PHYS 101 - ELEMENTS OF PHYSICS
- Ohm’s Law
Question Bank - Set 1
Liberty University
Question 1
Question
A resistor is connected to a voltage source of 12 V and a current of 2 A flows
through it. Calculate the resistance of the resistor.
Solution
Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor
is equal to the current (I) flowing through it multiplied by the resistance (R)
of the resistor. Mathematically, this is expressed as V=I×R.
Step 2: Given that the voltage (V) is 12 V and the current (I) is 2 A, we can
substitute these values into Ohm’s Law to find the resistance (R). This gives
us:
12 = 2 ×R
Step 3: Now, solve for Rby dividing both sides by 2:
R=12
2
R= 6 Ω
Step 4: Therefore, the resistance of the resistor is 6 Ohms.
Question 2
Question
A circuit consists of a resistor with resistance Rconnected to a battery with
voltage V. If the current in the circuit is given by I=V
2R, determine the power
dissipated by the resistor in terms of Vand R.
Solution
Step 1: Recall that the power dissipated by a resistor can be calculated using
the formula P=IV .
Step 2: Substitute the given expression for current Iinto the formula for
power P:
P=V
2R·V
Step 3: Simplify the expression:
P=V2
2R
Therefore, the power dissipated by the resistor in terms of Vand
Ris P=V22R.
Question 3
Question
An electrical circuit consists of a resistor with a resistance of 25 Ω connected to
a power supply with a voltage of 100 V. Calculate the current flowing through
the circuit.
Solution
To find the current flowing through the circuit, 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: Given data: Resistance, R= 25 Ω
Voltage, V= 100 V
Step 2: Substitute the given values into Ohm’s Law to find the current, I:
I=V
R
I=100 V
25 Ω
I= 4 A
Therefore, the current flowing through the circuit is 4 A.
Question 4
Question
A resistor is connected to a 12V battery and a current of 2.5A flows through it.
What is the resistance of the resistor?
2
Solution
Let’s use Ohm’s Law to find the resistance of the resistor. Ohm’s Law states
that V=I·R, where Vis the voltage across the resistor, Iis the current flowing
through the resistor, and Ris the resistance of the resistor.
Step 1: Write down Ohm’s Law formula
V=I·R
Substitute V= 12V and I= 2.5A into the formula.
12 = 2.5·R
Step 2: Solve for R
R=12
2.5= 4.8 Ω
Step 3: Answer: The resistance of the resistor is 4.8 Ω.
Question 5
Question
A resistor with a resistance of 15 Ω is connected to a battery that produces a
voltage of 12 V. Calculate the current flowing through the resistor.
Solution
Ohm’s Law 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, i.e.,
I=V
R.
Step 1: Given values are V= 12 Vand R= 15 Ω.
We need to calculate the current I.
Step 2: Substitute the given values into Ohm’s Law:
I=V
R=12
15 = 0.8A.
Therefore, the current flowing through the resistor is 0.8A.
Question 6
Question
A circuit consists of a resistor with resistance R= 50 Ω connected to a battery
with voltage V= 120 V. Calculate the current flowing through the circuit.
3
Solution
Step 1: Recall Ohm’s Law which states that the current Iflowing through
a conductor is directly proportional to the voltage Vapplied across it and
inversely proportional to the resistance Rof the conductor. This relationship
can be expressed as:
V=I·R
Step 2: We are given that V= 120 V and R= 50 Ω. We can rearrange
Ohm’s Law to solve for the current Ias:
I=V
R
Step 3: Substituting the given values into the formula, we get:
I=120 V
50 Ω
Step 4: Simplifying the expression, we find:
I= 2.4 A
Therefore, the current flowing through the circuit is 2.4 Amperes.
Question 7
Question
A circuit consists of a resistor with a resistance of 10 Ω connected to a battery
with a voltage of 12 V. Calculate the current flowing through the circuit.
Solution
Let’s use Ohm’s Law, which states that the current flowing through a conductor
between two points is directly proportional to the voltage across the two points
and inversely proportional to the resistance.
Step 1: Write down Ohm’s Law:
V=IR
where: V= voltage across the resistor (12 V), I= current flowing through the
resistor (to be determined), R= resistance of the resistor (10 Ω).
Step 2: Rearrange Ohm’s Law to solve for current I:
I=V
R
Step 3: Substitute the given values into the formula:
I=12 V
10 Ω
4
Step 4: Calculate the current flowing through the circuit:
I=12
10 = 1.2A
Therefore, the current flowing through the circuit is 1.2 A.
Question 8
Question
A resistor with resistance Ris connected to a battery with voltage V. If the
current flowing through the resistor is given by I=V/R, what happens to the
current if the resistance is doubled?
Solution
To find out what happens to the current when the resistance is doubled, we can
start by considering Ohm’s Law, which relates the voltage (V), current (I), and
resistance (R) in a circuit:
V=IR
We are given that I=V /R, so we can substitute this into the Ohm’s Law
equation:
V=V
RR
Step 1: Simplify the equation by multiplying V
Rby R.
V=V
Step 2: Since the equation is true for any values of Vand R, we can
conclude that when the resistance is doubled, the current flowing through the
resistor also doubles.
Question 9
Question
A resistor has a resistance of 10 Ω and a current of 2 Apassing through it.
Determine the voltage across the resistor.
5
Solution
Ohm’s Law relates the voltage (V), current (I), and resistance (R) in a circuit
with the equation V=I·R.
Step 1: Given values are R= 10 Ω and I= 2 A. We need to find V.
V=I·R
V= 2 A·10 Ω
V= 20 V
Step 2: Therefore, the voltage across the resistor is 20 V .
Question 10
Question
A resistor with resistance R= 50 Ω is connected to a battery with voltage
V= 12 V. Calculate the current flowing through the resistor.
Solution
Let’s use Ohm’s Law to find the current flowing through the resistor.
Ohm’s Law states that the current (I) flowing through a resistor is given by
the formula:
I=V
R
where: I= current in Amperes (A), V= voltage in Volts (V), and R= resistance
in Ohms (Ω).
Step 1: Write down the values given in the problem. V= 12 V (voltage)
R= 50 Ω (resistance)
Step 2: Substitute the values into Ohm’s Law formula to find the current
(I).
I=12 V
50 Ω
Step 3: Calculate the current flowing through the resistor.
I=12
50
I= 0.24 A
Therefore, the current flowing through the resistor is 0.24 A.
6
Question 11
Question
A circuit contains a resistor with resistance R= 15 Ω and a power supply that
provides a voltage of V= 120 V. If a current of I= 5 A flows through the
circuit, what is the power dissipated by the resistor?
Solution
Step 1: Recall Ohm’s Law, which states that the current flowing through a re-
sistor is directly proportional to the voltage across it and inversely proportional
to the resistance. Mathematically, Ohm’s Law is given by V=I×R, where V
is the voltage across the resistor, Iis the current flowing through the resistor,
and Ris the resistance of the resistor.
Step 2: Substitute the given values into Ohm’s Law to find the voltage across
the resistor. We have V=I×R= 5 A ×15 Ω = 75 V.
Step 3: The power dissipated by a resistor can be calculated using the for-
mula P=I2×R, where Pis the power dissipated, Iis the current flowing
through the resistor, and Ris the resistance of the resistor.
Step 4: Substitute the given values into the power formula to find the power
dissipated by the resistor. We have P=I2×R= (5 A)2×15 Ω = 25 A2×15 Ω =
375 W.
Step 5: Therefore, the power dissipated by the resistor in the circuit is 375 W.
Question 12
Question
A circuit consists of a resistor with resistance R= 30 Ω connected to a battery
with emf E= 12 V. Determine 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 Ithrough a circuit is equal to the voltage Vacross the
circuit divided by the resistance Rof the circuit: I=V
R.
Step 1: Determine the voltage across the circuit. Given that the emf of
the battery is E= 12 V, the voltage across the circuit is equal to the emf:
V=E= 12 V.
Step 2: Calculate the current flowing through the circuit. Using Ohm’s
Law, we have:
I=V
R=12
30 = 0.4 A
Therefore, the current flowing through the circuit is 0.4 A.
7
Question 13
Question
A resistor with a resistance of 15 Ω is connected to a battery that provides a
voltage of 120 V. Calculate the current flowing through the resistor.
Solution
Let’s use Ohm’s Law to calculate the current flowing through the resistor.
Ohm’s Law states:
V=I·R
where: V= voltage (in volts), I= current (in amperes), R= resistance (in
ohms).
Step 1: Given data: Voltage, V= 120 V, Resistance, R= 15 Ω. Substitute
the given values into Ohm’s Law to find the current, I:
I=V
R
Step 2: Calculate the current, I:
I=120
15 = 8 A
The current flowing through the resistor is 8 A .
Question 14
Question
A circuit consists of a resistor with resistance R= 10 Ω and a battery with an
emf of E= 12 V. If a current of I= 1.2Aflows through the circuit, what is the
potential difference across the resistor?
Solution
To find the potential difference across the resistor using Ohm’s Law, we can use
the formula V=IR, where Vis the potential difference across the resistor, Iis
the current flowing through the circuit, and Ris the resistance of the resistor.
Step 1: Given that I= 1.2Aand R= 10 Ω, we can substitute these values
into Ohm’s Law to find the potential difference V:
V=I·R= 1.2A·10 Ω = 12 V
Step 2: Therefore, the potential difference across the resistor in the circuit
is 12 V.
