PHYS 231 - UNIVERSITY PHYSICS I Ohm’s Law and its applications Question Bank Set 2

PHYS 231 - UNIVERSITY PHYSICS I

- Ohm’s Law and its applications

Question Bank - Set 2

Liberty University

Question 1

Question

A circuit consists of a resistor with a resistance of 10 Ω, a capacitor with a

capacitance of 0.1 F, and a battery with an electromotive force of 12 V. The

switch is closed at time t= 0. Calculate the current in the circuit at t= 0,

t=∞, and at any time t.

Solution

Step 1: Find the current at t= 0. The current at t= 0 is given by Ohm’s Law:

I(t= 0) = V

Given: Resistance, R= 10Ω Voltage, V= 12V

I(t= 0) = 12

10 = 1.2A

So, the current in the circuit at t= 0 is 1.2 A.

Step 2: Find the current at t=∞. At t=∞, the capacitor is fully charged

and acts like an open circuit. Therefore, the current at t=∞will be 0 A.

Step 3: Find the current at any time t. The total current in the circuit at any

time tis the sum of the currents through the resistor and capacitor. The current

through a capacitor is given by IC(t) = CdVC

dt , where Cis the capacitance and

VCis the voltage across the capacitor.

The current through a resistor is given by Ohm’s Law: IR(t) = VR(t)

R, where

VRis the voltage across the resistor.

Since the sum of the voltages across the resistor and capacitor should equal

the voltage of the battery: V=VR+VC

Now, differentiate the above equation with respect to time, t:dV

dt =dVR

dt +

dVC

But we know that VR=IR and VC=Q

C, where Qis the charge on the

capacitor, Iis the current in the circuit, and Cis the capacitance.

So, we have: dV

dt =RdI

dt +1

To find I(t), we need to solve this differential equation with the initial con-

dition I(t= 0) = 1.2A. This solution involves calculus and beyond the scope of

this problem due to its complexity.

Question 2

Question

A circuit consists of a battery with an EMF of 9V and an internal resistance of

3Ωconnected in series with a resistor of 6Ω. Find the current flowing through

the circuit.

Solution

Step 1: Calculate the total resistance in the circuit. The total resistance Rtotal

in the circuit is the sum of the internal resistance rand the resistance Rof the

external resistor.

Rtotal =r+R= 3Ω + 6Ω = 9Ω

Step 2: Apply Ohm’s Law to find the total current Itotal. Ohm’s Law states

that the current Iflowing through a circuit is given by the ratio of voltage V

applied across the circuit to the total resistance Rtotal in the circuit.

Itotal =V

Rtotal

Itotal =9V

9Ω = 1A

Step 3: Determine the current flowing through the circuit. The current

flowing through the circuit is equal to the total current Itotal calculated in step

I= 1A

Therefore, the current flowing through the circuit is 1A.

Question 3

Question

A circuit consists of a resistor with resistance R= 10 Ω and an electric potential

difference of V= 12 V. Determine the current flowing through the circuit.

Solution

Step 1: Recall Ohm’s Law, which states that the current (I) flowing through

a resistor is directly proportional to the voltage (V) across the resistor and

inversely proportional to the resistance (R) of the resistor. Mathematically,

Ohm’s Law is expressed as: V=IR.

Step 2: Rearrange Ohm’s Law to solve for the current I:I=V

Step 3: Substitute the given values V= 12 Vand R= 10 Ω into the equation

for current: I=12 V

10 Ω .

Step 4: Perform the division to find the current: I= 1.2A.

Therefore, the current flowing through the circuit is 1.2A.

Question 4

Question

A cylindrical wire made of copper has a resistance of 0.4 ohms. If the diameter

of the wire is doubled, what will be the new resistance of the wire? Assume the

resistivity of copper remains constant.

Solution

Step 1: Find the original cross-sectional area of the wire using the formula for

the area of a circle: A=πr2, where ris the radius of the wire. Given that the

wire is cylindrical, the diameter is doubled, so the radius will also be doubled.

Thus, the original radius r=d

2, where dis the original diameter.

Step 2: Calculate the original cross-sectional area.

A=π(d

2)2

=π(1

4d2)=π

4d2

Step 3: Find the new cross-sectional area after doubling the diameter. The

new radius r′= 2r= 2 ×d

2=d.

