Vis the voltage (potential
difference) across the resistor, Iis the current passing through the resistor, and
Ris the resistance of the resistor.
Step 2: Given in the question, R= 100 Ω and I= 0.1A. We can use Ohm’s
Law to find the voltage across the resistor: V=I·R.
Step 3: Substituting the given values into the formula, we get V= 0.1A·
100 Ω = 10 V.
Step 4: Since the voltmeter is connected in parallel to the resistor, it will
read the same voltage as the resistor. Therefore, the voltmeter will display
10 V .
Question 23
Question
A circuit consists of a resistor with resistance R= 15 Ω and an inductor with
inductance L= 0.04 H. The circuit is connected to a voltage source with an
emf of V= 20 V and a frequency of f= 50 Hz. Calculate the current flowing
through the circuit.
16
Solution
Step 1: Calculate the reactance of the inductor using the formula XL= 2πf L.
XL= 2π×50 ×0.04 = 4πΩ
Step 2: Calculate the total impedance of the circuit using the formula Z=
√R2+X2
L.
Z=√152+ (4π)2≈√225 + 39.48 ≈√264.48 ≈16.27 Ω
Step 3: Calculate the current flowing through the circuit using Ohm’s Law:
I=V
Z.
I=20
16.27 ≈1.23 A
Therefore, the current flowing through the circuit is approximately 1.23 A.
Question 24
Question
A circuit consists of a resistor with resistance R, an inductor with inductance
L, a capacitor with capacitance C, and an AC voltage source V=V0sin(ωt).
The instantaneous current i(t)through the circuit is given by the equation:
i(t) = i0sin(ωt +ϕ)
where i0= 3 A, ω= 50 rad/s, and ϕ=π
6rad.
Determine the voltage amplitude V0applied by the source.
Solution
Step 1: Ohm’s Law states that for a circuit, the voltage Vapplied across a com-
ponent is equal to the product of the current Iflowing through the component
and the resistance Rof the component.
V=IR
For the given circuit, the voltage across the resistor is VR=i(t)·R. Since the
current through the resistor is i(t)and the resistance of the resistor is R, we
have:
VR=i(t)·R
Step 2: In an AC circuit with a resistor, inductor, and capacitor in series,
the total voltage across the components will be the same as the voltage of the
AC source. Therefore, the voltages across the resistor, inductor, and capacitor
must sum up to the applied voltage V=V0sin(ωt).
V=VR+VL+VC
17
Now, let’s find the expressions for VLand VC.
Step 3: The voltage across an inductor in an AC circuit is given by:
VL=Ldi
dt
where Lis the inductance of the inductor. Since i(t) = i0sin(ωt +ϕ), we can
find di
dt by differentiating i(t)with respect to t.
di
dt =d
dt(i0sin(ωt +ϕ)) = i0ωcos(ωt +ϕ)
Step 4: Substituting into the expression for VL, we get:
VL=Ldi
dt =Li0ωcos(ωt +ϕ)
Step 5: The voltage across a capacitor in an AC circuit is given by:
VC=1
C∫i dt
where Cis the capacitance of the capacitor. Since i(t) = i0sin(ωt +ϕ), we can
find ∫i dt by integrating i(t)with respect to t.
∫i dt =∫i0sin(ωt +ϕ)dt =−i0
ωcos(ωt +ϕ)
Step 6: Substituting into the expression for VC, we have:
VC=1
C∫i dt =−i0
ωC cos(ωt +ϕ)
Step 7: Now, substituting VR,VL, and VCinto the equation V=VR+VL+
VC, we get:
V0sin(ωt) = i0Rsin(ωt +ϕ) + Li0ωcos(ωt +ϕ)−i0
ωC cos(ωt +ϕ)
Step 8: We know that the voltage amplitudes of sin(ωt),cos(ωt), and cos(ωt)
terms on both sides of the equation must be equal. So, the coefficient of sin(ωt)
term on the right side must be V0. Thus, we can determine V0by equating
coefficients:
V0=i0R
V0= 3 A×R
Question 25
Question
A circuit consists of a resistor with resistance R, an inductor with induc-
tance L, and a capacitor with capacitance Cconnected in series to an AC
voltage source with frequency f. The impedance of the circuit is given by
Z=√R2+(ωL −1
ωC )2, where ω= 2πf. Determine the angular frequency ω
at which the impedance of the circuit is minimized.
18
Solution
Step 1: To find the angular frequency ωat which the impedance of the circuit is
minimized, we need to find the minimum value of the expression for impedance
Z. This minimum occurs when the derivative of Zwith respect to ωis zero.
Step 2: Calculate the derivative of Zwith respect to ω:
dZ
dω =1
2√R2+(ωL −1
ωC )2·2·(ωL −1
ωC )(L−1
ω2C)
Step 3: Set the derivative equal to zero and solve for ω:
ωL −1
ωC
√R2+(ωL −1
ωC )2= 0
Step 4: Simplify the equation and solve for ω:
ωL −1
ωC = 0
ω2=1
LC
ω=√1
LC
Therefore, the angular frequency ωat which the impedance of the circuit is
minimized is ω=√1
LC .
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