PHYS 101 - ELEMENTS OF PHYSICS
- Ohm’s Law
Question Bank - Set 2
Liberty University
Question 1
Question
A resistor with a resistance of 10 Ω is connected to a 12 V battery. Calculate
the current flowing through the resistor.
Solution
Step 1: Recall Ohm’s Law, which states: 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: Given that the voltage Vis 12 V and the resistance Ris 10 Ω, we
can rearrange Ohm’s Law to solve for the current I:
I=V
R
Step 3: Plugging in the given values, we have:
I=12 V
10Ω
Step 4: Calculating the current flowing through the resistor:
I=12
10 A=1.2 A
Step 5: Therefore, the current flowing through the resistor is 1.2 A.
Question 2
Question
A circuit consists of a resistor with a resistance of 30 Ω connected to a power
supply that produces a voltage of 120 V. 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 ratio of the voltage (V) across the circuit to the resistance
(R) of the circuit. Mathematically, Ohm’s Law can be expressed as:
I=V
R
Step 2: Given that the voltage across the circuit (V) is 120 V and the
resistance of the resistor (R) is 30 Ω, we can substitute these values into Ohm’s
Law to find the current (I):
I=120
30
I= 4 Amperes
Step 3: Therefore, the current flowing through the circuit is 4 Amperes.
Question 3
Question
A resistor with resistance R= 150 Ω is connected to a battery with voltage
V= 12 V. Calculate the current that flows through the resistor.
Solution
To calculate the current that flows through the resistor, we can use Ohm’s Law,
which states 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
I=V
R
Step 2: Plug in the values V= 12 Vand R= 150 Ω:
I=12 V
150 Ω
2
Step 3: Calculate the current flowing through the resistor:
I=12
150
I= 0.08 A
Therefore, the current that flows through the resistor is 0.08 A.
Question 4
Question
A resistor is connected to a 12-volt battery and a current of 0.5 amperes flows
through it. Determine the resistance of the resistor.
Solution
Let’s use Ohm’s Law (V=IR) to find the resistance of the resistor.
Step 1: Write down the known values. The voltage across the resistor (V)
is 12 volts and the current flowing through it (I) is 0.5 amperes.
Step 2: Use Ohm’s Law to find the resistance. We can rearrange Ohm’s
Law to solve for resistance:
R=V
I
Step 3: Substitute the known values into the formula.
R=12
0.5
Step 4: Calculate the resistance.
R=12
0.5= 24 ohms
Step 5: Write the final answer. The resistance of the resistor is 24 ohms.
Question 5
Question
A resistor with a resistance of 12 Ω is connected to a 24 V battery. Determine
the current flowing through the resistor.
3
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: Given that the voltage V= 24 V and the resistance R= 12 Ω, we
can apply Ohm’s Law to find the current I.
V=IR
I=V
R
I=24
12
I= 2 A
Step 3: Therefore, the current flowing through the resistor is 2 A.
Question 6
Question
A circuit consists of a resistor with a resistance of 10 Ω and a battery with a
voltage of 12 V. If a current of 1.2Ais flowing through the circuit, determine
the power dissipated by the resistor.
Solution
Let’s first recall Ohm’s Law, which states that the current flowing through a
resistor is directly proportional to the voltage across it and inversely propor-
tional to the resistance. Mathematically, Ohm’s Law is represented as 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: Calculate the voltage across the resistor using Ohm’s Law. Given
that I= 1.2Aand R= 10 Ω, we can use Ohm’s Law to find V:
V=IR = 1.2A×10 Ω = 12 V
Step 2: Calculate the power dissipated by the resistor using the formula
P=IV . The power dissipated by a resistor can be calculated using the formula
P=IV , where: - Pis the power dissipated by the resistor, - Iis the current
flowing through the resistor, and - Vis the voltage across the resistor.
Substitute I= 1.2Aand V= 12 Vinto the formula to find the power:
P= 1.2A×12 V= 14.4W
Step 3: Answer: The power dissipated by the resistor in the circuit is
14.4W.
4
Question 7
Question
A circuit consists of a resistor with resistance Rconnected to a battery with
voltage V. If the current flowing through the circuit is I, prove Ohm’s Law,
V=IR, using Kirchhoff’s voltage law.
Solution
To prove Ohm’s Law using Kirchhoff’s voltage law, we consider the voltage
around the closed loop of the circuit.
Step 1: Start by considering the voltage drop across the resistor: Let VRbe
the voltage drop across the resistor R, which is given by Ohm’s Law: VR=IR.
This voltage drop occurs in the direction of the current flow.
Step 2: Consider the voltage rise across the battery: Let Vbattery be the
voltage rise across the battery. Since the current Iflows from the positive
terminal of the battery to the negative terminal, Vbattery =−V. This negative
sign indicates that the battery provides a voltage rise in the opposite direction
to the current flow.
Step 3: Apply Kirchhoff’s voltage law: According to Kirchhoff’s voltage
law, the sum of the voltages around a closed loop in a circuit must be zero.
Therefore, we have:
Vbattery +VR= 0
(−V)+(IR) = 0
Step 4: Substitute VR=IR into the equation:
−V+IR = 0
Step 5: Rearrange the equation to prove Ohm’s Law:
V=IR
Thus, we have proven Ohm’s Law, V=IR, using Kirchhoff’s voltage law in
the given circuit.
Question 8
Question
A resistor with a resistance of 5 Ω is connected to a 12 V battery. Determine
the current flowing through the resistor.
5
Solution
Ohm’s Law relates the voltage (V), current (I), and resistance (R) in a circuit
through the equation V=IR.
1. The given values are: Resistance, R= 5 Ω Voltage, V= 12 V
2. Using Ohm’s Law, we can rearrange the formula to solve for current:
I=V
R
3. Now, substitute the known values into the formula to find the current:
I=12
5
4. Thus, the current flowing through the resistor is:
I= 2.4 A
Question 9
Question
A circuit consists of a resistor with resistance R= 300 Ω connected to a power
supply with voltage V= 120 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: Identify the given values. The resistance of the resistor, R= 300 Ω,
and the voltage across the resistor, V= 120 V.
Step 2: Substitute the values into Ohm’s Law equation.
V=IR
120 = I×300
Step 3: Solve for the current, I.
I=120
300
I= 0.4A
Step 4: Answer: The current flowing through the circuit is 0.4 amps.
6
Question 10
Question
A resistor with a resistance of 20 Ω is connected to a 12 V battery. What is the
current flowing through the resistor?
Solution
Given: Resistance, R= 20 Ω Battery voltage, V= 12 V
We can use Ohm’s Law to find the current flowing through the resistor:
I=V
R
Step 1: Substituting the given values into Ohm’s Law:
I=12 V
20 Ω
Step 2: Calculating the current:
I= 0.6 A
Therefore, the current flowing through the resistor is 0.6 A.
Question 11
Question
A circuit consists of a resistor with a resistance of 100 Ω, a capacitor with a
capacitance of 0.1 F, and an inductor with an inductance of 0.2 H connected in
series to a voltage source. If the frequency of the source is 50 Hz, calculate the
current flowing through the circuit.
Solution
To calculate the current flowing through the circuit, we can use Ohm’s Law,
which states that V=I·Z, where Vis the voltage across the circuit, Iis the
current flowing through the circuit, and Zis the impedance of the circuit.
The impedance of a series RLC circuit is given by:
Z=pR2+ (XL−XC)2
where: - R= Resistance in the circuit (Ω) - XL= Inductive reactance, XL=
2πfL (Ω) - XC= Capacitive reactance, XC=1
2πfC (Ω) - f= Frequency of
the source (Hz) - L= Inductance of the inductor (H) - C= Capacitance of the
capacitor (F)
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Given: Resistance, R= 100 Ω Capacitance, C= 0.1 F Inductance, L= 0.2 H
Frequency, f= 50 Hz
Step 1: Calculate the inductive reactance XL
XL= 2πfL = 2π×50 ×0.2 = 62.83 Ω
Step 2: Calculate the capacitive reactance XC
XC=1
2πfC =1
2π×50 ×0.1= 31.83 Ω
Step 3: Calculate the impedance Z
Z=pR2+ (XL−XC)2=p1002+ (62.83 −31.83)2=√10000 + 961 = √10961 = 104.69 Ω
Step 4: Apply Ohm’s Law to find the current I
V=I·Z
I=V
Z=V
104.69
Since no value for the voltage source Vwas provided, the current flowing
through the circuit cannot be determined without this information.
Question 12
Question
A resistor with a resistance of 15 Ω is connected to a 9 Vbattery. What is the
current flowing through the resistor?
Solution
Let’s use Ohm’s Law, V=IR, where Vis the voltage across the resistor, Iis
the current flowing through the resistor, and Ris the resistance of the resistor.
We are given V= 9 Vand R= 15 Ω. We need to find I.
Step 1: Substitute the given values into Ohm’s Law equation.
9 = I×15
Step 2: Solve for I.
I=9
15 = 0.6A
Step 3: Answer: The current flowing through the resistor is 0.6A.
Question 13
Question
A 20 Ω resistor, a 30 Ω resistor, and a 40 Ω resistor are connected in series to
a 12V battery. What is the current flowing through the circuit?
8
Solution
Let’s denote the current flowing through the circuit as I.
Step 1: Calculate the total resistance of the circuit. The total resistance
Rtotal of resistors in series is the sum of individual resistances.