8
Question 15
Question
A circuit consists of a resistor with resistance R= 50 Ω connected to a battery
with voltage V= 12 V. What current flows through the circuit?
Solution
Step 1: Recall Ohm’s Law, which states that the current Iflowing through a
resistor is given by I=V
R, where Vis the voltage across the resistor and Ris
the resistance of the resistor.
Step 2: Substitute the given values into Ohm’s Law:
I=V
R=12 V
50 Ω
Step 3: Calculate the current flowing through the circuit:
I=12
50 = 0.24 A
Step 4: Therefore, the current flowing through the circuit is 0.24 A .
Question 16
Question
A circuit consists of a resistor with resistance R= 30 Ω and a voltage source
with V= 120 V. If a current of I= 3 A flows through the circuit, what is the
power dissipated by the resistor?
Solution
To find the power dissipated by the resistor, we can use the formula for power
in a circuit:
P=IV
where Pis power, Iis current, and Vis voltage.
Step 1: Calculate the power using P=IV .
P=IV
= (3 A)(120 V)
= 360 W
Step 2: Verify the result. Therefore, the power dissipated by the resistor
in the circuit is 360 W.
9
Question 17
Question
A circuit consists of a resistor with resistance R= 100 ohms connected to a
battery with voltage V= 12 volts. Determine 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)
across the circuit divided by the resistance (R) of the circuit. Mathematically,
this can be represented as:
I=V
R
Step 1: Substitute the given values into Ohm’s Law:
I=12 V
100 ohms
Step 2: Calculate the current flowing through the circuit:
I=12
100 = 0.12 Amperes
Step 3: Therefore, the current flowing through the circuit is 0.12 Amperes.
Question 18
Question
A circuit consists of a resistor, an inductor, and a capacitor connected in series.
The voltage across the circuit is given by V(t) = 10 sin(100t) volts. The cur-
rent in the circuit is given by I(t) = 2 sin100t−π
3amperes. Determine the
equivalent resistance in the circuit.
Solution
Step 1: We know that Ohm’s Law states that V=IR, where Vis voltage, Iis
current, and Ris resistance.
Step 2: The voltage and current in the circuit are given by V(t) = 10 sin(100t)
and I(t) = 2 sin100t−π
3respectively.
Step 3: By comparing the given voltage and current expressions with Ohm’s
Law, we can see that the voltage amplitude is 10 volts and the current amplitude
is 2 amperes.
Step 4: The RMS voltage (Vrms) is related to the voltage amplitude by
Vrms =Vamplitude
√2=10
√2volts.
10
Step 5: The RMS current (Irms) is related to the current amplitude by
Irms =Iamplitude
√2=2
√2amperes.
Step 6: The equivalent resistance (R) in the circuit can be calculated using
the formula R=Vrms
Irms =
10
√2
2
√2
=10
2= 5 ohms.
Therefore, the equivalent resistance in the circuit is 5 ohms .
Question 19
Question
A resistor has a resistance of 50 Ω and a current of 2 Apassing through it.
Calculate the voltage drop across the resistor.
Solution
Let’s use Ohm’s Law, which states that the voltage drop across a resistor is
equal to the product of its resistance and the current passing through it.
Step 1: Identify the given values.
Resistance, R= 50 Ω
Current, I= 2 A
Step 2: Write Ohm’s Law formula.
V=I·R
Step 3: Substitute the given values into the formula.
V= 2 A·50 Ω
Step 4: Perform the calculation.
V= 100 V
The voltage drop across the resistor is 100 V.
Question 20
Question
A resistor with a resistance of 10 Ω is connected to a 12 V battery. Calculate
the current flowing through the circuit.
11
Solution
Let’s use 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.
Step 1: Given values are V= 12 V and R= 10 Ω. We need to find the
current I.
Step 2: Use Ohm’s Law:
I=V
R
Step 3: Substitute the given values:
I=12
10
I= 1.2 A
Step 4: The current flowing through the circuit is 1.2 A.
Question 21
Question
A resistor has a resistance of 50 ohms. If a current of 2 amps passes 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 can be written as: V=I·R.
Step 2: Given that the resistance Ris 50 ohms and the current Iis 2 amps,
we can substitute these values into Ohm’s Law: V= 2 A ×50 Ω.
Step 3: Calculate the voltage drop across the resistor: V= 2 ×50 = 100
volts.
Step 4: Therefore, the voltage drop across the resistor when a current of 2
amps passes through it is 100 volts.
Question 22
Question
A resistor with a resistance of 120 Ω is connected to a 24 V battery. What is
the current passing through the resistor?
12
Solution
Step 1: We can use Ohm’s Law to find the current passing through the resistor.
Ohm’s law 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.
Rearranging the formula to solve for current, we have:
I=V
R
Step 2: Substituting the given values into the formula:
I=24V
120Ω
Step 3: Simplifying the expression:
I=1
5A= 0.2A
Step 4: Therefore, the current passing through the resistor is 0.2 A.
Question 23
Question
A resistor is connected to a 12 V battery, and a current of 3 A flows through
the resistor. If the resistance of the resistor is doubled, what will be the new
current flowing through the resistor?
Solution
Step 1: Use Ohm’s Law V=IR, where Vis the voltage, Iis the current, and
Ris the resistance. We are given V= 12 V and I= 3 A. Thus, we can find the
initial resistance R1using Ohm’s Law.
R1=V
I=12 V
3 A = 4 Ω
Step 2: When the resistance is doubled, the new resistance R2is 2R1.
R2= 2R1= 2 ×4 Ω = 8 Ω
Step 3: Using Ohm’s Law with the new resistance R2and the same voltage
V:
I2=V
R2
=12 V
8 Ω = 1.5 A
Answer: The new current flowing through the resistor when the resistance
is doubled is 1.5 A.
13
Question 24
Question
A circuit consists of a resistor, an inductor, and a capacitor connected in series
with a voltage source. The resistor has a resistance of 30 Ω, the inductor has an
inductance of 0.1 H, and the capacitor has a capacitance of 0.01 F. If the voltage
across the circuit is 12 V and the frequency of the source is 100 Hz, calculate the
current flowing through the circuit.
Solution
Step 1: Calculate the total impedance of the circuit. The total impedance
(Ztotal) of the circuit in series can be calculated using the formula:
Ztotal =pR2+ (XL−XC)2
Where: - Ris the resistance of the resistor, - XLis the reactance of the inductor,
and - XCis the reactance of the capacitor. The reactance of an inductor (XL)
is given by XL= 2πfL, and the reactance of a capacitor (XC) is given by
XC=1
2πfC .
Given: - R= 30 Ω, - L= 0.1 H, - C= 0.01 F, and - f= 100 Hz.
We can now substitute these values into the equations to find XLand XC
and then substitute all values into the formula for Ztotal.
Question 25
Question
A circuit consists of a resistor with resistance R= 20 Ω connected to a battery
with voltage V= 100 V. If a current Iflows through the circuit, determine the
power dissipated by the resistor in the circuit.
Solution
To find the power dissipated by the resistor, we can use Ohm’s Law, which
relates voltage (V), current (I), and resistance (R) in a circuit, along with the
formula for power (P=V I).
Step 1: Determine the current flowing through the circuit using Ohm’s Law
V=IR.
I=V
R=100 V
20 Ω = 5 A
Step 2: Calculate the power dissipated by the resistor using the formula
P=V I.
P=V·I= 100 V·5A= 500 W
Therefore, the power dissipated by the resistor in the circuit is 500 W.
14
Question 26
Question
A resistor with a resistance of 50 ohms is connected to a voltage source. If
a current of 2.5 A flows through the resistor, what is the voltage across the
resistor?
Solution
Ohm’s Law states that the voltage (V) across a resistor is equal to the current
(I) flowing through the resistor multiplied by the resistance (R) of the resistor.
Mathematically, Ohm’s Law can be represented as V=I·R.
Step 1: Write down the given values. The resistance of the resistor is
R= 50 ohms and the current flowing through it is I= 2.5 A.
Step 2: Use Ohm’s Law to find the voltage across the resistor. Plug in the
values of Rand Iinto Ohm’s Law equation:
V= 2.5 A ×50 Ω
Step 3: Calculate the voltage.
V= 125 V
Step 4: Write the final answer. The voltage across the resistor is 125 V .
Question 27
Question
A wire with resistance Ris connected to a battery of emf Eand negligible
internal resistance. When the wire carries a current of I, the power dissipated
in the wire is equal to the power delivered by the battery. Prove Ohm’s Law in
this scenario.
Solution
Let’s start by defining the different quantities involved. According to the prob-
lem statement,
Resistance of the wire: R
Electromotive force (emf) of the battery: E
Current flowing through the wire: I
15
The power dissipated in the wire can be calculated using the formula P=
I2R, where Iis the current flowing through the wire and Ris the resistance.
The power delivered by the battery is given by the product of the emf (E)
and the current (I), i.e., P=EI.
Given that the power dissipated in the wire is equal to the power delivered
by the battery, we have:
I2R=EI
Solving for I, we get:
I=E
R
This expression is a statement of Ohm’s Law, V=IR, where Vis the
voltage across the wire. Therefore, Ohm’s Law is verified in this scenario.
