PHYS 101 - ELEMENTS OF PHYSICS
- Ohm’s Law
Question Bank - Set 3
Liberty University
Question 1
Question
A circuit consists of a resistor with resistance R, an inductor with inductance L,
and a capacitor with capacitance Cconnected in series. The voltage across the
circuit is given by V(t) = V0sin(ωt), where V0,ω, and tare constants. Using
Ohm’s Law, derive an expression for the current I(t) in the circuit.
Solution
Step 1: Ohm’s Law states that the voltage across a circuit element is equal
to the product of the current flowing through it and its impedance. In this
case, the impedance ZRof the resistor R, impedance ZLof the inductor L, and
impedance ZCof the capacitor Care given by:
ZR=R, ZL=jωL, ZC=1
jωC
where jis the imaginary unit.
Step 2: The total impedance Ztotal of the circuit is the sum of the impedances
of the resistor, inductor, and capacitor since they are connected in series:
Ztotal =ZR+ZL+ZC=R+jωL +1
jωC
Step 3: The current I(t) through the circuit can be calculated using Ohm’s
Law where V(t) = V0sin(ωt) is the voltage across the circuit:
I(t) = V(t)
Ztotal
=V0sin(ωt)
R+jωL +1
jωC
Step 4: To simplify the expression, we can rationalize the denominator. Mul-
tiplying the numerator and denominator by the conjugate of the denominator:
I(t) = V0sin(ωt)
R+jωL +1
jωC ·R−jωL −1
jωC
R−jωL −1
jωC
Step 5: After simplifying and expanding, the expression for the current I(t)
becomes:
I(t) = V0sin(ωt)(R−jωL −1
jωC )
R2+ω2L2+1
ω2C2
Question 2
Question
A resistor with a resistance of 350 Ω is connected to a battery with a voltage of
12 V. What is the current flowing through the circuit?
Solution
To find the current flowing through the circuit, we can use Ohm’s Law, which
states that the current (I) flowing through a circuit is equal to the voltage (V)
divided by the resistance (R).
Step 1: Write down Ohm’s Law formula:
I=V
R
Step 2: Plug in the given values:
I=12 V
350 Ω
Step 3: Calculate the current:
I=12
350 A
I≈0.0343 A
Step 4: Therefore, the current flowing through the circuit is approximately
0.0343 A.
Question 3
Question
A resistor is connected to a voltage source of 10 V and a current of 2 A flows
through it. Determine the resistance of the resistor.
2
Solution
Let’s use Ohm’s Law, which states that the voltage across a resistor is equal
to the current flowing through it multiplied by the resistance. Mathematically,
Ohm’s Law can be written as:
V=IR
Where: - Vis the voltage across the resistor (in volts), - Iis the current
flowing through the resistor (in amperes), and - Ris the resistance of the resistor
(in ohms).
Step 1: Given that the voltage V= 10 V and the current I= 2 A, we can
substitute these values into Ohm’s Law to find the resistance:
10 V = 2 A ×R
Step 2: Solving for the resistance R, we get:
R=10 V
2 A = 5 Ω
Therefore, the resistance of the resistor is 5 Ω.
Question 4
Question
A circuit consists of a resistor, a capacitor, and an inductor connected in series.
The resistor has a resistance of 50 Ω, the capacitor has a capacitance of 0.1 F,
and the inductor has an inductance of 0.2 H. If a current of 2 A flows through
the circuit, calculate the voltage drop across each component using Ohm’s Law.
Solution
To calculate the voltage drop across each component, we can use Ohm’s Law,
V=IR, where Vis the voltage drop, Iis the current, and Ris the resistance.
Step 1: Calculate the voltage drop across the resistor Given: R=
50 Ω and I= 2 A Using Ohm’s Law: Vresistor =I·R= 2 A ×50 Ω = 100 V.
Therefore, the voltage drop across the resistor is 100 V.
Step 2: Calculate the voltage drop across the capacitor The rela-
tionship between voltage and current for a capacitor is V=Q
C, where Qis the
charge stored on the capacitor plates and Cis the capacitance. Since we are
given current, we can relate it to the charge using I=dQ
dt . Therefore, Q=RI dt.
Given: I= 2 A and C= 0.1 F Solving the integral: Q=R2dt = 2t+Q0Given
that Q0= 0 (initial charge): Q= 2tUsing V=Q
C:Vcapacitor =2t
0.1= 20tV
Step 3: Calculate the voltage drop across the inductor The relation-
ship between voltage and current for an inductor is V=Ldi
dt , where Lis the
inductance and di
dt is the rate of change of current. Given: L= 0.2 H and I= 2
3
A Taking derivative of Iwith respect to t:di
dt = 0 (since current is constant)
Using V=Ldi
dt :Vinductor = 0 V
Therefore, the voltage drop across the resistor is 100 V, across the capacitor
is 20tV, and across the inductor is 0 V.
Question 5
Question
A circuit consists of a resistor with resistance R= 20 Ω connected to a voltage
source of V= 100 V. Calculate the current flowing through the resistor.
Solution
To calculate the current flowing through the resistor, we can use Ohm’s Law,
which states that V=IR, where Vis the voltage across the resistor, Iis the
current flowing through the resistor, and Ris the resistance of the resistor.
Step 1: Substitute V= 100 V and R= 20 Ω into Ohm’s Law, V=IR, and
solve for I:
I=V
R
I=100 V
20 Ω
I= 5 A
Step 2: The current flowing through the resistor is I= 5 A.
Question 6
Question
A circuit contains a resistor with a resistance of 20 Ω and a current of 2 A
flowing through it. Determine the voltage drop across the resistor.
Solution
Step 1: Write down Ohm’s Law, which states that V=IR, where Vis the
voltage, Iis the current, and Ris the resistance of the resistor.
Step 2: Plug in the values given in the question. In this case, I= 2 A and
R= 20 Ω.
V= (2 A)(20 Ω)
Step 3: Calculate the voltage drop across the resistor.
V= 2 ×20 = 40 V
Step 4: Therefore, the voltage drop across the resistor is 40 V.
4
Question 7
Question
A circuit consists of a resistor, an inductor, and a capacitor connected in series.
The resistance of the resistor is 10 Ω, the inductance of the inductor is 5 H, and
the capacitance of the capacitor is 0.02 F. If the peak current flowing through
the circuit is 2 A, find the peak voltage across the circuit.
Solution
The peak voltage across the circuit can be found using Ohm’s Law, which states:
V=I·Z, where Vis the voltage, Iis the current, and Zis the impedance of
the circuit.
Step 1: Calculate the impedance of the circuit using the total resistance,
inductive reactance, and capacitive reactance. The impedance of the circuit is
given by the formula: Z=pR2+ (XL−XC)2, where Ris the resistance, XL
is the inductive reactance, and XCis the capacitive reactance. Given: Resistor
resistance R= 10 Ω, Inductive reactance XL= 2πfL, Capacitive reactance
XC=1
2πfC . Let’s choose a frequency, f= 1 Hz, to simplify calculations.
Substitute the given values into the formulas: XL= 2π·1·5 = 10 Ω, XC=
1
2π·1·0.02 = 7.96 Ω. Now, calculate the impedance: Z=p102+ (10 −7.96)2=
√100 + 4.082=√116.94 ≈10.81 Ω.
Step 2: Calculate the peak voltage across the circuit using Ohm’s Law.
Given peak current I= 2 A and impedance Z= 10.81 Ω, V=I·Z= 2·10.81 =
21.62 V.
Therefore, the peak voltage across the circuit is 21.62 V.
Question 8
Question
A circuit consists of a resistor with resistance R= 20 Ω and a battery with EMF
E= 12 V. If the current flowing through the circuit is I= 0.5A, calculate the
power dissipated in the resistor.
Solution
Step 1: To find the power dissipated in the resistor, we can use the formula for
power in a resistor: P=I2R.
Step 2: Substitute the given values into the formula. We have I= 0.5Aand
R= 20 Ω.
Step 3: Calculate the power:
P= (0.5A)2×20 Ω = 0.25 A2×20 Ω = 5 W
Step 4: Therefore, the power dissipated in the resistor is 5 W.
5
Question 9
Question
A resistor has a resistance of 10 Ω and a current of 2 A passing through it.
Determine the voltage drop across the resistor.
Solution
Step 1: Write down Ohm’s Law, which states that V=IR, where Vis the
voltage drop across the resistor, Iis the current passing through the resistor,
and Ris the resistance of the resistor.
Step 2: Substitute the given values into Ohm’s Law:
V= (2 A)(10Ω)
Step 3: Calculate the voltage drop:
V= 20 V
Therefore, the voltage drop across the resistor is 20 V.
Question 10
Question
A resistor with a resistance of 10 Ω is connected in series with a resistor with
an unknown resistance. A potential difference of 50 V is applied across the
combination of resistors, and a current of 2 A flows through the circuit. What
is the resistance of the unknown resistor?
Solution
Step 1: We can start by writing down Ohm’s Law, which relates voltage (V),
current (I), and resistance (R) using the equation V=IR. In this case, we
have the following values: - Total voltage across the resistors, Vtotal = 50 V, -
Total current through the circuit, Itotal = 2 A.
Step 2: The total voltage across the two resistors is the sum of the voltage
drops across each resistor:
Vtotal =V1+V2
Step 3: We can express the voltage drops in terms of the resistances and the
total current using Ohm’s Law:
V1=Itotal ·R1
V2=Itotal ·R2
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Step 4: We are given the resistance of one resistor (R1= 10 Ω) and the cur-
rent flowing through the circuit. We need to find the resistance of the unknown
resistor (R2).
Step 5: Substituting the known values into the equations, we get:
50 = 2 ·10 + 2 ·R2
Step 6: Solving for R2gives:
50 = 20 + 2R2
2R2= 30
R2= 15 Ω
Step 7: Therefore, the resistance of the unknown resistor is 15 Ω.
Question 11
Question
A circuit consists of a resistor with resistance R= 20 Ω connected to a battery
with emf E= 9 V. If the current flowing through the circuit is I= 0.4 A, what
is the potential difference across the resistor?
Solution
To find the potential difference across the resistor, we can use Ohm’s Law which
states that V=IR, where Vis the potential difference, Iis the current, and R
is the resistance.
Step 1: Write down Ohm’s Law:
V=IR
Step 2: Substitute the given values into the equation:
V= (0.4 A)(20 Ω)
Step 3: Perform the calculation:
V= 8 V
Step 4: Answer: The potential difference across the resistor is 8 volts.
Question 12
Question
A circuit consists of a resistor, a capacitor, and an inductor connected in series.
The values of the components are as follows: resistor R= 10 Ω, capacitor
C= 5 µF , and inductor L= 2 mH. If a sinusoidal voltage source with frequency
f= 1 kHz is connected to the circuit, determine the current flowing through
the circuit using Ohm’s Law.
7
Solution
Step 1: Calculate the total impedance of the circuit using the formula for
impedance in an RLC series circuit:
Z=qR2+ (XL−XC)2
where XL= 2πf L is the inductive reactance and XC=1
2πfC is the capacitive
reactance.
Step 2: Calculate the inductive reactance XL:
XL= 2πfL = 2 ·π·1000 ·0.002 = 12.57 Ω
Step 3: Calculate the capacitive reactance XC:
XC=1
2πf C =1
2·π·1000 ·5×10−6= 31.83 Ω
Step 4: Calculate the total impedance Z:
Z=p102+ (12.57 −31.83)2=√100 + 338.56 = √438.56 = 20.92 Ω
Step 5: Calculate the current Iflowing through the circuit using Ohm’s Law:
I=V
Z
where Vis the voltage of the sinusoidal source. Let us assume V= 10 Vfor
simplicity.
I=10
20.92 = 0.477 A
Therefore, the current flowing through the circuit is 0.477 A.
Question 13
Question
A resistor has a resistance of 500 Ω and a current of 0.8 A flowing through it.
Calculate the voltage drop across the resistor.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor
is equal to the product of the current (I) flowing through it and the resistance
(R) of the resistor. Mathematically, this relationship is represented as:
V=I×R
8
Step 2: Given that the resistance (R) is 500 Ω and the current (I) is 0.8 A,
we can substitute those values into Ohm’s Law to solve for the voltage:
V= 0.8 A ×500 Ω
Step 3: Multiply the current and resistance to find the voltage drop:
V= 0.8 A ×500 Ω = 400 V
Step 4: Therefore, the voltage drop across the resistor is 400 V.
Question 14
Question
A circuit contains a resistor with resistance R= 100 Ω connected to a battery
with voltage V= 12 V. If a current of I= 0.1 A flows through the circuit, what
is the power dissipated by the resistor?