Rtotal = 20Ω + 30Ω + 40Ω = 90Ω
Step 2: Apply Ohm’s Law to find the current. Ohm’s Law states that
V=IR, where Vis the voltage, Iis the current, and Ris the resistance.
Substitute the values into the equation:
I=V
Rtotal
=12V
90Ω = 0.1333 A
Therefore, the current flowing through the circuit is 0.1333 A.
Question 14
Question
A resistor has a resistance of 10 Ω and a current of 2 A flowing through it.
Determine the voltage drop across the resistor.
Solution
Step 1: Recall Ohm’s Law, which states that V=I×R, where Vis the voltage
drop 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: V= 2 A ×10 Ω.
Step 3: Calculate the voltage drop: V= 20 V.
Therefore, the voltage drop across the resistor is 20 V.
Question 15
Question
A circuit consists of a resistor with resistance R= 30 Ω, an inductor with
inductance L= 0.2 H, and a capacitor with capacitance C= 8 F. If a voltage of
V(t) = 24 sin(100t) volts is applied to the circuit, determine the current flowing
through the circuit at time t=π
200 seconds.
Solution
Step 1: Find the total impedance of the circuit. The impedance of a resistor
is ZR=R, the impedance of an inductor is ZL=jωL, and the impedance of
a capacitor is ZC=1
jωC , where ωis the angular frequency. Since the circuit
9
contains all three components in series, the total impedance Ztotal is the sum of
the individual impedances.
Ztotal =R+jωL +1
jωC
Step 2: Substitute the given values into the impedance expression. The
angular frequency is given by ω= 100 rad/s. Thus,
Ztotal = 30 + j(100)(0.2) + 1
j(100)(8)
Ztotal = 30 + j20 −j0.125 = 30 + j20 −j0.125
Step 3: Simplify the total impedance.
Ztotal = 30 + j20 −j0.125 = 30 + j20 + j0.125
Ztotal = 30 + j20.125 Ω
Step 4: Use Ohm’s Law to find the current flowing through the circuit at
t=π
200 seconds. Ohm’s Law states V(t) = I(t)Z, where V(t) is the voltage
across the circuit at time t,I(t) is the current flowing through the circuit at
time t, and Zis the total impedance of the circuit.
I(t) = V(t)
Ztotal
=24 sin(100t)
30 + j20.125
Step 5: Substitute t=π
200 seconds into the expression for current.
Iπ
200=24 sin 100 ×π
200
30 + j20.125 =24 sin π
2
30 + j20.125 =24
√302+ 20.1252
∠arctan 20.125
30 A
Question 16
Question
A resistor with a resistance of 240 Ω is connected to a battery with a voltage of
12 V. Calculate the current flowing through the resistor.
Solution
Let’s use Ohm’s Law, which states that the current passing through a conduc-
tor is directly proportional to the voltage across the conductor and inversely
proportional to the resistance of the conductor. Mathematically, Ohm’s Law is
represented as: 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 values are: - Resistance R= 240 Ω, - Voltage V= 12 V.
10
Step 2: We can rearrange Ohm’s Law to solve for current I:
I=V
R
Step 3: Substitute the values of Vand Rinto the formula:
I=12 V
240 Ω
Step 4: Calculate the current passing through the resistor:
I=1
20 A= 0.05 A
Therefore, the current flowing through the resistor is 0.05 A.
Question 17
Question
A wire of resistivity ρand length Lis connected to a voltage source of Vvolts,
creating a current Ithrough it. If the resistance of the wire is R, prove that
Ohm’s Law holds true for the wire, i.e., V=IR.
Solution
To prove Ohm’s Law for the wire, we will first calculate the resistance of the
wire and then show that V=IR.
Step 1: Calculate the resistance of the wire. The resistance of a wire is
given by the formula R=ρL
A, where ρis the resistivity of the material, Lis the
length of the wire, and Ais the cross-sectional area of the wire. Since the wire
is assumed to be uniform, we can write A= constant.
Step 2: Write Ohm’s Law for the wire. Ohm’s Law states that the volt-
age across a conductor is directly proportional to the current passing through
it, with the constant of proportionality being the resistance of the conductor.
Mathematically, this is represented as V=IR.
Step 3: Substitute the expression for resistance. Substitute the expression
for the resistance of the wire (R=ρL
A) into Ohm’s Law to get:
V=IρL
A
Step 4: Simplify the expression. Since Ais a constant for the uniform wire,
we can rewrite the equation as:
V=IρL
A=IR
Step 5: Conclusion. Therefore, we have shown that Ohm’s Law holds true
for the wire, as V=IR.
11
Question 18
Question
A resistor has a resistance of 10 Ω. If a current of 2 A passes through the
resistor, what is the voltage drop across it?
Solution
Let’s use Ohm’s Law, which states that the voltage drop (V) across a resistor
is equal to the product of the current (I) passing through it and the resistance
(R) of the resistor, i.e., V=IR.
Step 1: Given that the resistance R= 10 Ω and the current I= 2 A, we
can calculate the voltage drop V.
Voltage drop, V=I×R
V= 2 A ×10Ω
V= 20 V
Therefore, the voltage drop across the resistor is 20 V.
Question 19
Question
A resistor with a resistance of 30 Ω is connected to a battery with a voltage of
12 V. Calculate the current passing through the resistor.
Solution
Let’s 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.
Step 1: Write down Ohm’s Law formula.
V=IR
Step 2: Plug in the given values.
12 = I×30
Step 3: Solve for I.
I=12
30
I= 0.4 A
Therefore, the current passing through the resistor is 0.4 Amperes.
12
Question 20
Question
A resistor with resistance R= 10 Ω is connected to a power supply that delivers
a current of I= 2 A. 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 current (I) flowing through it multiplied by the resistance (R)
of the resistor. Mathematically, Ohm’s Law is represented as:
V=I×R
Step 2: Given that R= 10 Ω and I= 2 A, we can substitute these values
into Ohm’s Law to find the voltage drop across the resistor:
V= 2 A ×10 Ω
Step 3: Perform the calculation to find the voltage drop:
V= 20 V
Step 4: Therefore, the voltage drop across the resistor is 20 V .
Question 21
Question
An electrical circuit consists of a resistor with resistance R= 50 Ω connected to
a voltage source with voltage V= 120 V. Calculate the current flowing through
the circuit.
Solution
To calculate the current flowing through the circuit, we can use Ohm’s Law,
which states that the current Iin a circuit is equal to the voltage Vacross the
circuit divided by the resistance Rof the circuit.
Step 1: Write down Ohm’s Law formula:
I=V
R
Step 2: Substitute the given values into the formula:
I=120 V
50 Ω
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Step 3: Calculate the current:
I=120
50
I= 2.4 A
Therefore, the current flowing through the circuit is 2.4 A.
Question 22
Question
A resistor with a resistance of 8 Ω is connected to a 12 V battery. Calculate the
current flowing through the resistor.
Solution
Let’s use Ohm’s Law, which states that the current flowing through a resistor is
equal to the voltage across the resistor divided by the resistance of the resistor.
Mathematically, Ohm’s Law can be expressed as:
I=V
R
where: - Iis the current flowing through the resistor, - Vis the voltage across
the resistor, - Ris the resistance of the resistor.
Step 1: Given values: - Voltage, V= 12 V - Resistance, R= 8 Ω
Step 2: Substitute the given values into Ohm’s Law:
I=12
8
Step 3: Simplify the expression to find the current:
I= 1.5 A
Step 4: Therefore, the current flowing through the resistor is 1.5 A.
Question 23
Question
A resistor with resistance R1= 10 Ω is connected in series with a variable
resistor R2across a potential difference of V= 20 V. If the total current in the
circuit is 3 A, determine the resistance of R2.
14
Solution
Let’s denote the resistance of R2as R2and the total resistance in the circuit as
Rtotal. We can use Ohm’s Law, V=I·R, to determine the resistance of R2.
Step 1: Determine the total resistance in the circuit. The total resistance
in a series circuit is the sum of the individual resistances. So,
Rtotal =R1+R2= 10 Ω + R2Ω = (10 + R2) Ω
Step 2: Calculate the total resistance using Ohm’s Law. Given that the
potential difference V= 20 V and the total current I= 3 A, we can use Ohm’s
Law to find the total resistance:
Rtotal =V
I=20 V
3 A = 6.6 Ω
Step 3: Set up the equation to solve for R2. Since the total resistance is
equal to 10 + R2= 6.6, we can write the equation as:
10 + R2= 6.6
Step 4: Solve for R2. Subtracting 10 from both sides gives:
R2= 6.6−10 = −3.3 Ω
Therefore, the resistance of R2is −3.3 Ω.
Question 24
Question
A resistor with a resistance of 12 Ω is connected to a battery with a voltage of
48 V. Determine 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.
Step 1: Write down Ohm’s Law: I=V
R, where Iis the current, Vis the
voltage, and Ris the resistance.
Step 2: Substitute the given values into Ohm’s Law: I=48
12 .
Step 3: Calculate the current passing through the resistor: I= 4 A.
Therefore, the current passing through the resistor is 4 A.
Question 25
Question
A resistor with a resistance of 100 Ω is connected to a 12 V battery. Calculate
the current flowing through the resistor.
