PHYS 231 - UNIVERSITY PHYSICS I
- Ohm’s Law and its applications
Question Bank - Set 1
Liberty University
Question 1
Question
A circuit consists of a 12 V battery connected to a resistor. When a current
of 3 A flows through the circuit, the power dissipated in the resistor is 36 W.
Calculate the resistance of 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 flowing through
the resistor, and Ris the resistance of the resistor.
Step 2: Given that the power dissipated is 36 W and the current is 3 A, we
can substitute these values into the formula: 36 = 32×R.
Step 3: Simplifying the equation, we get 36 = 9R.
Step 4: Divide both sides by 9 to solve for the resistance: R=36
9= 4.
Step 5: Therefore, the resistance of the resistor is 4 ohms.
Question 2
Question
A resistor with a resistance of 5 Ωis connected to a 12 V battery in a circuit.
Calculate the current flowing through the resistor.
Solution
Step 1: Recall Ohm’s Law, which states that the current passing through a
resistor is directly proportional to the voltage across the resistor and inversely
proportional to the resistance of the resistor. Mathematically, Ohm’s Law is
expressed as:
V=IR
where: V= voltage across the resistor in volts, I= current flowing through the
resistor in amperes, and R= resistance of the resistor in ohms.
Step 2: Given information: Resistance of the resistor, R= 5 Ω Voltage across
the resistor, V= 12 V
Step 3: Substitute the given values into Ohm’s Law equation:
I=V
R
I=12 V
5 Ω
Step 4: Perform the division to calculate the current flowing through the
resistor:
I=12
5
I= 2.4A
Step 5: Therefore, the current flowing through the resistor is 2.4 amperes.
Question 3
Question
A circuit consists of a battery with a voltage of 12 V connected in series to three
resistors: R1= 4 Ω,R2= 6 Ω, and R3= 8 Ω. Calculate the current passing
through each resistor and the power dissipated by each resistor.
Solution
Step 1: First, we calculate the total resistance of the circuit by adding up the
individual resistances:
Rtotal =R1+R2+R3= 4 Ω + 6 Ω + 8 Ω = 18 Ω
Step 2: Next, we use Ohm’s Law to find the total current passing through
the circuit:
V=I·Rtotal
I=V
Rtotal
=12 V
18 Ω = 0.67 A
2
Step 3: To find the current passing through each resistor, we can use the
fact that in a series circuit, the current remains constant:
I=I1=I2=I3= 0.67 A
Step 4: Now, we can calculate the power dissipated by each resistor using
the formula P=I2·R: For R1:
P1=I2·R1= (0.67 A)2·4 Ω = 1.792 W
For R2:
P2=I2·R2= (0.67 A)2·6 Ω = 2.688 W
For R3:
P3=I2·R3= (0.67 A)2·8 Ω = 3.584 W
So, the current passing through each resistor is 0.67 A, and the power dissi-
pated by each resistor is 1.792 W, 2.688 W, and 3.584 W, respectively.
Question 4
Question
A circuit consists of a resistor with resistance R= 5 Ω and an unknown resistor
connected in series to a battery with emf E= 12 V. When a current of 2A
flows through the circuit, the potential difference across the unknown resistor
is found to be 6V. Determine the resistance of the unknown resistor.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor is
equal to the current (I) flowing through the resistor multiplied by the resistance
(R) of the resistor: V=IR.
Step 2: For the total circuit, the total potential difference provided by the
battery is equal to the sum of the potential differences across each resistor.
Therefore, the potential difference Vtotal across the total circuit is equal to the
potential difference across the 5 Ω resistor (V5) plus the potential difference
across the unknown resistor (Vunknown). Mathematically, this can be expressed
as: Vtotal =V5+Vunknown.
Step 3: We are given that the potential difference across the 5 Ω resistor is
6V and the total current in the circuit is 2A. Using Ohm’s Law for the 5 Ω
resistor, we have V5=I·R= 2 A×5 Ω = 10 V.
Step 4: Substitute V5= 10 V and Vtotal = 12 V into the equation from Step
2: 12 V= 10 V+Vunknown, we can solve for Vunknown.
Step 5: Vunknown = 12 V−10 V= 2 V.
Step 6: Finally, to find the resistance of the unknown resistor Runknown,
we can use Ohm’s Law with the current I= 2 A and the potential difference
Vunknown = 2 V: Runknown =Vunknown
I=2V
2A= 1 Ω.
Therefore, the resistance of the unknown resistor is 1 Ω.
3
Question 5
Question
A circuit consists of a resistor with resistance R= 10 Ω and an unknown resistor
connected in series to a 12 Vbattery. If the current passing through this circuit
is 0.5A, find the resistance of the unknown resistor.
Solution
Step 1: We know that in a series circuit, the total resistance Rtotal is the sum
of the individual resistances. Thus, Rtotal =R+Runknown.
Step 2: Ohm’s Law states that the voltage across a resistor equals the current
passing through it multiplied by the resistance. Therefore, the voltage across
the resistor with known resistance Ris VR=I·R= 0.5A·10 Ω = 5 V.
Step 3: Since the total voltage in a series circuit is equal to the sum of the
voltages across each component, we have Vtotal = 12 V. Thus, the voltage across
the unknown resistor is Vunknown =Vtotal −VR= 12 V−5V= 7 V.
Step 4: Applying Ohm’s Law again, we can find the resistance of the un-
known resistor: Runknown =Vunknown
I=7V
0.5A= 14 Ω.
Therefore, the resistance of the unknown resistor in the circuit is 14 Ω.
Question 6
Question
A circuit consists of a resistor with a resistance of 10 Ωand a battery with an
electromotive force of 12 V. If a current of 1.2 A flows through the circuit, what
is the potential difference across the resistor?
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 is represented by the formula V=IR.
Step 2: Substituting the given values, we have V= 1.2A×10 Ω.
Step 3: Calculate the potential difference using the formula:
V= 1.2A×10 Ω = 12 V
Step 4: Therefore, the potential difference across the resistor is 12 V.
4
Question 7
Question
A 12 V battery is connected to a series circuit containing three resistors: R1=
10 Ω,R2= 15 Ω, and R3= 20 Ω. Calculate the current passing through each
resistor in the circuit.
Solution
We can use Ohm’s Law, V=IR, to find the current passing through each
resistor in the circuit.
Step 1: Calculate the total resistance in the circuit. The total resistance in
a series circuit is the sum of the individual resistances:
Rtotal =R1+R2+R3= 10 Ω + 15 Ω + 20 Ω = 45 Ω
Step 2: Calculate the total current in the circuit. Using Ohm’s Law for the
total circuit:
Itotal =V
Rtotal
=12 V
45 Ω =4
15 A= 0.27 A
Step 3: Calculate the current passing through each resistor. Now, using
Ohm’s Law for individual resistors: For R1:
I1=V
R1
=12 V
10 Ω = 1.2A
For R2:
I2=V
R2
=12 V
15 Ω = 0.8A
For R3:
I3=V
R3
=12 V
20 Ω = 0.6A
Therefore, the current passing through R1is 1.2 A, through R2is 0.8 A, and
through R3is 0.6 A in the circuit.
Question 8
Question
A circuit consists of a resistor with resistance 10 Ω, an inductor with inductance
0.02 H, and a capacitor with capacitance 5 µF connected in series to a 12 V AC
power supply that operates at a frequency of 60 Hz.
Calculate the impedance of the circuit at this frequency.
5
Solution
Step 1: Calculate the impedance of each component.
• The impedance of a resistor is equal to its resistance. Therefore, the
impedance of the resistor is 10 Ω.
• The impedance of an inductor is given by ZL=jωL, where Lis the
inductance and ω= 2πf . Plugging in the values, we get:
ZL=j(2π·60)(0.02) = j2.4 Ω.
• The impedance of a capacitor is given by ZC=1
jωC , where Cis the
capacitance. Substituting the values, we get:
ZC=1
j(2π·60)(5 ×10−6)=−j53.05 Ω.
Step 2: Calculate the total impedance of the circuit.
Ztotal =ZR+ZL+ZC= 10 −j2.4−j53.05 Ω.
Step 3: Convert the total impedance to polar form.
Ztotal = 10 −j55.45 Ω.
Step 4: Calculate the magnitude of the total impedance.
|Ztotal|=√102+ (−55.45)2=√3113.7025 ≈55.78 Ω.
Therefore, the impedance of the circuit at a frequency of 60 Hz is approxi-
mately 55.78 Ω.
Question 9
Question
A circuit consists of a resistor of resistance R1, a resistor of resistance R2, and
a battery with emf Eand internal resistance r, all connected in series. The
resistance R1is twice as large as the resistance R2. The potential difference
across R2is found to be half the value of the emf of the battery. Find an
expression for the current Iin the circuit in terms of E,R1,R2, and r.
Given data:
R1= 2R2, VR2=1
2E
6
Solution
Step 1: Use Ohm’s Law to relate the potential differences and resistances in the
circuit.
E=VR1+Vr+VR2
IR1=IR2+Ir +E
2
Step 2: Substitute R1= 2R2into the equation.
I(2R2) = IR2+Ir +E
2
Step 3: Rearrange the equation in terms of I.
2IR2=IR2+Ir +E
2
2I−I=E
2R2−Ir
R2
I=E
2R2−Ir
R2
Step 4: Substitute R1= 2R2into the expression for I.
I=E
2R2−Ir
R2
Therefore, the current Iin the circuit in terms of E,R1,R2, and ris I=
E
2R2−Ir
R2.
Question 10
Question
A circuit consists of a resistor with resistance R= 20Ω and an unknown device
represented by a box with two terminals. When a current of 2.5A passes through
the circuit, a potential difference of 25 V is observed across the terminals of the
unknown device. Determine the resistance of the unknown device.
Solution
Step 1: Recall Ohm’s Law, which states that the potential difference (V) across
a device is equal to the current (I) passing through the device multiplied by the
resistance (R) of the device. The formula for Ohm’s Law is V=IR.
Step 2: We know that a potential difference of 25 V is observed across the
terminals of the unknown device when a current of 2.5A passes through the
7
circuit. Using Ohm’s Law, we can find the resistance of the unknown device.
Given V= 25 V and I= 2.5A, we have:
R=V
I=25 V
2.5A
Step 3: Calculate the resistance of the unknown device:
R= 10Ω
Step 4: Therefore, the resistance of the unknown device represented by the
box is 10Ω.
Question 11
Question
A wire with resistance Ris connected to a battery with emf Eand internal resis-
tance r. When a certain current Iflows through the wire, the power dissipated
is at its maximum value. Prove that the internal resistance rof the battery
must be equal to the resistance Rof the wire.
Solution
Step 1: Recall that the power dissipated in a circuit is given by P=I2R, where
Iis the current and Ris the resistance.
Step 2: The total resistance in the circuit is the sum of the resistance of the
wire (R) and the internal resistance of the battery (r), so the total resistance is
Rtotal =R+r.
Step 3: Using Ohm’s Law (V=IR), we can express the current Iin terms
of the total resistance Rtotal and the emf Eof the battery: I=E
Rtotal =E
R+r.
Step 4: Substitute the expression for Iinto the equation for power P=I2R
to get the power dissipated in terms of R,r, and E:P=(E
R+r)2·R.
Step 5: To find the maximum power dissipated, we can take the derivative
of Pwith respect to r, set it equal to 0, and solve for r.
Step 6: Differentiate Pwith respect to r:dP
dr =2E2R
(R+r)3−E2R
(R+r)2.
Step 7: Set dP
dr = 0 and solve for r:2E2R
(R+r)3−E2R
(R+r)2= 0.
Step 8: Simplifying the equation, we find that r=R. Thus, the internal
resistance of the battery must be equal to the resistance of the wire for the
power dissipated to be at its maximum value.
Question 12
Question
In a circuit, a resistor with resistance R1= 20 Ω and a resistor with resistance
R2= 30 Ω are connected in series with a battery that provides a voltage of
8
V= 60 V. Calculate the total current in the circuit.
Solution
Step 1: Calculate the total resistance in the circuit. The total resistance in
a series circuit is the sum of the individual resistances. Therefore, the total
resistance Rtotal is given by:
Rtotal =R1+R2= 20 Ω + 30 Ω = 50 Ω
Step 2: Apply Ohm’s Law to find the total current. Ohm’s Law states
that the current flowing through a circuit is inversely proportional to the total
resistance and directly proportional to the applied voltage. Mathematically,
Ohm’s Law is given by:
V=I·R
where Vis the voltage, Iis the current, and Ris the resistance.
Substitute the values we have into the equation:
I=V
Rtotal
=60 V
50 Ω = 1.2A
Therefore, the total current in the circuit is 1.2A.
Question 13
Question
A copper wire with a resistance of 2.5 Ωis connected in series with a resistor of
unknown resistance. When a potential difference of 12 V is applied across the
circuit, a current of 3 A is observed. Determine the resistance of the unknown
resistor.
Solution
Step 1: Use Ohm’s Law (V=IR) to find the total resistance of the circuit.
Given: Resistance of the copper wire, Rcopper = 2.5 Ω Potential difference,
V= 12 V Current, I= 3 A
The total resistance can be calculated as:
Rtotal =V
I
Rtotal =12 V
3A= 4 Ω
Step 2: Use the total resistance to find the resistance of the unknown resis-
tor. Let Runknown be the resistance of the unknown resistor. Since the resistors
9
are connected in series, the total resistance Rtotal is the sum of individual resis-
tances:
Rtotal =Rcopper +Runknown
4 Ω = 2.5 Ω + Runknown
Runknown = 4 Ω −2.5 Ω = 1.5 Ω
Therefore, the resistance of the unknown resistor is 1.5 Ω.
Question 14
Question
A circuit consists of three resistors connected in series. The first resistor has a
resistance of 4 Ω, the second resistor has a resistance of 6 Ω, and the third resistor
has an unknown resistance R. A potential difference of 24 Vis applied across
the circuit. If the current through the circuit is 2A, determine the resistance R
of the third resistor.
Solution
Step 1: Recall the formula for Ohm’s Law: V=IR, where Vis the potential
difference across the circuit, Iis the current flowing through the circuit, and Ris
the total resistance of the circuit. Step 2: Determine the total resistance Rtotal of
the circuit by summing the individual resistances: Rtotal =R1+R2+R. Step 3:
Substitute the given resistances into the total resistance formula: Rtotal = 4 Ω+
6 Ω + R= 10 Ω + R. Step 4: Use Ohm’s Law to find the total resistance of the
circuit: V=IRtotal. Substitute V= 24 Vand I= 2 A:24 V= 2 A×(10 Ω+R).
Step 5: Solve for the unknown resistance R:24 V= 20 Ω + 2R
2R= 24 V−20 Ω
2R= 4 V
R= 2 Ω.
Therefore, the resistance of the third resistor is 2 Ω.
Question 15
Question
A circuit consists of three resistors connected in series. The resistances of the
resistors are R1= 6 Ω,R2= 8 Ω, and R3= 10 Ω. If the total potential difference
across the circuit is V= 120 V, what is the current flowing through the circuit?
10
Solution
Step 1: Calculate the total resistance of the circuit using the formula for resistors
in series: Rtotal =R1+R2+R3
= 6 Ω + 8 Ω + 10 Ω
= 24 Ω.
Step 2: Use Ohm’s Law, V=IR, to find the current flowing through the
circuit:
I=V
Rtotal
=120 V
24 Ω
= 5 A.
Therefore, the current flowing through the circuit is 5A.
Question 16
Question
A copper wire with a resistance of 2.5 Ωis connected to a battery that provides
a current of 0.8 A. If the resistance of the wire is increased to 5 Ω, what is the
new current in the circuit?
Solution
Step 1: Calculate the initial voltage drop across the 2.5 Ωresistor using Ohm’s
Law, V=IR.
Initial Voltage = (0.8A)(2.5 Ω) = 2 V
Step 2: Since the battery provides a constant voltage, the total voltage drop
in the circuit remains 2 V when the resistance is increased to 5 Ω. Using Ohm’s
Law, we can find the new current in the circuit.
2V=I(5 Ω)
I=2V
5 Ω = 0.4A
Therefore, the new current in the circuit when the resistance is increased to
5Ωis 0.4 A.
11
Question 17
Question
A circuit consists of a resistor with a resistance of 30 Ω connected in series with a
battery that provides a voltage of 12 V. Determine the current flowing through
the circuit.
Solution
Step 1: Write down Ohm’s Law, which relates voltage (V), current (I), and
resistance (R) in a circuit:
V=I·R
Step 2: Given that the resistance R= 30 Ω and voltage V= 12 V, we can
use Ohm’s Law to solve for the current I:
12 V=I·30 Ω
Step 3: Solve for the current I:
I=12 V
30 Ω = 0.4A
Therefore, the current flowing through the circuit is 0.4A.
Question 18
Question
A resistor with resistance R1= 6 Ω and another resistor with resistance R2=
4 Ω are connected in parallel across a 12 Vbattery. Find the total current flowing
through the circuit.
Solution
Step 1: Calculate the equivalent resistance of the two resistors in parallel. The
formula to find the total resistance of two resistors in parallel is given by:
1
Rtotal
=1
R1
+1
R2
Substitute the given values: R1= 6 Ω and R2= 4 Ω:
1
Rtotal
=1
6+1
4
1
Rtotal
=2
12 +3
12
12
1
Rtotal
=5
12
Therefore, the equivalent resistance Rtotal is:
Rtotal =12
5Ω = 2.4 Ω
Step 2: Calculate the total current using Ohm’s Law. The total current
flowing through the circuit can be found using Ohm’s Law: I=V
R, where
V= 12 Vis the voltage of the battery and Rtotal = 2.4 Ω is the total resistance.
I=12
2.4= 5 A
Therefore, the total current flowing through the circuit is 5A.
Question 19
Question
A circuit consists of a resistor with resistance R= 30 Ω and an unknown resistor
connected in series. When a voltage of V= 12 Vis applied across the circuit,
a current of I= 0.4Aflows through it. What is the resistance of the unknown
resistor?
Solution
Step 1: Recall Ohm’s Law which states V=IR, where Vis the voltage across
a component, Iis the current flowing through it, and Ris the resistance of the
component.
Step 2: We first need to find the total resistance of the circuit. In a se-
ries circuit, the total resistance Rtotal is the sum of the individual resistances:
Rtotal =R1+R2+. . . +Rn.
Step 3: In this case, the total resistance Rtotal is given by Rtotal =R+
Runknown.
Step 4: We can now use Ohm’s Law to find the total resistance of the circuit.