Solution
Ohm’s Law states that the current through a conductor between two points is
directly proportional to the voltage across the two points. Mathematically, this
is represented by the equation V=IR, where Vis the voltage, Iis the current,
and Ris the resistance.
Step 1: Calculate the power dissipated by the resistor using the formula
P=IV .
P=IV
= (0.1 A)(12 V)
= 1.2 W
Step 2: Verify the calculated power using the formula P=I2R.
P=I2R
= (0.1 A)2×100 Ω
= 0.01 ×100
= 1 W
Therefore, the power dissipated by the resistor is 1.2 W.
Question 15
Question
A resistor with a resistance of 12 ohms is connected to a 24-volt battery. What
is the current flowing through the resistor?
9
Solution
Step 1: Recall Ohm’s Law, which states that the current (I) flowing through a
resistor is equal to the voltage (V) across the resistor divided by the resistance
(R) of the resistor. Mathematically, this can be represented as I=V
R.
Step 2: Given that the resistance Ris 12 ohms and the voltage Vis 24 volts,
we can substitute these values into Ohm’s Law to find the current I:
I=24 V
12Ω
Step 3: Simplifying the expression, we get:
I= 2 A
Step 4: Therefore, the current flowing through the resistor is 2 amperes.
Question 16
Question
A circuit consists of a resistor with a resistance of 30 Ω connected to a 12 V
battery. Calculate the current flowing through the circuit.
Solution
Step 1: Recall Ohm’s Law, which states that the current (I) flowing through a
circuit is equal to the voltage (V) across the circuit divided by the resistance
(R) of the circuit. Mathematically, this can be expressed as:
I=V
R
Step 2: Given that the voltage Vis 12 V and the resistance Ris 30 Ω,
substitute these values into Ohm’s Law to find the current I:
I=12 V
30Ω
Step 3: Calculate the current I:
I=12
30 A
I= 0.4 A
Step 4: Therefore, the current flowing through the circuit is 0.4 A.
10
Question 17
Question
A resistor with resistance 10 ohms is connected to a battery with a voltage of
12 volts. What is the current flowing through the resistor?
Solution
To find the current flowing through the resistor, we can use Ohm’s Law, which
states that V=IR, where: - Vis the voltage across the resistor, - Iis the
current flowing through the resistor, and - Ris the resistance of the resistor.
Step 1: Substitute the given values into Ohm’s Law.
V=IR
12 = I×10
Step 2: Solve for the current, I.
I=12
10
I= 1.2 amps
Therefore, the current flowing through the resistor is 1.2 amps.
Question 18
Question
A resistor has a resistance of 15 Ω and a current of 2 A passing through it.
What is the voltage drop across the resistor?
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor
is equal to the product of the current (I) passing through it and the resistance
(R) of the resistor. Mathematically, this is expressed as:
V=I×R
Step 2: Given that the resistance Ris 15 Ω and the current Iis 2 A, we can
substitute these values into Ohm’s Law to find the voltage drop (V) across the
resistor.
V= 2 A ×15 Ω
Step 3: Calculate the voltage drop across the resistor:
V= 30 V
Therefore, the voltage drop across the resistor is 30 V.
11
Question 19
Question
A certain circuit has a resistance of 30 Ω and a current of 0.5 A flowing through
it. If the voltage across the circuit is measured to be 15 V, what is the power
dissipated by the circuit?
Solution
To find the power dissipated by the circuit, we can use the formula for power in
a circuit: P=V I, where Pis the power, Vis the voltage, and Iis the current.
We are also given the value of resistance Rin the circuit, so we can use Ohm’s
Law, V=IR, to determine the current Iin the circuit.
Step 1: Calculate the current Iusing Ohm’s Law: V=IR
I=V
R=15 V
30 Ω = 0.5 A
Step 2: Substitute the values of Vand Iinto the power formula P=V I:
P=V×I= 15 V ×0.5 A = 7.5 W
Therefore, the power dissipated by the circuit is 7.5 W.
Question 20
Question
A resistor with a resistance of 6 Ω is connected to a 12 V battery. Determine
the current flowing through the resistor.
Solution
Let’s use Ohm’s Law, which states that the current passing through a conductor
between two points is directly proportional to the voltage across the two points
and inversely proportional to the resistance. Mathematically, Ohm’s Law can
be expressed as V=IR, where V= Voltage across the resistor, I= Current
passing through the resistor, and R= Resistance of the resistor.
Step 1: Given that the voltage across the resistor is V= 12 V and the
resistance of the resistor is R= 6 Ω, we can use Ohm’s Law to find the current
passing through the resistor.
I=V
R
Step 2: Substituting V= 12 V and R= 6 Ω into the formula, we get
I=12
6
12
I= 2 A
Step 3: Therefore, the current passing through the resistor is 2 A when
connected to a 12 V battery.
Question 21
Question
A circuit consists of a resistor, an inductor, and a capacitor connected in series.
The resistor has a resistance of 20 Ω, the inductor has an inductance of 0.01 H,
and the capacitor has a capacitance of 10 µF. If the frequency of the alternating
current in the circuit is 100 Hz, determine the current flowing through the circuit.
Solution
Let’s denote the resistance, inductive reactance, and capacitive reactance as R,
XL, and XCrespectively. The total impedance Zof the circuit can be calculated
using the formula:
Z=pR2+ (XL−XC)2
Step 1: Calculate the inductive reactance The inductive reactance XL
is given by:
XL= 2πf L
where fis the frequency of the AC current and Lis the inductance of the
inductor. Substitute f= 100 Hz and L= 0.01 H into the formula to get:
XL= 2π×100 ×0.01 = 2 Ω
Step 2: Calculate the capacitive reactance The capacitive reactance
XCis given by:
XC=1
2πf C
where fis the frequency of the AC current and Cis the capacitance of the
capacitor. Substitute f= 100 Hz and C= 10 µF = 10−5F into the formula to
get:
XC=1
2π×100 ×10−5= 1592.46 Ω
Step 3: Calculate the total impedance Substitute R= 20 Ω, XL= 2 Ω,
and XC= 1592.46 Ω into the formula for total impedance Z:
Z=p202+ (2 −1592.46)2=p400 + 1590.462≈1591.92 Ω
Step 4: Use Ohm’s Law The current Iflowing through the circuit can
be calculated using Ohm’s Law:
I=V
Z
where Vis the voltage across the circuit. Since the voltage Vhas not been
specified, we cannot calculate the current accurately.
13
Question 22
Question
A circuit consists of a resistor with resistance R= 15 Ω connected to a battery
with voltage V= 120 V. Calculate the current passing through the circuit.
Solution
To calculate the current passing through the circuit, we can use Ohm’s Law,
which states that V=IR, where Vis the voltage across the resistor, Iis the
current passing through the resistor, and Ris the resistance of the resistor.
Step 1: Write down Ohm’s Law equation:
V=IR
Step 2: Rearrange the equation to solve for current I:
I=V
R
Step 3: Substitute V= 120 Vand R= 15 Ω into the equation:
I=120 V
15 Ω
Step 4: Calculate the current passing through the circuit:
I=120 V
15 Ω = 8 A
Therefore, the current passing through the circuit is 8 A.
Question 23
Question
A resistor with resistance 25 Ω is connected to a potential difference of 50 V.
What is the current flowing through the resistor?
Solution
Let’s use Ohm’s Law, which states that V=IR, where Vis the potential
difference across the resistor, Iis the current flowing through the resistor, and
Ris the resistance of the resistor.
Step 1: Given data: Resistance of the resistor, R= 25 Ω Potential difference
across the resistor, V= 50 V
14
Step 2: Apply Ohm’s Law to find the current: From Ohm’s Law, V=IR,
we can rearrange the formula to solve for current I:
I=V
R
Step 3: Substitute the known values into the formula:
I=50
25 = 2 A
Step 4: Answer: The current flowing through the resistor is 2 A.
Question 24
Question
A resistor with resistance R= 100 Ω is connected to a voltage source with
V= 12 V. Calculate the current flowing through the resistor.
Solution
Step 1: Write Ohm’s Law: V=IR, where Vis the voltage, Iis the current,
and Ris the resistance.
Step 2: Rearrange Ohm’s Law to solve for current: I=V
R.
Step 3: Substitute the given values V= 12 V and R= 100 Ω into the formula
I=V
R.
Step 4: Calculate the current flowing through the resistor:
I=12 V
100 Ω = 0.12 A
Therefore, the current flowing through the resistor is 0.12 A.
Question 25
Question
A resistor has a resistance of 30 Ω and a current of 0.5Aflowing through it.
Determine the voltage drop across the resistor.
Solution
Ohm’s Law states that the voltage drop (V) across a resistor is equal to the
product of the resistance (R) and the current (I) flowing through it. Mathe-
matically, Ohm’s Law is represented as V=IR.
15
Step 1: Given the resistance R= 30 Ω and the current I= 0.5A, we can
use Ohm’s Law to find the voltage drop V.
V=IR
V= 0.5A×30 Ω
V= 15 V
Step 2: Therefore, the voltage drop across the resistor is 15 V.
Question 26
Question
A cylindrical resistor has a resistance of 10 Ω and a length of 2.0 meters. If the
resistivity of the material is 1.7×10−6Ω·m, what is the radius of the resistor?
Solution
Step 1: The resistance Rof a cylindrical resistor is given by the formula:
R=ρ·L
A
where ρis the resistivity of the material, Lis the length of the resistor, and A
is the cross-sectional area.
Step 2: We are given that R= 10 Ω, ρ= 1.7×10−6Ω·m, and L= 2.0 m.
We need to find the radius r.
Step 3: The cross-sectional area of a cylindrical resistor is given by the
formula:
A=πr2
Step 4: Substitute the expressions for Rand Ainto the formula for resis-
tance:
10 = 1.7×10−6·2
πr2
Step 5: Simplify the equation:
10 = 3.4×10−6
πr2
πr2=3.4×10−6
10
r2=3.4×10−6
10π
Step 6: Calculate the radius:
r=r3.4×10−6
10π≈0.00491 m
Step 7: Therefore, the radius of the cylindrical resistor is approximately
0.00491 meters.
16
Question 27
Question
A circuit consists of a resistor with a resistance of 50 Ω, a capacitor with a
capacitance of 0.01 F, and an inductor with an inductance of 0.05 H connected
in series to a voltage source of 12 V. Calculate the current flowing through the
circuit using Ohm’s Law.
Solution
To calculate the current flowing through the circuit, we can use Ohm’s Law,
which states V=IR, where: - Vis the voltage across the circuit, - Iis the
current flowing through the circuit, and - Ris the total resistance of the circuit.
Step 1: Calculate the total resistance of the circuit. The total resistance
Rtotal in a series circuit is the sum of the individual resistances:
Rtotal =Rresistor +Rinductor +Rcapacitor
Plugging in the given values:
Rtotal = 50 Ω + 0 Ω + 0 Ω = 50 Ω
Step 2: Use Ohm’s Law to find the current. Given V= 12 V, R= 50 Ω, we
can rearrange Ohm’s Law to solve for the current I:
I=V
R
I=12
50
I= 0.24 A
Therefore, the current flowing through the circuit is 0.24 A.
Question 28
Question
A copper wire has a resistance of 5 ohms. If a current of 2 amperes flows through
the wire, what is the voltage across the wire?
Solution
Step 1: Recall Ohm’s Law, which states that the voltage across a resistor is
equal to the product of the current flowing through it and the resistance of
the resistor: V=IR, where Vis the voltage (in volts), Iis the current (in
amperes), and Ris the resistance (in ohms).
17
Step 2: Given that the resistance of the copper wire is 5 ohms and the
current flowing through it is 2 amperes, we can use Ohm’s Law to find the
voltage across the wire:
V=I×R= 2 A ×5 Ω = 10 V
Step 3: Therefore, the voltage across the copper wire is 10 volts.
Question 29
Question
A resistor with resistance Ris connected to a voltage source such that a current I
flows through it. The power dissipated in the resistor is given by P=V I −1
2RI2.
Prove Ohm’s Law using this relationship.
Solution
To prove Ohm’s Law, we need to show that the relationship between voltage
(V), current (I), and resistance (R) is V=IR.