A′=π(d)2=πd2

Step 4: Use the formula for resistance Rin terms of resistivity ρ, length L,

and cross-sectional area A:

R=ρL

Step 5: Since the resistivity ρand the length Lof the wire remain constant,

the resistance Ris inversely proportional to the cross-sectional area A.

R′

R=A

A′=

4d2

πd2=1

Step 6: Find the new resistance of the wire.

R′=1

4R=1

4×0.4 Ω = 0.1 Ω

Therefore, the new resistance of the wire after doubling the diameter is 0.1

ohms.

Question 5

Question

A circuit consists of three resistors connected in series. The resistances of the

three resistors are R1= 5 Ω,R2= 7 Ω, and R3= 9 Ω. A battery of emf

E= 24 Vis connected to the circuit. Calculate: (a) The total resistance of the

circuit. (b) The current flowing through the circuit. (c) The potential difference

across each resistor.

Solution

(a) To find the total resistance of the circuit, we sum the individual resistances

of the resistors in series:

Rtotal =R1+R2+R3

Rtotal = 5 Ω + 7 Ω + 9 Ω

Rtotal = 21 Ω

Step 1: The total resistance of the circuit is 21 Ω.

(b) Using Ohm’s Law V=IR, we can find the current flowing through the

circuit:

I=E

Rtotal

=24 V

21 Ω

I≈1.14 A

Step 2: The current flowing through the circuit is approximately 1.14 A.

V=IR for each resistor:

For R1:

V1=I·R1= 1.14 A·5 Ω

V1≈5.71 V

For R2:

V2=I·R2= 1.14 A·7 Ω

V2≈7.98 V

For R3:

V3=I·R3= 1.14 A·9 Ω

V3≈10.26 V

Step 3: The potential difference across R1,R2, and R3are approximately

5.71 V,7.98 V, and 10.26 V, respectively.

Question 6

Question

A copper wire with a resistance of 10 Ω is connected in series with a carbon

resistor whose resistance varies with temperature as R(T) = R0(1 + αT ). If the

temperature coeﬀicient of resistance for the carbon resistor is α= 2 ×10−3K−1

and the initial resistance R0= 5 Ω, find the total resistance of the circuit at a

temperature of 100 ◦C.

Solution

Step 1: The total resistance of the circuit can be found by summing the in-

dividual resistances in series. Step 2: The resistance of the copper wire is

given as 10 Ω. Step 3: The resistance of the carbon resistor at 100 ◦C is

given by R(100) = 5(1 + 2 ×10−3×100). Step 4: Simplifying, we have

R(100) = 5(1 + 0.2) = 6 Ω. Step 5: Therefore, the total resistance of the

circuit at 100 ◦C is 10 Ω + 6 Ω = 16 Ω.

Question 7

Question

A circuit consists of a 12 V battery connected to two resistors in series. The

first resistor has a resistance of 4 Ωand the second resistor has a resistance of

6Ω. Find:

1. The current flowing through the circuit.

2. The voltage drop across each resistor.

Solution

1. To find the current flowing through the circuit, we can use Ohm’s Law

V=IR, where Vis the voltage, Iis the current, and Ris the total resistance

of the circuit. The total resistance Rtotal in a series circuit is the sum of the

individual resistances. Thus, Rtotal = 4 Ω + 6 Ω = 10 Ω. Given that V= 12 V,

we can rearrange Ohm’s Law to solve for the current I:

I=V

Rtotal

=12 V

10 Ω = 1.2A

2. To find the voltage drop across each resistor, we can use Ohm’s Law

V=IR.

• For the first resistor with R1= 4 Ω:

V1=I·R1= 1.2A·4 Ω = 4.8V

• For the second resistor with R2= 6 Ω:

V2=I·R2= 1.2A·6 Ω = 7.2V

Therefore,

1. The current flowing through the circuit is 1.2A.

2. The voltage drop across the 4 Ωresistor is 4.8V and the voltage drop

across the 6 Ωresistor is 7.2V.

Question 8

Question

A 10 Ωresistor is connected in series with a 5 Ωresistor across a 15 V battery.

Calculate the current flowing through each resistor.

Solution

Step 1: Calculate the total resistance in the circuit.

Rtotal =R1+R2

Rtotal = 10Ω + 5Ω

Rtotal = 15Ω

Step 2: Use Ohm’s Law to find the total current flowing in the circuit.

V=Itotal ·Rtotal

15 V=Itotal ·15 Ω

Itotal =15 V

15 Ω

Itotal = 1 A

Step 3: Calculate the current flowing through each resistor using the total

current.