15
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. Mathematically, this can be written as:
I=V
R
Step 1: Given that the voltage across the resistor is 12 V and the resistance
of the resistor is 100 Ω, we can substitute these values into Ohm’s Law to find
the current:
I=12 V
100 Ω
Step 2: Simplifying the expression gives:
I= 0.12 A
Therefore, the current flowing through the resistor is 0.12 A.
Question 26
Question
A circuit consists of a battery with a voltage of 12 V connected to a resistor
with a resistance of 8 Ω. Calculate the current flowing through the circuit.
Solution
Step 1: Recall Ohm’s Law which states V=IR, where Vis the voltage, Iis
the current, and Ris the resistance.
Step 2: Substitute the given values into Ohm’s Law: 12 = I×8.
Step 3: Solve for the current, I:
I=12
8= 1.5 A
Step 4: Therefore, the current flowing through the circuit is 1.5 A.
Question 27
Question
A circuit consists of a resistor with a resistance of 50 Ω and a voltage source
with a potential difference of 100 V. If a current of 2 Aflows through the circuit,
what is the power dissipated by the resistor?
16
Solution
Step 1: Recall 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.
V=IR
where Vis the voltage, Iis the current, and Ris the resistance.
Step 2: Given that V= 100 V,I= 2 A, and R= 50 Ω, we can use Ohm’s
Law to find the power dissipated by the resistor.
P=IV
Step 3: Substitute the values of Iand Vinto the formula for power:
P= (2 A)(100 V)
Step 4: Calculate the power dissipated by the resistor:
P= 200 W
Therefore, the power dissipated by the resistor in the circuit is 200 W.
Question 28
Question
A resistor with resistance R= 40Ω is connected to a battery with emf E= 12V.
Find the current passing through the circuit.
Solution
Step 1: Recall Ohm’s Law, which states that the current passing through a
resistor is given by I=E
R, where: - Iis the current passing through the resistor,
-Eis the emf of the battery, and - Ris the resistance of the resistor.
Step 2: Substitute the given values into Ohm’s Law:
I=12 V
40 Ω
Step 3: Calculate the current passing through the circuit:
I=3
10 A= 0.3A
Step 4: Therefore, the current passing through the circuit is 0.3A.
17
Question 29
Question
A circuit consists of a resistor with a resistance of 30 Ω and a battery with an
electromotive force of 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 V=IR, where Vis the voltage, Iis the current, and Ris the
resistance. In this case, we are given V= 12 V and R= 30 Ω. Let’s substitute
these values into Ohm’s Law to find the current I.
Step 1: Write down Ohm’s Law equation:
V=IR
Step 2: Substitute the given values:
12 = I×30
Step 3: Solve for the current I:
I=12
30 = 0.4 A
Step 4: Conclusion: The current flowing through the circuit is 0.4 A.
Question 30
Question
A resistor with resistance Ris connected to a battery, producing a current I
in the circuit. The power dissipated by the resistor is given by the equation
P=I2R. If the resistance Ris increased by a factor of 5, by what factor does
the power dissipated by the resistor change?
Solution
Let’s denote the original power dissipated by the resistor as Poriginal and the
new power dissipated by the resistor as Pnew.
Step 1: Express the original power dissipated in terms of Iand R.
Poriginal =I2R
Step 2: Find the new power dissipated when the resistance is increased by
a factor of 5. If the resistance Ris increased by a factor of 5, the new resistance
Rnew is 5R. Therefore, the new power dissipated is:
Pnew =I2·5R= 5(I2R)
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Step 3: Determine the factor by which the power dissipated by the resistor
changes. The factor by which the power dissipated by the resistor changes is:
Pnew
Poriginal
=5(I2R)
I2R= 5
So, when the resistance Ris increased by a factor of 5, the power dissipated
by the resistor changes by a factor of 5.
Question 31
Question
A resistor with a resistance of 10 Ω is connected to a battery with a voltage of
24 V. Calculate the current flowing through the circuit.
Solution
Step 1: Recall Ohm’s Law, which states that the current (I) flowing through a
conductor between two points is directly proportional to the voltage (V) across
the two points and inversely proportional to the resistance (R) of the conductor.
Mathematically, Ohm’s Law can be expressed as V=IR, where Iis the current
in the circuit, Vis the voltage across the circuit, and Ris the resistance of the
circuit.
Step 2: Given that the voltage Vis 24 V and the resistance Ris 10 Ω, we
can use Ohm’s Law to find the current Iwith the formula I=V
R.
Step 3: Substitute V= 24 V and R= 10 Ω into the formula I=V
R.
I=24 V
10 Ω
Step 4: Perform the division to find the current I.
I= 2.4 A
Step 5: Therefore, the current flowing through the circuit is 2.4 A.
Question 32
Question
A resistor has a resistance of 20 Ω. If a current of 0.5 A flows through it, what
is the voltage drop across the resistor?
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Solution
To find the voltage drop across the resistor, we can use Ohm’s Law, which states
that V=IR, where Vis the voltage drop across the resistor, Iis the current
flowing through the resistor, and Ris the resistance of the resistor.
Step 1: Given values: Resistance, R= 20Ω
Current, I= 0.5 A
Step 2: Apply Ohm’s Law to find the voltage drop:
V=IR
V= 0.5 A ×20 Ω
V= 10 V
The voltage drop across the resistor is 10 V.
Question 33
Question
A resistor with resistance R= 10 Ω is connected to a battery that provides
a potential difference V= 50 V. Calculate the current flowing through the
resistor.
Solution
Let’s use Ohm’s Law, which states that V=IR, where Vis the potential
difference (voltage), Iis the current, 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 the current I:
I=V
R
Step 3: Substitute the given values V= 50 V and R= 10 Ω into the
equation:
I=50 V
10 Ω
Step 4: Calculate the current:
I= 5 A
Step 5: Therefore, the current flowing through the resistor is 5 A .
20
Question 34
Question
A resistor with resistance R= 200 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 calculate the current flowing through the resistor.
Step 1: Recall Ohm’s Law: 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 2: Substitute the given values into Ohm’s Law:
12 = I·200
Step 3: Solve for I:
I=12
200 = 0.06 A
Step 4: Therefore, the current flowing through the resistor is 0.06 A.
Question 35
Question
A resistor with resistance R= 50 Ω is connected to a voltage source that pro-
vides V= 100 V across the resistor. Calculate the current passing through the
resistor.
Solution
Step 1: Recall Ohm’s Law, which states that the current passing through a
resistor is given by I=V
R, where Iis the current, Vis the voltage across the
resistor, and Ris the resistance of the resistor.
Step 2: Substitute the given values into the formula: I=100 V
50 Ω .
Step 3: Calculate the current passing through the resistor: I= 2 A.
Therefore, the current passing through the resistor is 2 A.
21
Question 2
Question
A circuit consists of a resistor with a resistance of 30 Ω connected to a power
supply that produces a voltage of 120 V. 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 ratio of the voltage (V) across the circuit to the resistance
(R) of the circuit. Mathematically, Ohm’s Law can be expressed as:
I=V
R
Step 2: Given that the voltage across the circuit (V) is 120 V and the
resistance of the resistor (R) is 30 Ω, we can substitute these values into Ohm’s
Law to find the current (I):
I=120
30
I= 4 Amperes
Step 3: Therefore, the current flowing through the circuit is 4 Amperes.
Question 3
Question
A resistor with resistance R= 150 Ω is connected to a battery with voltage
V= 12 V. Calculate the current that flows through the resistor.
Solution
To calculate the current that flows through the resistor, we can use Ohm’s Law,
which states 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
I=V
R
Step 2: Plug in the values V= 12 Vand R= 150 Ω:
I=12 V
150 Ω
2
Step 3: Calculate the current flowing through the resistor:
I=12
150
I= 0.08 A
Therefore, the current that flows through the resistor is 0.08 A.
Question 4
Question
A resistor is connected to a 12-volt battery and a current of 0.5 amperes flows
through it. Determine the resistance of the resistor.
Solution
Let’s use Ohm’s Law (V=IR) to find the resistance of the resistor.
Step 1: Write down the known values. The voltage across the resistor (V)
is 12 volts and the current flowing through it (I) is 0.5 amperes.
Step 2: Use Ohm’s Law to find the resistance. We can rearrange Ohm’s
Law to solve for resistance:
R=V
I
Step 3: Substitute the known values into the formula.
R=12
0.5
Step 4: Calculate the resistance.
R=12
0.5= 24 ohms
Step 5: Write the final answer. The resistance of the resistor is 24 ohms.
Question 5
Question
A resistor with a resistance of 12 Ω is connected to a 24 V battery. Determine
the current flowing through the resistor.
3
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: Given that the voltage V= 24 V and the resistance R= 12 Ω, we
can apply Ohm’s Law to find the current I.
V=IR
I=V
R
I=24
12
I= 2 A
Step 3: Therefore, the current flowing through the resistor is 2 A.
Question 6
Question
A circuit consists of a resistor with a resistance of 10 Ω and a battery with a
voltage of 12 V. If a current of 1.2Ais flowing through the circuit, determine
the power dissipated by the resistor.