Step 1: Start with the given expression for power:
P=V I −1
2RI2
Step 2: Differentiate both sides of the equation with respect to time t:
dP
dt =d(V I)
dt −d1
2RI2
dt
Step 3: Using the product rule for differentiation, we have:
dP
dt =VdI
dt +IdV
dt −1
2R·2I·dI
dt
Step 4: Simplify the equation:
dP
dt =VdI
dt +IdV
dt −RI dI
dt
Step 5: Recognize that dP
dt is the rate at which power is being dissipated,
which is equal to the rate at which energy is being supplied by the voltage
source. Therefore, the left-hand side simplifies to:
dP
dt =V I
Step 6: Substitute V I back into the equation:
V I =VdI
dt +IdV
dt −RI dI
dt
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Step 7: Rearrange the terms to get:
V I =IdV
dt −RI
Step 8: Since V=IR (Ohm’s Law), we know that dV
dt =RdI
dt . Substitute
this into the equation:
V I =I(RdI
dt −RI)
Step 9: Simplify the equation to:
V I =I(RdI
dt −RI)
V I =I2R−I2R
V I = 0
Step 10: Therefore, we have shown that V I = 0, which implies that V=
IR. This is Ohm’s Law.
Question 30
Question
A circuit consists of a resistor with resistance R= 100 Ω connected to a battery
with electromotive force (emf) E= 12 V. If the current flowing through the
circuit is I= 0.1 A, what is the internal resistance of the battery?
Solution
Ohm’s Law states that the potential difference across a resistor is equal to the
current flowing through it multiplied by the resistance. In this case, the potential
difference (voltage) across the resistor is equal to the emf of the battery.
Step 1: Calculate the potential difference across the resistor.
V=IR
V= 0.1 A ×100 Ω
V= 10 V
Step 2: Since the potential difference across the resistor is equal to the emf
of the battery, we have:
V=E
10 V = 12 V −I×r
10 V = 12 V −0.1 A ×r
10 V = 12 V −0.1 A ×r
19
Step 3: Solve for the internal resistance r.
0.1 A ×r= 2 V
r=2 V
0.1 A
r= 20 Ω
Therefore, the internal resistance of the battery is 20 Ω.
Question 31
Question
A resistor with a resistance of 10 ohms is connected to a 12-volt battery. What
is the current passing through the resistor?
Solution
Step 1: Recall Ohm’s Law, which states that the current passing through a
resistor is equal to the voltage across the resistor divided by the resistance of
the resistor. Mathematically, this can be represented as I=V
R, where Iis the
current in amperes (A), Vis the voltage in volts (V), and Ris the resistance in
ohms (Ω).
Step 2: Given that the resistance of the resistor is 10 ohms and the voltage
of the battery is 12 volts, we can plug these values into Ohm’s Law to calculate
the current passing through the resistor.
I=12 V
10 Ω
Step 3: Simplifying the expression, we find:
I= 1.2 A
Therefore, the current passing through the resistor is 1.2 amperes.
Question 32
Question
A circuit consists of a resistor, a capacitor, and an inductor in series. The
resistor has a resistance of 10 Ω, the capacitor has a reactance of 5 Ω, and the
inductor has a reactance of 8 Ω. If a voltage of 100 V is applied across the
circuit, what is the current flowing through the circuit?
20
Solution
Step 1: Calculate the total impedance of the circuit using the formula Ztotal =
pR2+ (XL−XC)2.
Ztotal =p102+ (8 −5)2
=√100 + 9
=√109
≈10.44 Ω
Step 2: Calculate the current flowing through the circuit using Ohm’s Law:
I=V
Ztotal .
I=100
10.44
≈9.57 A
Therefore, the current flowing through the circuit is approximately 9.57 A.
Question 33
Question
A circuit contains a resistor with resistance R= 30 Ω and a battery with voltage
V= 120 V. If a current of I= 4 Aflows through the circuit, what is the power
dissipated by the resistor?
Solution
Step 1: Recall Ohm’s Law, which relates voltage, current, and resistance: V=
I·R.
Step 2: Substitute the given values into Ohm’s Law to find the voltage across
the resistor:
V=I·R
120 V= 4 A·30 Ω
120 V= 120 V
Step 3: Since the voltage across the resistor is the same as the battery
voltage, the resistor is dissipating power. The power dissipated by a resistor
can be calculated using the formula P=I2·Ror P=V2
R.
Step 4: We will use the formula P=I2·Rto find the power dissipated by
the resistor:
P=I2·R
P= (4 A)2·30 Ω
P= 16 A2·30 Ω
P= 480 W
21
Step 5: Therefore, the power dissipated by the resistor in the circuit is
480 W.
Question 34
Question
A resistor has a resistance of 20 Ω and a current of 2 A passing through it.
Determine the voltage drop across the resistor using Ohm’s Law.
Solution
Let’s recall Ohm’s Law, which states that the voltage drop across a resistor
is equal to the product of the resistance and the current passing through it.
Mathematically, Ohm’s Law is represented as:
V=IR
where: V= voltage drop across the resistor (in volts), I= current passing
through the resistor (in amperes), and R= resistance of the resistor (in ohms).
Step 1: Given that the resistance Ris 20 Ω and the current Iis 2 A, we
can substitute these values into Ohm’s Law to find the voltage drop V:
V= (2 A)(20 Ω)
Step 2: Now, multiply the current and resistance to find the voltage drop:
V= 2 A ×20 Ω = 40 V
Therefore, the voltage drop across the resistor is 40 volts.
Question 35
Question
A resistor with a resistance of 10 Ω is connected to a voltage source that produces
a current of 2 A. Calculate the voltage across the resistor.
Solution
To calculate the voltage across the resistor, we can use Ohm’s Law, which states
that V=I×R, where Vis the voltage, Iis the current, and Ris the resistance
of the resistor.
Step 1: Identify the given values: The resistance of the resistor, R= 10 Ω,
and the current passing through it, I= 2 A.
Step 2: Apply Ohm’s Law to calculate the voltage:
V=I×R
22
Step 4: To simplify the expression, we can rationalize the denominator. Mul-
tiplying the numerator and denominator by the conjugate of the denominator:
I(t) = V0sin(ωt)
R+jωL +1
jωC ·R−jωL −1
jωC
R−jωL −1
jωC
Step 5: After simplifying and expanding, the expression for the current I(t)
becomes:
I(t) = V0sin(ωt)(R−jωL −1
jωC )
R2+ω2L2+1
ω2C2
Question 2
Question
A resistor with a resistance of 350 Ω is connected to a battery with a voltage of
12 V. What is the current flowing through the circuit?
Solution
To find the current flowing through the circuit, we can use Ohm’s Law, which
states that the current (I) flowing through a circuit is equal to the voltage (V)
divided by the resistance (R).
Step 1: Write down Ohm’s Law formula:
I=V
R
Step 2: Plug in the given values:
I=12 V
350 Ω
Step 3: Calculate the current:
I=12
350 A
I≈0.0343 A
Step 4: Therefore, the current flowing through the circuit is approximately
0.0343 A.
Question 3
Question
A resistor is connected to a voltage source of 10 V and a current of 2 A flows
through it. Determine the resistance of the resistor.
2
Solution
Let’s use Ohm’s Law, which states that the voltage across a resistor is equal
to the current flowing through it multiplied by the resistance. Mathematically,
Ohm’s Law can be written as:
V=IR
Where: - Vis the voltage across the resistor (in volts), - Iis the current
flowing through the resistor (in amperes), and - Ris the resistance of the resistor
(in ohms).
Step 1: Given that the voltage V= 10 V and the current I= 2 A, we can
substitute these values into Ohm’s Law to find the resistance:
10 V = 2 A ×R
Step 2: Solving for the resistance R, we get:
R=10 V
2 A = 5 Ω
Therefore, the resistance of the resistor is 5 Ω.
Question 4
Question
A circuit consists of a resistor, a capacitor, and an inductor connected in series.
The resistor has a resistance of 50 Ω, the capacitor has a capacitance of 0.1 F,
and the inductor has an inductance of 0.2 H. If a current of 2 A flows through
the circuit, calculate the voltage drop across each component using Ohm’s Law.
Solution
To calculate the voltage drop across each component, we can use Ohm’s Law,
V=IR, where Vis the voltage drop, Iis the current, and Ris the resistance.
Step 1: Calculate the voltage drop across the resistor Given: R=
50 Ω and I= 2 A Using Ohm’s Law: Vresistor =I·R= 2 A ×50 Ω = 100 V.
Therefore, the voltage drop across the resistor is 100 V.
Step 2: Calculate the voltage drop across the capacitor The rela-
tionship between voltage and current for a capacitor is V=Q
C, where Qis the
charge stored on the capacitor plates and Cis the capacitance. Since we are
given current, we can relate it to the charge using I=dQ
dt . Therefore, Q=RI dt.
Given: I= 2 A and C= 0.1 F Solving the integral: Q=R2dt = 2t+Q0Given
that Q0= 0 (initial charge): Q= 2tUsing V=Q
C:Vcapacitor =2t
0.1= 20tV
Step 3: Calculate the voltage drop across the inductor The relation-
ship between voltage and current for an inductor is V=Ldi
dt , where Lis the
inductance and di
dt is the rate of change of current. Given: L= 0.2 H and I= 2
3
A Taking derivative of Iwith respect to t:di
dt = 0 (since current is constant)
Using V=Ldi
dt :Vinductor = 0 V
Therefore, the voltage drop across the resistor is 100 V, across the capacitor
is 20tV, and across the inductor is 0 V.
Question 5
Question
A circuit consists of a resistor with resistance R= 20 Ω connected to a voltage
source of V= 100 V. Calculate the current flowing through the resistor.
Solution
To calculate the current flowing through the resistor, we can use Ohm’s Law,
which states that V=IR, where Vis the voltage across the resistor, Iis the
current flowing through the resistor, and Ris the resistance of the resistor.
Step 1: Substitute V= 100 V and R= 20 Ω into Ohm’s Law, V=IR, and
solve for I:
I=V
R
I=100 V
20 Ω
I= 5 A
Step 2: The current flowing through the resistor is I= 5 A.
Question 6
Question
A circuit contains a resistor with a resistance of 20 Ω and a current of 2 A
flowing through it. Determine the voltage drop across the resistor.
Solution
Step 1: Write down Ohm’s Law, which states that V=IR, where Vis the
voltage, Iis the current, and Ris the resistance of the resistor.
Step 2: Plug in the values given in the question. In this case, I= 2 A and
R= 20 Ω.
V= (2 A)(20 Ω)
Step 3: Calculate the voltage drop across the resistor.
V= 2 ×20 = 40 V
Step 4: Therefore, the voltage drop across the resistor is 40 V.
4
Question 7
Question
A circuit consists of a resistor, an inductor, and a capacitor connected in series.
The resistance of the resistor is 10 Ω, the inductance of the inductor is 5 H, and
the capacitance of the capacitor is 0.02 F. If the peak current flowing through
the circuit is 2 A, find the peak voltage across the circuit.
Solution
The peak voltage across the circuit can be found using Ohm’s Law, which states:
V=I·Z, where Vis the voltage, Iis the current, and Zis the impedance of
the circuit.
Step 1: Calculate the impedance of the circuit using the total resistance,
inductive reactance, and capacitive reactance. The impedance of the circuit is
given by the formula: Z=pR2+ (XL−XC)2, where Ris the resistance, XL
is the inductive reactance, and XCis the capacitive reactance. Given: Resistor
resistance R= 10 Ω, Inductive reactance XL= 2πfL, Capacitive reactance
XC=1
2πfC . Let’s choose a frequency, f= 1 Hz, to simplify calculations.
Substitute the given values into the formulas: XL= 2π·1·5 = 10 Ω, XC=
1
2π·1·0.02 = 7.96 Ω. Now, calculate the impedance: Z=p102+ (10 −7.96)2=
√100 + 4.082=√116.94 ≈10.81 Ω.
Step 2: Calculate the peak voltage across the circuit using Ohm’s Law.
Given peak current I= 2 A and impedance Z= 10.81 Ω, V=I·Z= 2·10.81 =
21.62 V.
Therefore, the peak voltage across the circuit is 21.62 V.
Question 8
Question
A circuit consists of a resistor with resistance R= 20 Ω and a battery with EMF
E= 12 V. If the current flowing through the circuit is I= 0.5A, calculate the
power dissipated in the resistor.
Solution
Step 1: To find the power dissipated in the resistor, we can use the formula for
power in a resistor: P=I2R.
Step 2: Substitute the given values into the formula. We have I= 0.5Aand
R= 20 Ω.
Step 3: Calculate the power:
P= (0.5A)2×20 Ω = 0.25 A2×20 Ω = 5 W
Step 4: Therefore, the power dissipated in the resistor is 5 W.
5
Question 9
Question
A resistor has a resistance of 10 Ω and a current of 2 A passing through it.
Determine the voltage drop across the resistor.
Solution
Step 1: Write down Ohm’s Law, which states that V=IR, where Vis the
voltage drop across the resistor, Iis the current passing through the resistor,
and Ris the resistance of the resistor.