Given that V= 12 Vand I= 0.4A, we have:
Rtotal =V
I
Step 5: Substituting the given values, we find:
R+Runknown =12 V
0.4A
Step 6: Simplifying the equation, we get:
30 Ω + Runknown = 30 Ω
13
Step 7: Subtracting 30 Ω from both sides gives:
Runknown = 0 Ω
Step 8: Therefore, the resistance of the unknown resistor in the circuit is
0 Ω .
Question 20
Question
A circuit consists of a resistor with resistance R, a capacitor with capacitance
C, and an inductor with inductance L. The total impedance of the circuit is
given by Z=√R2+ (XL−XC)2, where XL= 2πfL is the inductive reactance,
XC=1
2πf C is the capacitive reactance, and fis the frequency of the alternating
current in the circuit. Determine the frequency of the current that minimizes
the total impedance of the circuit.
Solution
Step 1: Substitute the expressions for XLand XCinto the equation for total
impedance Z.
Z=√R2+(2πfL −1
2πfC )2
Step 2: To minimize Z, we need to find the frequency fsuch that the
derivative of Zwith respect to fis zero.
dZ
df = 0
Step 3: Differentiate Zwith respect to fusing the chain rule and set the
derivative equal to zero.
dZ
df =1
2√R2+(2πfL −1
2πf C )2·2(2πL −1
2πC )= 0
Step 4: Solve for fby setting the expression inside the parentheses equal to
zero.
2πL −1
2πC = 0
Step 5: Solve for f.
2πL =1
2πC
4π2L=1
C
14
f=1
2π√LC
So, the frequency that minimizes the total impedance of the circuit is f=
1
2π√LC .
Question 21
Question
A circuit consists of a resistor with a resistance of 10 Ω, a capacitor with a
capacitance of 0.01 F, and a battery with an emf of 6 V. The switch in the
circuit is closed at t= 0. Calculate the current flowing through the circuit
when t→ ∞.
Solution
Step 1: Find the time constant, τ, of the circuit using the formula τ=RC
where Ris the resistance and Cis the capacitance.
τ= (10 Ω)(0.01 F) = 0.1s
Step 2: Determine the current, I(t), as a function of time tusing the equation
I(t) = I0e−t/τ
where I0is the initial current at t= 0. At t= 0, the current is given by Ohm’s
Law as I(0) = V
R.
I0=6V
10 Ω = 0.6A
So, the current as a function of time is:
I(t) = 0.6A·e−t/0.1s
Step 3: Calculate the current flowing through the circuit as t→ ∞ by
evaluating the limit of I(t)as tapproaches infinity.
lim
t→∞ I(t) = lim
t→∞ 0.6A·e−t/0.1s
lim
t→∞ I(t) = 0.6A·e−∞ = 0
Therefore, the current flowing through the circuit when t→ ∞ is 0.
Question 22
Question
A certain electrical device operates at a voltage of 120 V and draws a current
of 5 A. If the device has a resistance of 24 Ω, calculate the power consumed by
the device.
15
Solution
Step 1: Recall Ohm’s Law which states: V=I·R, where Vis the voltage across
the device, Iis the current passing through the device, and Ris the resistance
of the device.
Step 2: We are given that the voltage Vis 120 V, the current Iis 5 A, and
the resistance Ris 24 Ω. Therefore, we can rearrange Ohm’s Law to solve for
R:
R=V
I=120
5= 24 Ω
Step 3: Next, we calculate the power consumed by the device using the
formula: P=V·I, where Pis power in watts.
P=V·I= 120 ·5 = 600 W
Step 4: The power consumed by the device is 600 watts. Thus, the device
consumes 600 watts of power when operating at a voltage of 120 V and drawing
a current of 5 A with a resistance of 24 Ω.
Question 23
Question
An electric circuit consists of a resistor with resistance R, a capacitor with
capacitance C, and an inductor with inductance Lconnected in series to an
alternating voltage source with angular frequency ω. The voltage across the
resistor, capacitor, and inductor is VR=V0cos(ωt),VC=V0cos(ωt −π
2), and
VL=V0cos(ωt +π
2), respectively.
If the current in the circuit is given by I(t) = I0cos(ωt −ϕ), where I0is the
amplitude of the current and ϕis the phase angle between the current and the
voltage source, show that the amplitude I0of the current is given by
I0=V0
√R2+ (ωL −1
ωC )2
and the phase angle ϕis given by
tan ϕ=ωL −1
ωC
R
Solution
Step 1: Apply Ohm’s Law to find the relationship between the voltage VR,
current I, and resistance R.
VR=IR
16
Step 2: Substitute the expressions for VRand Iin terms of time to find the
expression for I0.
V0cos(ωt) = I0Rcos(ωt −ϕ)
Step 3: Rearrange the equation to solve for I0.
I0=V0
R
Step 4: Apply impedance of a series LRC circuit to find the total impedance
Z.
Z=√R2+(ωL −1
ωC )2
Step 5: Since Z=V0
I0, we have I0=V0
Z. Substitute Zinto the equation.
I0=V0
√R2+(ωL −1
ωC )2
Step 6: Finally, use the phase angle relationship tan ϕ=ωL−1
ωC
Rto determine
the phase angle ϕ.
Question 24
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
difference across the unknown device is measured to be 8 V. Find the resistance
of the unknown device.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor is
equal to the current (I) flowing through the resistor multiplied by the resistance
(R) of the resistor. Mathematically, this can be represented by the equation:
V=I·R.
Step 2: In this problem, the potential difference (V) across the unknown
device is 8 V, and the current (I) flowing through the circuit is 2 A. Therefore,
using Ohm’s Law, we can calculate the resistance (R) of the unknown device
as:
R=V
I=8V
2A= 4 Ω.
Step 3: Hence, the resistance of the unknown device in the circuit is 4 Ω.
17
Question 25
Question
A circuit consists of three resistors in series: R1= 10 Ω,R2= 20 Ω, and R3=
30 Ω. The circuit is connected to a 12 V battery. Calculate the current through
each resistor and the total power dissipated in the circuit.
Solution
Step 1: Calculate the total resistance in the circuit. The total resistance in a
series circuit is the sum of the individual resistances:
Rtotal =R1+R2+R3= 10 Ω + 20 Ω + 30 Ω = 60 Ω
Step 2: Calculate the total current in the circuit using Ohm’s Law (V=IR):
I=V
Rtotal
=12 V
60 Ω = 0.2A
Step 3: Calculate the current through each resistor. In a series circuit, the
current is the same through each resistor. Therefore, I=I1=I2=I3= 0.2A
Step 4: Calculate the power dissipated by each resistor using the formula
P=IV :
P1=I·V= 0.2A×12 V= 2.4W
P2=I·V= 0.2A×12 V= 2.4W
P3=I·V= 0.2A×12 V= 2.4W
Step 5: Calculate the total power dissipated in the circuit. The total power
is the sum of the power dissipated by each resistor:
Ptotal =P1+P2+P3= 2.4W+ 2.4W+ 2.4W= 7.2W
Therefore, the current through each resistor is 0.2A and the total power
dissipated in the circuit is 7.2W.
18
Solution
Step 1: Recall Ohm’s Law, which states that the current passing through a
resistor is directly proportional to the voltage across the resistor and inversely
proportional to the resistance of the resistor. Mathematically, Ohm’s Law is
expressed as:
V=IR
where: V= voltage across the resistor in volts, I= current flowing through the
resistor in amperes, and R= resistance of the resistor in ohms.
Step 2: Given information: Resistance of the resistor, R= 5 Ω Voltage across
the resistor, V= 12 V
Step 3: Substitute the given values into Ohm’s Law equation:
I=V
R
I=12 V
5 Ω
Step 4: Perform the division to calculate the current flowing through the
resistor:
I=12
5
I= 2.4A
Step 5: Therefore, the current flowing through the resistor is 2.4 amperes.
Question 3
Question
A circuit consists of a battery with a voltage of 12 V connected in series to three
resistors: R1= 4 Ω,R2= 6 Ω, and R3= 8 Ω. Calculate the current passing
through each resistor and the power dissipated by each resistor.
Solution
Step 1: First, we calculate the total resistance of the circuit by adding up the
individual resistances:
Rtotal =R1+R2+R3= 4 Ω + 6 Ω + 8 Ω = 18 Ω
Step 2: Next, we use Ohm’s Law to find the total current passing through
the circuit:
V=I·Rtotal
I=V
Rtotal
=12 V
18 Ω = 0.67 A
2
Step 3: To find the current passing through each resistor, we can use the
fact that in a series circuit, the current remains constant:
I=I1=I2=I3= 0.67 A
Step 4: Now, we can calculate the power dissipated by each resistor using
the formula P=I2·R: For R1:
P1=I2·R1= (0.67 A)2·4 Ω = 1.792 W
For R2:
P2=I2·R2= (0.67 A)2·6 Ω = 2.688 W
For R3:
P3=I2·R3= (0.67 A)2·8 Ω = 3.584 W
So, the current passing through each resistor is 0.67 A, and the power dissi-
pated by each resistor is 1.792 W, 2.688 W, and 3.584 W, respectively.
Question 4
Question
A circuit consists of a resistor with resistance R= 5 Ω and an unknown resistor
connected in series to a battery with emf E= 12 V. When a current of 2A
flows through the circuit, the potential difference across the unknown resistor
is found to be 6V. Determine the resistance of the unknown resistor.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor is
equal to the current (I) flowing through the resistor multiplied by the resistance
(R) of the resistor: V=IR.
Step 2: For the total circuit, the total potential difference provided by the
battery is equal to the sum of the potential differences across each resistor.
Therefore, the potential difference Vtotal across the total circuit is equal to the
potential difference across the 5 Ω resistor (V5) plus the potential difference
across the unknown resistor (Vunknown). Mathematically, this can be expressed
as: Vtotal =V5+Vunknown.
Step 3: We are given that the potential difference across the 5 Ω resistor is
6V and the total current in the circuit is 2A. Using Ohm’s Law for the 5 Ω
resistor, we have V5=I·R= 2 A×5 Ω = 10 V.
Step 4: Substitute V5= 10 V and Vtotal = 12 V into the equation from Step
2: 12 V= 10 V+Vunknown, we can solve for Vunknown.
Step 5: Vunknown = 12 V−10 V= 2 V.
Step 6: Finally, to find the resistance of the unknown resistor Runknown,
we can use Ohm’s Law with the current I= 2 A and the potential difference
Vunknown = 2 V: Runknown =Vunknown
I=2V
2A= 1 Ω.
Therefore, the resistance of the unknown resistor is 1 Ω.
3
Question 5
Question
A circuit consists of a resistor with resistance R= 10 Ω and an unknown resistor
connected in series to a 12 Vbattery. If the current passing through this circuit
is 0.5A, find the resistance of the unknown resistor.
Solution
Step 1: We know that in a series circuit, the total resistance Rtotal is the sum
of the individual resistances. Thus, Rtotal =R+Runknown.
Step 2: Ohm’s Law states that the voltage across a resistor equals the current
passing through it multiplied by the resistance. Therefore, the voltage across
the resistor with known resistance Ris VR=I·R= 0.5A·10 Ω = 5 V.
Step 3: Since the total voltage in a series circuit is equal to the sum of the
voltages across each component, we have Vtotal = 12 V. Thus, the voltage across
the unknown resistor is Vunknown =Vtotal −VR= 12 V−5V= 7 V.
Step 4: Applying Ohm’s Law again, we can find the resistance of the un-
known resistor: Runknown =Vunknown
I=7V
0.5A= 14 Ω.
Therefore, the resistance of the unknown resistor in the circuit is 14 Ω.
Question 6
Question
A circuit consists of a resistor with a resistance of 10 Ωand a battery with an
electromotive force of 12 V. If a current of 1.2 A flows through the circuit, what
is the potential difference across the resistor?
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 is represented by the formula V=IR.
Step 2: Substituting the given values, we have V= 1.2A×10 Ω.
Step 3: Calculate the potential difference using the formula:
V= 1.2A×10 Ω = 12 V
Step 4: Therefore, the potential difference across the resistor is 12 V.
4
Question 7
Question
A 12 V battery is connected to a series circuit containing three resistors: R1=
10 Ω,R2= 15 Ω, and R3= 20 Ω. Calculate the current passing through each
resistor in the circuit.
Solution
We can use Ohm’s Law, V=IR, to find the current passing through each
resistor in the circuit.
Step 1: Calculate the total resistance in the circuit. The total resistance in
a series circuit is the sum of the individual resistances:
Rtotal =R1+R2+R3= 10 Ω + 15 Ω + 20 Ω = 45 Ω
Step 2: Calculate the total current in the circuit. Using Ohm’s Law for the
total circuit:
Itotal =V
Rtotal
=12 V
45 Ω =4
15 A= 0.27 A
Step 3: Calculate the current passing through each resistor. Now, using
Ohm’s Law for individual resistors: For R1:
I1=V
R1
=12 V
10 Ω = 1.2A
For R2:
I2=V
R2
=12 V
15 Ω = 0.8A
For R3:
I3=V
R3
=12 V
20 Ω = 0.6A
Therefore, the current passing through R1is 1.2 A, through R2is 0.8 A, and
through R3is 0.6 A in the circuit.
Question 8
Question
A circuit consists of a resistor with resistance 10 Ω, an inductor with inductance
0.02 H, and a capacitor with capacitance 5 µF connected in series to a 12 V AC
power supply that operates at a frequency of 60 Hz.
Calculate the impedance of the circuit at this frequency.
5
Solution
Step 1: Calculate the impedance of each component.
• The impedance of a resistor is equal to its resistance. Therefore, the
impedance of the resistor is 10 Ω.
• The impedance of an inductor is given by ZL=jωL, where Lis the
inductance and ω= 2πf . Plugging in the values, we get:
ZL=j(2π·60)(0.02) = j2.4 Ω.
• The impedance of a capacitor is given by ZC=1
jωC , where Cis the
capacitance. Substituting the values, we get:
ZC=1
j(2π·60)(5 ×10−6)=−j53.05 Ω.
Step 2: Calculate the total impedance of the circuit.
Ztotal =ZR+ZL+ZC= 10 −j2.4−j53.05 Ω.
Step 3: Convert the total impedance to polar form.
Ztotal = 10 −j55.45 Ω.
Step 4: Calculate the magnitude of the total impedance.
|Ztotal|=√102+ (−55.45)2=√3113.7025 ≈55.78 Ω.
Therefore, the impedance of the circuit at a frequency of 60 Hz is approxi-
mately 55.78 Ω.
Question 9
Question
A circuit consists of a resistor of resistance R1, a resistor of resistance R2, and
a battery with emf Eand internal resistance r, all connected in series. The
resistance R1is twice as large as the resistance R2. The potential difference
across R2is found to be half the value of the emf of the battery. Find an
expression for the current Iin the circuit in terms of E,R1,R2, and r.
Given data:
R1= 2R2, VR2=1
2E
6
Solution
Step 1: Use Ohm’s Law to relate the potential differences and resistances in the
circuit.
E=VR1+Vr+VR2
IR1=IR2+Ir +E
2
Step 2: Substitute R1= 2R2into the equation.
I(2R2) = IR2+Ir +E
2
Step 3: Rearrange the equation in terms of I.
2IR2=IR2+Ir +E
2
2I−I=E
2R2−Ir
R2
I=E
2R2−Ir
R2
Step 4: Substitute R1= 2R2into the expression for I.
I=E
2R2−Ir
R2
Therefore, the current Iin the circuit in terms of E,R1,R2, and ris I=
E
2R2−Ir
R2.
Question 10
Question
A circuit consists of a resistor with resistance R= 20Ω and an unknown device
represented by a box with two terminals. When a current of 2.5A passes through
the circuit, a potential difference of 25 V is observed across the terminals of the
unknown device. Determine the resistance of the unknown device.
Solution
Step 1: Recall Ohm’s Law, which states that the potential difference (V) across
a device is equal to the current (I) passing through the device multiplied by the
resistance (R) of the device. The formula for Ohm’s Law is V=IR.
Step 2: We know that a potential difference of 25 V is observed across the
terminals of the unknown device when a current of 2.5A passes through the
7
circuit. Using Ohm’s Law, we can find the resistance of the unknown device.
Given V= 25 V and I= 2.5A, we have:
R=V
I=25 V
2.5A
Step 3: Calculate the resistance of the unknown device:
R= 10Ω
Step 4: Therefore, the resistance of the unknown device represented by the
box is 10Ω.
Question 11
Question
A wire with resistance Ris connected to a battery with emf Eand internal resis-
tance r. When a certain current Iflows through the wire, the power dissipated
is at its maximum value. Prove that the internal resistance rof the battery
must be equal to the resistance Rof the wire.
Solution
Step 1: Recall that the power dissipated in a circuit is given by P=I2R, where
Iis the current and Ris the resistance.
Step 2: The total resistance in the circuit is the sum of the resistance of the
wire (R) and the internal resistance of the battery (r), so the total resistance is
Rtotal =R+r.
Step 3: Using Ohm’s Law (V=IR), we can express the current Iin terms
of the total resistance Rtotal and the emf Eof the battery: I=E
Rtotal =E
R+r.
Step 4: Substitute the expression for Iinto the equation for power P=I2R
to get the power dissipated in terms of R,r, and E:P=(E
R+r)2·R.
Step 5: To find the maximum power dissipated, we can take the derivative
of Pwith respect to r, set it equal to 0, and solve for r.
Step 6: Differentiate Pwith respect to r:dP
dr =2E2R
(R+r)3−E2R
(R+r)2.
Step 7: Set dP
dr = 0 and solve for r:2E2R
(R+r)3−E2R
(R+r)2= 0.
Step 8: Simplifying the equation, we find that r=R. Thus, the internal
resistance of the battery must be equal to the resistance of the wire for the
power dissipated to be at its maximum value.
Question 12
Question
In a circuit, a resistor with resistance R1= 20 Ω and a resistor with resistance
R2= 30 Ω are connected in series with a battery that provides a voltage of
8
V= 60 V. Calculate the total current in the circuit.
Solution
Step 1: Calculate the total resistance in the circuit. The total resistance in
a series circuit is the sum of the individual resistances. Therefore, the total
resistance Rtotal is given by:
Rtotal =R1+R2= 20 Ω + 30 Ω = 50 Ω
Step 2: Apply Ohm’s Law to find the total current. Ohm’s Law states
that the current flowing through a circuit is inversely proportional to the total
resistance and directly proportional to the applied voltage. Mathematically,
Ohm’s Law is given by:
V=I·R
where Vis the voltage, Iis the current, and Ris the resistance.
Substitute the values we have into the equation:
I=V
Rtotal
=60 V
50 Ω = 1.2A
Therefore, the total current in the circuit is 1.2A.