Question 28
Question
A circuit consists of a resistor with resistance R= 20 Ω, a capacitor with capac-
itance C= 0.002 F, and an inductor with inductance L= 0.1H. If the voltage
across the circuit is given by V(t) = 10 sin(100t) volts, find the current I(t)
flowing through the circuit at time t=π
200 seconds.
Solution
Step 1: Determine the total impedance of the circuit. The total impedance Z
of the circuit is given by the formula:
Z=pR2+ (XL−XC)2
where XL=ωL is the inductive reactance, XC=1
ωC is the capacitive reactance,
and ω= 2πf is the angular frequency with fbeing the frequency of the voltage
source. Substitute the given values into the formula to solve for Z.
Step 2: Calculate the current in the circuit. Using Ohm’s Law V=IZ,
where Vis the voltage across the circuit, Iis the current flowing through the
circuit, and Zis the total impedance, solve for I.
Step 3: Substitute the given time t=π
200 seconds into the expression for
I(t) to find the current at that time.
Step 1: We have R= 20 Ω, C= 0.002 F,L= 0.1H, and V(t) = 10 sin(100t)
volts. The angular frequency is ω= 100 ×2π= 200π.
XL=ωL = 200π×0.1 = 20πΩ
XC=1
ωC =1
200π×0.002 =1
0.4π=5
2Ω
16
Therefore, the total impedance Zis:
Z=r202+ (20π−5
2)2
=r400 + (20π−5
2)2
=r400 + 400π2−20π×5 + 25
4
=r400 + 400π2−100π+25
4
=r400π2−100π+ 10025
4
=r(20π−5
2)2
= 20π−5
2
Step 2: The current Iin the circuit is given by:
I=V(t)
Z
=10 sin(100t)
20π−5
2
Step 3: Substitute t=π
200 into the expression for I(t):
Iπ
200=10 sin 100 ×π
200
20π−5
2
=10 sin π
2
20π−5
2
=10
20π−5
2
=10
20π−5
2
Therefore, at time t=π
200 seconds, the current flowing through the circuit
is 10
20π−5
2
Amperes.
Question 29
Question
A resistor with a resistance of 12 Ω is connected to a 9 V battery. Determine
the current passing through the resistor.
17
Solution
Let’s use Ohm’s Law to find the current passing through the resistor.
Step 1: Write down Ohm’s Law equation. Ohm’s Law states that V=I·R,
where Vis the voltage 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 equation. We are given
that V= 9 V and R= 12 Ω. We can rearrange Ohm’s Law to solve for I:
I=V
R
Substitute in the known values:
I=9 V
12 Ω
Step 3: Calculate the current passing through the resistor.
I=9
12
I= 0.75 A
So, the current passing through the resistor is 0.75 A.
Question 30
Question
A circuit consists of a resistor with resistance R= 50 Ω connected to a voltage
source with voltage V= 120 V. Determine the current flowing through the
circuit.
Solution
Step 1: Recall 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 2: Substitute the given values into Ohm’s Law to solve for the current:
V=IR
I=V
R
I=120 V
50 Ω
Step 3: Calculate the current:
I=120
50
I= 2.4A
Therefore, the current flowing through the circuit is 2.4 Amperes.
18
Question 31
Question
A copper wire with a resistance of 5 Ω is connected to a potential difference of
100 V. What is the current flowing through the wire?
Solution
Ohm’s Law states that the current flowing through a conductor is directly pro-
portional to the potential difference applied across it and inversely proportional
to the resistance of the conductor. Mathematically, Ohm’s Law is represented
as:
V=IR
where: - Vis the potential difference (in volts, V), - Iis the current (in amperes,
A), and - Ris the resistance (in ohms, Ω).
Given that the resistance Ris 5 Ω and the potential difference Vis 100 V,
we can rearrange Ohm’s Law to solve for the current I:
I=V
R
Step 1: Substitute V= 100 V and R= 5Ω into the equation:
I=100
5
Step 2: Calculate the current I:
I= 20 A
Therefore, the current flowing through the wire is 20 A.
Question 32
Question
A resistor with resistance 50 Ω is connected to a battery with voltage 12 V.
Calculate the current passing through the resistor.
Solution
Let’s use 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,
I=V
R.
Step 1: Write down the known values. The resistance of the resistor, R=
50 Ω, and the voltage of the battery, V= 12 V.
19
Step 2: Substitute the values into Ohm’s Law and solve for the current.
I=V
R=12 V
50 Ω = 0.24 A
Step 3: Answer: The current passing through the resistor is 0.24 A.
Question 33
Question
A resistor with resistance Ris connected to a battery with voltage V. If the
current passing through the resistor is I, prove Ohm’s Law, which states that
V=IR.
Solution
To prove Ohm’s Law, we start from the definition of resistance:
Step 1: The definition of resistance in terms of current and voltage is given
by R=V
I.
Step 2: Multiply both sides of the equation by I:
R·I=V
I·I
Step 3: Simplify the right side of the equation:
R·I=V
Step 4: Therefore, we have proved that V=IR, which is Ohm’s Law.
Question 34
Question
A resistor with a resistance of 100 Ω is connected to a 12 V battery. Determine
the current flowing through the resistor.
Solution
Step 1: Recall Ohm’s Law, which states that the current Iflowing through a
resistor is directly proportional to the voltage Vacross the resistor and inversely
proportional to the resistance Rof the resistor. Mathematically, Ohm’s Law is
represented by the formula: I=V
R.
Step 2: Given that the resistance Ris 100 Ω and the voltage Vis 12 V, we
can substitute these values into Ohm’s Law to find the current I:
I=12 V
100 Ω = 0.12 A
Step 3: Therefore, the current flowing through the resistor is 0.12 A.
20
Question 35
Question
A circuit consists of a resistor with resistance R, a capacitor with capacitance C,
and an inductor with inductance Lconnected in series. The alternating current
(AC) voltage across the circuit is given by V(t) = V0sin(ωt), where V0= 10 V
is the amplitude of the voltage and ω= 100 rad/s is the angular frequency.
Find the expression for the current I(t) flowing through the circuit in terms
of R,L,C,V0, and ω.
Solution
Step 1: By Ohm’s Law, the relationship between the current I(t), voltage V(t),
and resistance Ris given by V(t) = IR. In this circuit with an inductor and a
capacitor, we need to consider the impedance Zinstead of resistance:
Z=R+j(ωL −1
ωC )
where j=√−1.
Step 2: Using Ohm’s Law for impedance, we have:
V(t) = I(t)Z
Step 3: Substituting the given values into the equation for voltage, we have:
10 sin(100t) = I(t)[R+j(100L−1
100C)]
Step 4: Equating the real parts on both sides of the equation, we get:
10 sin(100t) = I(t)R
Step 5: Therefore, the expression for the current I(t) flowing through the
circuit is:
I(t) = 10 sin(100t)
R
21
Solution
Step 1: Recall that the power dissipated by a resistor can be calculated using
the formula P=IV .
Step 2: Substitute the given expression for current Iinto the formula for
power P:
P=V
2R·V
Step 3: Simplify the expression:
P=V2
2R
Therefore, the power dissipated by the resistor in terms of Vand
Ris P=V22R.
Question 3
Question
An electrical circuit consists of a resistor with a resistance of 25 Ω connected to
a power supply with a voltage of 100 V. Calculate the current flowing through
the circuit.
Solution
To find the current flowing through the circuit, 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: Given data: Resistance, R= 25 Ω
Voltage, V= 100 V
Step 2: Substitute the given values into Ohm’s Law to find the current, I:
I=V
R
I=100 V
25 Ω
I= 4 A
Therefore, the current flowing through the circuit is 4 A.
Question 4
Question
A resistor is connected to a 12V battery and a current of 2.5A flows through it.
What is the resistance of the resistor?
2
Solution
Let’s use Ohm’s Law to find the resistance of the resistor. Ohm’s Law states
that V=I·R, where Vis the voltage across the resistor, Iis the current flowing
through the resistor, and Ris the resistance of the resistor.
Step 1: Write down Ohm’s Law formula
V=I·R
Substitute V= 12V and I= 2.5A into the formula.
12 = 2.5·R
Step 2: Solve for R
R=12
2.5= 4.8 Ω
Step 3: Answer: The resistance of the resistor is 4.8 Ω.
Question 5
Question
A resistor with a resistance of 15 Ω is connected to a battery that produces a
voltage of 12 V. Calculate the current flowing through the resistor.
Solution
Ohm’s Law 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, i.e.,
I=V
R.
Step 1: Given values are V= 12 Vand R= 15 Ω.
We need to calculate the current I.
Step 2: Substitute the given values into Ohm’s Law:
I=V
R=12
15 = 0.8A.
Therefore, the current flowing through the resistor is 0.8A.
Question 6
Question
A circuit consists of a resistor with resistance R= 50 Ω connected to a battery
with voltage V= 120 V. Calculate the current flowing through the circuit.
3
Solution
Step 1: Recall Ohm’s Law which states that the current Iflowing through
a conductor is directly proportional to the voltage Vapplied across it and
inversely proportional to the resistance Rof the conductor. This relationship
can be expressed as:
V=I·R
Step 2: We are given that V= 120 V and R= 50 Ω. We can rearrange
Ohm’s Law to solve for the current Ias:
I=V
R
Step 3: Substituting the given values into the formula, we get:
I=120 V
50 Ω
Step 4: Simplifying the expression, we find:
I= 2.4 A
Therefore, the current flowing through the circuit is 2.4 Amperes.