Solution
Let’s first recall Ohm’s Law, which states that the current flowing through a
resistor is directly proportional to the voltage across it and inversely propor-
tional to the resistance. Mathematically, Ohm’s Law is represented as 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: Calculate the voltage across the resistor using Ohm’s Law. Given
that I= 1.2Aand R= 10 Ω, we can use Ohm’s Law to find V:
V=IR = 1.2A×10 Ω = 12 V
Step 2: Calculate the power dissipated by the resistor using the formula
P=IV . The power dissipated by a resistor can be calculated using the formula
P=IV , where: - Pis the power dissipated by the resistor, - Iis the current
flowing through the resistor, and - Vis the voltage across the resistor.
Substitute I= 1.2Aand V= 12 Vinto the formula to find the power:
P= 1.2A×12 V= 14.4W
Step 3: Answer: The power dissipated by the resistor in the circuit is
14.4W.
4
Question 7
Question
A circuit consists of a resistor with resistance Rconnected to a battery with
voltage V. If the current flowing through the circuit is I, prove Ohm’s Law,
V=IR, using Kirchhoff’s voltage law.
Solution
To prove Ohm’s Law using Kirchhoff’s voltage law, we consider the voltage
around the closed loop of the circuit.
Step 1: Start by considering the voltage drop across the resistor: Let VRbe
the voltage drop across the resistor R, which is given by Ohm’s Law: VR=IR.
This voltage drop occurs in the direction of the current flow.
Step 2: Consider the voltage rise across the battery: Let Vbattery be the
voltage rise across the battery. Since the current Iflows from the positive
terminal of the battery to the negative terminal, Vbattery =−V. This negative
sign indicates that the battery provides a voltage rise in the opposite direction
to the current flow.
Step 3: Apply Kirchhoff’s voltage law: According to Kirchhoff’s voltage
law, the sum of the voltages around a closed loop in a circuit must be zero.
Therefore, we have:
Vbattery +VR= 0
(−V)+(IR) = 0
Step 4: Substitute VR=IR into the equation:
−V+IR = 0
Step 5: Rearrange the equation to prove Ohm’s Law:
V=IR
Thus, we have proven Ohm’s Law, V=IR, using Kirchhoff’s voltage law in
the given circuit.
Question 8
Question
A resistor with a resistance of 5 Ω is connected to a 12 V battery. Determine
the current flowing through the resistor.
5
Solution
Ohm’s Law relates the voltage (V), current (I), and resistance (R) in a circuit
through the equation V=IR.
1. The given values are: Resistance, R= 5 Ω Voltage, V= 12 V
2. Using Ohm’s Law, we can rearrange the formula to solve for current:
I=V
R
3. Now, substitute the known values into the formula to find the current:
I=12
5
4. Thus, the current flowing through the resistor is:
I= 2.4 A
Question 9
Question
A circuit consists of a resistor with resistance R= 300 Ω connected to a power
supply with voltage V= 120 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: Identify the given values. The resistance of the resistor, R= 300 Ω,
and the voltage across the resistor, V= 120 V.
Step 2: Substitute the values into Ohm’s Law equation.
V=IR
120 = I×300
Step 3: Solve for the current, I.
I=120
300
I= 0.4A
Step 4: Answer: The current flowing through the circuit is 0.4 amps.
6
Question 10
Question
A resistor with a resistance of 20 Ω is connected to a 12 V battery. What is the
current flowing through the resistor?
Solution
Given: Resistance, R= 20 Ω Battery voltage, V= 12 V
We can use Ohm’s Law to find the current flowing through the resistor:
I=V
R
Step 1: Substituting the given values into Ohm’s Law:
I=12 V
20 Ω
Step 2: Calculating the current:
I= 0.6 A
Therefore, the current flowing through the resistor is 0.6 A.
Question 11
Question
A circuit consists of a resistor with a resistance of 100 Ω, a capacitor with a
capacitance of 0.1 F, and an inductor with an inductance of 0.2 H connected in
series to a voltage source. If the frequency of the source is 50 Hz, calculate the
current flowing through the circuit.
Solution
To calculate the current flowing through the circuit, we can use Ohm’s Law,
which states that V=I·Z, where Vis the voltage across the circuit, Iis the
current flowing through the circuit, and Zis the impedance of the circuit.
The impedance of a series RLC circuit is given by:
Z=pR2+ (XL−XC)2
where: - R= Resistance in the circuit (Ω) - XL= Inductive reactance, XL=
2πfL (Ω) - XC= Capacitive reactance, XC=1
2πfC (Ω) - f= Frequency of
the source (Hz) - L= Inductance of the inductor (H) - C= Capacitance of the
capacitor (F)
7
Given: Resistance, R= 100 Ω Capacitance, C= 0.1 F Inductance, L= 0.2 H
Frequency, f= 50 Hz
Step 1: Calculate the inductive reactance XL
XL= 2πfL = 2π×50 ×0.2 = 62.83 Ω
Step 2: Calculate the capacitive reactance XC
XC=1
2πfC =1
2π×50 ×0.1= 31.83 Ω
Step 3: Calculate the impedance Z
Z=pR2+ (XL−XC)2=p1002+ (62.83 −31.83)2=√10000 + 961 = √10961 = 104.69 Ω
Step 4: Apply Ohm’s Law to find the current I
V=I·Z
I=V
Z=V
104.69
Since no value for the voltage source Vwas provided, the current flowing
through the circuit cannot be determined without this information.
Question 12
Question
A resistor with a resistance of 15 Ω is connected to a 9 Vbattery. What is the
current flowing through the resistor?
Solution
Let’s use Ohm’s Law, V=IR, where Vis the voltage across the resistor, Iis
the current flowing through the resistor, and Ris the resistance of the resistor.
We are given V= 9 Vand R= 15 Ω. We need to find I.
Step 1: Substitute the given values into Ohm’s Law equation.
9 = I×15
Step 2: Solve for I.
I=9
15 = 0.6A
Step 3: Answer: The current flowing through the resistor is 0.6A.
Question 13
Question
A 20 Ω resistor, a 30 Ω resistor, and a 40 Ω resistor are connected in series to
a 12V battery. What is the current flowing through the circuit?
8
Solution
Let’s denote the current flowing through the circuit as I.
Step 1: Calculate the total resistance of the circuit. The total resistance
Rtotal of resistors in series is the sum of individual resistances.
Rtotal = 20Ω + 30Ω + 40Ω = 90Ω
Step 2: Apply Ohm’s Law to find the current. Ohm’s Law states that
V=IR, where Vis the voltage, Iis the current, and Ris the resistance.
Substitute the values into the equation:
I=V
Rtotal
=12V
90Ω = 0.1333 A
Therefore, the current flowing through the circuit is 0.1333 A.
Question 14
Question
A resistor has a resistance of 10 Ω and a current of 2 A flowing through it.
Determine the voltage drop across the resistor.
Solution
Step 1: Recall Ohm’s Law, which states that V=I×R, where Vis the voltage
drop 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: V= 2 A ×10 Ω.
Step 3: Calculate the voltage drop: V= 20 V.
Therefore, the voltage drop across the resistor is 20 V.
Question 15
Question
A circuit consists of a resistor with resistance R= 30 Ω, an inductor with
inductance L= 0.2 H, and a capacitor with capacitance C= 8 F. If a voltage of
V(t) = 24 sin(100t) volts is applied to the circuit, determine the current flowing
through the circuit at time t=π
200 seconds.
Solution
Step 1: Find the total impedance of the circuit. The impedance of a resistor
is ZR=R, the impedance of an inductor is ZL=jωL, and the impedance of
a capacitor is ZC=1
jωC , where ωis the angular frequency. Since the circuit
9
contains all three components in series, the total impedance Ztotal is the sum of
the individual impedances.
Ztotal =R+jωL +1
jωC
Step 2: Substitute the given values into the impedance expression. The
angular frequency is given by ω= 100 rad/s. Thus,
Ztotal = 30 + j(100)(0.2) + 1
j(100)(8)
Ztotal = 30 + j20 −j0.125 = 30 + j20 −j0.125
Step 3: Simplify the total impedance.
Ztotal = 30 + j20 −j0.125 = 30 + j20 + j0.125
Ztotal = 30 + j20.125 Ω
Step 4: Use Ohm’s Law to find the current flowing through the circuit at
t=π
200 seconds. Ohm’s Law states V(t) = I(t)Z, where V(t) is the voltage
across the circuit at time t,I(t) is the current flowing through the circuit at
time t, and Zis the total impedance of the circuit.
I(t) = V(t)
Ztotal
=24 sin(100t)
30 + j20.125
Step 5: Substitute t=π
200 seconds into the expression for current.
Iπ
200=24 sin 100 ×π
200
30 + j20.125 =24 sin π
2
30 + j20.125 =24
√302+ 20.1252
∠arctan 20.125
30 A
Question 16
Question
A resistor with a resistance of 240 Ω is connected to a battery with a voltage of
12 V. Calculate the current flowing through the resistor.
Solution
Let’s use Ohm’s Law, which states that the current passing through a conduc-
tor is directly proportional to the voltage across the conductor and inversely
proportional to the resistance of the conductor. Mathematically, Ohm’s Law is
represented as: 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 values are: - Resistance R= 240 Ω, - Voltage V= 12 V.
10
Step 2: We can rearrange Ohm’s Law to solve for current I:
I=V
R
Step 3: Substitute the values of Vand Rinto the formula:
I=12 V
240 Ω
Step 4: Calculate the current passing through the resistor:
I=1
20 A= 0.05 A
Therefore, the current flowing through the resistor is 0.05 A.