Question 13
Question
A copper wire with a resistance of 2.5 Ωis connected in series with a resistor of
unknown resistance. When a potential difference of 12 V is applied across the
circuit, a current of 3 A is observed. Determine the resistance of the unknown
resistor.
Solution
Step 1: Use Ohm’s Law (V=IR) to find the total resistance of the circuit.
Given: Resistance of the copper wire, Rcopper = 2.5 Ω Potential difference,
V= 12 V Current, I= 3 A
The total resistance can be calculated as:
Rtotal =V
I
Rtotal =12 V
3A= 4 Ω
Step 2: Use the total resistance to find the resistance of the unknown resis-
tor. Let Runknown be the resistance of the unknown resistor. Since the resistors
9
are connected in series, the total resistance Rtotal is the sum of individual resis-
tances:
Rtotal =Rcopper +Runknown
4 Ω = 2.5 Ω + Runknown
Runknown = 4 Ω −2.5 Ω = 1.5 Ω
Therefore, the resistance of the unknown resistor is 1.5 Ω.
Question 14
Question
A circuit consists of three resistors connected in series. The first resistor has a
resistance of 4 Ω, the second resistor has a resistance of 6 Ω, and the third resistor
has an unknown resistance R. A potential difference of 24 Vis applied across
the circuit. If the current through the circuit is 2A, determine the resistance R
of the third resistor.
Solution
Step 1: Recall the formula for Ohm’s Law: V=IR, where Vis the potential
difference across the circuit, Iis the current flowing through the circuit, and Ris
the total resistance of the circuit. Step 2: Determine the total resistance Rtotal of
the circuit by summing the individual resistances: Rtotal =R1+R2+R. Step 3:
Substitute the given resistances into the total resistance formula: Rtotal = 4 Ω+
6 Ω + R= 10 Ω + R. Step 4: Use Ohm’s Law to find the total resistance of the
circuit: V=IRtotal. Substitute V= 24 Vand I= 2 A:24 V= 2 A×(10 Ω+R).
Step 5: Solve for the unknown resistance R:24 V= 20 Ω + 2R
2R= 24 V−20 Ω
2R= 4 V
R= 2 Ω.
Therefore, the resistance of the third resistor is 2 Ω.
Question 15
Question
A circuit consists of three resistors connected in series. The resistances of the
resistors are R1= 6 Ω,R2= 8 Ω, and R3= 10 Ω. If the total potential difference
across the circuit is V= 120 V, what is the current flowing through the circuit?
10
Solution
Step 1: Calculate the total resistance of the circuit using the formula for resistors
in series: Rtotal =R1+R2+R3
= 6 Ω + 8 Ω + 10 Ω
= 24 Ω.
Step 2: Use Ohm’s Law, V=IR, to find the current flowing through the
circuit:
I=V
Rtotal
=120 V
24 Ω
= 5 A.
Therefore, the current flowing through the circuit is 5A.
Question 16
Question
A copper wire with a resistance of 2.5 Ωis connected to a battery that provides
a current of 0.8 A. If the resistance of the wire is increased to 5 Ω, what is the
new current in the circuit?
Solution
Step 1: Calculate the initial voltage drop across the 2.5 Ωresistor using Ohm’s
Law, V=IR.
Initial Voltage = (0.8A)(2.5 Ω) = 2 V
Step 2: Since the battery provides a constant voltage, the total voltage drop
in the circuit remains 2 V when the resistance is increased to 5 Ω. Using Ohm’s
Law, we can find the new current in the circuit.
2V=I(5 Ω)
I=2V
5 Ω = 0.4A
Therefore, the new current in the circuit when the resistance is increased to
5Ωis 0.4 A.
11
Question 17
Question
A circuit consists of a resistor with a resistance of 30 Ω connected in series with a
battery that provides a voltage of 12 V. Determine the current flowing through
the circuit.
Solution
Step 1: Write down Ohm’s Law, which relates voltage (V), current (I), and
resistance (R) in a circuit:
V=I·R
Step 2: Given that the resistance R= 30 Ω and voltage V= 12 V, we can
use Ohm’s Law to solve for the current I:
12 V=I·30 Ω
Step 3: Solve for the current I:
I=12 V
30 Ω = 0.4A
Therefore, the current flowing through the circuit is 0.4A.
Question 18
Question
A resistor with resistance R1= 6 Ω and another resistor with resistance R2=
4 Ω are connected in parallel across a 12 Vbattery. Find the total current flowing
through the circuit.
Solution
Step 1: Calculate the equivalent resistance of the two resistors in parallel. The
formula to find the total resistance of two resistors in parallel is given by:
1
Rtotal
=1
R1
+1
R2
Substitute the given values: R1= 6 Ω and R2= 4 Ω:
1
Rtotal
=1
6+1
4
1
Rtotal
=2
12 +3
12
12
1
Rtotal
=5
12
Therefore, the equivalent resistance Rtotal is:
Rtotal =12
5Ω = 2.4 Ω
Step 2: Calculate the total current using Ohm’s Law. The total current
flowing through the circuit can be found using Ohm’s Law: I=V
R, where
V= 12 Vis the voltage of the battery and Rtotal = 2.4 Ω is the total resistance.
I=12
2.4= 5 A
Therefore, the total current flowing through the circuit is 5A.
Question 19
Question
A circuit consists of a resistor with resistance R= 30 Ω and an unknown resistor
connected in series. When a voltage of V= 12 Vis applied across the circuit,
a current of I= 0.4Aflows through it. What is the resistance of the unknown
resistor?
Solution
Step 1: Recall Ohm’s Law which states V=IR, where Vis the voltage across
a component, Iis the current flowing through it, and Ris the resistance of the
component.
Step 2: We first need to find the total resistance of the circuit. In a se-
ries circuit, the total resistance Rtotal is the sum of the individual resistances:
Rtotal =R1+R2+. . . +Rn.
Step 3: In this case, the total resistance Rtotal is given by Rtotal =R+
Runknown.
Step 4: We can now use Ohm’s Law to find the total resistance of the circuit.
Given that V= 12 Vand I= 0.4A, we have:
Rtotal =V
I
Step 5: Substituting the given values, we find:
R+Runknown =12 V
0.4A
Step 6: Simplifying the equation, we get:
30 Ω + Runknown = 30 Ω
13
Step 7: Subtracting 30 Ω from both sides gives:
Runknown = 0 Ω
Step 8: Therefore, the resistance of the unknown resistor in the circuit is
0 Ω .
Question 20
Question
A circuit consists of a resistor with resistance R, a capacitor with capacitance
C, and an inductor with inductance L. The total impedance of the circuit is
given by Z=√R2+ (XL−XC)2, where XL= 2πfL is the inductive reactance,
XC=1
2πf C is the capacitive reactance, and fis the frequency of the alternating
current in the circuit. Determine the frequency of the current that minimizes
the total impedance of the circuit.
Solution
Step 1: Substitute the expressions for XLand XCinto the equation for total
impedance Z.
Z=√R2+(2πfL −1
2πfC )2
Step 2: To minimize Z, we need to find the frequency fsuch that the
derivative of Zwith respect to fis zero.
dZ
df = 0
Step 3: Differentiate Zwith respect to fusing the chain rule and set the
derivative equal to zero.
dZ
df =1
2√R2+(2πfL −1
2πf C )2·2(2πL −1
2πC )= 0
Step 4: Solve for fby setting the expression inside the parentheses equal to
zero.
2πL −1
2πC = 0
Step 5: Solve for f.
2πL =1
2πC
4π2L=1
C
14
f=1
2π√LC
So, the frequency that minimizes the total impedance of the circuit is f=
1
2π√LC .
Question 21
Question
A circuit consists of a resistor with a resistance of 10 Ω, a capacitor with a
capacitance of 0.01 F, and a battery with an emf of 6 V. The switch in the
circuit is closed at t= 0. Calculate the current flowing through the circuit
when t→ ∞.
Solution
Step 1: Find the time constant, τ, of the circuit using the formula τ=RC
where Ris the resistance and Cis the capacitance.
τ= (10 Ω)(0.01 F) = 0.1s
Step 2: Determine the current, I(t), as a function of time tusing the equation
I(t) = I0e−t/τ
where I0is the initial current at t= 0. At t= 0, the current is given by Ohm’s
Law as I(0) = V
R.
I0=6V
10 Ω = 0.6A
So, the current as a function of time is:
I(t) = 0.6A·e−t/0.1s
Step 3: Calculate the current flowing through the circuit as t→ ∞ by
evaluating the limit of I(t)as tapproaches infinity.
lim
t→∞ I(t) = lim
t→∞ 0.6A·e−t/0.1s
lim
t→∞ I(t) = 0.6A·e−∞ = 0
Therefore, the current flowing through the circuit when t→ ∞ is 0.
Question 22
Question
A certain electrical device operates at a voltage of 120 V and draws a current
of 5 A. If the device has a resistance of 24 Ω, calculate the power consumed by
the device.
15
Solution
Step 1: Recall Ohm’s Law which states: V=I·R, where Vis the voltage across
the device, Iis the current passing through the device, and Ris the resistance
of the device.
Step 2: We are given that the voltage Vis 120 V, the current Iis 5 A, and
the resistance Ris 24 Ω. Therefore, we can rearrange Ohm’s Law to solve for
R:
R=V
I=120
5= 24 Ω
Step 3: Next, we calculate the power consumed by the device using the
formula: P=V·I, where Pis power in watts.
P=V·I= 120 ·5 = 600 W
Step 4: The power consumed by the device is 600 watts. Thus, the device
consumes 600 watts of power when operating at a voltage of 120 V and drawing
a current of 5 A with a resistance of 24 Ω.
Question 23
Question
An electric circuit consists of a resistor with resistance R, a capacitor with
capacitance C, and an inductor with inductance Lconnected in series to an
alternating voltage source with angular frequency ω. The voltage across the
resistor, capacitor, and inductor is VR=V0cos(ωt),VC=V0cos(ωt −π
2), and
VL=V0cos(ωt +π
2), respectively.
If the current in the circuit is given by I(t) = I0cos(ωt −ϕ), where I0is the
amplitude of the current and ϕis the phase angle between the current and the
voltage source, show that the amplitude I0of the current is given by
I0=V0
√R2+ (ωL −1
ωC )2
and the phase angle ϕis given by
tan ϕ=ωL −1
ωC
R
Solution
Step 1: Apply Ohm’s Law to find the relationship between the voltage VR,
current I, and resistance R.
VR=IR
16
Step 2: Substitute the expressions for VRand Iin terms of time to find the
expression for I0.
V0cos(ωt) = I0Rcos(ωt −ϕ)
Step 3: Rearrange the equation to solve for I0.
I0=V0
R
Step 4: Apply impedance of a series LRC circuit to find the total impedance
Z.
Z=√R2+(ωL −1
ωC )2
Step 5: Since Z=V0
I0, we have I0=V0
Z. Substitute Zinto the equation.
I0=V0
√R2+(ωL −1
ωC )2
Step 6: Finally, use the phase angle relationship tan ϕ=ωL−1
ωC
Rto determine
the phase angle ϕ.
Question 24
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
difference across the unknown device is measured to be 8 V. Find the resistance
of the unknown device.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor is
equal to the current (I) flowing through the resistor multiplied by the resistance
(R) of the resistor. Mathematically, this can be represented by the equation:
V=I·R.
Step 2: In this problem, the potential difference (V) across the unknown
device is 8 V, and the current (I) flowing through the circuit is 2 A. Therefore,
using Ohm’s Law, we can calculate the resistance (R) of the unknown device
as:
R=V
I=8V
2A= 4 Ω.
Step 3: Hence, the resistance of the unknown device in the circuit is 4 Ω.
17
Question 25
Question
A circuit consists of three resistors in series: R1= 10 Ω,R2= 20 Ω, and R3=
30 Ω. The circuit is connected to a 12 V battery. Calculate the current through
each resistor and the total power dissipated in the circuit.
Solution
Step 1: Calculate the total resistance in the circuit. The total resistance in a
series circuit is the sum of the individual resistances:
Rtotal =R1+R2+R3= 10 Ω + 20 Ω + 30 Ω = 60 Ω
Step 2: Calculate the total current in the circuit using Ohm’s Law (V=IR):
I=V
Rtotal
=12 V
60 Ω = 0.2A
Step 3: Calculate the current through each resistor. In a series circuit, the
current is the same through each resistor. Therefore, I=I1=I2=I3= 0.2A
Step 4: Calculate the power dissipated by each resistor using the formula
P=IV :
P1=I·V= 0.2A×12 V= 2.4W
P2=I·V= 0.2A×12 V= 2.4W
P3=I·V= 0.2A×12 V= 2.4W
Step 5: Calculate the total power dissipated in the circuit. The total power
is the sum of the power dissipated by each resistor:
Ptotal =P1+P2+P3= 2.4W+ 2.4W+ 2.4W= 7.2W
Therefore, the current through each resistor is 0.2A and the total power
dissipated in the circuit is 7.2W.
18
Solution
Step 1: Recall Ohm’s Law, which states that the current passing through a
resistor is directly proportional to the voltage across the resistor and inversely
proportional to the resistance of the resistor. Mathematically, Ohm’s Law is
expressed as:
V=IR
where: V= voltage across the resistor in volts, I= current flowing through the
resistor in amperes, and R= resistance of the resistor in ohms.
Step 2: Given information: Resistance of the resistor, R= 5 Ω Voltage across
the resistor, V= 12 V
Step 3: Substitute the given values into Ohm’s Law equation:
I=V
R
I=12 V
5 Ω
Step 4: Perform the division to calculate the current flowing through the
resistor:
I=12
5
I= 2.4A
Step 5: Therefore, the current flowing through the resistor is 2.4 amperes.
Question 3
Question
A circuit consists of a battery with a voltage of 12 V connected in series to three
resistors: R1= 4 Ω,R2= 6 Ω, and R3= 8 Ω. Calculate the current passing
through each resistor and the power dissipated by each resistor.
Solution
Step 1: First, we calculate the total resistance of the circuit by adding up the
individual resistances:
Rtotal =R1+R2+R3= 4 Ω + 6 Ω + 8 Ω = 18 Ω
Step 2: Next, we use Ohm’s Law to find the total current passing through
the circuit:
V=I·Rtotal
I=V
Rtotal
=12 V
18 Ω = 0.67 A
2
Step 3: To find the current passing through each resistor, we can use the
fact that in a series circuit, the current remains constant:
I=I1=I2=I3= 0.67 A
Step 4: Now, we can calculate the power dissipated by each resistor using
the formula P=I2·R: For R1:
P1=I2·R1= (0.67 A)2·4 Ω = 1.792 W
For R2:
P2=I2·R2= (0.67 A)2·6 Ω = 2.688 W
For R3:
P3=I2·R3= (0.67 A)2·8 Ω = 3.584 W
So, the current passing through each resistor is 0.67 A, and the power dissi-
pated by each resistor is 1.792 W, 2.688 W, and 3.584 W, respectively.
Question 4
Question
A circuit consists of a resistor with resistance R= 5 Ω and an unknown resistor
connected in series to a battery with emf E= 12 V. When a current of 2A
flows through the circuit, the potential difference across the unknown resistor
is found to be 6V. Determine the resistance of the unknown resistor.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor is
equal to the current (I) flowing through the resistor multiplied by the resistance
(R) of the resistor: V=IR.
Step 2: For the total circuit, the total potential difference provided by the
battery is equal to the sum of the potential differences across each resistor.
Therefore, the potential difference Vtotal across the total circuit is equal to the
potential difference across the 5 Ω resistor (V5) plus the potential difference
across the unknown resistor (Vunknown). Mathematically, this can be expressed
as: Vtotal =V5+Vunknown.
Step 3: We are given that the potential difference across the 5 Ω resistor is
6V and the total current in the circuit is 2A. Using Ohm’s Law for the 5 Ω
resistor, we have V5=I·R= 2 A×5 Ω = 10 V.
Step 4: Substitute V5= 10 V and Vtotal = 12 V into the equation from Step
2: 12 V= 10 V+Vunknown, we can solve for Vunknown.
Step 5: Vunknown = 12 V−10 V= 2 V.
Step 6: Finally, to find the resistance of the unknown resistor Runknown,
we can use Ohm’s Law with the current I= 2 A and the potential difference
Vunknown = 2 V: Runknown =Vunknown
I=2V
2A= 1 Ω.
Therefore, the resistance of the unknown resistor is 1 Ω.
3
Question 5
Question
A circuit consists of a resistor with resistance R= 10 Ω and an unknown resistor
connected in series to a 12 Vbattery. If the current passing through this circuit
is 0.5A, find the resistance of the unknown resistor.
Solution
Step 1: We know that in a series circuit, the total resistance Rtotal is the sum
of the individual resistances. Thus, Rtotal =R+Runknown.
Step 2: Ohm’s Law states that the voltage across a resistor equals the current
passing through it multiplied by the resistance. Therefore, the voltage across
the resistor with known resistance Ris VR=I·R= 0.5A·10 Ω = 5 V.
Step 3: Since the total voltage in a series circuit is equal to the sum of the
voltages across each component, we have Vtotal = 12 V. Thus, the voltage across
the unknown resistor is Vunknown =Vtotal −VR= 12 V−5V= 7 V.
Step 4: Applying Ohm’s Law again, we can find the resistance of the un-
known resistor: Runknown =Vunknown
I=7V
0.5A= 14 Ω.
Therefore, the resistance of the unknown resistor in the circuit is 14 Ω.
Question 6
Question
A circuit consists of a resistor with a resistance of 10 Ωand a battery with an
electromotive force of 12 V. If a current of 1.2 A flows through the circuit, what
is the potential difference across the resistor?
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 is represented by the formula V=IR.
Step 2: Substituting the given values, we have V= 1.2A×10 Ω.
Step 3: Calculate the potential difference using the formula:
V= 1.2A×10 Ω = 12 V
Step 4: Therefore, the potential difference across the resistor is 12 V.
4
Question 7
Question
A 12 V battery is connected to a series circuit containing three resistors: R1=
10 Ω,R2= 15 Ω, and R3= 20 Ω. Calculate the current passing through each
resistor in the circuit.
Solution
We can use Ohm’s Law, V=IR, to find the current passing through each
resistor in the circuit.