I1=V

I1=15 V

10 Ω

I1= 1.5A

Step 4: Calculate the current flowing through the second resistor.

I2=V

I2=15 V

5 Ω

I2= 3 A

Therefore, the current flowing through the 10 Ωresistor is 1.5 A and the

current flowing through the 5 Ωresistor is 3 A.

Question 9

Question

A circuit consists of a resistor with resistance R, a capacitor with capacitance

C, and an AC voltage source of amplitude V0and frequency ω. The current

in the circuit is given by the expression I(t) = I0sin(ωt +ϕ), where I0is the

amplitude of the current and ϕis the phase angle. Suppose the phase angle

is such that tan(ϕ) = ωRC

1−ω2R2C2. Determine the impedance Zof the circuit in

terms of Rand C.

Solution

Step 1: By Ohm’s Law, the impedance Zof the circuit is given by the ratio

of the amplitude of the voltage across the circuit (V0) to the amplitude of the

current (I0):

Z=V0

Step 2: Since V0=I0Z, we need to find the expression for I0.

Step 3: The current is given by I(t) = I0sin(ωt +ϕ). This can also be

written in the form I(t) = I0sin(ϕ) cos(ωt) + I0cos(ϕ) sin(ωt).

Step 4: Comparing this expression to the general form I(t) = I0sin(ωt +ϕ),

we have sin(ϕ) = I0and cos(ϕ) = I0sin(ϕ).

Step 5: We can now express sin(ϕ)and cos(ϕ)in terms of ϕand find I0:

I0= tan(ϕ)

I0=sin(ϕ)

cos(ϕ)=ωRC

1−ω2R2C2

Step 6: Finally, we can find the impedance Z:

Z=V0

=V0

ωRC

1−ω2R2C2

=(V0)(1 −ω2R2C2)

ωRC

Therefore, the impedance Zof the circuit in terms of Rand Cis (V0)(1 −ω2R2C2)

ωRC .

Question 10

Question

A circuit consists of a resistor, an inductor, and a capacitor connected in series

with an AC voltage source. The impedance of the circuit is given by the equation

Z=R+j(XL−XC), where R= 10 Ω,XL= 20 Ω, and XC= 15 Ω. Calculate

the amplitude of the current flowing through the circuit if the AC voltage source

has an amplitude of 5V and a frequency of 50 Hz.

Solution

Step 1: Calculate the total impedance of the circuit using the given values.

Z=R+j(XL−XC)

Z= 10 + j(20 −15)

Z= 10 + j5

Z=√102+ 52∠arctan (5

10)

Z=√125∠arctan(0.5)

Z≈11.18 Ω∠26.57◦

Step 2: Calculate the amplitude of the current using Ohm’s Law.

V=I·Z

I=V

Given V= 5 V and Z= 11.18∠26.57◦Ω,

I=5

11.18∠26.57◦

I=5

11.18∠−26.57◦

I≈0.447 A∠−26.57◦

Therefore, the amplitude of the current flowing through the circuit is 0.447 A.

Question 11

Question

A circuit consists of a resistor with resistance R, a capacitor with capacitance C,

and an inductor with inductance Lconnected in series to an AC voltage source

with angular frequency ω. The voltage across the resistor leads the current by

an angle ϕ, the voltage across the capacitor lags the current by an angle θ, and

the voltage across the inductor leads the current by an angle λ. Show that the

impedances of the resistor, capacitor, and inductor are given respectively by:

ZR=R, ZC=1

iωC , ZL=iωL

and derive an expression for the phase angle ϕin terms of θand λ.

Solution

Step 1: Impedance of the Resistor The impedance of a resistor is simply the

resistance itself, so:

ZR=R

Step 2: Impedance of the Capacitor The impedance of a capacitor is given

by:

ZC=1

iωC

Step 3: Impedance of the Inductor The impedance of an inductor is given

by:

ZL=iωL

Step 4: Deriving the Expression for Phase Angle ϕFrom the given informa-

tion, we have:

Voltage across resistor: VR=IZR=IReiϕ

Voltage across capacitor: VC=IZC=I1

iωC eiθ

Voltage across inductor: VL=IZL=I(iωL)eiλ

The total voltage across the circuit is the sum of the voltages across the

resistor, capacitor, and inductor:

Vtotal =VR+VC+VL=IReiϕ +I1

iωC eiθ +I(iωL)eiλ

Comparing the total voltage to the current Imultiplied by the total impedance

Ztotal, we can write:

Vtotal =IZtotaleiϕ

Equating the expressions for Vtotal gives:

Ztotal =R+1

iωC +iωL =ZR+ZC+ZL

Therefore,

Ztotal =R+1

iωC +iωL

Comparing the real and imaginary parts of Ztotal to the real and imaginary

parts of ZR,ZC, and ZL, we can find the condition for the phase angle ϕin

terms of θand λ.