Question 17
Question
A wire of resistivity ρand length Lis connected to a voltage source of Vvolts,
creating a current Ithrough it. If the resistance of the wire is R, prove that
Ohm’s Law holds true for the wire, i.e., V=IR.
Solution
To prove Ohm’s Law for the wire, we will first calculate the resistance of the
wire and then show that V=IR.
Step 1: Calculate the resistance of the wire. The resistance of a wire is
given by the formula R=ρL
A, where ρis the resistivity of the material, Lis the
length of the wire, and Ais the cross-sectional area of the wire. Since the wire
is assumed to be uniform, we can write A= constant.
Step 2: Write Ohm’s Law for the wire. Ohm’s Law states that the volt-
age across a conductor is directly proportional to the current passing through
it, with the constant of proportionality being the resistance of the conductor.
Mathematically, this is represented as V=IR.
Step 3: Substitute the expression for resistance. Substitute the expression
for the resistance of the wire (R=ρL
A) into Ohm’s Law to get:
V=IρL
A
Step 4: Simplify the expression. Since Ais a constant for the uniform wire,
we can rewrite the equation as:
V=IρL
A=IR
Step 5: Conclusion. Therefore, we have shown that Ohm’s Law holds true
for the wire, as V=IR.
11
Question 18
Question
A resistor has a resistance of 10 Ω. If a current of 2 A passes through the
resistor, what is the voltage drop across it?
Solution
Let’s use Ohm’s Law, which states that the voltage drop (V) across a resistor
is equal to the product of the current (I) passing through it and the resistance
(R) of the resistor, i.e., V=IR.
Step 1: Given that the resistance R= 10 Ω and the current I= 2 A, we
can calculate the voltage drop V.
Voltage drop, V=I×R
V= 2 A ×10Ω
V= 20 V
Therefore, the voltage drop across the resistor is 20 V.
Question 19
Question
A resistor with a resistance of 30 Ω is connected to a battery with a voltage of
12 V. Calculate the current passing through the resistor.
Solution
Let’s 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.
Step 1: Write down Ohm’s Law formula.
V=IR
Step 2: Plug in the given values.
12 = I×30
Step 3: Solve for I.
I=12
30
I= 0.4 A
Therefore, the current passing through the resistor is 0.4 Amperes.
12
Question 20
Question
A resistor with resistance R= 10 Ω is connected to a power supply that delivers
a current of I= 2 A. 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 current (I) flowing through it multiplied by the resistance (R)
of the resistor. Mathematically, Ohm’s Law is represented as:
V=I×R
Step 2: Given that R= 10 Ω and I= 2 A, we can substitute these values
into Ohm’s Law to find the voltage drop across the resistor:
V= 2 A ×10 Ω
Step 3: Perform the calculation to find the voltage drop:
V= 20 V
Step 4: Therefore, the voltage drop across the resistor is 20 V .
Question 21
Question
An electrical circuit consists of a resistor with resistance R= 50 Ω connected to
a voltage source with voltage V= 120 V. Calculate the current flowing through
the circuit.
Solution
To calculate the current flowing through the circuit, we can use Ohm’s Law,
which states that the current Iin a circuit is equal to the voltage Vacross the
circuit divided by the resistance Rof the circuit.
Step 1: Write down Ohm’s Law formula:
I=V
R
Step 2: Substitute the given values into the formula:
I=120 V
50 Ω
13
Step 3: Calculate the current:
I=120
50
I= 2.4 A
Therefore, the current flowing through the circuit is 2.4 A.
Question 22
Question
A resistor with a resistance of 8 Ω is connected to a 12 V battery. Calculate the
current flowing through the resistor.
Solution
Let’s use Ohm’s Law, which states that the current flowing through a resistor is
equal to the voltage across the resistor divided by the resistance of the resistor.
Mathematically, Ohm’s Law can be expressed as:
I=V
R
where: - Iis the current flowing through the resistor, - Vis the voltage across
the resistor, - Ris the resistance of the resistor.
Step 1: Given values: - Voltage, V= 12 V - Resistance, R= 8 Ω
Step 2: Substitute the given values into Ohm’s Law:
I=12
8
Step 3: Simplify the expression to find the current:
I= 1.5 A
Step 4: Therefore, the current flowing through the resistor is 1.5 A.
Question 23
Question
A resistor with resistance R1= 10 Ω is connected in series with a variable
resistor R2across a potential difference of V= 20 V. If the total current in the
circuit is 3 A, determine the resistance of R2.
14
Solution
Let’s denote the resistance of R2as R2and the total resistance in the circuit as
Rtotal. We can use Ohm’s Law, V=I·R, to determine the resistance of R2.
Step 1: Determine the total resistance in the circuit. The total resistance
in a series circuit is the sum of the individual resistances. So,
Rtotal =R1+R2= 10 Ω + R2Ω = (10 + R2) Ω
Step 2: Calculate the total resistance using Ohm’s Law. Given that the
potential difference V= 20 V and the total current I= 3 A, we can use Ohm’s
Law to find the total resistance:
Rtotal =V
I=20 V
3 A = 6.6 Ω
Step 3: Set up the equation to solve for R2. Since the total resistance is
equal to 10 + R2= 6.6, we can write the equation as:
10 + R2= 6.6
Step 4: Solve for R2. Subtracting 10 from both sides gives:
R2= 6.6−10 = −3.3 Ω
Therefore, the resistance of R2is −3.3 Ω.
Question 24
Question
A resistor with a resistance of 12 Ω is connected to a battery with a voltage of
48 V. Determine 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.
Step 1: Write down Ohm’s Law: I=V
R, where Iis the current, Vis the
voltage, and Ris the resistance.
Step 2: Substitute the given values into Ohm’s Law: I=48
12 .
Step 3: Calculate the current passing through the resistor: I= 4 A.
Therefore, the current passing through the resistor is 4 A.
Question 25
Question
A resistor with a resistance of 100 Ω is connected to a 12 V battery. Calculate
the current flowing through the resistor.
15
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. Mathematically, this can be written as:
I=V
R
Step 1: Given that the voltage across the resistor is 12 V and the resistance
of the resistor is 100 Ω, we can substitute these values into Ohm’s Law to find
the current:
I=12 V
100 Ω
Step 2: Simplifying the expression gives:
I= 0.12 A
Therefore, the current flowing through the resistor is 0.12 A.
Question 26
Question
A circuit consists of a battery with a voltage of 12 V connected to a resistor
with a resistance of 8 Ω. Calculate the current flowing through the circuit.
Solution
Step 1: Recall Ohm’s Law which states V=IR, where Vis the voltage, Iis
the current, and Ris the resistance.
Step 2: Substitute the given values into Ohm’s Law: 12 = I×8.
Step 3: Solve for the current, I:
I=12
8= 1.5 A
Step 4: Therefore, the current flowing through the circuit is 1.5 A.
Question 27
Question
A circuit consists of a resistor with a resistance of 50 Ω and a voltage source
with a potential difference of 100 V. If a current of 2 Aflows through the circuit,
what is the power dissipated by the resistor?
16
Solution
Step 1: Recall 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.
V=IR
where Vis the voltage, Iis the current, and Ris the resistance.
Step 2: Given that V= 100 V,I= 2 A, and R= 50 Ω, we can use Ohm’s
Law to find the power dissipated by the resistor.
P=IV
Step 3: Substitute the values of Iand Vinto the formula for power:
P= (2 A)(100 V)
Step 4: Calculate the power dissipated by the resistor:
P= 200 W
Therefore, the power dissipated by the resistor in the circuit is 200 W.
Question 28
Question
A resistor with resistance R= 40Ω is connected to a battery with emf E= 12V.
Find the current passing through the circuit.
Solution
Step 1: Recall Ohm’s Law, which states that the current passing through a
resistor is given by I=E
R, where: - Iis the current passing through the resistor,
-Eis the emf of the battery, and - Ris the resistance of the resistor.
Step 2: Substitute the given values into Ohm’s Law:
I=12 V
40 Ω
Step 3: Calculate the current passing through the circuit:
I=3
10 A= 0.3A
Step 4: Therefore, the current passing through the circuit is 0.3A.
17
Question 29
Question
A circuit consists of a resistor with a resistance of 30 Ω and a battery with an
electromotive force of 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 V=IR, where Vis the voltage, Iis the current, and Ris the
resistance. In this case, we are given V= 12 V and R= 30 Ω. Let’s substitute
these values into Ohm’s Law to find the current I.
Step 1: Write down Ohm’s Law equation:
V=IR
Step 2: Substitute the given values:
12 = I×30
Step 3: Solve for the current I:
I=12
30 = 0.4 A
Step 4: Conclusion: The current flowing through the circuit is 0.4 A.
Question 30
Question
A resistor with resistance Ris connected to a battery, producing a current I
in the circuit. The power dissipated by the resistor is given by the equation
P=I2R. If the resistance Ris increased by a factor of 5, by what factor does
the power dissipated by the resistor change?
Solution
Let’s denote the original power dissipated by the resistor as Poriginal and the
new power dissipated by the resistor as Pnew.
Step 1: Express the original power dissipated in terms of Iand R.