Step 1: Calculate the total resistance in the circuit. The total resistance in
a series circuit is the sum of the individual resistances:
Rtotal =R1+R2+R3= 10 Ω + 15 Ω + 20 Ω = 45 Ω
Step 2: Calculate the total current in the circuit. Using Ohm’s Law for the
total circuit:
Itotal =V
Rtotal
=12 V
45 Ω =4
15 A= 0.27 A
Step 3: Calculate the current passing through each resistor. Now, using
Ohm’s Law for individual resistors: For R1:
I1=V
R1
=12 V
10 Ω = 1.2A
For R2:
I2=V
R2
=12 V
15 Ω = 0.8A
For R3:
I3=V
R3
=12 V
20 Ω = 0.6A
Therefore, the current passing through R1is 1.2 A, through R2is 0.8 A, and
through R3is 0.6 A in the circuit.
Question 8
Question
A circuit consists of a resistor with resistance 10 Ω, an inductor with inductance
0.02 H, and a capacitor with capacitance 5 µF connected in series to a 12 V AC
power supply that operates at a frequency of 60 Hz.
Calculate the impedance of the circuit at this frequency.
5
Solution
Step 1: Calculate the impedance of each component.
• The impedance of a resistor is equal to its resistance. Therefore, the
impedance of the resistor is 10 Ω.
• The impedance of an inductor is given by ZL=jωL, where Lis the
inductance and ω= 2πf . Plugging in the values, we get:
ZL=j(2π·60)(0.02) = j2.4 Ω.
• The impedance of a capacitor is given by ZC=1
jωC , where Cis the
capacitance. Substituting the values, we get:
ZC=1
j(2π·60)(5 ×10−6)=−j53.05 Ω.
Step 2: Calculate the total impedance of the circuit.
Ztotal =ZR+ZL+ZC= 10 −j2.4−j53.05 Ω.
Step 3: Convert the total impedance to polar form.
Ztotal = 10 −j55.45 Ω.
Step 4: Calculate the magnitude of the total impedance.
|Ztotal|=√102+ (−55.45)2=√3113.7025 ≈55.78 Ω.
Therefore, the impedance of the circuit at a frequency of 60 Hz is approxi-
mately 55.78 Ω.
Question 9
Question
A circuit consists of a resistor of resistance R1, a resistor of resistance R2, and
a battery with emf Eand internal resistance r, all connected in series. The
resistance R1is twice as large as the resistance R2. The potential difference
across R2is found to be half the value of the emf of the battery. Find an
expression for the current Iin the circuit in terms of E,R1,R2, and r.
Given data:
R1= 2R2, VR2=1
2E
6
Solution
Step 1: Use Ohm’s Law to relate the potential differences and resistances in the
circuit.
E=VR1+Vr+VR2
IR1=IR2+Ir +E
2
Step 2: Substitute R1= 2R2into the equation.
I(2R2) = IR2+Ir +E
2
Step 3: Rearrange the equation in terms of I.
2IR2=IR2+Ir +E
2
2I−I=E
2R2−Ir
R2
I=E
2R2−Ir
R2
Step 4: Substitute R1= 2R2into the expression for I.
I=E
2R2−Ir
R2
Therefore, the current Iin the circuit in terms of E,R1,R2, and ris I=
E
2R2−Ir
R2.
Question 10
Question
A circuit consists of a resistor with resistance R= 20Ω and an unknown device
represented by a box with two terminals. When a current of 2.5A passes through
the circuit, a potential difference of 25 V is observed across the terminals of the
unknown device. Determine the resistance of the unknown device.
Solution
Step 1: Recall Ohm’s Law, which states that the potential difference (V) across
a device is equal to the current (I) passing through the device multiplied by the
resistance (R) of the device. The formula for Ohm’s Law is V=IR.
Step 2: We know that a potential difference of 25 V is observed across the
terminals of the unknown device when a current of 2.5A passes through the
7
circuit. Using Ohm’s Law, we can find the resistance of the unknown device.
Given V= 25 V and I= 2.5A, we have:
R=V
I=25 V
2.5A
Step 3: Calculate the resistance of the unknown device:
R= 10Ω
Step 4: Therefore, the resistance of the unknown device represented by the
box is 10Ω.
Question 11
Question
A wire with resistance Ris connected to a battery with emf Eand internal resis-
tance r. When a certain current Iflows through the wire, the power dissipated
is at its maximum value. Prove that the internal resistance rof the battery
must be equal to the resistance Rof the wire.
Solution
Step 1: Recall that the power dissipated in a circuit is given by P=I2R, where
Iis the current and Ris the resistance.
Step 2: The total resistance in the circuit is the sum of the resistance of the
wire (R) and the internal resistance of the battery (r), so the total resistance is
Rtotal =R+r.
Step 3: Using Ohm’s Law (V=IR), we can express the current Iin terms
of the total resistance Rtotal and the emf Eof the battery: I=E
Rtotal =E
R+r.
Step 4: Substitute the expression for Iinto the equation for power P=I2R
to get the power dissipated in terms of R,r, and E:P=(E
R+r)2·R.
Step 5: To find the maximum power dissipated, we can take the derivative
of Pwith respect to r, set it equal to 0, and solve for r.
Step 6: Differentiate Pwith respect to r:dP
dr =2E2R
(R+r)3−E2R
(R+r)2.
Step 7: Set dP
dr = 0 and solve for r:2E2R
(R+r)3−E2R
(R+r)2= 0.
Step 8: Simplifying the equation, we find that r=R. Thus, the internal
resistance of the battery must be equal to the resistance of the wire for the
power dissipated to be at its maximum value.
Question 12
Question
In a circuit, a resistor with resistance R1= 20 Ω and a resistor with resistance
R2= 30 Ω are connected in series with a battery that provides a voltage of
8
V= 60 V. Calculate the total current in the circuit.
Solution
Step 1: Calculate the total resistance in the circuit. The total resistance in
a series circuit is the sum of the individual resistances. Therefore, the total
resistance Rtotal is given by:
Rtotal =R1+R2= 20 Ω + 30 Ω = 50 Ω
Step 2: Apply Ohm’s Law to find the total current. Ohm’s Law states
that the current flowing through a circuit is inversely proportional to the total
resistance and directly proportional to the applied voltage. Mathematically,
Ohm’s Law is given by:
V=I·R
where Vis the voltage, Iis the current, and Ris the resistance.
Substitute the values we have into the equation:
I=V
Rtotal
=60 V
50 Ω = 1.2A
Therefore, the total current in the circuit is 1.2A.
Question 13
Question
A copper wire with a resistance of 2.5 Ωis connected in series with a resistor of
unknown resistance. When a potential difference of 12 V is applied across the
circuit, a current of 3 A is observed. Determine the resistance of the unknown
resistor.
Solution
Step 1: Use Ohm’s Law (V=IR) to find the total resistance of the circuit.
Given: Resistance of the copper wire, Rcopper = 2.5 Ω Potential difference,
V= 12 V Current, I= 3 A
The total resistance can be calculated as:
Rtotal =V
I
Rtotal =12 V
3A= 4 Ω
Step 2: Use the total resistance to find the resistance of the unknown resis-
tor. Let Runknown be the resistance of the unknown resistor. Since the resistors
9
are connected in series, the total resistance Rtotal is the sum of individual resis-
tances:
Rtotal =Rcopper +Runknown
4 Ω = 2.5 Ω + Runknown
Runknown = 4 Ω −2.5 Ω = 1.5 Ω
Therefore, the resistance of the unknown resistor is 1.5 Ω.
Question 14
Question
A circuit consists of three resistors connected in series. The first resistor has a
resistance of 4 Ω, the second resistor has a resistance of 6 Ω, and the third resistor
has an unknown resistance R. A potential difference of 24 Vis applied across
the circuit. If the current through the circuit is 2A, determine the resistance R
of the third resistor.
Solution
Step 1: Recall the formula for Ohm’s Law: V=IR, where Vis the potential
difference across the circuit, Iis the current flowing through the circuit, and Ris
the total resistance of the circuit. Step 2: Determine the total resistance Rtotal of
the circuit by summing the individual resistances: Rtotal =R1+R2+R. Step 3:
Substitute the given resistances into the total resistance formula: Rtotal = 4 Ω+
6 Ω + R= 10 Ω + R. Step 4: Use Ohm’s Law to find the total resistance of the
circuit: V=IRtotal. Substitute V= 24 Vand I= 2 A:24 V= 2 A×(10 Ω+R).
Step 5: Solve for the unknown resistance R:24 V= 20 Ω + 2R
2R= 24 V−20 Ω
2R= 4 V
R= 2 Ω.
Therefore, the resistance of the third resistor is 2 Ω.
Question 15
Question
A circuit consists of three resistors connected in series. The resistances of the
resistors are R1= 6 Ω,R2= 8 Ω, and R3= 10 Ω. If the total potential difference
across the circuit is V= 120 V, what is the current flowing through the circuit?
10
Solution
Step 1: Calculate the total resistance of the circuit using the formula for resistors
in series: Rtotal =R1+R2+R3
= 6 Ω + 8 Ω + 10 Ω
= 24 Ω.
Step 2: Use Ohm’s Law, V=IR, to find the current flowing through the
circuit:
I=V
Rtotal
=120 V
24 Ω
= 5 A.
Therefore, the current flowing through the circuit is 5A.
Question 16
Question
A copper wire with a resistance of 2.5 Ωis connected to a battery that provides
a current of 0.8 A. If the resistance of the wire is increased to 5 Ω, what is the
new current in the circuit?
Solution
Step 1: Calculate the initial voltage drop across the 2.5 Ωresistor using Ohm’s
Law, V=IR.
Initial Voltage = (0.8A)(2.5 Ω) = 2 V
Step 2: Since the battery provides a constant voltage, the total voltage drop
in the circuit remains 2 V when the resistance is increased to 5 Ω. Using Ohm’s
Law, we can find the new current in the circuit.
2V=I(5 Ω)
I=2V
5 Ω = 0.4A
Therefore, the new current in the circuit when the resistance is increased to
5Ωis 0.4 A.
11
Question 17
Question
A circuit consists of a resistor with a resistance of 30 Ω connected in series with a
battery that provides a voltage of 12 V. Determine the current flowing through
the circuit.
Solution
Step 1: Write down Ohm’s Law, which relates voltage (V), current (I), and
resistance (R) in a circuit:
V=I·R
Step 2: Given that the resistance R= 30 Ω and voltage V= 12 V, we can
use Ohm’s Law to solve for the current I:
12 V=I·30 Ω
Step 3: Solve for the current I:
I=12 V
30 Ω = 0.4A
Therefore, the current flowing through the circuit is 0.4A.
Question 18
Question
A resistor with resistance R1= 6 Ω and another resistor with resistance R2=
4 Ω are connected in parallel across a 12 Vbattery. Find the total current flowing
through the circuit.
Solution
Step 1: Calculate the equivalent resistance of the two resistors in parallel. The
formula to find the total resistance of two resistors in parallel is given by:
1
Rtotal
=1
R1
+1
R2
Substitute the given values: R1= 6 Ω and R2= 4 Ω:
1
Rtotal
=1
6+1
4
1
Rtotal
=2
12 +3
12
12
1
Rtotal
=5
12
Therefore, the equivalent resistance Rtotal is:
Rtotal =12
5Ω = 2.4 Ω
Step 2: Calculate the total current using Ohm’s Law. The total current
flowing through the circuit can be found using Ohm’s Law: I=V
R, where
V= 12 Vis the voltage of the battery and Rtotal = 2.4 Ω is the total resistance.
I=12
2.4= 5 A
Therefore, the total current flowing through the circuit is 5A.
Question 19
Question
A circuit consists of a resistor with resistance R= 30 Ω and an unknown resistor
connected in series. When a voltage of V= 12 Vis applied across the circuit,
a current of I= 0.4Aflows through it. What is the resistance of the unknown
resistor?
Solution
Step 1: Recall Ohm’s Law which states V=IR, where Vis the voltage across
a component, Iis the current flowing through it, and Ris the resistance of the
component.
Step 2: We first need to find the total resistance of the circuit. In a se-
ries circuit, the total resistance Rtotal is the sum of the individual resistances:
Rtotal =R1+R2+. . . +Rn.
Step 3: In this case, the total resistance Rtotal is given by Rtotal =R+
Runknown.
Step 4: We can now use Ohm’s Law to find the total resistance of the circuit.
Given that V= 12 Vand I= 0.4A, we have:
Rtotal =V
I
Step 5: Substituting the given values, we find:
R+Runknown =12 V
0.4A
Step 6: Simplifying the equation, we get:
30 Ω + Runknown = 30 Ω
13
Step 7: Subtracting 30 Ω from both sides gives:
Runknown = 0 Ω
Step 8: Therefore, the resistance of the unknown resistor in the circuit is
0 Ω .
Question 20
Question
A circuit consists of a resistor with resistance R, a capacitor with capacitance
C, and an inductor with inductance L. The total impedance of the circuit is
given by Z=√R2+ (XL−XC)2, where XL= 2πfL is the inductive reactance,
XC=1
2πf C is the capacitive reactance, and fis the frequency of the alternating
current in the circuit. Determine the frequency of the current that minimizes
the total impedance of the circuit.
Solution
Step 1: Substitute the expressions for XLand XCinto the equation for total
impedance Z.
Z=√R2+(2πfL −1
2πfC )2
Step 2: To minimize Z, we need to find the frequency fsuch that the
derivative of Zwith respect to fis zero.
dZ
df = 0
Step 3: Differentiate Zwith respect to fusing the chain rule and set the
derivative equal to zero.
dZ
df =1
2√R2+(2πfL −1
2πf C )2·2(2πL −1
2πC )= 0
Step 4: Solve for fby setting the expression inside the parentheses equal to
zero.
2πL −1
2πC = 0
Step 5: Solve for f.
2πL =1
2πC
4π2L=1
C
14
f=1
2π√LC
So, the frequency that minimizes the total impedance of the circuit is f=
1
2π√LC .
Question 21
Question
A circuit consists of a resistor with a resistance of 10 Ω, a capacitor with a
capacitance of 0.01 F, and a battery with an emf of 6 V. The switch in the
circuit is closed at t= 0. Calculate the current flowing through the circuit
when t→ ∞.
Solution
Step 1: Find the time constant, τ, of the circuit using the formula τ=RC
where Ris the resistance and Cis the capacitance.
τ= (10 Ω)(0.01 F) = 0.1s
Step 2: Determine the current, I(t), as a function of time tusing the equation
I(t) = I0e−t/τ
where I0is the initial current at t= 0. At t= 0, the current is given by Ohm’s
Law as I(0) = V
R.
I0=6V
10 Ω = 0.6A
So, the current as a function of time is:
I(t) = 0.6A·e−t/0.1s
Step 3: Calculate the current flowing through the circuit as t→ ∞ by
evaluating the limit of I(t)as tapproaches infinity.
lim
t→∞ I(t) = lim
t→∞ 0.6A·e−t/0.1s
lim
t→∞ I(t) = 0.6A·e−∞ = 0
Therefore, the current flowing through the circuit when t→ ∞ is 0.
Question 22
Question
A certain electrical device operates at a voltage of 120 V and draws a current
of 5 A. If the device has a resistance of 24 Ω, calculate the power consumed by
the device.
15
Solution
Step 1: Recall Ohm’s Law which states: V=I·R, where Vis the voltage across
the device, Iis the current passing through the device, and Ris the resistance
of the device.
Step 2: We are given that the voltage Vis 120 V, the current Iis 5 A, and
the resistance Ris 24 Ω. Therefore, we can rearrange Ohm’s Law to solve for
R:
R=V
I=120
5= 24 Ω
Step 3: Next, we calculate the power consumed by the device using the
formula: P=V·I, where Pis power in watts.
P=V·I= 120 ·5 = 600 W
Step 4: The power consumed by the device is 600 watts. Thus, the device
consumes 600 watts of power when operating at a voltage of 120 V and drawing
a current of 5 A with a resistance of 24 Ω.
Question 23
Question
An electric circuit consists of a resistor with resistance R, a capacitor with
capacitance C, and an inductor with inductance Lconnected in series to an
alternating voltage source with angular frequency ω. The voltage across the
resistor, capacitor, and inductor is VR=V0cos(ωt),VC=V0cos(ωt −π
2), and
VL=V0cos(ωt +π
2), respectively.
If the current in the circuit is given by I(t) = I0cos(ωt −ϕ), where I0is the
amplitude of the current and ϕis the phase angle between the current and the
voltage source, show that the amplitude I0of the current is given by
I0=V0
√R2+ (ωL −1
ωC )2
and the phase angle ϕis given by
tan ϕ=ωL −1
ωC
R
Solution
Step 1: Apply Ohm’s Law to find the relationship between the voltage VR,
current I, and resistance R.
VR=IR
16
Step 2: Substitute the expressions for VRand Iin terms of time to find the
expression for I0.
V0cos(ωt) = I0Rcos(ωt −ϕ)
Step 3: Rearrange the equation to solve for I0.
I0=V0
R
Step 4: Apply impedance of a series LRC circuit to find the total impedance
Z.
Z=√R2+(ωL −1
ωC )2
Step 5: Since Z=V0
I0, we have I0=V0
Z. Substitute Zinto the equation.
I0=V0
√R2+(ωL −1
ωC )2
Step 6: Finally, use the phase angle relationship tan ϕ=ωL−1
ωC
Rto determine
the phase angle ϕ.
Question 24
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
difference across the unknown device is measured to be 8 V. Find the resistance
of the unknown device.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor is
equal to the current (I) flowing through the resistor multiplied by the resistance
(R) of the resistor. Mathematically, this can be represented by the equation:
V=I·R.
Step 2: In this problem, the potential difference (V) across the unknown
device is 8 V, and the current (I) flowing through the circuit is 2 A. Therefore,
using Ohm’s Law, we can calculate the resistance (R) of the unknown device
as:
R=V
I=8V
2A= 4 Ω.
Step 3: Hence, the resistance of the unknown device in the circuit is 4 Ω.
17
Question 25
Question
A circuit consists of three resistors in series: R1= 10 Ω,R2= 20 Ω, and R3=
30 Ω. The circuit is connected to a 12 V battery. Calculate the current through
each resistor and the total power dissipated in the circuit.
Solution
Step 1: Calculate the total resistance in the circuit. The total resistance in a
series circuit is the sum of the individual resistances:
Rtotal =R1+R2+R3= 10 Ω + 20 Ω + 30 Ω = 60 Ω
Step 2: Calculate the total current in the circuit using Ohm’s Law (V=IR):
I=V
Rtotal
=12 V
60 Ω = 0.2A
Step 3: Calculate the current through each resistor. In a series circuit, the
current is the same through each resistor. Therefore, I=I1=I2=I3= 0.2A
Step 4: Calculate the power dissipated by each resistor using the formula
P=IV :
P1=I·V= 0.2A×12 V= 2.4W
P2=I·V= 0.2A×12 V= 2.4W
P3=I·V= 0.2A×12 V= 2.4W
Step 5: Calculate the total power dissipated in the circuit. The total power
is the sum of the power dissipated by each resistor:
Ptotal =P1+P2+P3= 2.4W+ 2.4W+ 2.4W= 7.2W
Therefore, the current through each resistor is 0.2A and the total power
dissipated in the circuit is 7.2W.