Step 2: Substitute the given values into Ohm’s Law:
V= (2 A)(10Ω)
Step 3: Calculate the voltage drop:
V= 20 V
Therefore, the voltage drop across the resistor is 20 V.
Question 10
Question
A resistor with a resistance of 10 Ω is connected in series with a resistor with
an unknown resistance. A potential difference of 50 V is applied across the
combination of resistors, and a current of 2 A flows through the circuit. What
is the resistance of the unknown resistor?
Solution
Step 1: We can start by writing down Ohm’s Law, which relates voltage (V),
current (I), and resistance (R) using the equation V=IR. In this case, we
have the following values: - Total voltage across the resistors, Vtotal = 50 V, -
Total current through the circuit, Itotal = 2 A.
Step 2: The total voltage across the two resistors is the sum of the voltage
drops across each resistor:
Vtotal =V1+V2
Step 3: We can express the voltage drops in terms of the resistances and the
total current using Ohm’s Law:
V1=Itotal ·R1
V2=Itotal ·R2
6
Step 4: We are given the resistance of one resistor (R1= 10 Ω) and the cur-
rent flowing through the circuit. We need to find the resistance of the unknown
resistor (R2).
Step 5: Substituting the known values into the equations, we get:
50 = 2 ·10 + 2 ·R2
Step 6: Solving for R2gives:
50 = 20 + 2R2
2R2= 30
R2= 15 Ω
Step 7: Therefore, the resistance of the unknown resistor is 15 Ω.
Question 11
Question
A circuit consists of a resistor with resistance R= 20 Ω connected to a battery
with emf E= 9 V. If the current flowing through the circuit is I= 0.4 A, what
is the potential difference across the resistor?
Solution
To find the potential difference across the resistor, we can use Ohm’s Law which
states that V=IR, where Vis the potential difference, Iis the current, and R
is the resistance.
Step 1: Write down Ohm’s Law:
V=IR
Step 2: Substitute the given values into the equation:
V= (0.4 A)(20 Ω)
Step 3: Perform the calculation:
V= 8 V
Step 4: Answer: The potential difference across the resistor is 8 volts.
Question 12
Question
A circuit consists of a resistor, a capacitor, and an inductor connected in series.
The values of the components are as follows: resistor R= 10 Ω, capacitor
C= 5 µF , and inductor L= 2 mH. If a sinusoidal voltage source with frequency
f= 1 kHz is connected to the circuit, determine the current flowing through
the circuit using Ohm’s Law.
7
Solution
Step 1: Calculate the total impedance of the circuit using the formula for
impedance in an RLC series circuit:
Z=qR2+ (XL−XC)2
where XL= 2πf L is the inductive reactance and XC=1
2πfC is the capacitive
reactance.
Step 2: Calculate the inductive reactance XL:
XL= 2πfL = 2 ·π·1000 ·0.002 = 12.57 Ω
Step 3: Calculate the capacitive reactance XC:
XC=1
2πf C =1
2·π·1000 ·5×10−6= 31.83 Ω
Step 4: Calculate the total impedance Z:
Z=p102+ (12.57 −31.83)2=√100 + 338.56 = √438.56 = 20.92 Ω
Step 5: Calculate the current Iflowing through the circuit using Ohm’s Law:
I=V
Z
where Vis the voltage of the sinusoidal source. Let us assume V= 10 Vfor
simplicity.
I=10
20.92 = 0.477 A
Therefore, the current flowing through the circuit is 0.477 A.
Question 13
Question
A resistor has a resistance of 500 Ω and a current of 0.8 A flowing through it.
Calculate the voltage drop across the resistor.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor
is equal to the product of the current (I) flowing through it and the resistance
(R) of the resistor. Mathematically, this relationship is represented as:
V=I×R
8
Step 2: Given that the resistance (R) is 500 Ω and the current (I) is 0.8 A,
we can substitute those values into Ohm’s Law to solve for the voltage:
V= 0.8 A ×500 Ω
Step 3: Multiply the current and resistance to find the voltage drop:
V= 0.8 A ×500 Ω = 400 V
Step 4: Therefore, the voltage drop across the resistor is 400 V.
Question 14
Question
A circuit contains a resistor with resistance R= 100 Ω connected to a battery
with voltage V= 12 V. If a current of I= 0.1 A flows through the circuit, what
is the power dissipated by the resistor?
Solution
Ohm’s Law states that the current through a conductor between two points is
directly proportional to the voltage across the two points. Mathematically, this
is represented by the equation V=IR, where Vis the voltage, Iis the current,
and Ris the resistance.
Step 1: Calculate the power dissipated by the resistor using the formula
P=IV .
P=IV
= (0.1 A)(12 V)
= 1.2 W
Step 2: Verify the calculated power using the formula P=I2R.
P=I2R
= (0.1 A)2×100 Ω
= 0.01 ×100
= 1 W
Therefore, the power dissipated by the resistor is 1.2 W.
Question 15
Question
A resistor with a resistance of 12 ohms is connected to a 24-volt battery. What
is the current flowing through the resistor?
9
Solution
Step 1: Recall Ohm’s Law, which states that the current (I) flowing through a
resistor is equal to the voltage (V) across the resistor divided by the resistance
(R) of the resistor. Mathematically, this can be represented as I=V
R.
Step 2: Given that the resistance Ris 12 ohms and the voltage Vis 24 volts,
we can substitute these values into Ohm’s Law to find the current I:
I=24 V
12Ω
Step 3: Simplifying the expression, we get:
I= 2 A
Step 4: Therefore, the current flowing through the resistor is 2 amperes.
Question 16
Question
A circuit consists of a resistor with a resistance of 30 Ω connected to a 12 V
battery. Calculate the current flowing through the circuit.
Solution
Step 1: Recall Ohm’s Law, which states that the current (I) flowing through a
circuit is equal to the voltage (V) across the circuit divided by the resistance
(R) of the circuit. Mathematically, this can be expressed as:
I=V
R
Step 2: Given that the voltage Vis 12 V and the resistance Ris 30 Ω,
substitute these values into Ohm’s Law to find the current I:
I=12 V
30Ω
Step 3: Calculate the current I:
I=12
30 A
I= 0.4 A
Step 4: Therefore, the current flowing through the circuit is 0.4 A.
10
Question 17
Question
A resistor with resistance 10 ohms is connected to a battery with a voltage of
12 volts. What is the current flowing through the resistor?
Solution
To find the current flowing through the resistor, we can use Ohm’s Law, which
states that V=IR, where: - Vis the voltage across the resistor, - Iis the
current flowing through the resistor, and - Ris the resistance of the resistor.
Step 1: Substitute the given values into Ohm’s Law.
V=IR
12 = I×10
Step 2: Solve for the current, I.
I=12
10
I= 1.2 amps
Therefore, the current flowing through the resistor is 1.2 amps.
Question 18
Question
A resistor has a resistance of 15 Ω and a current of 2 A passing through it.
What is the voltage drop across the resistor?
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor
is equal to the product of the current (I) passing through it and the resistance
(R) of the resistor. Mathematically, this is expressed as:
V=I×R
Step 2: Given that the resistance Ris 15 Ω and the current Iis 2 A, we can
substitute these values into Ohm’s Law to find the voltage drop (V) across the
resistor.
V= 2 A ×15 Ω
Step 3: Calculate the voltage drop across the resistor:
V= 30 V
Therefore, the voltage drop across the resistor is 30 V.
11
Question 19
Question
A certain circuit has a resistance of 30 Ω and a current of 0.5 A flowing through
it. If the voltage across the circuit is measured to be 15 V, what is the power
dissipated by the circuit?
Solution
To find the power dissipated by the circuit, we can use the formula for power in
a circuit: P=V I, where Pis the power, Vis the voltage, and Iis the current.
We are also given the value of resistance Rin the circuit, so we can use Ohm’s
Law, V=IR, to determine the current Iin the circuit.
Step 1: Calculate the current Iusing Ohm’s Law: V=IR
I=V
R=15 V
30 Ω = 0.5 A
Step 2: Substitute the values of Vand Iinto the power formula P=V I:
P=V×I= 15 V ×0.5 A = 7.5 W
Therefore, the power dissipated by the circuit is 7.5 W.
Question 20
Question
A resistor with a resistance of 6 Ω is connected to a 12 V battery. Determine
the current flowing through the resistor.
Solution
Let’s use Ohm’s Law, which states that the current passing through a conductor
between two points is directly proportional to the voltage across the two points
and inversely proportional to the resistance. Mathematically, Ohm’s Law can
be expressed as V=IR, where V= Voltage across the resistor, I= Current
passing through the resistor, and R= Resistance of the resistor.
Step 1: Given that the voltage across the resistor is V= 12 V and the
resistance of the resistor is R= 6 Ω, we can use Ohm’s Law to find the current
passing through the resistor.
I=V
R
Step 2: Substituting V= 12 V and R= 6 Ω into the formula, we get
I=12
6
12
I= 2 A
Step 3: Therefore, the current passing through the resistor is 2 A when
connected to a 12 V battery.
Question 21
Question
A circuit consists of a resistor, an inductor, and a capacitor connected in series.
The resistor has a resistance of 20 Ω, the inductor has an inductance of 0.01 H,
and the capacitor has a capacitance of 10 µF. If the frequency of the alternating
current in the circuit is 100 Hz, determine the current flowing through the circuit.
Solution
Let’s denote the resistance, inductive reactance, and capacitive reactance as R,
XL, and XCrespectively. The total impedance Zof the circuit can be calculated
using the formula:
Z=pR2+ (XL−XC)2
Step 1: Calculate the inductive reactance The inductive reactance XL
is given by:
XL= 2πf L
where fis the frequency of the AC current and Lis the inductance of the
inductor. Substitute f= 100 Hz and L= 0.01 H into the formula to get:
XL= 2π×100 ×0.01 = 2 Ω
Step 2: Calculate the capacitive reactance The capacitive reactance
XCis given by:
XC=1
2πf C
where fis the frequency of the AC current and Cis the capacitance of the
capacitor. Substitute f= 100 Hz and C= 10 µF = 10−5F into the formula to
get:
XC=1
2π×100 ×10−5= 1592.46 Ω
Step 3: Calculate the total impedance Substitute R= 20 Ω, XL= 2 Ω,
and XC= 1592.46 Ω into the formula for total impedance Z:
Z=p202+ (2 −1592.46)2=p400 + 1590.462≈1591.92 Ω
Step 4: Use Ohm’s Law The current Iflowing through the circuit can
be calculated using Ohm’s Law:
I=V
Z
where Vis the voltage across the circuit. Since the voltage Vhas not been
specified, we cannot calculate the current accurately.
13
Question 22
Question
A circuit consists of a resistor with resistance R= 15 Ω connected to a battery
with voltage V= 120 V. Calculate the current passing through the circuit.
Solution
To calculate the current passing through the circuit, we can use Ohm’s Law,
which states that V=IR, where Vis the voltage across the resistor, Iis the
current passing through the resistor, and Ris the resistance of the resistor.
Step 1: Write down Ohm’s Law equation:
V=IR
Step 2: Rearrange the equation to solve for current I:
I=V
R
Step 3: Substitute V= 120 Vand R= 15 Ω into the equation:
I=120 V
15 Ω
Step 4: Calculate the current passing through the circuit:
I=120 V
15 Ω = 8 A
Therefore, the current passing through the circuit is 8 A.
Question 23
Question
A resistor with resistance 25 Ω is connected to a potential difference of 50 V.
What is the current flowing through the resistor?
Solution
Let’s use Ohm’s Law, which states that V=IR, where Vis the potential
difference across the resistor, Iis the current flowing through the resistor, and
Ris the resistance of the resistor.
Step 1: Given data: Resistance of the resistor, R= 25 Ω Potential difference
across the resistor, V= 50 V
14
Step 2: Apply Ohm’s Law to find the current: From Ohm’s Law, V=IR,
we can rearrange the formula to solve for current I:
I=V
R
Step 3: Substitute the known values into the formula:
I=50
25 = 2 A
Step 4: Answer: The current flowing through the resistor is 2 A.
Question 24
Question
A resistor with resistance R= 100 Ω is connected to a voltage source with
V= 12 V. Calculate the current flowing through the resistor.
Solution
Step 1: Write Ohm’s Law: V=IR, where Vis the voltage, Iis the current,
and Ris the resistance.
Step 2: Rearrange Ohm’s Law to solve for current: I=V
R.
Step 3: Substitute the given values V= 12 V and R= 100 Ω into the formula
I=V
R.