Question 7
Question
A circuit consists of a resistor with a resistance of 10 Ω connected to a battery
with a voltage of 12 V. Calculate the current flowing through the circuit.
Solution
Let’s use Ohm’s Law, which states that the current flowing through a conductor
between two points is directly proportional to the voltage across the two points
and inversely proportional to the resistance.
Step 1: Write down Ohm’s Law:
V=IR
where: V= voltage across the resistor (12 V), I= current flowing through the
resistor (to be determined), R= resistance of the resistor (10 Ω).
Step 2: Rearrange Ohm’s Law to solve for current I:
I=V
R
Step 3: Substitute the given values into the formula:
I=12 V
10 Ω
4
Step 4: Calculate the current flowing through the circuit:
I=12
10 = 1.2A
Therefore, the current flowing through the circuit is 1.2 A.
Question 8
Question
A resistor with resistance Ris connected to a battery with voltage V. If the
current flowing through the resistor is given by I=V/R, what happens to the
current if the resistance is doubled?
Solution
To find out what happens to the current when the resistance is doubled, we can
start by considering Ohm’s Law, which relates the voltage (V), current (I), and
resistance (R) in a circuit:
V=IR
We are given that I=V /R, so we can substitute this into the Ohm’s Law
equation:
V=V
RR
Step 1: Simplify the equation by multiplying V
Rby R.
V=V
Step 2: Since the equation is true for any values of Vand R, we can
conclude that when the resistance is doubled, the current flowing through the
resistor also doubles.
Question 9
Question
A resistor has a resistance of 10 Ω and a current of 2 Apassing through it.
Determine the voltage across the resistor.
5
Solution
Ohm’s Law relates the voltage (V), current (I), and resistance (R) in a circuit
with the equation V=I·R.
Step 1: Given values are R= 10 Ω and I= 2 A. We need to find V.
V=I·R
V= 2 A·10 Ω
V= 20 V
Step 2: Therefore, the voltage across the resistor is 20 V .
Question 10
Question
A resistor with resistance R= 50 Ω is connected to a battery with voltage
V= 12 V. Calculate the current flowing through the resistor.
Solution
Let’s use Ohm’s Law to find the current flowing through the resistor.
Ohm’s Law states that the current (I) flowing through a resistor is given by
the formula:
I=V
R
where: I= current in Amperes (A), V= voltage in Volts (V), and R= resistance
in Ohms (Ω).
Step 1: Write down the values given in the problem. V= 12 V (voltage)
R= 50 Ω (resistance)
Step 2: Substitute the values into Ohm’s Law formula to find the current
(I).
I=12 V
50 Ω
Step 3: Calculate the current flowing through the resistor.
I=12
50
I= 0.24 A
Therefore, the current flowing through the resistor is 0.24 A.
6
Question 11
Question
A circuit contains a resistor with resistance R= 15 Ω and a power supply that
provides a voltage of V= 120 V. If a current of I= 5 A flows through the
circuit, what is the power dissipated by the resistor?
Solution
Step 1: Recall Ohm’s Law, which states that the current flowing through a re-
sistor is directly proportional to the voltage across it and inversely proportional
to the resistance. Mathematically, Ohm’s Law is given by V=I×R, where V
is the voltage across the resistor, Iis the current flowing through the resistor,
and Ris the resistance of the resistor.
Step 2: Substitute the given values into Ohm’s Law to find the voltage across
the resistor. We have V=I×R= 5 A ×15 Ω = 75 V.
Step 3: The power dissipated by a resistor can be calculated using the for-
mula P=I2×R, where Pis the power dissipated, Iis the current flowing
through the resistor, and Ris the resistance of the resistor.
Step 4: Substitute the given values into the power formula to find the power
dissipated by the resistor. We have P=I2×R= (5 A)2×15 Ω = 25 A2×15 Ω =
375 W.
Step 5: Therefore, the power dissipated by the resistor in the circuit is 375 W.
Question 12
Question
A circuit consists of a resistor with resistance R= 30 Ω connected to a battery
with emf E= 12 V. Determine 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 Ithrough a circuit is equal to the voltage Vacross the
circuit divided by the resistance Rof the circuit: I=V
R.
Step 1: Determine the voltage across the circuit. Given that the emf of
the battery is E= 12 V, the voltage across the circuit is equal to the emf:
V=E= 12 V.
Step 2: Calculate the current flowing through the circuit. Using Ohm’s
Law, we have:
I=V
R=12
30 = 0.4 A
Therefore, the current flowing through the circuit is 0.4 A.
7
Question 13
Question
A resistor with a resistance of 15 Ω is connected to a battery that provides a
voltage of 120 V. Calculate the current flowing through the resistor.
Solution
Let’s use Ohm’s Law to calculate the current flowing through the resistor.
Ohm’s Law states:
V=I·R
where: V= voltage (in volts), I= current (in amperes), R= resistance (in
ohms).
Step 1: Given data: Voltage, V= 120 V, Resistance, R= 15 Ω. Substitute
the given values into Ohm’s Law to find the current, I:
I=V
R
Step 2: Calculate the current, I:
I=120
15 = 8 A
The current flowing through the resistor is 8 A .
Question 14
Question
A circuit consists of a resistor with resistance R= 10 Ω and a battery with an
emf of E= 12 V. If a current of I= 1.2Aflows through the circuit, what is the
potential difference across the resistor?
Solution
To find the potential difference across the resistor using Ohm’s Law, we can use
the formula V=IR, where Vis the potential difference across the resistor, Iis
the current flowing through the circuit, and Ris the resistance of the resistor.
Step 1: Given that I= 1.2Aand R= 10 Ω, we can substitute these values
into Ohm’s Law to find the potential difference V:
V=I·R= 1.2A·10 Ω = 12 V
Step 2: Therefore, the potential difference across the resistor in the circuit
is 12 V.
8
Question 15
Question
A circuit consists of a resistor with resistance R= 50 Ω connected to a battery
with voltage V= 12 V. What current flows through the circuit?
Solution
Step 1: Recall Ohm’s Law, which states that the current Iflowing through a
resistor is given by I=V
R, where Vis the voltage across the resistor and Ris
the resistance of the resistor.
Step 2: Substitute the given values into Ohm’s Law:
I=V
R=12 V
50 Ω
Step 3: Calculate the current flowing through the circuit:
I=12
50 = 0.24 A
Step 4: Therefore, the current flowing through the circuit is 0.24 A .
Question 16
Question
A circuit consists of a resistor with resistance R= 30 Ω and a voltage source
with V= 120 V. If a current of I= 3 A flows through the circuit, what is the
power dissipated by the resistor?
Solution
To find the power dissipated by the resistor, we can use the formula for power
in a circuit:
P=IV
where Pis power, Iis current, and Vis voltage.
Step 1: Calculate the power using P=IV .
P=IV
= (3 A)(120 V)
= 360 W
Step 2: Verify the result. Therefore, the power dissipated by the resistor
in the circuit is 360 W.
9
Question 17
Question
A circuit consists of a resistor with resistance R= 100 ohms connected to a
battery with voltage V= 12 volts. Determine 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)
across the circuit divided by the resistance (R) of the circuit. Mathematically,
this can be represented as:
I=V
R
Step 1: Substitute the given values into Ohm’s Law:
I=12 V
100 ohms
Step 2: Calculate the current flowing through the circuit:
I=12
100 = 0.12 Amperes
Step 3: Therefore, the current flowing through the circuit is 0.12 Amperes.
Question 18
Question
A circuit consists of a resistor, an inductor, and a capacitor connected in series.
The voltage across the circuit is given by V(t) = 10 sin(100t) volts. The cur-
rent in the circuit is given by I(t) = 2 sin100t−π
3amperes. Determine the
equivalent resistance in the circuit.
Solution
Step 1: We know that Ohm’s Law states that V=IR, where Vis voltage, Iis
current, and Ris resistance.
Step 2: The voltage and current in the circuit are given by V(t) = 10 sin(100t)
and I(t) = 2 sin100t−π
3respectively.
Step 3: By comparing the given voltage and current expressions with Ohm’s
Law, we can see that the voltage amplitude is 10 volts and the current amplitude
is 2 amperes.
Step 4: The RMS voltage (Vrms) is related to the voltage amplitude by
Vrms =Vamplitude
√2=10
√2volts.
10
Step 5: The RMS current (Irms) is related to the current amplitude by
Irms =Iamplitude
√2=2
√2amperes.
Step 6: The equivalent resistance (R) in the circuit can be calculated using
the formula R=Vrms
Irms =
10
√2
2
√2
=10
2= 5 ohms.
Therefore, the equivalent resistance in the circuit is 5 ohms .
Question 19
Question
A resistor has a resistance of 50 Ω and a current of 2 Apassing through it.
Calculate the voltage drop across the resistor.
Solution
Let’s use Ohm’s Law, which states that the voltage drop across a resistor is
equal to the product of its resistance and the current passing through it.
Step 1: Identify the given values.
Resistance, R= 50 Ω
Current, I= 2 A
Step 2: Write Ohm’s Law formula.