18
Solution
Step 1: Recall Ohm’s Law, which states that the current passing through a
resistor is directly proportional to the voltage across the resistor and inversely
proportional to the resistance of the resistor. Mathematically, Ohm’s Law is
expressed as:
V=IR
where: V= voltage across the resistor in volts, I= current flowing through the
resistor in amperes, and R= resistance of the resistor in ohms.
Step 2: Given information: Resistance of the resistor, R= 5 Ω Voltage across
the resistor, V= 12 V
Step 3: Substitute the given values into Ohm’s Law equation:
I=V
R
I=12 V
5 Ω
Step 4: Perform the division to calculate the current flowing through the
resistor:
I=12
5
I= 2.4A
Step 5: Therefore, the current flowing through the resistor is 2.4 amperes.
Question 3
Question
A circuit consists of a battery with a voltage of 12 V connected in series to three
resistors: R1= 4 Ω,R2= 6 Ω, and R3= 8 Ω. Calculate the current passing
through each resistor and the power dissipated by each resistor.
Solution
Step 1: First, we calculate the total resistance of the circuit by adding up the
individual resistances:
Rtotal =R1+R2+R3= 4 Ω + 6 Ω + 8 Ω = 18 Ω
Step 2: Next, we use Ohm’s Law to find the total current passing through
the circuit:
V=I·Rtotal
I=V
Rtotal
=12 V
18 Ω = 0.67 A
2
Step 3: To find the current passing through each resistor, we can use the
fact that in a series circuit, the current remains constant:
I=I1=I2=I3= 0.67 A
Step 4: Now, we can calculate the power dissipated by each resistor using
the formula P=I2·R: For R1:
P1=I2·R1= (0.67 A)2·4 Ω = 1.792 W
For R2:
P2=I2·R2= (0.67 A)2·6 Ω = 2.688 W
For R3:
P3=I2·R3= (0.67 A)2·8 Ω = 3.584 W
So, the current passing through each resistor is 0.67 A, and the power dissi-
pated by each resistor is 1.792 W, 2.688 W, and 3.584 W, respectively.
Question 4
Question
A circuit consists of a resistor with resistance R= 5 Ω and an unknown resistor
connected in series to a battery with emf E= 12 V. When a current of 2A
flows through the circuit, the potential difference across the unknown resistor
is found to be 6V. Determine the resistance of the unknown resistor.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor is
equal to the current (I) flowing through the resistor multiplied by the resistance
(R) of the resistor: V=IR.
Step 2: For the total circuit, the total potential difference provided by the
battery is equal to the sum of the potential differences across each resistor.
Therefore, the potential difference Vtotal across the total circuit is equal to the
potential difference across the 5 Ω resistor (V5) plus the potential difference
across the unknown resistor (Vunknown). Mathematically, this can be expressed
as: Vtotal =V5+Vunknown.
Step 3: We are given that the potential difference across the 5 Ω resistor is
6V and the total current in the circuit is 2A. Using Ohm’s Law for the 5 Ω
resistor, we have V5=I·R= 2 A×5 Ω = 10 V.
Step 4: Substitute V5= 10 V and Vtotal = 12 V into the equation from Step
2: 12 V= 10 V+Vunknown, we can solve for Vunknown.
Step 5: Vunknown = 12 V−10 V= 2 V.
Step 6: Finally, to find the resistance of the unknown resistor Runknown,
we can use Ohm’s Law with the current I= 2 A and the potential difference
Vunknown = 2 V: Runknown =Vunknown
I=2V
2A= 1 Ω.
Therefore, the resistance of the unknown resistor is 1 Ω.
3
Question 5
Question
A circuit consists of a resistor with resistance R= 10 Ω and an unknown resistor
connected in series to a 12 Vbattery. If the current passing through this circuit
is 0.5A, find the resistance of the unknown resistor.
Solution
Step 1: We know that in a series circuit, the total resistance Rtotal is the sum
of the individual resistances. Thus, Rtotal =R+Runknown.
Step 2: Ohm’s Law states that the voltage across a resistor equals the current
passing through it multiplied by the resistance. Therefore, the voltage across
the resistor with known resistance Ris VR=I·R= 0.5A·10 Ω = 5 V.
Step 3: Since the total voltage in a series circuit is equal to the sum of the
voltages across each component, we have Vtotal = 12 V. Thus, the voltage across
the unknown resistor is Vunknown =Vtotal −VR= 12 V−5V= 7 V.
Step 4: Applying Ohm’s Law again, we can find the resistance of the un-
known resistor: Runknown =Vunknown
I=7V
0.5A= 14 Ω.
Therefore, the resistance of the unknown resistor in the circuit is 14 Ω.
Question 6
Question
A circuit consists of a resistor with a resistance of 10 Ωand a battery with an
electromotive force of 12 V. If a current of 1.2 A flows through the circuit, what
is the potential difference across the resistor?
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 is represented by the formula V=IR.
Step 2: Substituting the given values, we have V= 1.2A×10 Ω.
Step 3: Calculate the potential difference using the formula:
V= 1.2A×10 Ω = 12 V
Step 4: Therefore, the potential difference across the resistor is 12 V.
4
Question 7
Question
A 12 V battery is connected to a series circuit containing three resistors: R1=
10 Ω,R2= 15 Ω, and R3= 20 Ω. Calculate the current passing through each
resistor in the circuit.
Solution
We can use Ohm’s Law, V=IR, to find the current passing through each
resistor in the circuit.
Step 1: Calculate the total resistance in the circuit. The total resistance in
a series circuit is the sum of the individual resistances:
Rtotal =R1+R2+R3= 10 Ω + 15 Ω + 20 Ω = 45 Ω
Step 2: Calculate the total current in the circuit. Using Ohm’s Law for the
total circuit:
Itotal =V
Rtotal
=12 V
45 Ω =4
15 A= 0.27 A
Step 3: Calculate the current passing through each resistor. Now, using
Ohm’s Law for individual resistors: For R1:
I1=V
R1
=12 V
10 Ω = 1.2A
For R2:
I2=V
R2
=12 V
15 Ω = 0.8A
For R3:
I3=V
R3
=12 V
20 Ω = 0.6A
Therefore, the current passing through R1is 1.2 A, through R2is 0.8 A, and
through R3is 0.6 A in the circuit.
Question 8
Question
A circuit consists of a resistor with resistance 10 Ω, an inductor with inductance
0.02 H, and a capacitor with capacitance 5 µF connected in series to a 12 V AC
power supply that operates at a frequency of 60 Hz.
Calculate the impedance of the circuit at this frequency.
5
Solution
Step 1: Calculate the impedance of each component.
• The impedance of a resistor is equal to its resistance. Therefore, the
impedance of the resistor is 10 Ω.
• The impedance of an inductor is given by ZL=jωL, where Lis the
inductance and ω= 2πf . Plugging in the values, we get:
ZL=j(2π·60)(0.02) = j2.4 Ω.
• The impedance of a capacitor is given by ZC=1
jωC , where Cis the
capacitance. Substituting the values, we get:
ZC=1
j(2π·60)(5 ×10−6)=−j53.05 Ω.
Step 2: Calculate the total impedance of the circuit.
Ztotal =ZR+ZL+ZC= 10 −j2.4−j53.05 Ω.
Step 3: Convert the total impedance to polar form.
Ztotal = 10 −j55.45 Ω.
Step 4: Calculate the magnitude of the total impedance.
|Ztotal|=√102+ (−55.45)2=√3113.7025 ≈55.78 Ω.
Therefore, the impedance of the circuit at a frequency of 60 Hz is approxi-
mately 55.78 Ω.
Question 9
Question
A circuit consists of a resistor of resistance R1, a resistor of resistance R2, and
a battery with emf Eand internal resistance r, all connected in series. The
resistance R1is twice as large as the resistance R2. The potential difference
across R2is found to be half the value of the emf of the battery. Find an
expression for the current Iin the circuit in terms of E,R1,R2, and r.
Given data:
R1= 2R2, VR2=1
2E
6
Solution
Step 1: Use Ohm’s Law to relate the potential differences and resistances in the
circuit.
E=VR1+Vr+VR2
IR1=IR2+Ir +E
2
Step 2: Substitute R1= 2R2into the equation.
I(2R2) = IR2+Ir +E
2
Step 3: Rearrange the equation in terms of I.
2IR2=IR2+Ir +E
2
2I−I=E
2R2−Ir
R2
I=E
2R2−Ir
R2
Step 4: Substitute R1= 2R2into the expression for I.
I=E
2R2−Ir
R2
Therefore, the current Iin the circuit in terms of E,R1,R2, and ris I=
E
2R2−Ir
R2.
Question 10
Question
A circuit consists of a resistor with resistance R= 20Ω and an unknown device
represented by a box with two terminals. When a current of 2.5A passes through
the circuit, a potential difference of 25 V is observed across the terminals of the
unknown device. Determine the resistance of the unknown device.
Solution
Step 1: Recall Ohm’s Law, which states that the potential difference (V) across
a device is equal to the current (I) passing through the device multiplied by the
resistance (R) of the device. The formula for Ohm’s Law is V=IR.
Step 2: We know that a potential difference of 25 V is observed across the
terminals of the unknown device when a current of 2.5A passes through the
7
circuit. Using Ohm’s Law, we can find the resistance of the unknown device.
Given V= 25 V and I= 2.5A, we have:
R=V
I=25 V
2.5A
Step 3: Calculate the resistance of the unknown device:
R= 10Ω
Step 4: Therefore, the resistance of the unknown device represented by the
box is 10Ω.
Question 11
Question
A wire with resistance Ris connected to a battery with emf Eand internal resis-
tance r. When a certain current Iflows through the wire, the power dissipated
is at its maximum value. Prove that the internal resistance rof the battery
must be equal to the resistance Rof the wire.
Solution
Step 1: Recall that the power dissipated in a circuit is given by P=I2R, where
Iis the current and Ris the resistance.
Step 2: The total resistance in the circuit is the sum of the resistance of the
wire (R) and the internal resistance of the battery (r), so the total resistance is
Rtotal =R+r.
Step 3: Using Ohm’s Law (V=IR), we can express the current Iin terms
of the total resistance Rtotal and the emf Eof the battery: I=E
Rtotal =E
R+r.
Step 4: Substitute the expression for Iinto the equation for power P=I2R
to get the power dissipated in terms of R,r, and E:P=(E
R+r)2·R.
Step 5: To find the maximum power dissipated, we can take the derivative
of Pwith respect to r, set it equal to 0, and solve for r.
Step 6: Differentiate Pwith respect to r:dP
dr =2E2R
(R+r)3−E2R
(R+r)2.
Step 7: Set dP
dr = 0 and solve for r:2E2R
(R+r)3−E2R
(R+r)2= 0.
Step 8: Simplifying the equation, we find that r=R. Thus, the internal
resistance of the battery must be equal to the resistance of the wire for the
power dissipated to be at its maximum value.
Question 12
Question
In a circuit, a resistor with resistance R1= 20 Ω and a resistor with resistance
R2= 30 Ω are connected in series with a battery that provides a voltage of
8
V= 60 V. Calculate the total current in the circuit.
Solution
Step 1: Calculate the total resistance in the circuit. The total resistance in
a series circuit is the sum of the individual resistances. Therefore, the total
resistance Rtotal is given by:
Rtotal =R1+R2= 20 Ω + 30 Ω = 50 Ω
Step 2: Apply Ohm’s Law to find the total current. Ohm’s Law states
that the current flowing through a circuit is inversely proportional to the total
resistance and directly proportional to the applied voltage. Mathematically,
Ohm’s Law is given by:
V=I·R
where Vis the voltage, Iis the current, and Ris the resistance.
Substitute the values we have into the equation:
I=V
Rtotal
=60 V
50 Ω = 1.2A
Therefore, the total current in the circuit is 1.2A.
Question 13
Question
A copper wire with a resistance of 2.5 Ωis connected in series with a resistor of
unknown resistance. When a potential difference of 12 V is applied across the
circuit, a current of 3 A is observed. Determine the resistance of the unknown
resistor.
Solution
Step 1: Use Ohm’s Law (V=IR) to find the total resistance of the circuit.
Given: Resistance of the copper wire, Rcopper = 2.5 Ω Potential difference,
V= 12 V Current, I= 3 A
The total resistance can be calculated as:
Rtotal =V
I
Rtotal =12 V
3A= 4 Ω
Step 2: Use the total resistance to find the resistance of the unknown resis-
tor. Let Runknown be the resistance of the unknown resistor. Since the resistors
9
are connected in series, the total resistance Rtotal is the sum of individual resis-
tances:
Rtotal =Rcopper +Runknown
4 Ω = 2.5 Ω + Runknown
Runknown = 4 Ω −2.5 Ω = 1.5 Ω
Therefore, the resistance of the unknown resistor is 1.5 Ω.
Question 14
Question
A circuit consists of three resistors connected in series. The first resistor has a
resistance of 4 Ω, the second resistor has a resistance of 6 Ω, and the third resistor
has an unknown resistance R. A potential difference of 24 Vis applied across
the circuit. If the current through the circuit is 2A, determine the resistance R
of the third resistor.
Solution
Step 1: Recall the formula for Ohm’s Law: V=IR, where Vis the potential
difference across the circuit, Iis the current flowing through the circuit, and Ris
the total resistance of the circuit. Step 2: Determine the total resistance Rtotal of
the circuit by summing the individual resistances: Rtotal =R1+R2+R. Step 3:
Substitute the given resistances into the total resistance formula: Rtotal = 4 Ω+
6 Ω + R= 10 Ω + R. Step 4: Use Ohm’s Law to find the total resistance of the
circuit: V=IRtotal. Substitute V= 24 Vand I= 2 A:24 V= 2 A×(10 Ω+R).
Step 5: Solve for the unknown resistance R:24 V= 20 Ω + 2R
2R= 24 V−20 Ω
2R= 4 V
R= 2 Ω.
Therefore, the resistance of the third resistor is 2 Ω.
Question 15
Question
A circuit consists of three resistors connected in series. The resistances of the
resistors are R1= 6 Ω,R2= 8 Ω, and R3= 10 Ω. If the total potential difference
across the circuit is V= 120 V, what is the current flowing through the circuit?
10
Solution
Step 1: Calculate the total resistance of the circuit using the formula for resistors
in series: Rtotal =R1+R2+R3
= 6 Ω + 8 Ω + 10 Ω
= 24 Ω.
Step 2: Use Ohm’s Law, V=IR, to find the current flowing through the
circuit:
I=V
Rtotal
=120 V
24 Ω
= 5 A.
Therefore, the current flowing through the circuit is 5A.
Question 16
Question
A copper wire with a resistance of 2.5 Ωis connected to a battery that provides
a current of 0.8 A. If the resistance of the wire is increased to 5 Ω, what is the
new current in the circuit?
Solution
Step 1: Calculate the initial voltage drop across the 2.5 Ωresistor using Ohm’s
Law, V=IR.
Initial Voltage = (0.8A)(2.5 Ω) = 2 V
Step 2: Since the battery provides a constant voltage, the total voltage drop
in the circuit remains 2 V when the resistance is increased to 5 Ω. Using Ohm’s
Law, we can find the new current in the circuit.
2V=I(5 Ω)
I=2V
5 Ω = 0.4A
Therefore, the new current in the circuit when the resistance is increased to
5Ωis 0.4 A.
11
Question 17
Question
A circuit consists of a resistor with a resistance of 30 Ω connected in series with a
battery that provides a voltage of 12 V. Determine the current flowing through
the circuit.
Solution
Step 1: Write down Ohm’s Law, which relates voltage (V), current (I), and
resistance (R) in a circuit:
V=I·R
Step 2: Given that the resistance R= 30 Ω and voltage V= 12 V, we can
use Ohm’s Law to solve for the current I:
12 V=I·30 Ω
Step 3: Solve for the current I:
I=12 V
30 Ω = 0.4A
Therefore, the current flowing through the circuit is 0.4A.
Question 18
Question
A resistor with resistance R1= 6 Ω and another resistor with resistance R2=
4 Ω are connected in parallel across a 12 Vbattery. Find the total current flowing
through the circuit.
Solution
Step 1: Calculate the equivalent resistance of the two resistors in parallel. The
formula to find the total resistance of two resistors in parallel is given by:
1
Rtotal
=1
R1
+1
R2
Substitute the given values: R1= 6 Ω and R2= 4 Ω:
1
Rtotal
=1
6+1
4
1
Rtotal
=2
12 +3
12
12
1
Rtotal
=5
12
Therefore, the equivalent resistance Rtotal is:
Rtotal =12
5Ω = 2.4 Ω
Step 2: Calculate the total current using Ohm’s Law. The total current
flowing through the circuit can be found using Ohm’s Law: I=V
R, where
V= 12 Vis the voltage of the battery and Rtotal = 2.4 Ω is the total resistance.
I=12
2.4= 5 A
Therefore, the total current flowing through the circuit is 5A.
Question 19
Question
A circuit consists of a resistor with resistance R= 30 Ω and an unknown resistor
connected in series. When a voltage of V= 12 Vis applied across the circuit,
a current of I= 0.4Aflows through it. What is the resistance of the unknown
resistor?
Solution
Step 1: Recall Ohm’s Law which states V=IR, where Vis the voltage across
a component, Iis the current flowing through it, and Ris the resistance of the
component.
Step 2: We first need to find the total resistance of the circuit. In a se-
ries circuit, the total resistance Rtotal is the sum of the individual resistances:
Rtotal =R1+R2+. . . +Rn.
Step 3: In this case, the total resistance Rtotal is given by Rtotal =R+
Runknown.
Step 4: We can now use Ohm’s Law to find the total resistance of the circuit.
Given that V= 12 Vand I= 0.4A, we have:
Rtotal =V
I
Step 5: Substituting the given values, we find:
R+Runknown =12 V
0.4A
Step 6: Simplifying the equation, we get:
30 Ω + Runknown = 30 Ω
13
Step 7: Subtracting 30 Ω from both sides gives:
Runknown = 0 Ω
Step 8: Therefore, the resistance of the unknown resistor in the circuit is
0 Ω .