Step 4: Calculate the current flowing through the resistor:
I=12 V
100 Ω = 0.12 A
Therefore, the current flowing through the resistor is 0.12 A.
Question 25
Question
A resistor has a resistance of 30 Ω and a current of 0.5Aflowing through it.
Determine the voltage drop across the resistor.
Solution
Ohm’s Law states that the voltage drop (V) across a resistor is equal to the
product of the resistance (R) and the current (I) flowing through it. Mathe-
matically, Ohm’s Law is represented as V=IR.
15
Step 1: Given the resistance R= 30 Ω and the current I= 0.5A, we can
use Ohm’s Law to find the voltage drop V.
V=IR
V= 0.5A×30 Ω
V= 15 V
Step 2: Therefore, the voltage drop across the resistor is 15 V.
Question 26
Question
A cylindrical resistor has a resistance of 10 Ω and a length of 2.0 meters. If the
resistivity of the material is 1.7×10−6Ω·m, what is the radius of the resistor?
Solution
Step 1: The resistance Rof a cylindrical resistor is given by the formula:
R=ρ·L
A
where ρis the resistivity of the material, Lis the length of the resistor, and A
is the cross-sectional area.
Step 2: We are given that R= 10 Ω, ρ= 1.7×10−6Ω·m, and L= 2.0 m.
We need to find the radius r.
Step 3: The cross-sectional area of a cylindrical resistor is given by the
formula:
A=πr2
Step 4: Substitute the expressions for Rand Ainto the formula for resis-
tance:
10 = 1.7×10−6·2
πr2
Step 5: Simplify the equation:
10 = 3.4×10−6
πr2
πr2=3.4×10−6
10
r2=3.4×10−6
10π
Step 6: Calculate the radius:
r=r3.4×10−6
10π≈0.00491 m
Step 7: Therefore, the radius of the cylindrical resistor is approximately
0.00491 meters.
16
Question 27
Question
A circuit consists of a resistor with a resistance of 50 Ω, a capacitor with a
capacitance of 0.01 F, and an inductor with an inductance of 0.05 H connected
in series to a voltage source of 12 V. Calculate the current flowing through the
circuit using Ohm’s Law.
Solution
To calculate the current flowing through the circuit, we can use Ohm’s Law,
which states V=IR, where: - Vis the voltage across the circuit, - Iis the
current flowing through the circuit, and - Ris the total resistance of the circuit.
Step 1: Calculate the total resistance of the circuit. The total resistance
Rtotal in a series circuit is the sum of the individual resistances:
Rtotal =Rresistor +Rinductor +Rcapacitor
Plugging in the given values:
Rtotal = 50 Ω + 0 Ω + 0 Ω = 50 Ω
Step 2: Use Ohm’s Law to find the current. Given V= 12 V, R= 50 Ω, we
can rearrange Ohm’s Law to solve for the current I:
I=V
R
I=12
50
I= 0.24 A
Therefore, the current flowing through the circuit is 0.24 A.
Question 28
Question
A copper wire has a resistance of 5 ohms. If a current of 2 amperes flows through
the wire, what is the voltage across the wire?
Solution
Step 1: Recall Ohm’s Law, which states that the voltage across a resistor is
equal to the product of the current flowing through it and the resistance of
the resistor: V=IR, where Vis the voltage (in volts), Iis the current (in
amperes), and Ris the resistance (in ohms).
17
Step 2: Given that the resistance of the copper wire is 5 ohms and the
current flowing through it is 2 amperes, we can use Ohm’s Law to find the
voltage across the wire:
V=I×R= 2 A ×5 Ω = 10 V
Step 3: Therefore, the voltage across the copper wire is 10 volts.
Question 29
Question
A resistor with resistance Ris connected to a voltage source such that a current I
flows through it. The power dissipated in the resistor is given by P=V I −1
2RI2.
Prove Ohm’s Law using this relationship.
Solution
To prove Ohm’s Law, we need to show that the relationship between voltage
(V), current (I), and resistance (R) is V=IR.
Step 1: Start with the given expression for power:
P=V I −1
2RI2
Step 2: Differentiate both sides of the equation with respect to time t:
dP
dt =d(V I)
dt −d1
2RI2
dt
Step 3: Using the product rule for differentiation, we have:
dP
dt =VdI
dt +IdV
dt −1
2R·2I·dI
dt
Step 4: Simplify the equation:
dP
dt =VdI
dt +IdV
dt −RI dI
dt
Step 5: Recognize that dP
dt is the rate at which power is being dissipated,
which is equal to the rate at which energy is being supplied by the voltage
source. Therefore, the left-hand side simplifies to:
dP
dt =V I
Step 6: Substitute V I back into the equation:
V I =VdI
dt +IdV
dt −RI dI
dt
18
Step 7: Rearrange the terms to get:
V I =IdV
dt −RI
Step 8: Since V=IR (Ohm’s Law), we know that dV
dt =RdI
dt . Substitute
this into the equation:
V I =I(RdI
dt −RI)
Step 9: Simplify the equation to:
V I =I(RdI
dt −RI)
V I =I2R−I2R
V I = 0
Step 10: Therefore, we have shown that V I = 0, which implies that V=
IR. This is Ohm’s Law.
Question 30
Question
A circuit consists of a resistor with resistance R= 100 Ω connected to a battery
with electromotive force (emf) E= 12 V. If the current flowing through the
circuit is I= 0.1 A, what is the internal resistance of the battery?
Solution
Ohm’s Law states that the potential difference across a resistor is equal to the
current flowing through it multiplied by the resistance. In this case, the potential
difference (voltage) across the resistor is equal to the emf of the battery.
Step 1: Calculate the potential difference across the resistor.
V=IR
V= 0.1 A ×100 Ω
V= 10 V
Step 2: Since the potential difference across the resistor is equal to the emf
of the battery, we have:
V=E
10 V = 12 V −I×r
10 V = 12 V −0.1 A ×r
10 V = 12 V −0.1 A ×r
19
Step 3: Solve for the internal resistance r.
0.1 A ×r= 2 V
r=2 V
0.1 A
r= 20 Ω
Therefore, the internal resistance of the battery is 20 Ω.
Question 31
Question
A resistor with a resistance of 10 ohms is connected to a 12-volt battery. What
is the current passing through the resistor?
Solution
Step 1: Recall Ohm’s Law, which states that the current passing through a
resistor is equal to the voltage across the resistor divided by the resistance of
the resistor. Mathematically, this can be represented as I=V
R, where Iis the
current in amperes (A), Vis the voltage in volts (V), and Ris the resistance in
ohms (Ω).
Step 2: Given that the resistance of the resistor is 10 ohms and the voltage
of the battery is 12 volts, we can plug these values into Ohm’s Law to calculate
the current passing through the resistor.
I=12 V
10 Ω
Step 3: Simplifying the expression, we find:
I= 1.2 A
Therefore, the current passing through the resistor is 1.2 amperes.
Question 32
Question
A circuit consists of a resistor, a capacitor, and an inductor in series. The
resistor has a resistance of 10 Ω, the capacitor has a reactance of 5 Ω, and the
inductor has a reactance of 8 Ω. If a voltage of 100 V is applied across the
circuit, what is the current flowing through the circuit?
20
Solution
Step 1: Calculate the total impedance of the circuit using the formula Ztotal =
pR2+ (XL−XC)2.
Ztotal =p102+ (8 −5)2
=√100 + 9
=√109
≈10.44 Ω
Step 2: Calculate the current flowing through the circuit using Ohm’s Law:
I=V
Ztotal .
I=100
10.44
≈9.57 A
Therefore, the current flowing through the circuit is approximately 9.57 A.
Question 33
Question
A circuit contains a resistor with resistance R= 30 Ω and a battery with voltage
V= 120 V. If a current of I= 4 Aflows through the circuit, what is the power
dissipated by the resistor?
Solution
Step 1: Recall Ohm’s Law, which relates voltage, current, and resistance: V=
I·R.
Step 2: Substitute the given values into Ohm’s Law to find the voltage across
the resistor:
V=I·R
120 V= 4 A·30 Ω
120 V= 120 V
Step 3: Since the voltage across the resistor is the same as the battery
voltage, the resistor is dissipating power. The power dissipated by a resistor
can be calculated using the formula P=I2·Ror P=V2
R.
Step 4: We will use the formula P=I2·Rto find the power dissipated by
the resistor:
P=I2·R
P= (4 A)2·30 Ω
P= 16 A2·30 Ω
P= 480 W
21
Step 5: Therefore, the power dissipated by the resistor in the circuit is
480 W.
Question 34
Question
A resistor has a resistance of 20 Ω and a current of 2 A passing through it.
Determine the voltage drop across the resistor using Ohm’s Law.
Solution
Let’s recall Ohm’s Law, which states that the voltage drop across a resistor
is equal to the product of the resistance and the current passing through it.
Mathematically, Ohm’s Law is represented as:
V=IR
where: V= voltage drop across the resistor (in volts), I= current passing
through the resistor (in amperes), and R= resistance of the resistor (in ohms).
Step 1: Given that the resistance Ris 20 Ω and the current Iis 2 A, we
can substitute these values into Ohm’s Law to find the voltage drop V:
V= (2 A)(20 Ω)
Step 2: Now, multiply the current and resistance to find the voltage drop:
V= 2 A ×20 Ω = 40 V
Therefore, the voltage drop across the resistor is 40 volts.
Question 35
Question
A resistor with a resistance of 10 Ω is connected to a voltage source that produces
a current of 2 A. Calculate the voltage across the resistor.
Solution
To calculate the voltage across the resistor, we can use Ohm’s Law, which states
that V=I×R, where Vis the voltage, Iis the current, and Ris the resistance
of the resistor.
Step 1: Identify the given values: The resistance of the resistor, R= 10 Ω,
and the current passing through it, I= 2 A.
Step 2: Apply Ohm’s Law to calculate the voltage:
V=I×R
22
Solution
Let’s use Ohm’s Law, which states that the voltage across a resistor is equal
to the current flowing through it multiplied by the resistance. Mathematically,
Ohm’s Law can be written as:
V=IR
Where: - Vis the voltage across the resistor (in volts), - Iis the current
flowing through the resistor (in amperes), and - Ris the resistance of the resistor
(in ohms).
Step 1: Given that the voltage V= 10 V and the current I= 2 A, we can
substitute these values into Ohm’s Law to find the resistance:
10 V = 2 A ×R
Step 2: Solving for the resistance R, we get:
R=10 V
2 A = 5 Ω
Therefore, the resistance of the resistor is 5 Ω.
Question 4
Question
A circuit consists of a resistor, a capacitor, and an inductor connected in series.
The resistor has a resistance of 50 Ω, the capacitor has a capacitance of 0.1 F,
and the inductor has an inductance of 0.2 H. If a current of 2 A flows through
the circuit, calculate the voltage drop across each component using Ohm’s Law.
Solution
To calculate the voltage drop across each component, we can use Ohm’s Law,
V=IR, where Vis the voltage drop, Iis the current, and Ris the resistance.
Step 1: Calculate the voltage drop across the resistor Given: R=
50 Ω and I= 2 A Using Ohm’s Law: Vresistor =I·R= 2 A ×50 Ω = 100 V.
Therefore, the voltage drop across the resistor is 100 V.
Step 2: Calculate the voltage drop across the capacitor The rela-
tionship between voltage and current for a capacitor is V=Q
C, where Qis the
charge stored on the capacitor plates and Cis the capacitance. Since we are
given current, we can relate it to the charge using I=dQ
dt . Therefore, Q=RI dt.
Given: I= 2 A and C= 0.1 F Solving the integral: Q=R2dt = 2t+Q0Given
that Q0= 0 (initial charge): Q= 2tUsing V=Q
C:Vcapacitor =2t
0.1= 20tV
Step 3: Calculate the voltage drop across the inductor The relation-
ship between voltage and current for an inductor is V=Ldi
dt , where Lis the
inductance and di
dt is the rate of change of current. Given: L= 0.2 H and I= 2
3
A Taking derivative of Iwith respect to t:di
dt = 0 (since current is constant)
Using V=Ldi
dt :Vinductor = 0 V
Therefore, the voltage drop across the resistor is 100 V, across the capacitor
is 20tV, and across the inductor is 0 V.
Question 5
Question
A circuit consists of a resistor with resistance R= 20 Ω connected to a voltage
source of V= 100 V. Calculate the current flowing through the resistor.
Solution
To calculate the current flowing through the resistor, we can use Ohm’s Law,
which states that V=IR, where Vis the voltage across the resistor, Iis the
current flowing through the resistor, and Ris the resistance of the resistor.