Question 12

Question

A circuit consists of a 12 V battery connected in series with a resistor of resis-

tance 4 Ω. Calculate the current flowing through the circuit.

Solution

Step 1: Recall Ohm’s Law, which states that the current Iflowing through

a resistor is given by the equation I=V

R, where Vis the voltage across the

resistor and Ris the resistance of the resistor.

Step 2: We are given that the voltage Vis 12 V and the resistance Ris 4 Ω.

Substituting these values into Ohm’s Law, we get:

I=12 V

4 Ω

Step 3: Simplifying the expression, we find:

I= 3 A

Step 4: Therefore, the current flowing through the circuit is 3 A.

Question 13

Question

A circuit consists of a resistor with resistance R, a capacitor with capacitance

C, and an inductor with inductance L, all connected in series to an AC voltage

source. The impedance of the circuit is given by Z=R+j(ωL −1

ωC ), where

ωis the angular frequency of the AC source. If the impedance of the circuit is

Z= 8 + j6 Ω, determine the values of R,L, and C.

Solution

Step 1: We are given that the impedance of the circuit is Z= 8 + j6 Ω. Com-

paring this with the general form Z=R+j(ωL −1

ωC ), we can write:

R= 8 Ω

ωL −1

ωC = 6

Step 2: We know that for an AC circuit with angular frequency ω, the

reactance of the inductor is ωL and the reactance of the capacitor is 1

ωC . From

the given impedance, we can identify the values of R,L, and Cas follows:

R= 8 Ω

ωL = 6

ωC = 0

Step 3: Since the reactance of the capacitor is 0, this implies that the term

ωC is 0. Therefore, ωC =∞or C= 0. However, physically it is not possible

for the capacitance to be 0. Therefore, the correct interpretation is that the

impedance of the capacitor is much larger than the impedance of the inductor:

ωC =∞=⇒C= 0

Step 4: Now, we can substitute R= 8 and C= 0 into the equation ωL = 6

to solve for L:

ωL = 6 =⇒ω·L= 6 =⇒L=6

Step 5: Since we are not given a specific value for ω, the values of R,L, and

Ccan only be determined relative to ω. Therefore, the values of R,L, and C

in terms of ωare:

R= 8 Ω

L=6

ωH

C= 0 F

Question 14

Question

A circuit consists of a resistor with resistance R, a capacitor with capacitance

C, and an inductor with inductance L, all connected in series to an AC voltage

source. The circuit operates at a frequency of ω= 100 Hz. The peak voltage

of the AC source is V0= 10 V. Given that the impedance of the circuit is

Z=√R2+ (ωL −1

ωC )2, calculate the impedance of the circuit in ohms.

Solution

Step 1: Recall that the impedance of a circuit in an AC circuit is given by the

formula Z=√R2+ (ωL −1

ωC )2.

Step 2: Substitute the given values into the formula to find the impedance:

Z=√R2+ (ωL −1

ωC )2

Z=√R2+ ((100)(L)−1

(100)(C))2

Step 3: Since we have not been given specific values for R,L, and C, we

cannot calculate the impedance further without additional information.

Question 15

Question

A circuit consists of a resistor with resistance Rconnected across a battery with

voltage V. The current flowing through the resistor is I. If the resistance of

the resistor is doubled while the battery voltage remains constant, how will the

current through the resistor change?

Solution

Let’s denote the original resistance as R, the final resistance as 2R, the original

current as I, and the battery voltage as V.

Step 1: Recall Ohm’s Law, which states that the current flowing through

a resistor is given by I=V

Rwhere Vis the voltage across the resistor and Ris

the resistance of the resistor.

Step 2: For the original circuit, the current Iis given by I=V

Step 3: Now, consider the circuit with the resistance doubled. The current

I′in this circuit is given by I′=V

2R.

Step 4: Comparing Iand I′, we see that I′=V

2R=1

2·V

R=1

2·I.