Question 20
Question
A circuit consists of a resistor with resistance R, a capacitor with capacitance
C, and an inductor with inductance L. The total impedance of the circuit is
given by Z=√R2+ (XL−XC)2, where XL= 2πfL is the inductive reactance,
XC=1
2πf C is the capacitive reactance, and fis the frequency of the alternating
current in the circuit. Determine the frequency of the current that minimizes
the total impedance of the circuit.
Solution
Step 1: Substitute the expressions for XLand XCinto the equation for total
impedance Z.
Z=√R2+(2πfL −1
2πfC )2
Step 2: To minimize Z, we need to find the frequency fsuch that the
derivative of Zwith respect to fis zero.
dZ
df = 0
Step 3: Differentiate Zwith respect to fusing the chain rule and set the
derivative equal to zero.
dZ
df =1
2√R2+(2πfL −1
2πf C )2·2(2πL −1
2πC )= 0
Step 4: Solve for fby setting the expression inside the parentheses equal to
zero.
2πL −1
2πC = 0
Step 5: Solve for f.
2πL =1
2πC
4π2L=1
C
14
f=1
2π√LC
So, the frequency that minimizes the total impedance of the circuit is f=
1
2π√LC .
Question 21
Question
A circuit consists of a resistor with a resistance of 10 Ω, a capacitor with a
capacitance of 0.01 F, and a battery with an emf of 6 V. The switch in the
circuit is closed at t= 0. Calculate the current flowing through the circuit
when t→ ∞.
Solution
Step 1: Find the time constant, τ, of the circuit using the formula τ=RC
where Ris the resistance and Cis the capacitance.
τ= (10 Ω)(0.01 F) = 0.1s
Step 2: Determine the current, I(t), as a function of time tusing the equation
I(t) = I0e−t/τ
where I0is the initial current at t= 0. At t= 0, the current is given by Ohm’s
Law as I(0) = V
R.
I0=6V
10 Ω = 0.6A
So, the current as a function of time is:
I(t) = 0.6A·e−t/0.1s
Step 3: Calculate the current flowing through the circuit as t→ ∞ by
evaluating the limit of I(t)as tapproaches infinity.
lim
t→∞ I(t) = lim
t→∞ 0.6A·e−t/0.1s
lim
t→∞ I(t) = 0.6A·e−∞ = 0
Therefore, the current flowing through the circuit when t→ ∞ is 0.
Question 22
Question
A certain electrical device operates at a voltage of 120 V and draws a current
of 5 A. If the device has a resistance of 24 Ω, calculate the power consumed by
the device.
15
Solution
Step 1: Recall Ohm’s Law which states: V=I·R, where Vis the voltage across
the device, Iis the current passing through the device, and Ris the resistance
of the device.
Step 2: We are given that the voltage Vis 120 V, the current Iis 5 A, and
the resistance Ris 24 Ω. Therefore, we can rearrange Ohm’s Law to solve for
R:
R=V
I=120
5= 24 Ω
Step 3: Next, we calculate the power consumed by the device using the
formula: P=V·I, where Pis power in watts.
P=V·I= 120 ·5 = 600 W
Step 4: The power consumed by the device is 600 watts. Thus, the device
consumes 600 watts of power when operating at a voltage of 120 V and drawing
a current of 5 A with a resistance of 24 Ω.
Question 23
Question
An electric circuit consists of a resistor with resistance R, a capacitor with
capacitance C, and an inductor with inductance Lconnected in series to an
alternating voltage source with angular frequency ω. The voltage across the
resistor, capacitor, and inductor is VR=V0cos(ωt),VC=V0cos(ωt −π
2), and
VL=V0cos(ωt +π
2), respectively.
If the current in the circuit is given by I(t) = I0cos(ωt −ϕ), where I0is the
amplitude of the current and ϕis the phase angle between the current and the
voltage source, show that the amplitude I0of the current is given by
I0=V0
√R2+ (ωL −1
ωC )2
and the phase angle ϕis given by
tan ϕ=ωL −1
ωC
R
Solution
Step 1: Apply Ohm’s Law to find the relationship between the voltage VR,
current I, and resistance R.
VR=IR
16
Step 2: Substitute the expressions for VRand Iin terms of time to find the
expression for I0.
V0cos(ωt) = I0Rcos(ωt −ϕ)
Step 3: Rearrange the equation to solve for I0.
I0=V0
R
Step 4: Apply impedance of a series LRC circuit to find the total impedance
Z.
Z=√R2+(ωL −1
ωC )2
Step 5: Since Z=V0
I0, we have I0=V0
Z. Substitute Zinto the equation.
I0=V0
√R2+(ωL −1
ωC )2
Step 6: Finally, use the phase angle relationship tan ϕ=ωL−1
ωC
Rto determine
the phase angle ϕ.
Question 24
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
difference across the unknown device is measured to be 8 V. Find the resistance
of the unknown device.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor is
equal to the current (I) flowing through the resistor multiplied by the resistance
(R) of the resistor. Mathematically, this can be represented by the equation:
V=I·R.
Step 2: In this problem, the potential difference (V) across the unknown
device is 8 V, and the current (I) flowing through the circuit is 2 A. Therefore,
using Ohm’s Law, we can calculate the resistance (R) of the unknown device
as:
R=V
I=8V
2A= 4 Ω.
Step 3: Hence, the resistance of the unknown device in the circuit is 4 Ω.
17
Question 25
Question
A circuit consists of three resistors in series: R1= 10 Ω,R2= 20 Ω, and R3=
30 Ω. The circuit is connected to a 12 V battery. Calculate the current through
each resistor and the total power dissipated in the circuit.
Solution
Step 1: Calculate the total resistance in the circuit. The total resistance in a
series circuit is the sum of the individual resistances:
Rtotal =R1+R2+R3= 10 Ω + 20 Ω + 30 Ω = 60 Ω
Step 2: Calculate the total current in the circuit using Ohm’s Law (V=IR):
I=V
Rtotal
=12 V
60 Ω = 0.2A
Step 3: Calculate the current through each resistor. In a series circuit, the
current is the same through each resistor. Therefore, I=I1=I2=I3= 0.2A
Step 4: Calculate the power dissipated by each resistor using the formula
P=IV :
P1=I·V= 0.2A×12 V= 2.4W
P2=I·V= 0.2A×12 V= 2.4W
P3=I·V= 0.2A×12 V= 2.4W
Step 5: Calculate the total power dissipated in the circuit. The total power
is the sum of the power dissipated by each resistor:
Ptotal =P1+P2+P3= 2.4W+ 2.4W+ 2.4W= 7.2W
Therefore, the current through each resistor is 0.2A and the total power
dissipated in the circuit is 7.2W.
18
Solution
Step 1: Recall Ohm’s Law, which states that the current passing through a
resistor is directly proportional to the voltage across the resistor and inversely
proportional to the resistance of the resistor. Mathematically, Ohm’s Law is
expressed as:
V=IR
where: V= voltage across the resistor in volts, I= current flowing through the
resistor in amperes, and R= resistance of the resistor in ohms.
Step 2: Given information: Resistance of the resistor, R= 5 Ω Voltage across
the resistor, V= 12 V
Step 3: Substitute the given values into Ohm’s Law equation:
I=V
R
I=12 V
5 Ω
Step 4: Perform the division to calculate the current flowing through the
resistor:
I=12
5
I= 2.4A
Step 5: Therefore, the current flowing through the resistor is 2.4 amperes.
Question 3
Question
A circuit consists of a battery with a voltage of 12 V connected in series to three
resistors: R1= 4 Ω,R2= 6 Ω, and R3= 8 Ω. Calculate the current passing
through each resistor and the power dissipated by each resistor.
Solution
Step 1: First, we calculate the total resistance of the circuit by adding up the
individual resistances:
Rtotal =R1+R2+R3= 4 Ω + 6 Ω + 8 Ω = 18 Ω
Step 2: Next, we use Ohm’s Law to find the total current passing through
the circuit:
V=I·Rtotal
I=V
Rtotal
=12 V
18 Ω = 0.67 A
2
Step 3: To find the current passing through each resistor, we can use the
fact that in a series circuit, the current remains constant:
I=I1=I2=I3= 0.67 A
Step 4: Now, we can calculate the power dissipated by each resistor using
the formula P=I2·R: For R1:
P1=I2·R1= (0.67 A)2·4 Ω = 1.792 W
For R2:
P2=I2·R2= (0.67 A)2·6 Ω = 2.688 W
For R3:
P3=I2·R3= (0.67 A)2·8 Ω = 3.584 W
So, the current passing through each resistor is 0.67 A, and the power dissi-
pated by each resistor is 1.792 W, 2.688 W, and 3.584 W, respectively.
Question 4
Question
A circuit consists of a resistor with resistance R= 5 Ω and an unknown resistor
connected in series to a battery with emf E= 12 V. When a current of 2A
flows through the circuit, the potential difference across the unknown resistor
is found to be 6V. Determine the resistance of the unknown resistor.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor is
equal to the current (I) flowing through the resistor multiplied by the resistance
(R) of the resistor: V=IR.
Step 2: For the total circuit, the total potential difference provided by the
battery is equal to the sum of the potential differences across each resistor.
Therefore, the potential difference Vtotal across the total circuit is equal to the
potential difference across the 5 Ω resistor (V5) plus the potential difference
across the unknown resistor (Vunknown). Mathematically, this can be expressed
as: Vtotal =V5+Vunknown.
Step 3: We are given that the potential difference across the 5 Ω resistor is
6V and the total current in the circuit is 2A. Using Ohm’s Law for the 5 Ω
resistor, we have V5=I·R= 2 A×5 Ω = 10 V.
Step 4: Substitute V5= 10 V and Vtotal = 12 V into the equation from Step
2: 12 V= 10 V+Vunknown, we can solve for Vunknown.
Step 5: Vunknown = 12 V−10 V= 2 V.
Step 6: Finally, to find the resistance of the unknown resistor Runknown,
we can use Ohm’s Law with the current I= 2 A and the potential difference
Vunknown = 2 V: Runknown =Vunknown
I=2V
2A= 1 Ω.
Therefore, the resistance of the unknown resistor is 1 Ω.
3
Question 5
Question
A circuit consists of a resistor with resistance R= 10 Ω and an unknown resistor
connected in series to a 12 Vbattery. If the current passing through this circuit
is 0.5A, find the resistance of the unknown resistor.
Solution
Step 1: We know that in a series circuit, the total resistance Rtotal is the sum
of the individual resistances. Thus, Rtotal =R+Runknown.
Step 2: Ohm’s Law states that the voltage across a resistor equals the current
passing through it multiplied by the resistance. Therefore, the voltage across
the resistor with known resistance Ris VR=I·R= 0.5A·10 Ω = 5 V.
Step 3: Since the total voltage in a series circuit is equal to the sum of the
voltages across each component, we have Vtotal = 12 V. Thus, the voltage across
the unknown resistor is Vunknown =Vtotal −VR= 12 V−5V= 7 V.
Step 4: Applying Ohm’s Law again, we can find the resistance of the un-
known resistor: Runknown =Vunknown
I=7V
0.5A= 14 Ω.
Therefore, the resistance of the unknown resistor in the circuit is 14 Ω.
Question 6
Question
A circuit consists of a resistor with a resistance of 10 Ωand a battery with an
electromotive force of 12 V. If a current of 1.2 A flows through the circuit, what
is the potential difference across the resistor?
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 is represented by the formula V=IR.
Step 2: Substituting the given values, we have V= 1.2A×10 Ω.
Step 3: Calculate the potential difference using the formula:
V= 1.2A×10 Ω = 12 V
Step 4: Therefore, the potential difference across the resistor is 12 V.
4
Question 7
Question
A 12 V battery is connected to a series circuit containing three resistors: R1=
10 Ω,R2= 15 Ω, and R3= 20 Ω. Calculate the current passing through each
resistor in the circuit.
Solution
We can use Ohm’s Law, V=IR, to find the current passing through each
resistor in the circuit.
Step 1: Calculate the total resistance in the circuit. The total resistance in
a series circuit is the sum of the individual resistances:
Rtotal =R1+R2+R3= 10 Ω + 15 Ω + 20 Ω = 45 Ω
Step 2: Calculate the total current in the circuit. Using Ohm’s Law for the
total circuit:
Itotal =V
Rtotal
=12 V
45 Ω =4
15 A= 0.27 A
Step 3: Calculate the current passing through each resistor. Now, using
Ohm’s Law for individual resistors: For R1:
I1=V
R1
=12 V
10 Ω = 1.2A
For R2:
I2=V
R2
=12 V
15 Ω = 0.8A
For R3:
I3=V
R3
=12 V
20 Ω = 0.6A
Therefore, the current passing through R1is 1.2 A, through R2is 0.8 A, and
through R3is 0.6 A in the circuit.
Question 8
Question
A circuit consists of a resistor with resistance 10 Ω, an inductor with inductance
0.02 H, and a capacitor with capacitance 5 µF connected in series to a 12 V AC
power supply that operates at a frequency of 60 Hz.
Calculate the impedance of the circuit at this frequency.
5
Solution
Step 1: Calculate the impedance of each component.
• The impedance of a resistor is equal to its resistance. Therefore, the
impedance of the resistor is 10 Ω.
• The impedance of an inductor is given by ZL=jωL, where Lis the
inductance and ω= 2πf . Plugging in the values, we get:
ZL=j(2π·60)(0.02) = j2.4 Ω.
• The impedance of a capacitor is given by ZC=1
jωC , where Cis the
capacitance. Substituting the values, we get:
ZC=1
j(2π·60)(5 ×10−6)=−j53.05 Ω.
Step 2: Calculate the total impedance of the circuit.
Ztotal =ZR+ZL+ZC= 10 −j2.4−j53.05 Ω.
Step 3: Convert the total impedance to polar form.
Ztotal = 10 −j55.45 Ω.
Step 4: Calculate the magnitude of the total impedance.
|Ztotal|=√102+ (−55.45)2=√3113.7025 ≈55.78 Ω.
Therefore, the impedance of the circuit at a frequency of 60 Hz is approxi-
mately 55.78 Ω.
Question 9
Question
A circuit consists of a resistor of resistance R1, a resistor of resistance R2, and
a battery with emf Eand internal resistance r, all connected in series. The
resistance R1is twice as large as the resistance R2. The potential difference
across R2is found to be half the value of the emf of the battery. Find an
expression for the current Iin the circuit in terms of E,R1,R2, and r.
Given data:
R1= 2R2, VR2=1
2E
6
Solution
Step 1: Use Ohm’s Law to relate the potential differences and resistances in the
circuit.
E=VR1+Vr+VR2
IR1=IR2+Ir +E
2
Step 2: Substitute R1= 2R2into the equation.
I(2R2) = IR2+Ir +E
2
Step 3: Rearrange the equation in terms of I.
2IR2=IR2+Ir +E
2
2I−I=E
2R2−Ir
R2
I=E
2R2−Ir
R2
Step 4: Substitute R1= 2R2into the expression for I.
I=E
2R2−Ir
R2
Therefore, the current Iin the circuit in terms of E,R1,R2, and ris I=
E
2R2−Ir
R2.
Question 10
Question
A circuit consists of a resistor with resistance R= 20Ω and an unknown device
represented by a box with two terminals. When a current of 2.5A passes through
the circuit, a potential difference of 25 V is observed across the terminals of the
unknown device. Determine the resistance of the unknown device.
Solution
Step 1: Recall Ohm’s Law, which states that the potential difference (V) across
a device is equal to the current (I) passing through the device multiplied by the
resistance (R) of the device. The formula for Ohm’s Law is V=IR.
Step 2: We know that a potential difference of 25 V is observed across the
terminals of the unknown device when a current of 2.5A passes through the
7
circuit. Using Ohm’s Law, we can find the resistance of the unknown device.
Given V= 25 V and I= 2.5A, we have:
R=V
I=25 V
2.5A
Step 3: Calculate the resistance of the unknown device:
R= 10Ω
Step 4: Therefore, the resistance of the unknown device represented by the
box is 10Ω.
Question 11
Question
A wire with resistance Ris connected to a battery with emf Eand internal resis-
tance r. When a certain current Iflows through the wire, the power dissipated
is at its maximum value. Prove that the internal resistance rof the battery
must be equal to the resistance Rof the wire.
Solution
Step 1: Recall that the power dissipated in a circuit is given by P=I2R, where
Iis the current and Ris the resistance.
Step 2: The total resistance in the circuit is the sum of the resistance of the
wire (R) and the internal resistance of the battery (r), so the total resistance is
Rtotal =R+r.
Step 3: Using Ohm’s Law (V=IR), we can express the current Iin terms
of the total resistance Rtotal and the emf Eof the battery: I=E
Rtotal =E
R+r.
Step 4: Substitute the expression for Iinto the equation for power P=I2R
to get the power dissipated in terms of R,r, and E:P=(E
R+r)2·R.
Step 5: To find the maximum power dissipated, we can take the derivative
of Pwith respect to r, set it equal to 0, and solve for r.
Step 6: Differentiate Pwith respect to r:dP
dr =2E2R
(R+r)3−E2R
(R+r)2.
Step 7: Set dP
dr = 0 and solve for r:2E2R
(R+r)3−E2R
(R+r)2= 0.
Step 8: Simplifying the equation, we find that r=R. Thus, the internal
resistance of the battery must be equal to the resistance of the wire for the
power dissipated to be at its maximum value.
Question 12
Question
In a circuit, a resistor with resistance R1= 20 Ω and a resistor with resistance
R2= 30 Ω are connected in series with a battery that provides a voltage of
8
V= 60 V. Calculate the total current in the circuit.
Solution
Step 1: Calculate the total resistance in the circuit. The total resistance in
a series circuit is the sum of the individual resistances. Therefore, the total
resistance Rtotal is given by:
Rtotal =R1+R2= 20 Ω + 30 Ω = 50 Ω
Step 2: Apply Ohm’s Law to find the total current. Ohm’s Law states
that the current flowing through a circuit is inversely proportional to the total
resistance and directly proportional to the applied voltage. Mathematically,
Ohm’s Law is given by:
V=I·R
where Vis the voltage, Iis the current, and Ris the resistance.
Substitute the values we have into the equation:
I=V
Rtotal
=60 V
50 Ω = 1.2A
Therefore, the total current in the circuit is 1.2A.
Question 13
Question
A copper wire with a resistance of 2.5 Ωis connected in series with a resistor of
unknown resistance. When a potential difference of 12 V is applied across the
circuit, a current of 3 A is observed. Determine the resistance of the unknown
resistor.
Solution
Step 1: Use Ohm’s Law (V=IR) to find the total resistance of the circuit.