Step 1: Substitute V= 100 V and R= 20 Ω into Ohm’s Law, V=IR, and
solve for I:
I=V
R
I=100 V
20 Ω
I= 5 A
Step 2: The current flowing through the resistor is I= 5 A.
Question 6
Question
A circuit contains a resistor with a resistance of 20 Ω and a current of 2 A
flowing through it. Determine the voltage drop across the resistor.
Solution
Step 1: Write down Ohm’s Law, which states that V=IR, where Vis the
voltage, Iis the current, and Ris the resistance of the resistor.
Step 2: Plug in the values given in the question. In this case, I= 2 A and
R= 20 Ω.
V= (2 A)(20 Ω)
Step 3: Calculate the voltage drop across the resistor.
V= 2 ×20 = 40 V
Step 4: Therefore, the voltage drop across the resistor is 40 V.
4
Question 7
Question
A circuit consists of a resistor, an inductor, and a capacitor connected in series.
The resistance of the resistor is 10 Ω, the inductance of the inductor is 5 H, and
the capacitance of the capacitor is 0.02 F. If the peak current flowing through
the circuit is 2 A, find the peak voltage across the circuit.
Solution
The peak voltage across the circuit can be found using Ohm’s Law, which states:
V=I·Z, where Vis the voltage, Iis the current, and Zis the impedance of
the circuit.
Step 1: Calculate the impedance of the circuit using the total resistance,
inductive reactance, and capacitive reactance. The impedance of the circuit is
given by the formula: Z=pR2+ (XL−XC)2, where Ris the resistance, XL
is the inductive reactance, and XCis the capacitive reactance. Given: Resistor
resistance R= 10 Ω, Inductive reactance XL= 2πfL, Capacitive reactance
XC=1
2πfC . Let’s choose a frequency, f= 1 Hz, to simplify calculations.
Substitute the given values into the formulas: XL= 2π·1·5 = 10 Ω, XC=
1
2π·1·0.02 = 7.96 Ω. Now, calculate the impedance: Z=p102+ (10 −7.96)2=
√100 + 4.082=√116.94 ≈10.81 Ω.
Step 2: Calculate the peak voltage across the circuit using Ohm’s Law.
Given peak current I= 2 A and impedance Z= 10.81 Ω, V=I·Z= 2·10.81 =
21.62 V.
Therefore, the peak voltage across the circuit is 21.62 V.
Question 8
Question
A circuit consists of a resistor with resistance R= 20 Ω and a battery with EMF
E= 12 V. If the current flowing through the circuit is I= 0.5A, calculate the
power dissipated in the resistor.
Solution
Step 1: To find the power dissipated in the resistor, we can use the formula for
power in a resistor: P=I2R.
Step 2: Substitute the given values into the formula. We have I= 0.5Aand
R= 20 Ω.
Step 3: Calculate the power:
P= (0.5A)2×20 Ω = 0.25 A2×20 Ω = 5 W
Step 4: Therefore, the power dissipated in the resistor is 5 W.
5
Question 9
Question
A resistor has a resistance of 10 Ω and a current of 2 A passing through it.
Determine the voltage drop across the resistor.
Solution
Step 1: Write down Ohm’s Law, which states that V=IR, where Vis the
voltage drop across the resistor, Iis the current passing through the resistor,
and Ris the resistance of the resistor.
Step 2: Substitute the given values into Ohm’s Law:
V= (2 A)(10Ω)
Step 3: Calculate the voltage drop:
V= 20 V
Therefore, the voltage drop across the resistor is 20 V.
Question 10
Question
A resistor with a resistance of 10 Ω is connected in series with a resistor with
an unknown resistance. A potential difference of 50 V is applied across the
combination of resistors, and a current of 2 A flows through the circuit. What
is the resistance of the unknown resistor?
Solution
Step 1: We can start by writing down Ohm’s Law, which relates voltage (V),
current (I), and resistance (R) using the equation V=IR. In this case, we
have the following values: - Total voltage across the resistors, Vtotal = 50 V, -
Total current through the circuit, Itotal = 2 A.
Step 2: The total voltage across the two resistors is the sum of the voltage
drops across each resistor:
Vtotal =V1+V2
Step 3: We can express the voltage drops in terms of the resistances and the
total current using Ohm’s Law:
V1=Itotal ·R1
V2=Itotal ·R2
6
Step 4: We are given the resistance of one resistor (R1= 10 Ω) and the cur-
rent flowing through the circuit. We need to find the resistance of the unknown
resistor (R2).
Step 5: Substituting the known values into the equations, we get:
50 = 2 ·10 + 2 ·R2
Step 6: Solving for R2gives:
50 = 20 + 2R2
2R2= 30
R2= 15 Ω
Step 7: Therefore, the resistance of the unknown resistor is 15 Ω.
Question 11
Question
A circuit consists of a resistor with resistance R= 20 Ω connected to a battery
with emf E= 9 V. If the current flowing through the circuit is I= 0.4 A, what
is the potential difference across the resistor?
Solution
To find the potential difference across the resistor, we can use Ohm’s Law which
states that V=IR, where Vis the potential difference, Iis the current, and R
is the resistance.
Step 1: Write down Ohm’s Law:
V=IR
Step 2: Substitute the given values into the equation:
V= (0.4 A)(20 Ω)
Step 3: Perform the calculation:
V= 8 V
Step 4: Answer: The potential difference across the resistor is 8 volts.
Question 12
Question
A circuit consists of a resistor, a capacitor, and an inductor connected in series.
The values of the components are as follows: resistor R= 10 Ω, capacitor
C= 5 µF , and inductor L= 2 mH. If a sinusoidal voltage source with frequency
f= 1 kHz is connected to the circuit, determine the current flowing through
the circuit using Ohm’s Law.
7
Solution
Step 1: Calculate the total impedance of the circuit using the formula for
impedance in an RLC series circuit:
Z=qR2+ (XL−XC)2
where XL= 2πf L is the inductive reactance and XC=1
2πfC is the capacitive
reactance.
Step 2: Calculate the inductive reactance XL:
XL= 2πfL = 2 ·π·1000 ·0.002 = 12.57 Ω
Step 3: Calculate the capacitive reactance XC:
XC=1
2πf C =1
2·π·1000 ·5×10−6= 31.83 Ω
Step 4: Calculate the total impedance Z:
Z=p102+ (12.57 −31.83)2=√100 + 338.56 = √438.56 = 20.92 Ω
Step 5: Calculate the current Iflowing through the circuit using Ohm’s Law:
I=V
Z
where Vis the voltage of the sinusoidal source. Let us assume V= 10 Vfor
simplicity.
I=10
20.92 = 0.477 A
Therefore, the current flowing through the circuit is 0.477 A.
Question 13
Question
A resistor has a resistance of 500 Ω and a current of 0.8 A flowing through it.
Calculate the voltage drop across the resistor.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor
is equal to the product of the current (I) flowing through it and the resistance
(R) of the resistor. Mathematically, this relationship is represented as:
V=I×R
8
Step 2: Given that the resistance (R) is 500 Ω and the current (I) is 0.8 A,
we can substitute those values into Ohm’s Law to solve for the voltage:
V= 0.8 A ×500 Ω
Step 3: Multiply the current and resistance to find the voltage drop:
V= 0.8 A ×500 Ω = 400 V
Step 4: Therefore, the voltage drop across the resistor is 400 V.
Question 14
Question
A circuit contains a resistor with resistance R= 100 Ω connected to a battery
with voltage V= 12 V. If a current of I= 0.1 A flows through the circuit, what
is the power dissipated by the resistor?
Solution
Ohm’s Law states that the current through a conductor between two points is
directly proportional to the voltage across the two points. Mathematically, this
is represented by the equation V=IR, where Vis the voltage, Iis the current,
and Ris the resistance.
Step 1: Calculate the power dissipated by the resistor using the formula
P=IV .
P=IV
= (0.1 A)(12 V)
= 1.2 W
Step 2: Verify the calculated power using the formula P=I2R.
P=I2R
= (0.1 A)2×100 Ω
= 0.01 ×100
= 1 W
Therefore, the power dissipated by the resistor is 1.2 W.
Question 15
Question
A resistor with a resistance of 12 ohms is connected to a 24-volt battery. What
is the current flowing through the resistor?
9
Solution
Step 1: Recall Ohm’s Law, which states that the current (I) flowing through a
resistor is equal to the voltage (V) across the resistor divided by the resistance
(R) of the resistor. Mathematically, this can be represented as I=V
R.
Step 2: Given that the resistance Ris 12 ohms and the voltage Vis 24 volts,
we can substitute these values into Ohm’s Law to find the current I:
I=24 V
12Ω
Step 3: Simplifying the expression, we get:
I= 2 A
Step 4: Therefore, the current flowing through the resistor is 2 amperes.
Question 16
Question
A circuit consists of a resistor with a resistance of 30 Ω connected to a 12 V
battery. Calculate the current flowing through the circuit.
Solution
Step 1: Recall Ohm’s Law, which states that the current (I) flowing through a
circuit is equal to the voltage (V) across the circuit divided by the resistance
(R) of the circuit. Mathematically, this can be expressed as:
I=V
R
Step 2: Given that the voltage Vis 12 V and the resistance Ris 30 Ω,
substitute these values into Ohm’s Law to find the current I:
I=12 V
30Ω
Step 3: Calculate the current I:
I=12
30 A
I= 0.4 A
Step 4: Therefore, the current flowing through the circuit is 0.4 A.
10
Question 17
Question
A resistor with resistance 10 ohms is connected to a battery with a voltage of
12 volts. What is the current flowing through the resistor?
Solution
To find the current flowing through the resistor, we can use Ohm’s Law, which
states that V=IR, where: - Vis the voltage across the resistor, - Iis the
current flowing through the resistor, and - Ris the resistance of the resistor.
Step 1: Substitute the given values into Ohm’s Law.
V=IR
12 = I×10
Step 2: Solve for the current, I.
I=12
10
I= 1.2 amps
Therefore, the current flowing through the resistor is 1.2 amps.
Question 18
Question
A resistor has a resistance of 15 Ω and a current of 2 A passing through it.
What is the voltage drop across the resistor?
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor
is equal to the product of the current (I) passing through it and the resistance
(R) of the resistor. Mathematically, this is expressed as:
V=I×R
Step 2: Given that the resistance Ris 15 Ω and the current Iis 2 A, we can
substitute these values into Ohm’s Law to find the voltage drop (V) across the
resistor.
V= 2 A ×15 Ω
Step 3: Calculate the voltage drop across the resistor:
V= 30 V
Therefore, the voltage drop across the resistor is 30 V.
11
Question 19
Question
A certain circuit has a resistance of 30 Ω and a current of 0.5 A flowing through
it. If the voltage across the circuit is measured to be 15 V, what is the power
dissipated by the circuit?
Solution
To find the power dissipated by the circuit, we can use the formula for power in
a circuit: P=V I, where Pis the power, Vis the voltage, and Iis the current.
We are also given the value of resistance Rin the circuit, so we can use Ohm’s
Law, V=IR, to determine the current Iin the circuit.
Step 1: Calculate the current Iusing Ohm’s Law: V=IR
I=V
R=15 V
30 Ω = 0.5 A
Step 2: Substitute the values of Vand Iinto the power formula P=V I:
P=V×I= 15 V ×0.5 A = 7.5 W
Therefore, the power dissipated by the circuit is 7.5 W.
Question 20
Question
A resistor with a resistance of 6 Ω is connected to a 12 V battery. Determine
the current flowing through the resistor.
Solution
Let’s use Ohm’s Law, which states that the current passing through a conductor
between two points is directly proportional to the voltage across the two points
and inversely proportional to the resistance. Mathematically, Ohm’s Law can
be expressed as V=IR, where V= Voltage across the resistor, I= Current
passing through the resistor, and R= Resistance of the resistor.
Step 1: Given that the voltage across the resistor is V= 12 V and the
resistance of the resistor is R= 6 Ω, we can use Ohm’s Law to find the current
passing through the resistor.
I=V
R
Step 2: Substituting V= 12 V and R= 6 Ω into the formula, we get
I=12
6
12
I= 2 A
Step 3: Therefore, the current passing through the resistor is 2 A when
connected to a 12 V battery.
Question 21
Question
A circuit consists of a resistor, an inductor, and a capacitor connected in series.
The resistor has a resistance of 20 Ω, the inductor has an inductance of 0.01 H,
and the capacitor has a capacitance of 10 µF. If the frequency of the alternating
current in the circuit is 100 Hz, determine the current flowing through the circuit.