Poriginal =I2R
Step 2: Find the new power dissipated when the resistance is increased by
a factor of 5. If the resistance Ris increased by a factor of 5, the new resistance
Rnew is 5R. Therefore, the new power dissipated is:
Pnew =I2·5R= 5(I2R)
18
Step 3: Determine the factor by which the power dissipated by the resistor
changes. The factor by which the power dissipated by the resistor changes is:
Pnew
Poriginal
=5(I2R)
I2R= 5
So, when the resistance Ris increased by a factor of 5, the power dissipated
by the resistor changes by a factor of 5.
Question 31
Question
A resistor with a resistance of 10 Ω is connected to a battery with a voltage of
24 V. Calculate the current flowing through the circuit.
Solution
Step 1: Recall Ohm’s Law, which states that the current (I) flowing through a
conductor between two points is directly proportional to the voltage (V) across
the two points and inversely proportional to the resistance (R) of the conductor.
Mathematically, Ohm’s Law can be expressed as V=IR, where Iis the current
in the circuit, Vis the voltage across the circuit, and Ris the resistance of the
circuit.
Step 2: Given that the voltage Vis 24 V and the resistance Ris 10 Ω, we
can use Ohm’s Law to find the current Iwith the formula I=V
R.
Step 3: Substitute V= 24 V and R= 10 Ω into the formula I=V
R.
I=24 V
10 Ω
Step 4: Perform the division to find the current I.
I= 2.4 A
Step 5: Therefore, the current flowing through the circuit is 2.4 A.
Question 32
Question
A resistor has a resistance of 20 Ω. If a current of 0.5 A flows through it, what
is the voltage drop across the resistor?
19
Solution
To find the voltage drop across the resistor, we can use Ohm’s Law, which states
that V=IR, where Vis the voltage drop across the resistor, Iis the current
flowing through the resistor, and Ris the resistance of the resistor.
Step 1: Given values: Resistance, R= 20Ω
Current, I= 0.5 A
Step 2: Apply Ohm’s Law to find the voltage drop:
V=IR
V= 0.5 A ×20 Ω
V= 10 V
The voltage drop across the resistor is 10 V.
Question 33
Question
A resistor with resistance R= 10 Ω is connected to a battery that provides
a potential difference V= 50 V. Calculate the current flowing through the
resistor.
Solution
Let’s use Ohm’s Law, which states that V=IR, where Vis the potential
difference (voltage), Iis the current, 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 the current I:
I=V
R
Step 3: Substitute the given values V= 50 V and R= 10 Ω into the
equation:
I=50 V
10 Ω
Step 4: Calculate the current:
I= 5 A
Step 5: Therefore, the current flowing through the resistor is 5 A .
20
Question 34
Question
A resistor with resistance R= 200 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 calculate the current flowing through the resistor.
Step 1: Recall Ohm’s Law: 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 2: Substitute the given values into Ohm’s Law:
12 = I·200
Step 3: Solve for I:
I=12
200 = 0.06 A
Step 4: Therefore, the current flowing through the resistor is 0.06 A.
Question 35
Question
A resistor with resistance R= 50 Ω is connected to a voltage source that pro-
vides V= 100 V across the resistor. Calculate the current passing through the
resistor.
Solution
Step 1: Recall Ohm’s Law, which states that the current passing through a
resistor is given by I=V
R, where Iis the current, Vis the voltage across the
resistor, and Ris the resistance of the resistor.
Step 2: Substitute the given values into the formula: I=100 V
50 Ω .
Step 3: Calculate the current passing through the resistor: I= 2 A.
Therefore, the current passing through the resistor is 2 A.
21
Question 2
Question
A circuit consists of a resistor with a resistance of 30 Ω connected to a power
supply that produces a voltage of 120 V. 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 ratio of the voltage (V) across the circuit to the resistance
(R) of the circuit. Mathematically, Ohm’s Law can be expressed as:
I=V
R
Step 2: Given that the voltage across the circuit (V) is 120 V and the
resistance of the resistor (R) is 30 Ω, we can substitute these values into Ohm’s
Law to find the current (I):
I=120
30
I= 4 Amperes
Step 3: Therefore, the current flowing through the circuit is 4 Amperes.
Question 3
Question
A resistor with resistance R= 150 Ω is connected to a battery with voltage
V= 12 V. Calculate the current that flows through the resistor.
Solution
To calculate the current that flows through the resistor, we can use Ohm’s Law,
which states 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
I=V
R
Step 2: Plug in the values V= 12 Vand R= 150 Ω:
I=12 V
150 Ω
2
Step 3: Calculate the current flowing through the resistor:
I=12
150
I= 0.08 A
Therefore, the current that flows through the resistor is 0.08 A.
Question 4
Question
A resistor is connected to a 12-volt battery and a current of 0.5 amperes flows
through it. Determine the resistance of the resistor.
Solution
Let’s use Ohm’s Law (V=IR) to find the resistance of the resistor.
Step 1: Write down the known values. The voltage across the resistor (V)
is 12 volts and the current flowing through it (I) is 0.5 amperes.
Step 2: Use Ohm’s Law to find the resistance. We can rearrange Ohm’s
Law to solve for resistance:
R=V
I
Step 3: Substitute the known values into the formula.
R=12
0.5
Step 4: Calculate the resistance.
R=12
0.5= 24 ohms
Step 5: Write the final answer. The resistance of the resistor is 24 ohms.
Question 5
Question
A resistor with a resistance of 12 Ω is connected to a 24 V battery. Determine
the current flowing through the resistor.
3
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: Given that the voltage V= 24 V and the resistance R= 12 Ω, we
can apply Ohm’s Law to find the current I.
V=IR
I=V
R
I=24
12
I= 2 A
Step 3: Therefore, the current flowing through the resistor is 2 A.
Question 6
Question
A circuit consists of a resistor with a resistance of 10 Ω and a battery with a
voltage of 12 V. If a current of 1.2Ais flowing through the circuit, determine
the power dissipated by the resistor.
Solution
Let’s first recall Ohm’s Law, which states that the current flowing through a
resistor is directly proportional to the voltage across it and inversely propor-
tional to the resistance. Mathematically, Ohm’s Law is represented as 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: Calculate the voltage across the resistor using Ohm’s Law. Given
that I= 1.2Aand R= 10 Ω, we can use Ohm’s Law to find V:
V=IR = 1.2A×10 Ω = 12 V
Step 2: Calculate the power dissipated by the resistor using the formula
P=IV . The power dissipated by a resistor can be calculated using the formula
P=IV , where: - Pis the power dissipated by the resistor, - Iis the current
flowing through the resistor, and - Vis the voltage across the resistor.
Substitute I= 1.2Aand V= 12 Vinto the formula to find the power:
P= 1.2A×12 V= 14.4W
Step 3: Answer: The power dissipated by the resistor in the circuit is
14.4W.
4
Question 7
Question
A circuit consists of a resistor with resistance Rconnected to a battery with
voltage V. If the current flowing through the circuit is I, prove Ohm’s Law,
V=IR, using Kirchhoff’s voltage law.
Solution
To prove Ohm’s Law using Kirchhoff’s voltage law, we consider the voltage
around the closed loop of the circuit.
Step 1: Start by considering the voltage drop across the resistor: Let VRbe
the voltage drop across the resistor R, which is given by Ohm’s Law: VR=IR.
This voltage drop occurs in the direction of the current flow.
Step 2: Consider the voltage rise across the battery: Let Vbattery be the
voltage rise across the battery. Since the current Iflows from the positive
terminal of the battery to the negative terminal, Vbattery =−V. This negative
sign indicates that the battery provides a voltage rise in the opposite direction
to the current flow.
Step 3: Apply Kirchhoff’s voltage law: According to Kirchhoff’s voltage
law, the sum of the voltages around a closed loop in a circuit must be zero.
Therefore, we have:
Vbattery +VR= 0
(−V)+(IR) = 0
Step 4: Substitute VR=IR into the equation:
−V+IR = 0
Step 5: Rearrange the equation to prove Ohm’s Law:
V=IR
Thus, we have proven Ohm’s Law, V=IR, using Kirchhoff’s voltage law in
the given circuit.
Question 8
Question
A resistor with a resistance of 5 Ω is connected to a 12 V battery. Determine
the current flowing through the resistor.
5
Solution
Ohm’s Law relates the voltage (V), current (I), and resistance (R) in a circuit
through the equation V=IR.
1. The given values are: Resistance, R= 5 Ω Voltage, V= 12 V
2. Using Ohm’s Law, we can rearrange the formula to solve for current:
I=V
R
3. Now, substitute the known values into the formula to find the current:
I=12
5
4. Thus, the current flowing through the resistor is:
I= 2.4 A
Question 9
Question
A circuit consists of a resistor with resistance R= 300 Ω connected to a power
supply with voltage V= 120 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: Identify the given values. The resistance of the resistor, R= 300 Ω,
and the voltage across the resistor, V= 120 V.
Step 2: Substitute the values into Ohm’s Law equation.
V=IR
120 = I×300
Step 3: Solve for the current, I.
I=120
300
I= 0.4A
Step 4: Answer: The current flowing through the circuit is 0.4 amps.
6
Question 10
Question
A resistor with a resistance of 20 Ω is connected to a 12 V battery. What is the
current flowing through the resistor?
Solution
Given: Resistance, R= 20 Ω Battery voltage, V= 12 V
We can use Ohm’s Law to find the current flowing through the resistor:
I=V
R
Step 1: Substituting the given values into Ohm’s Law:
I=12 V
20 Ω
Step 2: Calculating the current:
I= 0.6 A
Therefore, the current flowing through the resistor is 0.6 A.