Given: Resistance of the copper wire, Rcopper = 2.5 Ω Potential difference,
V= 12 V Current, I= 3 A
The total resistance can be calculated as:
Rtotal =V
I
Rtotal =12 V
3A= 4 Ω
Step 2: Use the total resistance to find the resistance of the unknown resis-
tor. Let Runknown be the resistance of the unknown resistor. Since the resistors
9
are connected in series, the total resistance Rtotal is the sum of individual resis-
tances:
Rtotal =Rcopper +Runknown
4 Ω = 2.5 Ω + Runknown
Runknown = 4 Ω −2.5 Ω = 1.5 Ω
Therefore, the resistance of the unknown resistor is 1.5 Ω.
Question 14
Question
A circuit consists of three resistors connected in series. The first resistor has a
resistance of 4 Ω, the second resistor has a resistance of 6 Ω, and the third resistor
has an unknown resistance R. A potential difference of 24 Vis applied across
the circuit. If the current through the circuit is 2A, determine the resistance R
of the third resistor.
Solution
Step 1: Recall the formula for Ohm’s Law: V=IR, where Vis the potential
difference across the circuit, Iis the current flowing through the circuit, and Ris
the total resistance of the circuit. Step 2: Determine the total resistance Rtotal of
the circuit by summing the individual resistances: Rtotal =R1+R2+R. Step 3:
Substitute the given resistances into the total resistance formula: Rtotal = 4 Ω+
6 Ω + R= 10 Ω + R. Step 4: Use Ohm’s Law to find the total resistance of the
circuit: V=IRtotal. Substitute V= 24 Vand I= 2 A:24 V= 2 A×(10 Ω+R).
Step 5: Solve for the unknown resistance R:24 V= 20 Ω + 2R
2R= 24 V−20 Ω
2R= 4 V
R= 2 Ω.
Therefore, the resistance of the third resistor is 2 Ω.
Question 15
Question
A circuit consists of three resistors connected in series. The resistances of the
resistors are R1= 6 Ω,R2= 8 Ω, and R3= 10 Ω. If the total potential difference
across the circuit is V= 120 V, what is the current flowing through the circuit?
10
Solution
Step 1: Calculate the total resistance of the circuit using the formula for resistors
in series: Rtotal =R1+R2+R3
= 6 Ω + 8 Ω + 10 Ω
= 24 Ω.
Step 2: Use Ohm’s Law, V=IR, to find the current flowing through the
circuit:
I=V
Rtotal
=120 V
24 Ω
= 5 A.
Therefore, the current flowing through the circuit is 5A.
Question 16
Question
A copper wire with a resistance of 2.5 Ωis connected to a battery that provides
a current of 0.8 A. If the resistance of the wire is increased to 5 Ω, what is the
new current in the circuit?
Solution
Step 1: Calculate the initial voltage drop across the 2.5 Ωresistor using Ohm’s
Law, V=IR.
Initial Voltage = (0.8A)(2.5 Ω) = 2 V
Step 2: Since the battery provides a constant voltage, the total voltage drop
in the circuit remains 2 V when the resistance is increased to 5 Ω. Using Ohm’s
Law, we can find the new current in the circuit.
2V=I(5 Ω)
I=2V
5 Ω = 0.4A
Therefore, the new current in the circuit when the resistance is increased to
5Ωis 0.4 A.
11
Question 17
Question
A circuit consists of a resistor with a resistance of 30 Ω connected in series with a
battery that provides a voltage of 12 V. Determine the current flowing through
the circuit.
Solution
Step 1: Write down Ohm’s Law, which relates voltage (V), current (I), and
resistance (R) in a circuit:
V=I·R
Step 2: Given that the resistance R= 30 Ω and voltage V= 12 V, we can
use Ohm’s Law to solve for the current I:
12 V=I·30 Ω
Step 3: Solve for the current I:
I=12 V
30 Ω = 0.4A
Therefore, the current flowing through the circuit is 0.4A.
Question 18
Question
A resistor with resistance R1= 6 Ω and another resistor with resistance R2=
4 Ω are connected in parallel across a 12 Vbattery. Find the total current flowing
through the circuit.
Solution
Step 1: Calculate the equivalent resistance of the two resistors in parallel. The
formula to find the total resistance of two resistors in parallel is given by:
1
Rtotal
=1
R1
+1
R2
Substitute the given values: R1= 6 Ω and R2= 4 Ω:
1
Rtotal
=1
6+1
4
1
Rtotal
=2
12 +3
12
12
1
Rtotal
=5
12
Therefore, the equivalent resistance Rtotal is:
Rtotal =12
5Ω = 2.4 Ω
Step 2: Calculate the total current using Ohm’s Law. The total current
flowing through the circuit can be found using Ohm’s Law: I=V
R, where
V= 12 Vis the voltage of the battery and Rtotal = 2.4 Ω is the total resistance.
I=12
2.4= 5 A
Therefore, the total current flowing through the circuit is 5A.
Question 19
Question
A circuit consists of a resistor with resistance R= 30 Ω and an unknown resistor
connected in series. When a voltage of V= 12 Vis applied across the circuit,
a current of I= 0.4Aflows through it. What is the resistance of the unknown
resistor?
Solution
Step 1: Recall Ohm’s Law which states V=IR, where Vis the voltage across
a component, Iis the current flowing through it, and Ris the resistance of the
component.
Step 2: We first need to find the total resistance of the circuit. In a se-
ries circuit, the total resistance Rtotal is the sum of the individual resistances:
Rtotal =R1+R2+. . . +Rn.
Step 3: In this case, the total resistance Rtotal is given by Rtotal =R+
Runknown.
Step 4: We can now use Ohm’s Law to find the total resistance of the circuit.
Given that V= 12 Vand I= 0.4A, we have:
Rtotal =V
I
Step 5: Substituting the given values, we find:
R+Runknown =12 V
0.4A
Step 6: Simplifying the equation, we get:
30 Ω + Runknown = 30 Ω
13
Step 7: Subtracting 30 Ω from both sides gives:
Runknown = 0 Ω
Step 8: Therefore, the resistance of the unknown resistor in the circuit is
0 Ω .
Question 20
Question
A circuit consists of a resistor with resistance R, a capacitor with capacitance
C, and an inductor with inductance L. The total impedance of the circuit is
given by Z=√R2+ (XL−XC)2, where XL= 2πfL is the inductive reactance,
XC=1
2πf C is the capacitive reactance, and fis the frequency of the alternating
current in the circuit. Determine the frequency of the current that minimizes
the total impedance of the circuit.
Solution
Step 1: Substitute the expressions for XLand XCinto the equation for total
impedance Z.
Z=√R2+(2πfL −1
2πfC )2
Step 2: To minimize Z, we need to find the frequency fsuch that the
derivative of Zwith respect to fis zero.
dZ
df = 0
Step 3: Differentiate Zwith respect to fusing the chain rule and set the
derivative equal to zero.
dZ
df =1
2√R2+(2πfL −1
2πf C )2·2(2πL −1
2πC )= 0
Step 4: Solve for fby setting the expression inside the parentheses equal to
zero.
2πL −1
2πC = 0
Step 5: Solve for f.
2πL =1
2πC
4π2L=1
C
14
f=1
2π√LC
So, the frequency that minimizes the total impedance of the circuit is f=
1
2π√LC .
Question 21
Question
A circuit consists of a resistor with a resistance of 10 Ω, a capacitor with a
capacitance of 0.01 F, and a battery with an emf of 6 V. The switch in the
circuit is closed at t= 0. Calculate the current flowing through the circuit
when t→ ∞.
Solution
Step 1: Find the time constant, τ, of the circuit using the formula τ=RC
where Ris the resistance and Cis the capacitance.
τ= (10 Ω)(0.01 F) = 0.1s
Step 2: Determine the current, I(t), as a function of time tusing the equation
I(t) = I0e−t/τ
where I0is the initial current at t= 0. At t= 0, the current is given by Ohm’s
Law as I(0) = V
R.
I0=6V
10 Ω = 0.6A
So, the current as a function of time is:
I(t) = 0.6A·e−t/0.1s
Step 3: Calculate the current flowing through the circuit as t→ ∞ by
evaluating the limit of I(t)as tapproaches infinity.
lim
t→∞ I(t) = lim
t→∞ 0.6A·e−t/0.1s
lim
t→∞ I(t) = 0.6A·e−∞ = 0
Therefore, the current flowing through the circuit when t→ ∞ is 0.
Question 22
Question
A certain electrical device operates at a voltage of 120 V and draws a current
of 5 A. If the device has a resistance of 24 Ω, calculate the power consumed by
the device.
15
Solution
Step 1: Recall Ohm’s Law which states: V=I·R, where Vis the voltage across
the device, Iis the current passing through the device, and Ris the resistance
of the device.
Step 2: We are given that the voltage Vis 120 V, the current Iis 5 A, and
the resistance Ris 24 Ω. Therefore, we can rearrange Ohm’s Law to solve for
R:
R=V
I=120
5= 24 Ω
Step 3: Next, we calculate the power consumed by the device using the
formula: P=V·I, where Pis power in watts.
P=V·I= 120 ·5 = 600 W
Step 4: The power consumed by the device is 600 watts. Thus, the device
consumes 600 watts of power when operating at a voltage of 120 V and drawing
a current of 5 A with a resistance of 24 Ω.
Question 23
Question
An electric circuit consists of a resistor with resistance R, a capacitor with
capacitance C, and an inductor with inductance Lconnected in series to an
alternating voltage source with angular frequency ω. The voltage across the
resistor, capacitor, and inductor is VR=V0cos(ωt),VC=V0cos(ωt −π
2), and
VL=V0cos(ωt +π
2), respectively.
If the current in the circuit is given by I(t) = I0cos(ωt −ϕ), where I0is the
amplitude of the current and ϕis the phase angle between the current and the
voltage source, show that the amplitude I0of the current is given by
I0=V0
√R2+ (ωL −1
ωC )2
and the phase angle ϕis given by
tan ϕ=ωL −1
ωC
R
Solution
Step 1: Apply Ohm’s Law to find the relationship between the voltage VR,
current I, and resistance R.
VR=IR
16
Step 2: Substitute the expressions for VRand Iin terms of time to find the
expression for I0.
V0cos(ωt) = I0Rcos(ωt −ϕ)
Step 3: Rearrange the equation to solve for I0.
I0=V0
R
Step 4: Apply impedance of a series LRC circuit to find the total impedance
Z.
Z=√R2+(ωL −1
ωC )2
Step 5: Since Z=V0
I0, we have I0=V0
Z. Substitute Zinto the equation.
I0=V0
√R2+(ωL −1
ωC )2
Step 6: Finally, use the phase angle relationship tan ϕ=ωL−1
ωC
Rto determine
the phase angle ϕ.
Question 24
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
difference across the unknown device is measured to be 8 V. Find the resistance
of the unknown device.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor is
equal to the current (I) flowing through the resistor multiplied by the resistance
(R) of the resistor. Mathematically, this can be represented by the equation:
V=I·R.
Step 2: In this problem, the potential difference (V) across the unknown
device is 8 V, and the current (I) flowing through the circuit is 2 A. Therefore,
using Ohm’s Law, we can calculate the resistance (R) of the unknown device
as:
R=V
I=8V
2A= 4 Ω.
Step 3: Hence, the resistance of the unknown device in the circuit is 4 Ω.
17
Question 25
Question
A circuit consists of three resistors in series: R1= 10 Ω,R2= 20 Ω, and R3=
30 Ω. The circuit is connected to a 12 V battery. Calculate the current through
each resistor and the total power dissipated in the circuit.
Solution
Step 1: Calculate the total resistance in the circuit. The total resistance in a
series circuit is the sum of the individual resistances:
Rtotal =R1+R2+R3= 10 Ω + 20 Ω + 30 Ω = 60 Ω
Step 2: Calculate the total current in the circuit using Ohm’s Law (V=IR):
I=V
Rtotal
=12 V
60 Ω = 0.2A
Step 3: Calculate the current through each resistor. In a series circuit, the
current is the same through each resistor. Therefore, I=I1=I2=I3= 0.2A
Step 4: Calculate the power dissipated by each resistor using the formula
P=IV :
P1=I·V= 0.2A×12 V= 2.4W
P2=I·V= 0.2A×12 V= 2.4W
P3=I·V= 0.2A×12 V= 2.4W
Step 5: Calculate the total power dissipated in the circuit. The total power
is the sum of the power dissipated by each resistor:
Ptotal =P1+P2+P3= 2.4W+ 2.4W+ 2.4W= 7.2W
Therefore, the current through each resistor is 0.2A and the total power
dissipated in the circuit is 7.2W.
18
Solution
Step 1: Recall Ohm’s Law, which states that the current passing through a
resistor is directly proportional to the voltage across the resistor and inversely
proportional to the resistance of the resistor. Mathematically, Ohm’s Law is
expressed as:
V=IR
where: V= voltage across the resistor in volts, I= current flowing through the
resistor in amperes, and R= resistance of the resistor in ohms.
Step 2: Given information: Resistance of the resistor, R= 5 Ω Voltage across
the resistor, V= 12 V
Step 3: Substitute the given values into Ohm’s Law equation:
I=V
R
I=12 V
5 Ω
Step 4: Perform the division to calculate the current flowing through the
resistor:
I=12
5
I= 2.4A
Step 5: Therefore, the current flowing through the resistor is 2.4 amperes.
Question 3
Question
A circuit consists of a battery with a voltage of 12 V connected in series to three
resistors: R1= 4 Ω,R2= 6 Ω, and R3= 8 Ω. Calculate the current passing
through each resistor and the power dissipated by each resistor.
Solution
Step 1: First, we calculate the total resistance of the circuit by adding up the
individual resistances:
Rtotal =R1+R2+R3= 4 Ω + 6 Ω + 8 Ω = 18 Ω
Step 2: Next, we use Ohm’s Law to find the total current passing through
the circuit:
V=I·Rtotal
I=V
Rtotal
=12 V
18 Ω = 0.67 A
2
Step 3: To find the current passing through each resistor, we can use the
fact that in a series circuit, the current remains constant:
I=I1=I2=I3= 0.67 A
Step 4: Now, we can calculate the power dissipated by each resistor using
the formula P=I2·R: For R1:
P1=I2·R1= (0.67 A)2·4 Ω = 1.792 W
For R2:
P2=I2·R2= (0.67 A)2·6 Ω = 2.688 W
For R3:
P3=I2·R3= (0.67 A)2·8 Ω = 3.584 W
So, the current passing through each resistor is 0.67 A, and the power dissi-
pated by each resistor is 1.792 W, 2.688 W, and 3.584 W, respectively.
Question 4
Question
A circuit consists of a resistor with resistance R= 5 Ω and an unknown resistor
connected in series to a battery with emf E= 12 V. When a current of 2A
flows through the circuit, the potential difference across the unknown resistor
is found to be 6V. Determine the resistance of the unknown resistor.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor is
equal to the current (I) flowing through the resistor multiplied by the resistance
(R) of the resistor: V=IR.
Step 2: For the total circuit, the total potential difference provided by the
battery is equal to the sum of the potential differences across each resistor.
Therefore, the potential difference Vtotal across the total circuit is equal to the
potential difference across the 5 Ω resistor (V5) plus the potential difference
across the unknown resistor (Vunknown). Mathematically, this can be expressed
as: Vtotal =V5+Vunknown.
Step 3: We are given that the potential difference across the 5 Ω resistor is
6V and the total current in the circuit is 2A. Using Ohm’s Law for the 5 Ω
resistor, we have V5=I·R= 2 A×5 Ω = 10 V.
Step 4: Substitute V5= 10 V and Vtotal = 12 V into the equation from Step
2: 12 V= 10 V+Vunknown, we can solve for Vunknown.
Step 5: Vunknown = 12 V−10 V= 2 V.
Step 6: Finally, to find the resistance of the unknown resistor Runknown,
we can use Ohm’s Law with the current I= 2 A and the potential difference
Vunknown = 2 V: Runknown =Vunknown
I=2V
2A= 1 Ω.
Therefore, the resistance of the unknown resistor is 1 Ω.
3
Question 5
Question
A circuit consists of a resistor with resistance R= 10 Ω and an unknown resistor
connected in series to a 12 Vbattery. If the current passing through this circuit
is 0.5A, find the resistance of the unknown resistor.
Solution
Step 1: We know that in a series circuit, the total resistance Rtotal is the sum
of the individual resistances. Thus, Rtotal =R+Runknown.
Step 2: Ohm’s Law states that the voltage across a resistor equals the current
passing through it multiplied by the resistance. Therefore, the voltage across
the resistor with known resistance Ris VR=I·R= 0.5A·10 Ω = 5 V.
Step 3: Since the total voltage in a series circuit is equal to the sum of the
voltages across each component, we have Vtotal = 12 V. Thus, the voltage across
the unknown resistor is Vunknown =Vtotal −VR= 12 V−5V= 7 V.
Step 4: Applying Ohm’s Law again, we can find the resistance of the un-
known resistor: Runknown =Vunknown
I=7V
0.5A= 14 Ω.
Therefore, the resistance of the unknown resistor in the circuit is 14 Ω.
Question 6
Question
A circuit consists of a resistor with a resistance of 10 Ωand a battery with an
electromotive force of 12 V. If a current of 1.2 A flows through the circuit, what
is the potential difference across the resistor?
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 is represented by the formula V=IR.
Step 2: Substituting the given values, we have V= 1.2A×10 Ω.
Step 3: Calculate the potential difference using the formula:
V= 1.2A×10 Ω = 12 V
Step 4: Therefore, the potential difference across the resistor is 12 V.
4
Question 7
Question
A 12 V battery is connected to a series circuit containing three resistors: R1=
10 Ω,R2= 15 Ω, and R3= 20 Ω. Calculate the current passing through each
resistor in the circuit.
Solution
We can use Ohm’s Law, V=IR, to find the current passing through each
resistor in the circuit.
Step 1: Calculate the total resistance in the circuit. The total resistance in
a series circuit is the sum of the individual resistances:
Rtotal =R1+R2+R3= 10 Ω + 15 Ω + 20 Ω = 45 Ω
Step 2: Calculate the total current in the circuit. Using Ohm’s Law for the
total circuit:
Itotal =V
Rtotal
=12 V
45 Ω =4
15 A= 0.27 A
Step 3: Calculate the current passing through each resistor. Now, using
Ohm’s Law for individual resistors: For R1:
I1=V
R1
=12 V
10 Ω = 1.2A
For R2:
I2=V
R2
=12 V
15 Ω = 0.8A
For R3:
I3=V
R3
=12 V
20 Ω = 0.6A
Therefore, the current passing through R1is 1.2 A, through R2is 0.8 A, and
through R3is 0.6 A in the circuit.