V=I·R
Step 3: Substitute the given values into the formula.
V= 2 A·50 Ω
Step 4: Perform the calculation.
V= 100 V
The voltage drop across the resistor is 100 V.
Question 20
Question
A resistor with a resistance of 10 Ω is connected to a 12 V battery. Calculate
the current flowing through the circuit.
11
Solution
Let’s use 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.
Step 1: Given values are V= 12 V and R= 10 Ω. We need to find the
current I.
Step 2: Use Ohm’s Law:
I=V
R
Step 3: Substitute the given values:
I=12
10
I= 1.2 A
Step 4: The current flowing through the circuit is 1.2 A.
Question 21
Question
A resistor has a resistance of 50 ohms. If a current of 2 amps passes 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 can be written as: V=I·R.
Step 2: Given that the resistance Ris 50 ohms and the current Iis 2 amps,
we can substitute these values into Ohm’s Law: V= 2 A ×50 Ω.
Step 3: Calculate the voltage drop across the resistor: V= 2 ×50 = 100
volts.
Step 4: Therefore, the voltage drop across the resistor when a current of 2
amps passes through it is 100 volts.
Question 22
Question
A resistor with a resistance of 120 Ω is connected to a 24 V battery. What is
the current passing through the resistor?
12
Solution
Step 1: We can use Ohm’s Law to find the current passing through the resistor.
Ohm’s law 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.
Rearranging the formula to solve for current, we have:
I=V
R
Step 2: Substituting the given values into the formula:
I=24V
120Ω
Step 3: Simplifying the expression:
I=1
5A= 0.2A
Step 4: Therefore, the current passing through the resistor is 0.2 A.
Question 23
Question
A resistor is connected to a 12 V battery, and a current of 3 A flows through
the resistor. If the resistance of the resistor is doubled, what will be the new
current flowing through the resistor?
Solution
Step 1: Use Ohm’s Law V=IR, where Vis the voltage, Iis the current, and
Ris the resistance. We are given V= 12 V and I= 3 A. Thus, we can find the
initial resistance R1using Ohm’s Law.
R1=V
I=12 V
3 A = 4 Ω
Step 2: When the resistance is doubled, the new resistance R2is 2R1.
R2= 2R1= 2 ×4 Ω = 8 Ω
Step 3: Using Ohm’s Law with the new resistance R2and the same voltage
V:
I2=V
R2
=12 V
8 Ω = 1.5 A
Answer: The new current flowing through the resistor when the resistance
is doubled is 1.5 A.
13
Question 24
Question
A circuit consists of a resistor, an inductor, and a capacitor connected in series
with a voltage source. The resistor has a resistance of 30 Ω, the inductor has an
inductance of 0.1 H, and the capacitor has a capacitance of 0.01 F. If the voltage
across the circuit is 12 V and the frequency of the source is 100 Hz, calculate the
current flowing through the circuit.
Solution
Step 1: Calculate the total impedance of the circuit. The total impedance
(Ztotal) of the circuit in series can be calculated using the formula:
Ztotal =pR2+ (XL−XC)2
Where: - Ris the resistance of the resistor, - XLis the reactance of the inductor,
and - XCis the reactance of the capacitor. The reactance of an inductor (XL)
is given by XL= 2πfL, and the reactance of a capacitor (XC) is given by
XC=1
2πfC .
Given: - R= 30 Ω, - L= 0.1 H, - C= 0.01 F, and - f= 100 Hz.
We can now substitute these values into the equations to find XLand XC
and then substitute all values into the formula for Ztotal.
Question 25
Question
A circuit consists of a resistor with resistance R= 20 Ω connected to a battery
with voltage V= 100 V. If a current Iflows through the circuit, determine the
power dissipated by the resistor in the circuit.
Solution
To find the power dissipated by the resistor, we can use Ohm’s Law, which
relates voltage (V), current (I), and resistance (R) in a circuit, along with the
formula for power (P=V I).
Step 1: Determine the current flowing through the circuit using Ohm’s Law
V=IR.
I=V
R=100 V
20 Ω = 5 A
Step 2: Calculate the power dissipated by the resistor using the formula
P=V I.
P=V·I= 100 V·5A= 500 W
Therefore, the power dissipated by the resistor in the circuit is 500 W.
14
Question 26
Question
A resistor with a resistance of 50 ohms is connected to a voltage source. If
a current of 2.5 A flows through the resistor, what is the voltage across the
resistor?
Solution
Ohm’s Law states that the voltage (V) across a resistor is equal to the current
(I) flowing through the resistor multiplied by the resistance (R) of the resistor.
Mathematically, Ohm’s Law can be represented as V=I·R.
Step 1: Write down the given values. The resistance of the resistor is
R= 50 ohms and the current flowing through it is I= 2.5 A.
Step 2: Use Ohm’s Law to find the voltage across the resistor. Plug in the
values of Rand Iinto Ohm’s Law equation:
V= 2.5 A ×50 Ω
Step 3: Calculate the voltage.
V= 125 V
Step 4: Write the final answer. The voltage across the resistor is 125 V .
Question 27
Question
A wire with resistance Ris connected to a battery of emf Eand negligible
internal resistance. When the wire carries a current of I, the power dissipated
in the wire is equal to the power delivered by the battery. Prove Ohm’s Law in
this scenario.
Solution
Let’s start by defining the different quantities involved. According to the prob-
lem statement,
Resistance of the wire: R
Electromotive force (emf) of the battery: E
Current flowing through the wire: I
15
The power dissipated in the wire can be calculated using the formula P=
I2R, where Iis the current flowing through the wire and Ris the resistance.
The power delivered by the battery is given by the product of the emf (E)
and the current (I), i.e., P=EI.
Given that the power dissipated in the wire is equal to the power delivered
by the battery, we have:
I2R=EI
Solving for I, we get:
I=E
R
This expression is a statement of Ohm’s Law, V=IR, where Vis the
voltage across the wire. Therefore, Ohm’s Law is verified in this scenario.
Question 28
Question
A circuit consists of a resistor with resistance R= 20 Ω, a capacitor with capac-
itance C= 0.002 F, and an inductor with inductance L= 0.1H. If the voltage
across the circuit is given by V(t) = 10 sin(100t) volts, find the current I(t)
flowing through the circuit at time t=π
200 seconds.
Solution
Step 1: Determine the total impedance of the circuit. The total impedance Z
of the circuit is given by the formula:
Z=pR2+ (XL−XC)2
where XL=ωL is the inductive reactance, XC=1
ωC is the capacitive reactance,
and ω= 2πf is the angular frequency with fbeing the frequency of the voltage
source. Substitute the given values into the formula to solve for Z.
Step 2: Calculate the current in the circuit. Using Ohm’s Law V=IZ,
where Vis the voltage across the circuit, Iis the current flowing through the
circuit, and Zis the total impedance, solve for I.
Step 3: Substitute the given time t=π
200 seconds into the expression for
I(t) to find the current at that time.
Step 1: We have R= 20 Ω, C= 0.002 F,L= 0.1H, and V(t) = 10 sin(100t)
volts. The angular frequency is ω= 100 ×2π= 200π.
XL=ωL = 200π×0.1 = 20πΩ
XC=1
ωC =1
200π×0.002 =1
0.4π=5
2Ω
16
Therefore, the total impedance Zis:
Z=r202+ (20π−5
2)2
=r400 + (20π−5
2)2
=r400 + 400π2−20π×5 + 25
4
=r400 + 400π2−100π+25
4
=r400π2−100π+ 10025
4
=r(20π−5
2)2
= 20π−5
2
Step 2: The current Iin the circuit is given by:
I=V(t)
Z
=10 sin(100t)
20π−5
2
Step 3: Substitute t=π
200 into the expression for I(t):
Iπ
200=10 sin 100 ×π
200
20π−5
2
=10 sin π
2
20π−5
2
=10
20π−5
2
=10
20π−5
2
Therefore, at time t=π
200 seconds, the current flowing through the circuit
is 10
20π−5
2
Amperes.
Question 29
Question
A resistor with a resistance of 12 Ω is connected to a 9 V battery. Determine
the current passing through the resistor.
17
Solution
Let’s use Ohm’s Law to find the current passing through the resistor.
Step 1: Write down Ohm’s Law equation. Ohm’s Law states that V=I·R,
where Vis the voltage 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 equation. We are given
that V= 9 V and R= 12 Ω. We can rearrange Ohm’s Law to solve for I:
I=V
R
Substitute in the known values:
I=9 V
12 Ω
Step 3: Calculate the current passing through the resistor.
I=9
12
I= 0.75 A
So, the current passing through the resistor is 0.75 A.
Question 30
Question
A circuit consists of a resistor with resistance R= 50 Ω connected to a voltage
source with voltage V= 120 V. Determine the current flowing through the
circuit.
Solution
Step 1: Recall 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 2: Substitute the given values into Ohm’s Law to solve for the current:
V=IR
I=V
R
I=120 V
50 Ω
Step 3: Calculate the current:
I=120
50
I= 2.4A
Therefore, the current flowing through the circuit is 2.4 Amperes.
18
Question 31
Question
A copper wire with a resistance of 5 Ω is connected to a potential difference of
100 V. What is the current flowing through the wire?
Solution
Ohm’s Law states that the current flowing through a conductor is directly pro-
portional to the potential difference applied across it and inversely proportional
to the resistance of the conductor. Mathematically, Ohm’s Law is represented
as:
V=IR
where: - Vis the potential difference (in volts, V), - Iis the current (in amperes,
A), and - Ris the resistance (in ohms, Ω).