Solution
Let’s denote the resistance, inductive reactance, and capacitive reactance as R,
XL, and XCrespectively. The total impedance Zof the circuit can be calculated
using the formula:
Z=pR2+ (XL−XC)2
Step 1: Calculate the inductive reactance The inductive reactance XL
is given by:
XL= 2πf L
where fis the frequency of the AC current and Lis the inductance of the
inductor. Substitute f= 100 Hz and L= 0.01 H into the formula to get:
XL= 2π×100 ×0.01 = 2 Ω
Step 2: Calculate the capacitive reactance The capacitive reactance
XCis given by:
XC=1
2πf C
where fis the frequency of the AC current and Cis the capacitance of the
capacitor. Substitute f= 100 Hz and C= 10 µF = 10−5F into the formula to
get:
XC=1
2π×100 ×10−5= 1592.46 Ω
Step 3: Calculate the total impedance Substitute R= 20 Ω, XL= 2 Ω,
and XC= 1592.46 Ω into the formula for total impedance Z:
Z=p202+ (2 −1592.46)2=p400 + 1590.462≈1591.92 Ω
Step 4: Use Ohm’s Law The current Iflowing through the circuit can
be calculated using Ohm’s Law:
I=V
Z
where Vis the voltage across the circuit. Since the voltage Vhas not been
specified, we cannot calculate the current accurately.
13
Question 22
Question
A circuit consists of a resistor with resistance R= 15 Ω connected to a battery
with voltage V= 120 V. Calculate the current passing through the circuit.
Solution
To calculate the current passing through the circuit, we can use Ohm’s Law,
which states that V=IR, where Vis the voltage across the resistor, Iis the
current passing through the resistor, and Ris the resistance of the resistor.
Step 1: Write down Ohm’s Law equation:
V=IR
Step 2: Rearrange the equation to solve for current I:
I=V
R
Step 3: Substitute V= 120 Vand R= 15 Ω into the equation:
I=120 V
15 Ω
Step 4: Calculate the current passing through the circuit:
I=120 V
15 Ω = 8 A
Therefore, the current passing through the circuit is 8 A.
Question 23
Question
A resistor with resistance 25 Ω is connected to a potential difference of 50 V.
What is the current flowing through the resistor?
Solution
Let’s use Ohm’s Law, which states that V=IR, where Vis the potential
difference across the resistor, Iis the current flowing through the resistor, and
Ris the resistance of the resistor.
Step 1: Given data: Resistance of the resistor, R= 25 Ω Potential difference
across the resistor, V= 50 V
14
Step 2: Apply Ohm’s Law to find the current: From Ohm’s Law, V=IR,
we can rearrange the formula to solve for current I:
I=V
R
Step 3: Substitute the known values into the formula:
I=50
25 = 2 A
Step 4: Answer: The current flowing through the resistor is 2 A.
Question 24
Question
A resistor with resistance R= 100 Ω is connected to a voltage source with
V= 12 V. Calculate the current flowing through the resistor.
Solution
Step 1: Write Ohm’s Law: V=IR, where Vis the voltage, Iis the current,
and Ris the resistance.
Step 2: Rearrange Ohm’s Law to solve for current: I=V
R.
Step 3: Substitute the given values V= 12 V and R= 100 Ω into the formula
I=V
R.
Step 4: Calculate the current flowing through the resistor:
I=12 V
100 Ω = 0.12 A
Therefore, the current flowing through the resistor is 0.12 A.
Question 25
Question
A resistor has a resistance of 30 Ω and a current of 0.5Aflowing through it.
Determine the voltage drop across the resistor.
Solution
Ohm’s Law states that the voltage drop (V) across a resistor is equal to the
product of the resistance (R) and the current (I) flowing through it. Mathe-
matically, Ohm’s Law is represented as V=IR.
15
Step 1: Given the resistance R= 30 Ω and the current I= 0.5A, we can
use Ohm’s Law to find the voltage drop V.
V=IR
V= 0.5A×30 Ω
V= 15 V
Step 2: Therefore, the voltage drop across the resistor is 15 V.
Question 26
Question
A cylindrical resistor has a resistance of 10 Ω and a length of 2.0 meters. If the
resistivity of the material is 1.7×10−6Ω·m, what is the radius of the resistor?
Solution
Step 1: The resistance Rof a cylindrical resistor is given by the formula:
R=ρ·L
A
where ρis the resistivity of the material, Lis the length of the resistor, and A
is the cross-sectional area.
Step 2: We are given that R= 10 Ω, ρ= 1.7×10−6Ω·m, and L= 2.0 m.
We need to find the radius r.
Step 3: The cross-sectional area of a cylindrical resistor is given by the
formula:
A=πr2
Step 4: Substitute the expressions for Rand Ainto the formula for resis-
tance:
10 = 1.7×10−6·2
πr2
Step 5: Simplify the equation:
10 = 3.4×10−6
πr2
πr2=3.4×10−6
10
r2=3.4×10−6
10π
Step 6: Calculate the radius:
r=r3.4×10−6
10π≈0.00491 m
Step 7: Therefore, the radius of the cylindrical resistor is approximately
0.00491 meters.
16
Question 27
Question
A circuit consists of a resistor with a resistance of 50 Ω, a capacitor with a
capacitance of 0.01 F, and an inductor with an inductance of 0.05 H connected
in series to a voltage source of 12 V. Calculate the current flowing through the
circuit using Ohm’s Law.
Solution
To calculate the current flowing through the circuit, we can use Ohm’s Law,
which states V=IR, where: - Vis the voltage across the circuit, - Iis the
current flowing through the circuit, and - Ris the total resistance of the circuit.
Step 1: Calculate the total resistance of the circuit. The total resistance
Rtotal in a series circuit is the sum of the individual resistances:
Rtotal =Rresistor +Rinductor +Rcapacitor
Plugging in the given values:
Rtotal = 50 Ω + 0 Ω + 0 Ω = 50 Ω
Step 2: Use Ohm’s Law to find the current. Given V= 12 V, R= 50 Ω, we
can rearrange Ohm’s Law to solve for the current I:
I=V
R
I=12
50
I= 0.24 A
Therefore, the current flowing through the circuit is 0.24 A.
Question 28
Question
A copper wire has a resistance of 5 ohms. If a current of 2 amperes flows through
the wire, what is the voltage across the wire?
Solution
Step 1: Recall Ohm’s Law, which states that the voltage across a resistor is
equal to the product of the current flowing through it and the resistance of
the resistor: V=IR, where Vis the voltage (in volts), Iis the current (in
amperes), and Ris the resistance (in ohms).
17
Step 2: Given that the resistance of the copper wire is 5 ohms and the
current flowing through it is 2 amperes, we can use Ohm’s Law to find the
voltage across the wire:
V=I×R= 2 A ×5 Ω = 10 V
Step 3: Therefore, the voltage across the copper wire is 10 volts.
Question 29
Question
A resistor with resistance Ris connected to a voltage source such that a current I
flows through it. The power dissipated in the resistor is given by P=V I −1
2RI2.
Prove Ohm’s Law using this relationship.
Solution
To prove Ohm’s Law, we need to show that the relationship between voltage
(V), current (I), and resistance (R) is V=IR.
Step 1: Start with the given expression for power:
P=V I −1
2RI2
Step 2: Differentiate both sides of the equation with respect to time t:
dP
dt =d(V I)
dt −d1
2RI2
dt
Step 3: Using the product rule for differentiation, we have:
dP
dt =VdI
dt +IdV
dt −1
2R·2I·dI
dt
Step 4: Simplify the equation:
dP
dt =VdI
dt +IdV
dt −RI dI
dt
Step 5: Recognize that dP
dt is the rate at which power is being dissipated,
which is equal to the rate at which energy is being supplied by the voltage
source. Therefore, the left-hand side simplifies to:
dP
dt =V I
Step 6: Substitute V I back into the equation:
V I =VdI
dt +IdV
dt −RI dI
dt
18
Step 7: Rearrange the terms to get:
V I =IdV
dt −RI
Step 8: Since V=IR (Ohm’s Law), we know that dV
dt =RdI
dt . Substitute
this into the equation:
V I =I(RdI
dt −RI)
Step 9: Simplify the equation to:
V I =I(RdI
dt −RI)
V I =I2R−I2R
V I = 0
Step 10: Therefore, we have shown that V I = 0, which implies that V=
IR. This is Ohm’s Law.
Question 30
Question
A circuit consists of a resistor with resistance R= 100 Ω connected to a battery
with electromotive force (emf) E= 12 V. If the current flowing through the
circuit is I= 0.1 A, what is the internal resistance of the battery?
Solution
Ohm’s Law states that the potential difference across a resistor is equal to the
current flowing through it multiplied by the resistance. In this case, the potential
difference (voltage) across the resistor is equal to the emf of the battery.
Step 1: Calculate the potential difference across the resistor.
V=IR
V= 0.1 A ×100 Ω
V= 10 V
Step 2: Since the potential difference across the resistor is equal to the emf
of the battery, we have:
V=E
10 V = 12 V −I×r
10 V = 12 V −0.1 A ×r
10 V = 12 V −0.1 A ×r
19
Step 3: Solve for the internal resistance r.
0.1 A ×r= 2 V
r=2 V
0.1 A
r= 20 Ω
Therefore, the internal resistance of the battery is 20 Ω.
Question 31
Question
A resistor with a resistance of 10 ohms is connected to a 12-volt battery. What
is the current passing through the resistor?
Solution
Step 1: Recall Ohm’s Law, which states that the current passing through a
resistor is equal to the voltage across the resistor divided by the resistance of
the resistor. Mathematically, this can be represented as I=V
R, where Iis the
current in amperes (A), Vis the voltage in volts (V), and Ris the resistance in
ohms (Ω).
Step 2: Given that the resistance of the resistor is 10 ohms and the voltage
of the battery is 12 volts, we can plug these values into Ohm’s Law to calculate
the current passing through the resistor.
I=12 V
10 Ω
Step 3: Simplifying the expression, we find:
I= 1.2 A
Therefore, the current passing through the resistor is 1.2 amperes.
Question 32
Question
A circuit consists of a resistor, a capacitor, and an inductor in series. The
resistor has a resistance of 10 Ω, the capacitor has a reactance of 5 Ω, and the
inductor has a reactance of 8 Ω. If a voltage of 100 V is applied across the
circuit, what is the current flowing through the circuit?
20
Solution
Step 1: Calculate the total impedance of the circuit using the formula Ztotal =
pR2+ (XL−XC)2.
Ztotal =p102+ (8 −5)2
=√100 + 9
=√109
≈10.44 Ω
Step 2: Calculate the current flowing through the circuit using Ohm’s Law:
I=V
Ztotal .
I=100
10.44
≈9.57 A
Therefore, the current flowing through the circuit is approximately 9.57 A.
Question 33
Question
A circuit contains a resistor with resistance R= 30 Ω and a battery with voltage
V= 120 V. If a current of I= 4 Aflows through the circuit, what is the power
dissipated by the resistor?
Solution
Step 1: Recall Ohm’s Law, which relates voltage, current, and resistance: V=
I·R.
Step 2: Substitute the given values into Ohm’s Law to find the voltage across
the resistor:
V=I·R
120 V= 4 A·30 Ω
120 V= 120 V
Step 3: Since the voltage across the resistor is the same as the battery
voltage, the resistor is dissipating power. The power dissipated by a resistor
can be calculated using the formula P=I2·Ror P=V2
R.
Step 4: We will use the formula P=I2·Rto find the power dissipated by
the resistor:
P=I2·R
P= (4 A)2·30 Ω
P= 16 A2·30 Ω
P= 480 W
21
Step 5: Therefore, the power dissipated by the resistor in the circuit is
480 W.
Question 34
Question
A resistor has a resistance of 20 Ω and a current of 2 A passing through it.
Determine the voltage drop across the resistor using Ohm’s Law.
Solution
Let’s recall Ohm’s Law, which states that the voltage drop across a resistor
is equal to the product of the resistance and the current passing through it.
Mathematically, Ohm’s Law is represented as:
V=IR
where: V= voltage drop across the resistor (in volts), I= current passing
through the resistor (in amperes), and R= resistance of the resistor (in ohms).
Step 1: Given that the resistance Ris 20 Ω and the current Iis 2 A, we
can substitute these values into Ohm’s Law to find the voltage drop V:
V= (2 A)(20 Ω)
Step 2: Now, multiply the current and resistance to find the voltage drop:
V= 2 A ×20 Ω = 40 V
Therefore, the voltage drop across the resistor is 40 volts.
Question 35
Question
A resistor with a resistance of 10 Ω is connected to a voltage source that produces
a current of 2 A. Calculate the voltage across the resistor.