Question 8
Question
A circuit consists of a resistor with resistance 10 Ω, an inductor with inductance
0.02 H, and a capacitor with capacitance 5 µF connected in series to a 12 V AC
power supply that operates at a frequency of 60 Hz.
Calculate the impedance of the circuit at this frequency.
5
Solution
Step 1: Calculate the impedance of each component.
• The impedance of a resistor is equal to its resistance. Therefore, the
impedance of the resistor is 10 Ω.
• The impedance of an inductor is given by ZL=jωL, where Lis the
inductance and ω= 2πf . Plugging in the values, we get:
ZL=j(2π·60)(0.02) = j2.4 Ω.
• The impedance of a capacitor is given by ZC=1
jωC , where Cis the
capacitance. Substituting the values, we get:
ZC=1
j(2π·60)(5 ×10−6)=−j53.05 Ω.
Step 2: Calculate the total impedance of the circuit.
Ztotal =ZR+ZL+ZC= 10 −j2.4−j53.05 Ω.
Step 3: Convert the total impedance to polar form.
Ztotal = 10 −j55.45 Ω.
Step 4: Calculate the magnitude of the total impedance.
|Ztotal|=√102+ (−55.45)2=√3113.7025 ≈55.78 Ω.
Therefore, the impedance of the circuit at a frequency of 60 Hz is approxi-
mately 55.78 Ω.
Question 9
Question
A circuit consists of a resistor of resistance R1, a resistor of resistance R2, and
a battery with emf Eand internal resistance r, all connected in series. The
resistance R1is twice as large as the resistance R2. The potential difference
across R2is found to be half the value of the emf of the battery. Find an
expression for the current Iin the circuit in terms of E,R1,R2, and r.
Given data:
R1= 2R2, VR2=1
2E
6
Solution
Step 1: Use Ohm’s Law to relate the potential differences and resistances in the
circuit.
E=VR1+Vr+VR2
IR1=IR2+Ir +E
2
Step 2: Substitute R1= 2R2into the equation.
I(2R2) = IR2+Ir +E
2
Step 3: Rearrange the equation in terms of I.
2IR2=IR2+Ir +E
2
2I−I=E
2R2−Ir
R2
I=E
2R2−Ir
R2
Step 4: Substitute R1= 2R2into the expression for I.
I=E
2R2−Ir
R2
Therefore, the current Iin the circuit in terms of E,R1,R2, and ris I=
E
2R2−Ir
R2.
Question 10
Question
A circuit consists of a resistor with resistance R= 20Ω and an unknown device
represented by a box with two terminals. When a current of 2.5A passes through
the circuit, a potential difference of 25 V is observed across the terminals of the
unknown device. Determine the resistance of the unknown device.
Solution
Step 1: Recall Ohm’s Law, which states that the potential difference (V) across
a device is equal to the current (I) passing through the device multiplied by the
resistance (R) of the device. The formula for Ohm’s Law is V=IR.
Step 2: We know that a potential difference of 25 V is observed across the
terminals of the unknown device when a current of 2.5A passes through the
7
circuit. Using Ohm’s Law, we can find the resistance of the unknown device.
Given V= 25 V and I= 2.5A, we have:
R=V
I=25 V
2.5A
Step 3: Calculate the resistance of the unknown device:
R= 10Ω
Step 4: Therefore, the resistance of the unknown device represented by the
box is 10Ω.
Question 11
Question
A wire with resistance Ris connected to a battery with emf Eand internal resis-
tance r. When a certain current Iflows through the wire, the power dissipated
is at its maximum value. Prove that the internal resistance rof the battery
must be equal to the resistance Rof the wire.
Solution
Step 1: Recall that the power dissipated in a circuit is given by P=I2R, where
Iis the current and Ris the resistance.
Step 2: The total resistance in the circuit is the sum of the resistance of the
wire (R) and the internal resistance of the battery (r), so the total resistance is
Rtotal =R+r.
Step 3: Using Ohm’s Law (V=IR), we can express the current Iin terms
of the total resistance Rtotal and the emf Eof the battery: I=E
Rtotal =E
R+r.
Step 4: Substitute the expression for Iinto the equation for power P=I2R
to get the power dissipated in terms of R,r, and E:P=(E
R+r)2·R.
Step 5: To find the maximum power dissipated, we can take the derivative
of Pwith respect to r, set it equal to 0, and solve for r.
Step 6: Differentiate Pwith respect to r:dP
dr =2E2R
(R+r)3−E2R
(R+r)2.
Step 7: Set dP
dr = 0 and solve for r:2E2R
(R+r)3−E2R
(R+r)2= 0.
Step 8: Simplifying the equation, we find that r=R. Thus, the internal
resistance of the battery must be equal to the resistance of the wire for the
power dissipated to be at its maximum value.
Question 12
Question
In a circuit, a resistor with resistance R1= 20 Ω and a resistor with resistance
R2= 30 Ω are connected in series with a battery that provides a voltage of
8
V= 60 V. Calculate the total current in the circuit.
Solution
Step 1: Calculate the total resistance in the circuit. The total resistance in
a series circuit is the sum of the individual resistances. Therefore, the total
resistance Rtotal is given by:
Rtotal =R1+R2= 20 Ω + 30 Ω = 50 Ω
Step 2: Apply Ohm’s Law to find the total current. Ohm’s Law states
that the current flowing through a circuit is inversely proportional to the total
resistance and directly proportional to the applied voltage. Mathematically,
Ohm’s Law is given by:
V=I·R
where Vis the voltage, Iis the current, and Ris the resistance.
Substitute the values we have into the equation:
I=V
Rtotal
=60 V
50 Ω = 1.2A
Therefore, the total current in the circuit is 1.2A.
Question 13
Question
A copper wire with a resistance of 2.5 Ωis connected in series with a resistor of
unknown resistance. When a potential difference of 12 V is applied across the
circuit, a current of 3 A is observed. Determine the resistance of the unknown
resistor.
Solution
Step 1: Use Ohm’s Law (V=IR) to find the total resistance of the circuit.
Given: Resistance of the copper wire, Rcopper = 2.5 Ω Potential difference,
V= 12 V Current, I= 3 A
The total resistance can be calculated as:
Rtotal =V
I
Rtotal =12 V
3A= 4 Ω
Step 2: Use the total resistance to find the resistance of the unknown resis-
tor. Let Runknown be the resistance of the unknown resistor. Since the resistors
9
are connected in series, the total resistance Rtotal is the sum of individual resis-
tances:
Rtotal =Rcopper +Runknown
4 Ω = 2.5 Ω + Runknown
Runknown = 4 Ω −2.5 Ω = 1.5 Ω
Therefore, the resistance of the unknown resistor is 1.5 Ω.
Question 14
Question
A circuit consists of three resistors connected in series. The first resistor has a
resistance of 4 Ω, the second resistor has a resistance of 6 Ω, and the third resistor
has an unknown resistance R. A potential difference of 24 Vis applied across
the circuit. If the current through the circuit is 2A, determine the resistance R
of the third resistor.
Solution
Step 1: Recall the formula for Ohm’s Law: V=IR, where Vis the potential
difference across the circuit, Iis the current flowing through the circuit, and Ris
the total resistance of the circuit. Step 2: Determine the total resistance Rtotal of
the circuit by summing the individual resistances: Rtotal =R1+R2+R. Step 3:
Substitute the given resistances into the total resistance formula: Rtotal = 4 Ω+
6 Ω + R= 10 Ω + R. Step 4: Use Ohm’s Law to find the total resistance of the
circuit: V=IRtotal. Substitute V= 24 Vand I= 2 A:24 V= 2 A×(10 Ω+R).
Step 5: Solve for the unknown resistance R:24 V= 20 Ω + 2R
2R= 24 V−20 Ω
2R= 4 V
R= 2 Ω.
Therefore, the resistance of the third resistor is 2 Ω.
Question 15
Question
A circuit consists of three resistors connected in series. The resistances of the
resistors are R1= 6 Ω,R2= 8 Ω, and R3= 10 Ω. If the total potential difference
across the circuit is V= 120 V, what is the current flowing through the circuit?
10
Solution
Step 1: Calculate the total resistance of the circuit using the formula for resistors
in series: Rtotal =R1+R2+R3
= 6 Ω + 8 Ω + 10 Ω
= 24 Ω.
Step 2: Use Ohm’s Law, V=IR, to find the current flowing through the
circuit:
I=V
Rtotal
=120 V
24 Ω
= 5 A.
Therefore, the current flowing through the circuit is 5A.
Question 16
Question
A copper wire with a resistance of 2.5 Ωis connected to a battery that provides
a current of 0.8 A. If the resistance of the wire is increased to 5 Ω, what is the
new current in the circuit?
Solution
Step 1: Calculate the initial voltage drop across the 2.5 Ωresistor using Ohm’s
Law, V=IR.
Initial Voltage = (0.8A)(2.5 Ω) = 2 V
Step 2: Since the battery provides a constant voltage, the total voltage drop
in the circuit remains 2 V when the resistance is increased to 5 Ω. Using Ohm’s
Law, we can find the new current in the circuit.
2V=I(5 Ω)
I=2V
5 Ω = 0.4A
Therefore, the new current in the circuit when the resistance is increased to
5Ωis 0.4 A.
11
Question 17
Question
A circuit consists of a resistor with a resistance of 30 Ω connected in series with a
battery that provides a voltage of 12 V. Determine the current flowing through
the circuit.
Solution
Step 1: Write down Ohm’s Law, which relates voltage (V), current (I), and
resistance (R) in a circuit:
V=I·R
Step 2: Given that the resistance R= 30 Ω and voltage V= 12 V, we can
use Ohm’s Law to solve for the current I:
12 V=I·30 Ω
Step 3: Solve for the current I:
I=12 V
30 Ω = 0.4A
Therefore, the current flowing through the circuit is 0.4A.
Question 18
Question
A resistor with resistance R1= 6 Ω and another resistor with resistance R2=
4 Ω are connected in parallel across a 12 Vbattery. Find the total current flowing
through the circuit.
Solution
Step 1: Calculate the equivalent resistance of the two resistors in parallel. The
formula to find the total resistance of two resistors in parallel is given by:
1
Rtotal
=1
R1
+1
R2
Substitute the given values: R1= 6 Ω and R2= 4 Ω:
1
Rtotal
=1
6+1
4
1
Rtotal
=2
12 +3
12
12
1
Rtotal
=5
12
Therefore, the equivalent resistance Rtotal is:
Rtotal =12
5Ω = 2.4 Ω
Step 2: Calculate the total current using Ohm’s Law. The total current
flowing through the circuit can be found using Ohm’s Law: I=V
R, where
V= 12 Vis the voltage of the battery and Rtotal = 2.4 Ω is the total resistance.
I=12
2.4= 5 A
Therefore, the total current flowing through the circuit is 5A.
Question 19
Question
A circuit consists of a resistor with resistance R= 30 Ω and an unknown resistor
connected in series. When a voltage of V= 12 Vis applied across the circuit,
a current of I= 0.4Aflows through it. What is the resistance of the unknown
resistor?
Solution
Step 1: Recall Ohm’s Law which states V=IR, where Vis the voltage across
a component, Iis the current flowing through it, and Ris the resistance of the
component.
Step 2: We first need to find the total resistance of the circuit. In a se-
ries circuit, the total resistance Rtotal is the sum of the individual resistances:
Rtotal =R1+R2+. . . +Rn.
Step 3: In this case, the total resistance Rtotal is given by Rtotal =R+
Runknown.
Step 4: We can now use Ohm’s Law to find the total resistance of the circuit.
Given that V= 12 Vand I= 0.4A, we have:
Rtotal =V
I
Step 5: Substituting the given values, we find:
R+Runknown =12 V
0.4A
Step 6: Simplifying the equation, we get:
30 Ω + Runknown = 30 Ω
13
Step 7: Subtracting 30 Ω from both sides gives:
Runknown = 0 Ω
Step 8: Therefore, the resistance of the unknown resistor in the circuit is
0 Ω .
Question 20
Question
A circuit consists of a resistor with resistance R, a capacitor with capacitance
C, and an inductor with inductance L. The total impedance of the circuit is
given by Z=√R2+ (XL−XC)2, where XL= 2πfL is the inductive reactance,
XC=1
2πf C is the capacitive reactance, and fis the frequency of the alternating
current in the circuit. Determine the frequency of the current that minimizes
the total impedance of the circuit.
Solution
Step 1: Substitute the expressions for XLand XCinto the equation for total
impedance Z.
Z=√R2+(2πfL −1
2πfC )2
Step 2: To minimize Z, we need to find the frequency fsuch that the
derivative of Zwith respect to fis zero.
dZ
df = 0
Step 3: Differentiate Zwith respect to fusing the chain rule and set the
derivative equal to zero.
dZ
df =1
2√R2+(2πfL −1
2πf C )2·2(2πL −1
2πC )= 0
Step 4: Solve for fby setting the expression inside the parentheses equal to
zero.
2πL −1
2πC = 0
Step 5: Solve for f.
2πL =1
2πC
4π2L=1
C
14
f=1
2π√LC
So, the frequency that minimizes the total impedance of the circuit is f=
1
2π√LC .
Question 21
Question
A circuit consists of a resistor with a resistance of 10 Ω, a capacitor with a
capacitance of 0.01 F, and a battery with an emf of 6 V. The switch in the
circuit is closed at t= 0. Calculate the current flowing through the circuit
when t→ ∞.
Solution
Step 1: Find the time constant, τ, of the circuit using the formula τ=RC
where Ris the resistance and Cis the capacitance.
τ= (10 Ω)(0.01 F) = 0.1s
Step 2: Determine the current, I(t), as a function of time tusing the equation
I(t) = I0e−t/τ
where I0is the initial current at t= 0. At t= 0, the current is given by Ohm’s
Law as I(0) = V
R.
I0=6V
10 Ω = 0.6A
So, the current as a function of time is:
I(t) = 0.6A·e−t/0.1s
Step 3: Calculate the current flowing through the circuit as t→ ∞ by
evaluating the limit of I(t)as tapproaches infinity.
lim
t→∞ I(t) = lim
t→∞ 0.6A·e−t/0.1s
lim
t→∞ I(t) = 0.6A·e−∞ = 0
Therefore, the current flowing through the circuit when t→ ∞ is 0.
Question 22
Question
A certain electrical device operates at a voltage of 120 V and draws a current
of 5 A. If the device has a resistance of 24 Ω, calculate the power consumed by
the device.
15
Solution
Step 1: Recall Ohm’s Law which states: V=I·R, where Vis the voltage across
the device, Iis the current passing through the device, and Ris the resistance
of the device.
Step 2: We are given that the voltage Vis 120 V, the current Iis 5 A, and
the resistance Ris 24 Ω. Therefore, we can rearrange Ohm’s Law to solve for
R:
R=V
I=120
5= 24 Ω
Step 3: Next, we calculate the power consumed by the device using the
formula: P=V·I, where Pis power in watts.
P=V·I= 120 ·5 = 600 W
Step 4: The power consumed by the device is 600 watts. Thus, the device
consumes 600 watts of power when operating at a voltage of 120 V and drawing
a current of 5 A with a resistance of 24 Ω.
Question 23
Question
An electric circuit consists of a resistor with resistance R, a capacitor with
capacitance C, and an inductor with inductance Lconnected in series to an
alternating voltage source with angular frequency ω. The voltage across the
resistor, capacitor, and inductor is VR=V0cos(ωt),VC=V0cos(ωt −π
2), and
VL=V0cos(ωt +π
2), respectively.
If the current in the circuit is given by I(t) = I0cos(ωt −ϕ), where I0is the
amplitude of the current and ϕis the phase angle between the current and the
voltage source, show that the amplitude I0of the current is given by
I0=V0
√R2+ (ωL −1
ωC )2
and the phase angle ϕis given by
tan ϕ=ωL −1
ωC
R
Solution
Step 1: Apply Ohm’s Law to find the relationship between the voltage VR,
current I, and resistance R.
VR=IR
16
Step 2: Substitute the expressions for VRand Iin terms of time to find the
expression for I0.
V0cos(ωt) = I0Rcos(ωt −ϕ)
Step 3: Rearrange the equation to solve for I0.
I0=V0
R
Step 4: Apply impedance of a series LRC circuit to find the total impedance
Z.
Z=√R2+(ωL −1
ωC )2
Step 5: Since Z=V0
I0, we have I0=V0
Z. Substitute Zinto the equation.
I0=V0
√R2+(ωL −1
ωC )2
Step 6: Finally, use the phase angle relationship tan ϕ=ωL−1
ωC
Rto determine
the phase angle ϕ.
Question 24
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
difference across the unknown device is measured to be 8 V. Find the resistance
of the unknown device.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor is
equal to the current (I) flowing through the resistor multiplied by the resistance
(R) of the resistor. Mathematically, this can be represented by the equation:
V=I·R.
Step 2: In this problem, the potential difference (V) across the unknown
device is 8 V, and the current (I) flowing through the circuit is 2 A. Therefore,
using Ohm’s Law, we can calculate the resistance (R) of the unknown device
as:
R=V
I=8V
2A= 4 Ω.
Step 3: Hence, the resistance of the unknown device in the circuit is 4 Ω.
17
Question 25
Question
A circuit consists of three resistors in series: R1= 10 Ω,R2= 20 Ω, and R3=
30 Ω. The circuit is connected to a 12 V battery. Calculate the current through
each resistor and the total power dissipated in the circuit.
Solution
Step 1: Calculate the total resistance in the circuit. The total resistance in a
series circuit is the sum of the individual resistances:
Rtotal =R1+R2+R3= 10 Ω + 20 Ω + 30 Ω = 60 Ω
Step 2: Calculate the total current in the circuit using Ohm’s Law (V=IR):
I=V
Rtotal
=12 V
60 Ω = 0.2A
Step 3: Calculate the current through each resistor. In a series circuit, the
current is the same through each resistor. Therefore, I=I1=I2=I3= 0.2A
Step 4: Calculate the power dissipated by each resistor using the formula
P=IV :
P1=I·V= 0.2A×12 V= 2.4W
P2=I·V= 0.2A×12 V= 2.4W
P3=I·V= 0.2A×12 V= 2.4W
Step 5: Calculate the total power dissipated in the circuit. The total power
is the sum of the power dissipated by each resistor:
Ptotal =P1+P2+P3= 2.4W+ 2.4W+ 2.4W= 7.2W
Therefore, the current through each resistor is 0.2A and the total power
dissipated in the circuit is 7.2W.
18