Given that the resistance Ris 5 Ω and the potential difference Vis 100 V,
we can rearrange Ohm’s Law to solve for the current I:
I=V
R
Step 1: Substitute V= 100 V and R= 5Ω into the equation:
I=100
5
Step 2: Calculate the current I:
I= 20 A
Therefore, the current flowing through the wire is 20 A.
Question 32
Question
A resistor with resistance 50 Ω is connected to a battery with voltage 12 V.
Calculate the current passing through the resistor.
Solution
Let’s use 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,
I=V
R.
Step 1: Write down the known values. The resistance of the resistor, R=
50 Ω, and the voltage of the battery, V= 12 V.
19
Step 2: Substitute the values into Ohm’s Law and solve for the current.
I=V
R=12 V
50 Ω = 0.24 A
Step 3: Answer: The current passing through the resistor is 0.24 A.
Question 33
Question
A resistor with resistance Ris connected to a battery with voltage V. If the
current passing through the resistor is I, prove Ohm’s Law, which states that
V=IR.
Solution
To prove Ohm’s Law, we start from the definition of resistance:
Step 1: The definition of resistance in terms of current and voltage is given
by R=V
I.
Step 2: Multiply both sides of the equation by I:
R·I=V
I·I
Step 3: Simplify the right side of the equation:
R·I=V
Step 4: Therefore, we have proved that V=IR, which is Ohm’s Law.
Question 34
Question
A resistor with a resistance of 100 Ω is connected to a 12 V battery. Determine
the current flowing through the resistor.
Solution
Step 1: Recall Ohm’s Law, which states that the current Iflowing through a
resistor is directly proportional to the voltage Vacross the resistor and inversely
proportional to the resistance Rof the resistor. Mathematically, Ohm’s Law is
represented by the formula: I=V
R.
Step 2: Given that the resistance Ris 100 Ω and the voltage Vis 12 V, we
can substitute these values into Ohm’s Law to find the current I:
I=12 V
100 Ω = 0.12 A
Step 3: Therefore, the current flowing through the resistor is 0.12 A.
20
Question 35
Question
A circuit consists of a resistor with resistance R, a capacitor with capacitance C,
and an inductor with inductance Lconnected in series. The alternating current
(AC) voltage across the circuit is given by V(t) = V0sin(ωt), where V0= 10 V
is the amplitude of the voltage and ω= 100 rad/s is the angular frequency.
Find the expression for the current I(t) flowing through the circuit in terms
of R,L,C,V0, and ω.
Solution
Step 1: By Ohm’s Law, the relationship between the current I(t), voltage V(t),
and resistance Ris given by V(t) = IR. In this circuit with an inductor and a
capacitor, we need to consider the impedance Zinstead of resistance:
Z=R+j(ωL −1
ωC )
where j=√−1.
Step 2: Using Ohm’s Law for impedance, we have:
V(t) = I(t)Z
Step 3: Substituting the given values into the equation for voltage, we have:
10 sin(100t) = I(t)[R+j(100L−1
100C)]
Step 4: Equating the real parts on both sides of the equation, we get:
10 sin(100t) = I(t)R
Step 5: Therefore, the expression for the current I(t) flowing through the
circuit is:
I(t) = 10 sin(100t)
R
21
Solution
Step 1: Recall that the power dissipated by a resistor can be calculated using
the formula P=IV .
Step 2: Substitute the given expression for current Iinto the formula for
power P:
P=V
2R·V
Step 3: Simplify the expression:
P=V2
2R
Therefore, the power dissipated by the resistor in terms of Vand
Ris P=V22R.
Question 3
Question
An electrical circuit consists of a resistor with a resistance of 25 Ω connected to
a power supply with a voltage of 100 V. Calculate the current flowing through
the circuit.
Solution
To find the current flowing through the circuit, 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: Given data: Resistance, R= 25 Ω
Voltage, V= 100 V
Step 2: Substitute the given values into Ohm’s Law to find the current, I:
I=V
R
I=100 V
25 Ω
I= 4 A
Therefore, the current flowing through the circuit is 4 A.
Question 4
Question
A resistor is connected to a 12V battery and a current of 2.5A flows through it.
What is the resistance of the resistor?
2
Solution
Let’s use Ohm’s Law to find the resistance of the resistor. Ohm’s Law states
that V=I·R, where Vis the voltage across the resistor, Iis the current flowing
through the resistor, and Ris the resistance of the resistor.
Step 1: Write down Ohm’s Law formula
V=I·R
Substitute V= 12V and I= 2.5A into the formula.
12 = 2.5·R
Step 2: Solve for R
R=12
2.5= 4.8 Ω
Step 3: Answer: The resistance of the resistor is 4.8 Ω.
Question 5
Question
A resistor with a resistance of 15 Ω is connected to a battery that produces a
voltage of 12 V. Calculate the current flowing through the resistor.
Solution
Ohm’s Law 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, i.e.,
I=V
R.
Step 1: Given values are V= 12 Vand R= 15 Ω.
We need to calculate the current I.
Step 2: Substitute the given values into Ohm’s Law:
I=V
R=12
15 = 0.8A.
Therefore, the current flowing through the resistor is 0.8A.
Question 6
Question
A circuit consists of a resistor with resistance R= 50 Ω connected to a battery
with voltage V= 120 V. Calculate the current flowing through the circuit.
3
Solution
Step 1: Recall Ohm’s Law which states that the current Iflowing through
a conductor is directly proportional to the voltage Vapplied across it and
inversely proportional to the resistance Rof the conductor. This relationship
can be expressed as:
V=I·R
Step 2: We are given that V= 120 V and R= 50 Ω. We can rearrange
Ohm’s Law to solve for the current Ias:
I=V
R
Step 3: Substituting the given values into the formula, we get:
I=120 V
50 Ω
Step 4: Simplifying the expression, we find:
I= 2.4 A
Therefore, the current flowing through the circuit is 2.4 Amperes.
Question 7
Question
A circuit consists of a resistor with a resistance of 10 Ω connected to a battery
with a voltage of 12 V. Calculate the current flowing through the circuit.
Solution
Let’s use Ohm’s Law, which states that the current flowing through a conductor
between two points is directly proportional to the voltage across the two points
and inversely proportional to the resistance.
Step 1: Write down Ohm’s Law:
V=IR
where: V= voltage across the resistor (12 V), I= current flowing through the
resistor (to be determined), R= resistance of the resistor (10 Ω).
Step 2: Rearrange Ohm’s Law to solve for current I:
I=V
R
Step 3: Substitute the given values into the formula:
I=12 V
10 Ω
4
Step 4: Calculate the current flowing through the circuit:
I=12
10 = 1.2A
Therefore, the current flowing through the circuit is 1.2 A.
Question 8
Question
A resistor with resistance Ris connected to a battery with voltage V. If the
current flowing through the resistor is given by I=V/R, what happens to the
current if the resistance is doubled?
Solution
To find out what happens to the current when the resistance is doubled, we can
start by considering Ohm’s Law, which relates the voltage (V), current (I), and
resistance (R) in a circuit:
V=IR
We are given that I=V /R, so we can substitute this into the Ohm’s Law
equation:
V=V
RR
Step 1: Simplify the equation by multiplying V
Rby R.
V=V
Step 2: Since the equation is true for any values of Vand R, we can
conclude that when the resistance is doubled, the current flowing through the
resistor also doubles.
Question 9
Question
A resistor has a resistance of 10 Ω and a current of 2 Apassing through it.
Determine the voltage across the resistor.
5
Solution
Ohm’s Law relates the voltage (V), current (I), and resistance (R) in a circuit
with the equation V=I·R.
Step 1: Given values are R= 10 Ω and I= 2 A. We need to find V.
V=I·R
V= 2 A·10 Ω
V= 20 V
Step 2: Therefore, the voltage across the resistor is 20 V .
Question 10
Question
A resistor with resistance R= 50 Ω is connected to a battery with voltage
V= 12 V. Calculate the current flowing through the resistor.
Solution
Let’s use Ohm’s Law to find the current flowing through the resistor.
Ohm’s Law states that the current (I) flowing through a resistor is given by
the formula:
I=V
R
where: I= current in Amperes (A), V= voltage in Volts (V), and R= resistance
in Ohms (Ω).
Step 1: Write down the values given in the problem. V= 12 V (voltage)
R= 50 Ω (resistance)
Step 2: Substitute the values into Ohm’s Law formula to find the current
(I).
I=12 V
50 Ω
Step 3: Calculate the current flowing through the resistor.
I=12
50
I= 0.24 A
Therefore, the current flowing through the resistor is 0.24 A.
6
Question 11
Question
A circuit contains a resistor with resistance R= 15 Ω and a power supply that
provides a voltage of V= 120 V. If a current of I= 5 A flows through the
circuit, what is the power dissipated by the resistor?
Solution
Step 1: Recall Ohm’s Law, which states that the current flowing through a re-
sistor is directly proportional to the voltage across it and inversely proportional
to the resistance. Mathematically, Ohm’s Law is given by V=I×R, where V
is the voltage across the resistor, Iis the current flowing through the resistor,
and Ris the resistance of the resistor.