Question 11
Question
A circuit consists of a resistor with a resistance of 100 Ω, a capacitor with a
capacitance of 0.1 F, and an inductor with an inductance of 0.2 H connected in
series to a voltage source. If the frequency of the source is 50 Hz, calculate the
current flowing through the circuit.
Solution
To calculate the current flowing through the circuit, we can use Ohm’s Law,
which states that V=I·Z, where Vis the voltage across the circuit, Iis the
current flowing through the circuit, and Zis the impedance of the circuit.
The impedance of a series RLC circuit is given by:
Z=pR2+ (XL−XC)2
where: - R= Resistance in the circuit (Ω) - XL= Inductive reactance, XL=
2πfL (Ω) - XC= Capacitive reactance, XC=1
2πfC (Ω) - f= Frequency of
the source (Hz) - L= Inductance of the inductor (H) - C= Capacitance of the
capacitor (F)
7
Given: Resistance, R= 100 Ω Capacitance, C= 0.1 F Inductance, L= 0.2 H
Frequency, f= 50 Hz
Step 1: Calculate the inductive reactance XL
XL= 2πfL = 2π×50 ×0.2 = 62.83 Ω
Step 2: Calculate the capacitive reactance XC
XC=1
2πfC =1
2π×50 ×0.1= 31.83 Ω
Step 3: Calculate the impedance Z
Z=pR2+ (XL−XC)2=p1002+ (62.83 −31.83)2=√10000 + 961 = √10961 = 104.69 Ω
Step 4: Apply Ohm’s Law to find the current I
V=I·Z
I=V
Z=V
104.69
Since no value for the voltage source Vwas provided, the current flowing
through the circuit cannot be determined without this information.
Question 12
Question
A resistor with a resistance of 15 Ω is connected to a 9 Vbattery. What is the
current flowing through the resistor?
Solution
Let’s use Ohm’s Law, V=IR, where Vis the voltage across the resistor, Iis
the current flowing through the resistor, and Ris the resistance of the resistor.
We are given V= 9 Vand R= 15 Ω. We need to find I.
Step 1: Substitute the given values into Ohm’s Law equation.
9 = I×15
Step 2: Solve for I.
I=9
15 = 0.6A
Step 3: Answer: The current flowing through the resistor is 0.6A.
Question 13
Question
A 20 Ω resistor, a 30 Ω resistor, and a 40 Ω resistor are connected in series to
a 12V battery. What is the current flowing through the circuit?
8
Solution
Let’s denote the current flowing through the circuit as I.
Step 1: Calculate the total resistance of the circuit. The total resistance
Rtotal of resistors in series is the sum of individual resistances.
Rtotal = 20Ω + 30Ω + 40Ω = 90Ω
Step 2: Apply Ohm’s Law to find the current. Ohm’s Law states that
V=IR, where Vis the voltage, Iis the current, and Ris the resistance.
Substitute the values into the equation:
I=V
Rtotal
=12V
90Ω = 0.1333 A
Therefore, the current flowing through the circuit is 0.1333 A.
Question 14
Question
A resistor has a resistance of 10 Ω and a current of 2 A flowing through it.
Determine the voltage drop across the resistor.
Solution
Step 1: Recall Ohm’s Law, which states that V=I×R, where Vis the voltage
drop 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: V= 2 A ×10 Ω.
Step 3: Calculate the voltage drop: V= 20 V.
Therefore, the voltage drop across the resistor is 20 V.
Question 15
Question
A circuit consists of a resistor with resistance R= 30 Ω, an inductor with
inductance L= 0.2 H, and a capacitor with capacitance C= 8 F. If a voltage of
V(t) = 24 sin(100t) volts is applied to the circuit, determine the current flowing
through the circuit at time t=π
200 seconds.
Solution
Step 1: Find the total impedance of the circuit. The impedance of a resistor
is ZR=R, the impedance of an inductor is ZL=jωL, and the impedance of
a capacitor is ZC=1
jωC , where ωis the angular frequency. Since the circuit
9
contains all three components in series, the total impedance Ztotal is the sum of
the individual impedances.
Ztotal =R+jωL +1
jωC
Step 2: Substitute the given values into the impedance expression. The
angular frequency is given by ω= 100 rad/s. Thus,
Ztotal = 30 + j(100)(0.2) + 1
j(100)(8)
Ztotal = 30 + j20 −j0.125 = 30 + j20 −j0.125
Step 3: Simplify the total impedance.
Ztotal = 30 + j20 −j0.125 = 30 + j20 + j0.125
Ztotal = 30 + j20.125 Ω
Step 4: Use Ohm’s Law to find the current flowing through the circuit at
t=π
200 seconds. Ohm’s Law states V(t) = I(t)Z, where V(t) is the voltage
across the circuit at time t,I(t) is the current flowing through the circuit at
time t, and Zis the total impedance of the circuit.
I(t) = V(t)
Ztotal
=24 sin(100t)
30 + j20.125
Step 5: Substitute t=π
200 seconds into the expression for current.
Iπ
200=24 sin 100 ×π
200
30 + j20.125 =24 sin π
2
30 + j20.125 =24
√302+ 20.1252
∠arctan 20.125
30 A
Question 16
Question
A resistor with a resistance of 240 Ω is connected to a battery with a voltage of
12 V. Calculate the current flowing through the resistor.
Solution
Let’s use Ohm’s Law, which states that the current passing through a conduc-
tor is directly proportional to the voltage across the conductor and inversely
proportional to the resistance of the conductor. Mathematically, Ohm’s Law is
represented as: 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 values are: - Resistance R= 240 Ω, - Voltage V= 12 V.
10
Step 2: We can rearrange Ohm’s Law to solve for current I:
I=V
R
Step 3: Substitute the values of Vand Rinto the formula:
I=12 V
240 Ω
Step 4: Calculate the current passing through the resistor:
I=1
20 A= 0.05 A
Therefore, the current flowing through the resistor is 0.05 A.
Question 17
Question
A wire of resistivity ρand length Lis connected to a voltage source of Vvolts,
creating a current Ithrough it. If the resistance of the wire is R, prove that
Ohm’s Law holds true for the wire, i.e., V=IR.
Solution
To prove Ohm’s Law for the wire, we will first calculate the resistance of the
wire and then show that V=IR.
Step 1: Calculate the resistance of the wire. The resistance of a wire is
given by the formula R=ρL
A, where ρis the resistivity of the material, Lis the
length of the wire, and Ais the cross-sectional area of the wire. Since the wire
is assumed to be uniform, we can write A= constant.
Step 2: Write Ohm’s Law for the wire. Ohm’s Law states that the volt-
age across a conductor is directly proportional to the current passing through
it, with the constant of proportionality being the resistance of the conductor.
Mathematically, this is represented as V=IR.
Step 3: Substitute the expression for resistance. Substitute the expression
for the resistance of the wire (R=ρL
A) into Ohm’s Law to get:
V=IρL
A
Step 4: Simplify the expression. Since Ais a constant for the uniform wire,
we can rewrite the equation as:
V=IρL
A=IR
Step 5: Conclusion. Therefore, we have shown that Ohm’s Law holds true
for the wire, as V=IR.
11
Question 18
Question
A resistor has a resistance of 10 Ω. If a current of 2 A passes through the
resistor, what is the voltage drop across it?
Solution
Let’s use Ohm’s Law, which states that the voltage drop (V) across a resistor
is equal to the product of the current (I) passing through it and the resistance
(R) of the resistor, i.e., V=IR.
Step 1: Given that the resistance R= 10 Ω and the current I= 2 A, we
can calculate the voltage drop V.
Voltage drop, V=I×R
V= 2 A ×10Ω
V= 20 V
Therefore, the voltage drop across the resistor is 20 V.
Question 19
Question
A resistor with a resistance of 30 Ω is connected to a battery with a voltage of
12 V. Calculate the current passing through the resistor.
Solution
Let’s 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.
Step 1: Write down Ohm’s Law formula.
V=IR
Step 2: Plug in the given values.
12 = I×30
Step 3: Solve for I.
I=12
30
I= 0.4 A
Therefore, the current passing through the resistor is 0.4 Amperes.
12
Question 20
Question
A resistor with resistance R= 10 Ω is connected to a power supply that delivers
a current of I= 2 A. 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 current (I) flowing through it multiplied by the resistance (R)
of the resistor. Mathematically, Ohm’s Law is represented as:
V=I×R
Step 2: Given that R= 10 Ω and I= 2 A, we can substitute these values
into Ohm’s Law to find the voltage drop across the resistor:
V= 2 A ×10 Ω
Step 3: Perform the calculation to find the voltage drop:
V= 20 V
Step 4: Therefore, the voltage drop across the resistor is 20 V .
Question 21
Question
An electrical circuit consists of a resistor with resistance R= 50 Ω connected to
a voltage source with voltage V= 120 V. Calculate the current flowing through
the circuit.
Solution
To calculate the current flowing through the circuit, we can use Ohm’s Law,
which states that the current Iin a circuit is equal to the voltage Vacross the
circuit divided by the resistance Rof the circuit.
Step 1: Write down Ohm’s Law formula:
I=V
R
Step 2: Substitute the given values into the formula:
I=120 V
50 Ω
13
Step 3: Calculate the current:
I=120
50
I= 2.4 A
Therefore, the current flowing through the circuit is 2.4 A.