Solution
To calculate the voltage across the resistor, we can use Ohm’s Law, which states
that V=I×R, where Vis the voltage, Iis the current, and Ris the resistance
of the resistor.
Step 1: Identify the given values: The resistance of the resistor, R= 10 Ω,
and the current passing through it, I= 2 A.
Step 2: Apply Ohm’s Law to calculate the voltage:
V=I×R
22
Step 4: To simplify the expression, we can rationalize the denominator. Mul-
tiplying the numerator and denominator by the conjugate of the denominator:
I(t) = V0sin(ωt)
R+jωL +1
jωC ·R−jωL −1
jωC
R−jωL −1
jωC
Step 5: After simplifying and expanding, the expression for the current I(t)
becomes:
I(t) = V0sin(ωt)(R−jωL −1
jωC )
R2+ω2L2+1
ω2C2
Question 2
Question
A resistor with a resistance of 350 Ω is connected to a battery with a voltage of
12 V. What is the current flowing through the circuit?
Solution
To find the current flowing through the circuit, we can use Ohm’s Law, which
states that the current (I) flowing through a circuit is equal to the voltage (V)
divided by the resistance (R).
Step 1: Write down Ohm’s Law formula:
I=V
R
Step 2: Plug in the given values:
I=12 V
350 Ω
Step 3: Calculate the current:
I=12
350 A
I≈0.0343 A
Step 4: Therefore, the current flowing through the circuit is approximately
0.0343 A.
Question 3
Question
A resistor is connected to a voltage source of 10 V and a current of 2 A flows
through it. Determine the resistance of the resistor.
2
Solution
Let’s use Ohm’s Law, which states that the voltage across a resistor is equal
to the current flowing through it multiplied by the resistance. Mathematically,
Ohm’s Law can be written as:
V=IR
Where: - Vis the voltage across the resistor (in volts), - Iis the current
flowing through the resistor (in amperes), and - Ris the resistance of the resistor
(in ohms).
Step 1: Given that the voltage V= 10 V and the current I= 2 A, we can
substitute these values into Ohm’s Law to find the resistance:
10 V = 2 A ×R
Step 2: Solving for the resistance R, we get:
R=10 V
2 A = 5 Ω
Therefore, the resistance of the resistor is 5 Ω.
Question 4
Question
A circuit consists of a resistor, a capacitor, and an inductor connected in series.
The resistor has a resistance of 50 Ω, the capacitor has a capacitance of 0.1 F,
and the inductor has an inductance of 0.2 H. If a current of 2 A flows through
the circuit, calculate the voltage drop across each component using Ohm’s Law.
Solution
To calculate the voltage drop across each component, we can use Ohm’s Law,
V=IR, where Vis the voltage drop, Iis the current, and Ris the resistance.
Step 1: Calculate the voltage drop across the resistor Given: R=
50 Ω and I= 2 A Using Ohm’s Law: Vresistor =I·R= 2 A ×50 Ω = 100 V.
Therefore, the voltage drop across the resistor is 100 V.
Step 2: Calculate the voltage drop across the capacitor The rela-
tionship between voltage and current for a capacitor is V=Q
C, where Qis the
charge stored on the capacitor plates and Cis the capacitance. Since we are
given current, we can relate it to the charge using I=dQ
dt . Therefore, Q=RI dt.
Given: I= 2 A and C= 0.1 F Solving the integral: Q=R2dt = 2t+Q0Given
that Q0= 0 (initial charge): Q= 2tUsing V=Q
C:Vcapacitor =2t
0.1= 20tV
Step 3: Calculate the voltage drop across the inductor The relation-
ship between voltage and current for an inductor is V=Ldi
dt , where Lis the
inductance and di
dt is the rate of change of current. Given: L= 0.2 H and I= 2
3
A Taking derivative of Iwith respect to t:di
dt = 0 (since current is constant)
Using V=Ldi
dt :Vinductor = 0 V
Therefore, the voltage drop across the resistor is 100 V, across the capacitor
is 20tV, and across the inductor is 0 V.
Question 5
Question
A circuit consists of a resistor with resistance R= 20 Ω connected to a voltage
source of V= 100 V. Calculate the current flowing through the resistor.
Solution
To calculate the current flowing through the resistor, we can use Ohm’s Law,
which states that V=IR, where Vis the voltage across the resistor, Iis the
current flowing through the resistor, and Ris the resistance of the resistor.
Step 1: Substitute V= 100 V and R= 20 Ω into Ohm’s Law, V=IR, and
solve for I:
I=V
R
I=100 V
20 Ω
I= 5 A
Step 2: The current flowing through the resistor is I= 5 A.
Question 6
Question
A circuit contains a resistor with a resistance of 20 Ω and a current of 2 A
flowing through it. Determine the voltage drop across the resistor.
Solution
Step 1: Write down Ohm’s Law, which states that V=IR, where Vis the
voltage, Iis the current, and Ris the resistance of the resistor.
Step 2: Plug in the values given in the question. In this case, I= 2 A and
R= 20 Ω.
V= (2 A)(20 Ω)
Step 3: Calculate the voltage drop across the resistor.
V= 2 ×20 = 40 V
Step 4: Therefore, the voltage drop across the resistor is 40 V.
4
Question 7
Question
A circuit consists of a resistor, an inductor, and a capacitor connected in series.
The resistance of the resistor is 10 Ω, the inductance of the inductor is 5 H, and
the capacitance of the capacitor is 0.02 F. If the peak current flowing through
the circuit is 2 A, find the peak voltage across the circuit.
Solution
The peak voltage across the circuit can be found using Ohm’s Law, which states:
V=I·Z, where Vis the voltage, Iis the current, and Zis the impedance of
the circuit.
Step 1: Calculate the impedance of the circuit using the total resistance,
inductive reactance, and capacitive reactance. The impedance of the circuit is
given by the formula: Z=pR2+ (XL−XC)2, where Ris the resistance, XL
is the inductive reactance, and XCis the capacitive reactance. Given: Resistor
resistance R= 10 Ω, Inductive reactance XL= 2πfL, Capacitive reactance
XC=1
2πfC . Let’s choose a frequency, f= 1 Hz, to simplify calculations.
Substitute the given values into the formulas: XL= 2π·1·5 = 10 Ω, XC=
1
2π·1·0.02 = 7.96 Ω. Now, calculate the impedance: Z=p102+ (10 −7.96)2=
√100 + 4.082=√116.94 ≈10.81 Ω.
Step 2: Calculate the peak voltage across the circuit using Ohm’s Law.
Given peak current I= 2 A and impedance Z= 10.81 Ω, V=I·Z= 2·10.81 =
21.62 V.
Therefore, the peak voltage across the circuit is 21.62 V.
Question 8
Question
A circuit consists of a resistor with resistance R= 20 Ω and a battery with EMF
E= 12 V. If the current flowing through the circuit is I= 0.5A, calculate the
power dissipated in the resistor.
Solution
Step 1: To find the power dissipated in the resistor, we can use the formula for
power in a resistor: P=I2R.
Step 2: Substitute the given values into the formula. We have I= 0.5Aand
R= 20 Ω.
Step 3: Calculate the power:
P= (0.5A)2×20 Ω = 0.25 A2×20 Ω = 5 W
Step 4: Therefore, the power dissipated in the resistor is 5 W.
5
Question 9
Question
A resistor has a resistance of 10 Ω and a current of 2 A passing through it.
Determine the voltage drop across the resistor.
Solution
Step 1: Write down Ohm’s Law, which states that V=IR, where Vis the
voltage drop across the resistor, Iis the current passing through the resistor,
and Ris the resistance of the resistor.
Step 2: Substitute the given values into Ohm’s Law:
V= (2 A)(10Ω)
Step 3: Calculate the voltage drop:
V= 20 V
Therefore, the voltage drop across the resistor is 20 V.
Question 10
Question
A resistor with a resistance of 10 Ω is connected in series with a resistor with
an unknown resistance. A potential difference of 50 V is applied across the
combination of resistors, and a current of 2 A flows through the circuit. What
is the resistance of the unknown resistor?
Solution
Step 1: We can start by writing down Ohm’s Law, which relates voltage (V),
current (I), and resistance (R) using the equation V=IR. In this case, we
have the following values: - Total voltage across the resistors, Vtotal = 50 V, -
Total current through the circuit, Itotal = 2 A.
Step 2: The total voltage across the two resistors is the sum of the voltage
drops across each resistor:
Vtotal =V1+V2
Step 3: We can express the voltage drops in terms of the resistances and the
total current using Ohm’s Law:
V1=Itotal ·R1
V2=Itotal ·R2
6
Step 4: We are given the resistance of one resistor (R1= 10 Ω) and the cur-
rent flowing through the circuit. We need to find the resistance of the unknown
resistor (R2).
Step 5: Substituting the known values into the equations, we get:
50 = 2 ·10 + 2 ·R2
Step 6: Solving for R2gives:
50 = 20 + 2R2
2R2= 30
R2= 15 Ω
Step 7: Therefore, the resistance of the unknown resistor is 15 Ω.
Question 11
Question
A circuit consists of a resistor with resistance R= 20 Ω connected to a battery
with emf E= 9 V. If the current flowing through the circuit is I= 0.4 A, what
is the potential difference across the resistor?
Solution
To find the potential difference across the resistor, we can use Ohm’s Law which
states that V=IR, where Vis the potential difference, Iis the current, and R
is the resistance.
Step 1: Write down Ohm’s Law:
V=IR
Step 2: Substitute the given values into the equation:
V= (0.4 A)(20 Ω)
Step 3: Perform the calculation:
V= 8 V
Step 4: Answer: The potential difference across the resistor is 8 volts.
Question 12
Question
A circuit consists of a resistor, a capacitor, and an inductor connected in series.
The values of the components are as follows: resistor R= 10 Ω, capacitor
C= 5 µF , and inductor L= 2 mH. If a sinusoidal voltage source with frequency
f= 1 kHz is connected to the circuit, determine the current flowing through
the circuit using Ohm’s Law.
7
Solution
Step 1: Calculate the total impedance of the circuit using the formula for
impedance in an RLC series circuit:
Z=qR2+ (XL−XC)2
where XL= 2πf L is the inductive reactance and XC=1
2πfC is the capacitive
reactance.
Step 2: Calculate the inductive reactance XL:
XL= 2πfL = 2 ·π·1000 ·0.002 = 12.57 Ω
Step 3: Calculate the capacitive reactance XC:
XC=1
2πf C =1
2·π·1000 ·5×10−6= 31.83 Ω
Step 4: Calculate the total impedance Z:
Z=p102+ (12.57 −31.83)2=√100 + 338.56 = √438.56 = 20.92 Ω
Step 5: Calculate the current Iflowing through the circuit using Ohm’s Law:
I=V
Z
where Vis the voltage of the sinusoidal source. Let us assume V= 10 Vfor
simplicity.
I=10
20.92 = 0.477 A
Therefore, the current flowing through the circuit is 0.477 A.
Question 13
Question
A resistor has a resistance of 500 Ω and a current of 0.8 A flowing through it.
Calculate the voltage drop across the resistor.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor
is equal to the product of the current (I) flowing through it and the resistance
(R) of the resistor. Mathematically, this relationship is represented as:
V=I×R
8
Step 2: Given that the resistance (R) is 500 Ω and the current (I) is 0.8 A,
we can substitute those values into Ohm’s Law to solve for the voltage:
V= 0.8 A ×500 Ω
Step 3: Multiply the current and resistance to find the voltage drop:
V= 0.8 A ×500 Ω = 400 V
Step 4: Therefore, the voltage drop across the resistor is 400 V.
Question 14
Question
A circuit contains a resistor with resistance R= 100 Ω connected to a battery
with voltage V= 12 V. If a current of I= 0.1 A flows through the circuit, what
is the power dissipated by the resistor?
Solution
Ohm’s Law states that the current through a conductor between two points is
directly proportional to the voltage across the two points. Mathematically, this
is represented by the equation V=IR, where Vis the voltage, Iis the current,
and Ris the resistance.
Step 1: Calculate the power dissipated by the resistor using the formula
P=IV .
P=IV
= (0.1 A)(12 V)
= 1.2 W
Step 2: Verify the calculated power using the formula P=I2R.
P=I2R
= (0.1 A)2×100 Ω
= 0.01 ×100
= 1 W
Therefore, the power dissipated by the resistor is 1.2 W.
Question 15
Question
A resistor with a resistance of 12 ohms is connected to a 24-volt battery. What
is the current flowing through the resistor?
9
V= 2 A×10 Ω
V= 20 V
Step 3: Conclusion: The voltage across the resistor is 20 V.
23