Step 2: Substitute the given values into Ohm’s Law to find the voltage across
the resistor. We have V=I×R= 5 A ×15 Ω = 75 V.
Step 3: The power dissipated by a resistor can be calculated using the for-
mula P=I2×R, where Pis the power dissipated, Iis the current flowing
through the resistor, and Ris the resistance of the resistor.
Step 4: Substitute the given values into the power formula to find the power
dissipated by the resistor. We have P=I2×R= (5 A)2×15 Ω = 25 A2×15 Ω =
375 W.
Step 5: Therefore, the power dissipated by the resistor in the circuit is 375 W.
Question 12
Question
A circuit consists of a resistor with resistance R= 30 Ω connected to a battery
with emf E= 12 V. Determine 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 Ithrough a circuit is equal to the voltage Vacross the
circuit divided by the resistance Rof the circuit: I=V
R.
Step 1: Determine the voltage across the circuit. Given that the emf of
the battery is E= 12 V, the voltage across the circuit is equal to the emf:
V=E= 12 V.
Step 2: Calculate the current flowing through the circuit. Using Ohm’s
Law, we have:
I=V
R=12
30 = 0.4 A
Therefore, the current flowing through the circuit is 0.4 A.
7
Question 13
Question
A resistor with a resistance of 15 Ω is connected to a battery that provides a
voltage of 120 V. Calculate the current flowing through the resistor.
Solution
Let’s use Ohm’s Law to calculate the current flowing through the resistor.
Ohm’s Law states:
V=I·R
where: V= voltage (in volts), I= current (in amperes), R= resistance (in
ohms).
Step 1: Given data: Voltage, V= 120 V, Resistance, R= 15 Ω. Substitute
the given values into Ohm’s Law to find the current, I:
I=V
R
Step 2: Calculate the current, I:
I=120
15 = 8 A
The current flowing through the resistor is 8 A .
Question 14
Question
A circuit consists of a resistor with resistance R= 10 Ω and a battery with an
emf of E= 12 V. If a current of I= 1.2Aflows through the circuit, what is the
potential difference across the resistor?
Solution
To find the potential difference across the resistor using Ohm’s Law, we can use
the formula V=IR, where Vis the potential difference across the resistor, Iis
the current flowing through the circuit, and Ris the resistance of the resistor.
Step 1: Given that I= 1.2Aand R= 10 Ω, we can substitute these values
into Ohm’s Law to find the potential difference V:
V=I·R= 1.2A·10 Ω = 12 V
Step 2: Therefore, the potential difference across the resistor in the circuit
is 12 V.
8
Question 15
Question
A circuit consists of a resistor with resistance R= 50 Ω connected to a battery
with voltage V= 12 V. What current flows through the circuit?
Solution
Step 1: Recall Ohm’s Law, which states that the current Iflowing through a
resistor is given by I=V
R, where Vis the voltage across the resistor and Ris
the resistance of the resistor.
Step 2: Substitute the given values into Ohm’s Law:
I=V
R=12 V
50 Ω
Step 3: Calculate the current flowing through the circuit:
I=12
50 = 0.24 A
Step 4: Therefore, the current flowing through the circuit is 0.24 A .
Question 16
Question
A circuit consists of a resistor with resistance R= 30 Ω and a voltage source
with V= 120 V. If a current of I= 3 A flows through the circuit, what is the
power dissipated by the resistor?
Solution
To find the power dissipated by the resistor, we can use the formula for power
in a circuit:
P=IV
where Pis power, Iis current, and Vis voltage.
Step 1: Calculate the power using P=IV .
P=IV
= (3 A)(120 V)
= 360 W
Step 2: Verify the result. Therefore, the power dissipated by the resistor
in the circuit is 360 W.
9
Question 17
Question
A circuit consists of a resistor with resistance R= 100 ohms connected to a
battery with voltage V= 12 volts. Determine 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)
across the circuit divided by the resistance (R) of the circuit. Mathematically,
this can be represented as:
I=V
R
Step 1: Substitute the given values into Ohm’s Law:
I=12 V
100 ohms
Step 2: Calculate the current flowing through the circuit:
I=12
100 = 0.12 Amperes
Step 3: Therefore, the current flowing through the circuit is 0.12 Amperes.
Question 18
Question
A circuit consists of a resistor, an inductor, and a capacitor connected in series.
The voltage across the circuit is given by V(t) = 10 sin(100t) volts. The cur-
rent in the circuit is given by I(t) = 2 sin100t−π
3amperes. Determine the
equivalent resistance in the circuit.
Solution
Step 1: We know that Ohm’s Law states that V=IR, where Vis voltage, Iis
current, and Ris resistance.
Step 2: The voltage and current in the circuit are given by V(t) = 10 sin(100t)
and I(t) = 2 sin100t−π
3respectively.
Step 3: By comparing the given voltage and current expressions with Ohm’s
Law, we can see that the voltage amplitude is 10 volts and the current amplitude
is 2 amperes.
Step 4: The RMS voltage (Vrms) is related to the voltage amplitude by
Vrms =Vamplitude
√2=10
√2volts.
10
Step 5: The RMS current (Irms) is related to the current amplitude by
Irms =Iamplitude
√2=2
√2amperes.
Step 6: The equivalent resistance (R) in the circuit can be calculated using
the formula R=Vrms
Irms =
10
√2
2
√2
=10
2= 5 ohms.
Therefore, the equivalent resistance in the circuit is 5 ohms .
Question 19
Question
A resistor has a resistance of 50 Ω and a current of 2 Apassing through it.
Calculate the voltage drop across the resistor.
Solution
Let’s use Ohm’s Law, which states that the voltage drop across a resistor is
equal to the product of its resistance and the current passing through it.
Step 1: Identify the given values.
Resistance, R= 50 Ω
Current, I= 2 A
Step 2: Write Ohm’s Law formula.
V=I·R
Step 3: Substitute the given values into the formula.
V= 2 A·50 Ω
Step 4: Perform the calculation.
V= 100 V
The voltage drop across the resistor is 100 V.
Question 20
Question
A resistor with a resistance of 10 Ω is connected to a 12 V battery. Calculate
the current flowing through the circuit.
11
Solution
Let’s use 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.
Step 1: Given values are V= 12 V and R= 10 Ω. We need to find the
current I.
Step 2: Use Ohm’s Law:
I=V
R
Step 3: Substitute the given values:
I=12
10
I= 1.2 A
Step 4: The current flowing through the circuit is 1.2 A.
Question 21
Question
A resistor has a resistance of 50 ohms. If a current of 2 amps passes 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 can be written as: V=I·R.
Step 2: Given that the resistance Ris 50 ohms and the current Iis 2 amps,
we can substitute these values into Ohm’s Law: V= 2 A ×50 Ω.
Step 3: Calculate the voltage drop across the resistor: V= 2 ×50 = 100
volts.
Step 4: Therefore, the voltage drop across the resistor when a current of 2
amps passes through it is 100 volts.
Question 22
Question
A resistor with a resistance of 120 Ω is connected to a 24 V battery. What is
the current passing through the resistor?
12
Solution
Step 1: We can use Ohm’s Law to find the current passing through the resistor.
Ohm’s law 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.
Rearranging the formula to solve for current, we have:
I=V
R
Step 2: Substituting the given values into the formula:
I=24V
120Ω
Step 3: Simplifying the expression:
I=1
5A= 0.2A
Step 4: Therefore, the current passing through the resistor is 0.2 A.
Question 23
Question
A resistor is connected to a 12 V battery, and a current of 3 A flows through
the resistor. If the resistance of the resistor is doubled, what will be the new
current flowing through the resistor?
Solution
Step 1: Use Ohm’s Law V=IR, where Vis the voltage, Iis the current, and
Ris the resistance. We are given V= 12 V and I= 3 A. Thus, we can find the
initial resistance R1using Ohm’s Law.
R1=V
I=12 V
3 A = 4 Ω
Step 2: When the resistance is doubled, the new resistance R2is 2R1.
R2= 2R1= 2 ×4 Ω = 8 Ω
Step 3: Using Ohm’s Law with the new resistance R2and the same voltage
V:
I2=V
R2
=12 V
8 Ω = 1.5 A
Answer: The new current flowing through the resistor when the resistance
is doubled is 1.5 A.
13
Question 24
Question
A circuit consists of a resistor, an inductor, and a capacitor connected in series
with a voltage source. The resistor has a resistance of 30 Ω, the inductor has an
inductance of 0.1 H, and the capacitor has a capacitance of 0.01 F. If the voltage
across the circuit is 12 V and the frequency of the source is 100 Hz, calculate the
current flowing through the circuit.
Solution
Step 1: Calculate the total impedance of the circuit. The total impedance
(Ztotal) of the circuit in series can be calculated using the formula:
Ztotal =pR2+ (XL−XC)2
Where: - Ris the resistance of the resistor, - XLis the reactance of the inductor,
and - XCis the reactance of the capacitor. The reactance of an inductor (XL)
is given by XL= 2πfL, and the reactance of a capacitor (XC) is given by
XC=1
2πfC .
Given: - R= 30 Ω, - L= 0.1 H, - C= 0.01 F, and - f= 100 Hz.
We can now substitute these values into the equations to find XLand XC
and then substitute all values into the formula for Ztotal.