Question 22
Question
A resistor with a resistance of 8 Ω is connected to a 12 V battery. Calculate the
current flowing through the resistor.
Solution
Let’s use Ohm’s Law, which states that the current flowing through a resistor is
equal to the voltage across the resistor divided by the resistance of the resistor.
Mathematically, Ohm’s Law can be expressed as:
I=V
R
where: - Iis the current flowing through the resistor, - Vis the voltage across
the resistor, - Ris the resistance of the resistor.
Step 1: Given values: - Voltage, V= 12 V - Resistance, R= 8 Ω
Step 2: Substitute the given values into Ohm’s Law:
I=12
8
Step 3: Simplify the expression to find the current:
I= 1.5 A
Step 4: Therefore, the current flowing through the resistor is 1.5 A.
Question 23
Question
A resistor with resistance R1= 10 Ω is connected in series with a variable
resistor R2across a potential difference of V= 20 V. If the total current in the
circuit is 3 A, determine the resistance of R2.
14
Solution
Let’s denote the resistance of R2as R2and the total resistance in the circuit as
Rtotal. We can use Ohm’s Law, V=I·R, to determine the resistance of R2.
Step 1: Determine the total resistance in the circuit. The total resistance
in a series circuit is the sum of the individual resistances. So,
Rtotal =R1+R2= 10 Ω + R2Ω = (10 + R2) Ω
Step 2: Calculate the total resistance using Ohm’s Law. Given that the
potential difference V= 20 V and the total current I= 3 A, we can use Ohm’s
Law to find the total resistance:
Rtotal =V
I=20 V
3 A = 6.6 Ω
Step 3: Set up the equation to solve for R2. Since the total resistance is
equal to 10 + R2= 6.6, we can write the equation as:
10 + R2= 6.6
Step 4: Solve for R2. Subtracting 10 from both sides gives:
R2= 6.6−10 = −3.3 Ω
Therefore, the resistance of R2is −3.3 Ω.
Question 24
Question
A resistor with a resistance of 12 Ω is connected to a battery with a voltage of
48 V. Determine 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.
Step 1: Write down Ohm’s Law: I=V
R, where Iis the current, Vis the
voltage, and Ris the resistance.
Step 2: Substitute the given values into Ohm’s Law: I=48
12 .
Step 3: Calculate the current passing through the resistor: I= 4 A.
Therefore, the current passing through the resistor is 4 A.
Question 25
Question
A resistor with a resistance of 100 Ω is connected to a 12 V battery. Calculate
the current flowing through the resistor.
15
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. Mathematically, this can be written as:
I=V
R
Step 1: Given that the voltage across the resistor is 12 V and the resistance
of the resistor is 100 Ω, we can substitute these values into Ohm’s Law to find
the current:
I=12 V
100 Ω
Step 2: Simplifying the expression gives:
I= 0.12 A
Therefore, the current flowing through the resistor is 0.12 A.
Question 26
Question
A circuit consists of a battery with a voltage of 12 V connected to a resistor
with a resistance of 8 Ω. Calculate the current flowing through the circuit.
Solution
Step 1: Recall Ohm’s Law which states V=IR, where Vis the voltage, Iis
the current, and Ris the resistance.
Step 2: Substitute the given values into Ohm’s Law: 12 = I×8.
Step 3: Solve for the current, I:
I=12
8= 1.5 A
Step 4: Therefore, the current flowing through the circuit is 1.5 A.
Question 27
Question
A circuit consists of a resistor with a resistance of 50 Ω and a voltage source
with a potential difference of 100 V. If a current of 2 Aflows through the circuit,
what is the power dissipated by the resistor?
16
Solution
Step 1: Recall 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.
V=IR
where Vis the voltage, Iis the current, and Ris the resistance.
Step 2: Given that V= 100 V,I= 2 A, and R= 50 Ω, we can use Ohm’s
Law to find the power dissipated by the resistor.
P=IV
Step 3: Substitute the values of Iand Vinto the formula for power:
P= (2 A)(100 V)
Step 4: Calculate the power dissipated by the resistor:
P= 200 W
Therefore, the power dissipated by the resistor in the circuit is 200 W.
Question 28
Question
A resistor with resistance R= 40Ω is connected to a battery with emf E= 12V.
Find the current passing through the circuit.
Solution
Step 1: Recall Ohm’s Law, which states that the current passing through a
resistor is given by I=E
R, where: - Iis the current passing through the resistor,
-Eis the emf of the battery, and - Ris the resistance of the resistor.
Step 2: Substitute the given values into Ohm’s Law:
I=12 V
40 Ω
Step 3: Calculate the current passing through the circuit:
I=3
10 A= 0.3A
Step 4: Therefore, the current passing through the circuit is 0.3A.
17
Question 29
Question
A circuit consists of a resistor with a resistance of 30 Ω and a battery with an
electromotive force of 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 V=IR, where Vis the voltage, Iis the current, and Ris the
resistance. In this case, we are given V= 12 V and R= 30 Ω. Let’s substitute
these values into Ohm’s Law to find the current I.
Step 1: Write down Ohm’s Law equation:
V=IR
Step 2: Substitute the given values:
12 = I×30
Step 3: Solve for the current I:
I=12
30 = 0.4 A
Step 4: Conclusion: The current flowing through the circuit is 0.4 A.
Question 30
Question
A resistor with resistance Ris connected to a battery, producing a current I
in the circuit. The power dissipated by the resistor is given by the equation
P=I2R. If the resistance Ris increased by a factor of 5, by what factor does
the power dissipated by the resistor change?
Solution
Let’s denote the original power dissipated by the resistor as Poriginal and the
new power dissipated by the resistor as Pnew.
Step 1: Express the original power dissipated in terms of Iand R.
Poriginal =I2R
Step 2: Find the new power dissipated when the resistance is increased by
a factor of 5. If the resistance Ris increased by a factor of 5, the new resistance
Rnew is 5R. Therefore, the new power dissipated is:
Pnew =I2·5R= 5(I2R)
18
Step 3: Determine the factor by which the power dissipated by the resistor
changes. The factor by which the power dissipated by the resistor changes is:
Pnew
Poriginal
=5(I2R)
I2R= 5
So, when the resistance Ris increased by a factor of 5, the power dissipated
by the resistor changes by a factor of 5.
Question 31
Question
A resistor with a resistance of 10 Ω is connected to a battery with a voltage of
24 V. Calculate the current flowing through the circuit.
Solution
Step 1: Recall Ohm’s Law, which states that the current (I) flowing through a
conductor between two points is directly proportional to the voltage (V) across
the two points and inversely proportional to the resistance (R) of the conductor.
Mathematically, Ohm’s Law can be expressed as V=IR, where Iis the current
in the circuit, Vis the voltage across the circuit, and Ris the resistance of the
circuit.
Step 2: Given that the voltage Vis 24 V and the resistance Ris 10 Ω, we
can use Ohm’s Law to find the current Iwith the formula I=V
R.
Step 3: Substitute V= 24 V and R= 10 Ω into the formula I=V
R.
I=24 V
10 Ω
Step 4: Perform the division to find the current I.
I= 2.4 A
Step 5: Therefore, the current flowing through the circuit is 2.4 A.
Question 32
Question
A resistor has a resistance of 20 Ω. If a current of 0.5 A flows through it, what
is the voltage drop across the resistor?
19
Solution
To find the voltage drop across the resistor, we can use Ohm’s Law, which states
that V=IR, where Vis the voltage drop across the resistor, Iis the current
flowing through the resistor, and Ris the resistance of the resistor.
Step 1: Given values: Resistance, R= 20Ω
Current, I= 0.5 A
Step 2: Apply Ohm’s Law to find the voltage drop:
V=IR
V= 0.5 A ×20 Ω
V= 10 V
The voltage drop across the resistor is 10 V.
Question 33
Question
A resistor with resistance R= 10 Ω is connected to a battery that provides
a potential difference V= 50 V. Calculate the current flowing through the
resistor.
Solution
Let’s use Ohm’s Law, which states that V=IR, where Vis the potential
difference (voltage), Iis the current, 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 the current I:
I=V
R
Step 3: Substitute the given values V= 50 V and R= 10 Ω into the
equation:
I=50 V
10 Ω
Step 4: Calculate the current:
I= 5 A
Step 5: Therefore, the current flowing through the resistor is 5 A .
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Question 34
Question
A resistor with resistance R= 200 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 calculate the current flowing through the resistor.
Step 1: Recall Ohm’s Law: 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 2: Substitute the given values into Ohm’s Law:
12 = I·200
Step 3: Solve for I:
I=12
200 = 0.06 A
Step 4: Therefore, the current flowing through the resistor is 0.06 A.
Question 35
Question
A resistor with resistance R= 50 Ω is connected to a voltage source that pro-
vides V= 100 V across the resistor. Calculate the current passing through the
resistor.
Solution
Step 1: Recall Ohm’s Law, which states that the current passing through a
resistor is given by I=V
R, where Iis the current, Vis the voltage across the
resistor, and Ris the resistance of the resistor.
Step 2: Substitute the given values into the formula: I=100 V
50 Ω .
Step 3: Calculate the current passing through the resistor: I= 2 A.
Therefore, the current passing through the resistor is 2 A.
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