Question 25
Question
A circuit consists of a resistor with resistance R= 20 Ω connected to a battery
with voltage V= 100 V. If a current Iflows through the circuit, determine the
power dissipated by the resistor in the circuit.
Solution
To find the power dissipated by the resistor, we can use Ohm’s Law, which
relates voltage (V), current (I), and resistance (R) in a circuit, along with the
formula for power (P=V I).
Step 1: Determine the current flowing through the circuit using Ohm’s Law
V=IR.
I=V
R=100 V
20 Ω = 5 A
Step 2: Calculate the power dissipated by the resistor using the formula
P=V I.
P=V·I= 100 V·5A= 500 W
Therefore, the power dissipated by the resistor in the circuit is 500 W.
14
Question 26
Question
A resistor with a resistance of 50 ohms is connected to a voltage source. If
a current of 2.5 A flows through the resistor, what is the voltage across the
resistor?
Solution
Ohm’s Law states that the voltage (V) across a resistor is equal to the current
(I) flowing through the resistor multiplied by the resistance (R) of the resistor.
Mathematically, Ohm’s Law can be represented as V=I·R.
Step 1: Write down the given values. The resistance of the resistor is
R= 50 ohms and the current flowing through it is I= 2.5 A.
Step 2: Use Ohm’s Law to find the voltage across the resistor. Plug in the
values of Rand Iinto Ohm’s Law equation:
V= 2.5 A ×50 Ω
Step 3: Calculate the voltage.
V= 125 V
Step 4: Write the final answer. The voltage across the resistor is 125 V .
Question 27
Question
A wire with resistance Ris connected to a battery of emf Eand negligible
internal resistance. When the wire carries a current of I, the power dissipated
in the wire is equal to the power delivered by the battery. Prove Ohm’s Law in
this scenario.
Solution
Let’s start by defining the different quantities involved. According to the prob-
lem statement,
Resistance of the wire: R
Electromotive force (emf) of the battery: E
Current flowing through the wire: I
15
The power dissipated in the wire can be calculated using the formula P=
I2R, where Iis the current flowing through the wire and Ris the resistance.
The power delivered by the battery is given by the product of the emf (E)
and the current (I), i.e., P=EI.
Given that the power dissipated in the wire is equal to the power delivered
by the battery, we have:
I2R=EI
Solving for I, we get:
I=E
R
This expression is a statement of Ohm’s Law, V=IR, where Vis the
voltage across the wire. Therefore, Ohm’s Law is verified in this scenario.
Question 28
Question
A circuit consists of a resistor with resistance R= 20 Ω, a capacitor with capac-
itance C= 0.002 F, and an inductor with inductance L= 0.1H. If the voltage
across the circuit is given by V(t) = 10 sin(100t) volts, find the current I(t)
flowing through the circuit at time t=π
200 seconds.
Solution
Step 1: Determine the total impedance of the circuit. The total impedance Z
of the circuit is given by the formula:
Z=pR2+ (XL−XC)2
where XL=ωL is the inductive reactance, XC=1
ωC is the capacitive reactance,
and ω= 2πf is the angular frequency with fbeing the frequency of the voltage
source. Substitute the given values into the formula to solve for Z.
Step 2: Calculate the current in the circuit. Using Ohm’s Law V=IZ,
where Vis the voltage across the circuit, Iis the current flowing through the
circuit, and Zis the total impedance, solve for I.
Step 3: Substitute the given time t=π
200 seconds into the expression for
I(t) to find the current at that time.
Step 1: We have R= 20 Ω, C= 0.002 F,L= 0.1H, and V(t) = 10 sin(100t)
volts. The angular frequency is ω= 100 ×2π= 200π.
XL=ωL = 200π×0.1 = 20πΩ
XC=1
ωC =1
200π×0.002 =1
0.4π=5
2Ω
16
Therefore, the total impedance Zis:
Z=r202+ (20π−5
2)2
=r400 + (20π−5
2)2
=r400 + 400π2−20π×5 + 25
4
=r400 + 400π2−100π+25
4
=r400π2−100π+ 10025
4
=r(20π−5
2)2
= 20π−5
2
Step 2: The current Iin the circuit is given by:
I=V(t)
Z
=10 sin(100t)
20π−5
2
Step 3: Substitute t=π
200 into the expression for I(t):
Iπ
200=10 sin 100 ×π
200
20π−5
2
=10 sin π
2
20π−5
2
=10
20π−5
2
=10
20π−5
2
Therefore, at time t=π
200 seconds, the current flowing through the circuit
is 10
20π−5
2
Amperes.
Question 29
Question
A resistor with a resistance of 12 Ω is connected to a 9 V battery. Determine
the current passing through the resistor.
17
Solution
Let’s use Ohm’s Law to find the current passing through the resistor.
Step 1: Write down Ohm’s Law equation. Ohm’s Law states that V=I·R,
where Vis the voltage 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 equation. We are given
that V= 9 V and R= 12 Ω. We can rearrange Ohm’s Law to solve for I:
I=V
R
Substitute in the known values:
I=9 V
12 Ω
Step 3: Calculate the current passing through the resistor.
I=9
12
I= 0.75 A
So, the current passing through the resistor is 0.75 A.
Question 30
Question
A circuit consists of a resistor with resistance R= 50 Ω connected to a voltage
source with voltage V= 120 V. Determine the current flowing through the
circuit.
Solution
Step 1: Recall 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 2: Substitute the given values into Ohm’s Law to solve for the current:
V=IR
I=V
R
I=120 V
50 Ω
Step 3: Calculate the current:
I=120
50
I= 2.4A
Therefore, the current flowing through the circuit is 2.4 Amperes.
18
Question 31
Question
A copper wire with a resistance of 5 Ω is connected to a potential difference of
100 V. What is the current flowing through the wire?
Solution
Ohm’s Law states that the current flowing through a conductor is directly pro-
portional to the potential difference applied across it and inversely proportional
to the resistance of the conductor. Mathematically, Ohm’s Law is represented
as:
V=IR
where: - Vis the potential difference (in volts, V), - Iis the current (in amperes,
A), and - Ris the resistance (in ohms, Ω).
Given that the resistance Ris 5 Ω and the potential difference Vis 100 V,
we can rearrange Ohm’s Law to solve for the current I:
I=V
R
Step 1: Substitute V= 100 V and R= 5Ω into the equation:
I=100
5
Step 2: Calculate the current I:
I= 20 A
Therefore, the current flowing through the wire is 20 A.
Question 32
Question
A resistor with resistance 50 Ω is connected to a battery with voltage 12 V.
Calculate the current passing through the resistor.
Solution
Let’s use 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,
I=V
R.
Step 1: Write down the known values. The resistance of the resistor, R=
50 Ω, and the voltage of the battery, V= 12 V.
19
Step 2: Substitute the values into Ohm’s Law and solve for the current.
I=V
R=12 V
50 Ω = 0.24 A
Step 3: Answer: The current passing through the resistor is 0.24 A.
Question 33
Question
A resistor with resistance Ris connected to a battery with voltage V. If the
current passing through the resistor is I, prove Ohm’s Law, which states that
V=IR.
Solution
To prove Ohm’s Law, we start from the definition of resistance:
Step 1: The definition of resistance in terms of current and voltage is given
by R=V
I.
Step 2: Multiply both sides of the equation by I:
R·I=V
I·I
Step 3: Simplify the right side of the equation:
R·I=V
Step 4: Therefore, we have proved that V=IR, which is Ohm’s Law.
Question 34
Question
A resistor with a resistance of 100 Ω is connected to a 12 V battery. Determine
the current flowing through the resistor.
Solution
Step 1: Recall Ohm’s Law, which states that the current Iflowing through a
resistor is directly proportional to the voltage Vacross the resistor and inversely
proportional to the resistance Rof the resistor. Mathematically, Ohm’s Law is
represented by the formula: I=V
R.
Step 2: Given that the resistance Ris 100 Ω and the voltage Vis 12 V, we
can substitute these values into Ohm’s Law to find the current I:
I=12 V
100 Ω = 0.12 A
Step 3: Therefore, the current flowing through the resistor is 0.12 A.
20
Question 35
Question
A circuit consists of a resistor with resistance R, a capacitor with capacitance C,
and an inductor with inductance Lconnected in series. The alternating current
(AC) voltage across the circuit is given by V(t) = V0sin(ωt), where V0= 10 V
is the amplitude of the voltage and ω= 100 rad/s is the angular frequency.
Find the expression for the current I(t) flowing through the circuit in terms
of R,L,C,V0, and ω.
Solution
Step 1: By Ohm’s Law, the relationship between the current I(t), voltage V(t),
and resistance Ris given by V(t) = IR. In this circuit with an inductor and a
capacitor, we need to consider the impedance Zinstead of resistance:
Z=R+j(ωL −1
ωC )
where j=√−1.
Step 2: Using Ohm’s Law for impedance, we have:
V(t) = I(t)Z
Step 3: Substituting the given values into the equation for voltage, we have:
10 sin(100t) = I(t)[R+j(100L−1
100C)]
Step 4: Equating the real parts on both sides of the equation, we get:
10 sin(100t) = I(t)R
Step 5: Therefore, the expression for the current I(t) flowing through the
circuit is:
I(t) = 10 sin(100t)
R
21