Step 5: Therefore, if the resistance of the resistor is doubled while the

battery voltage remains constant, the current through the resistor will be halved.

Question 16

Question

A circuit consists of a resistor with resistance R, a capacitor with capacitance

C, and a battery with emf Econnected in series. Initially, the capacitor is

uncharged. At t= 0, the switch is closed. After a long time t, the charge on

the capacitor is Qf. Show that the steady-state current in the circuit is given

by Iss =E

Solution

Step 1: After a long time t, the capacitor is fully charged. The potential dif-

ference across the capacitor terminals is equal to the emf of the battery, which

means VC=E.

Step 2: The current Iin the circuit can be calculated using the loop rule

(Kirchhoff’s voltage law):

E=IR +Q

where Qis the charge on the capacitor at time t. Since the capacitor is fully

charged (Q=Qf), the current in the circuit is given by Iss:

Iss =E

Therefore, the steady-state current in the circuit is Iss =E

Question 17

Question

A resistor with resistance R1= 10 Ω is connected in series with a resistor with

resistance R2= 20 Ω. The combination is connected to a 12 Vbattery. Calcu-

late: (a) the total resistance of the circuit, (b) the current passing through the

circuit, and (c) the voltage across each resistor.

Solution

(a) To find the total resistance of the circuit in series, we simply add the resis-

tances of the two resistors:

Rtotal =R1+R2= 10 Ω + 20 Ω = 30 Ω

(b) The current passing through the circuit can be found using Ohm’s Law,

V=IR. Since the total resistance is 30 Ω and the voltage across the circuit is

12 V:

I=V

Rtotal

=12 V

30 Ω = 0.4A

The voltage across R1is:

VR1=I×R1= 0.4A×10 Ω = 4 V

The voltage across R2is:

VR2=I×R2= 0.4A×20 Ω = 8 V

Question 18

Question

A circuit consists of three resistors connected in series with a 12 V battery. The

resistors have resistance values of 4 Ω, 6 Ω, and 8 Ω. Calculate the total current

in the circuit.

Solution

Step 1: Calculate the total resistance in the circuit using Ohm’s law, which

states that the total resistance in a series circuit is the sum of the individual

resistances.

Rtotal =R1+R2+R3= 4 Ω + 6 Ω + 8 Ω = 18 Ω

Step 2: Use Ohm’s law (V=IR) to find the total current in the circuit.

Given that the battery voltage is 12 V:

I=V

Rtotal

=12 V

18 Ω = 0.67 A

Therefore, the total current in the circuit is 0.67 A.

Question 19

Question

A circuit consists of a 12 V battery connected in series with a resistor and

an unknown device. When the current through the circuit is 2 A, the voltage

across the unknown device is measured to be 8 V. Determine the resistance of

the unknown device.

Solution

Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor

is equal to the current (I) flowing through it multiplied by the resistance (R):

V=IR

Step 2: In this circuit, the voltage across the unknown device is 8 V and the

current through the circuit is 2 A. Therefore, we have:

8 = 2R

Step 3: Solve for Rby dividing both sides by 2:

R=8

2= 4 ohms

Step 4: Therefore, the resistance of the unknown device is 4 ohms.

Question 20

Question

A resistor with a resistance of 25 Ω is connected to a battery with an emf of 12

V. If the current in the circuit is 0.4 A, determine the power dissipated by the

resistor.

Solution

Given: Resistance, R= 25 Ω EMF of the battery, ε= 12 VCurrent in the

circuit, I= 0.4A

We can determine the power dissipated by the resistor using the formula

P=I2R, where Pis the power, Iis the current, and Ris the resistance.

Step 1: Find the power dissipated by the resistor.

Power, P=I2R= (0.4A)2×25 Ω

P= 0.16 A2×25 Ω = 4 W

Therefore, the power dissipated by the resistor is 4 W.

Question 21

Question

A 10 V battery is connected in a circuit with two resistors in series. The first

resistor has a resistance of 4 Ωand the second resistor has a resistance of 6 Ω.

Determine the current passing through each 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, and Ris

the resistance.

Step 2: To find the total resistance in the circuit, we sum the resistances in

series. The total resistance, Rtotal, is given by Rtotal =R1+R2.

Step 3: Substitute the given values into the equation to find the total resis-

tance:

Rtotal = 4 Ω + 6 Ω = 10 Ω

Step 4: Now, use Ohm’s Law to find the total current passing through the

circuit. The total current, Itotal, is given by Itotal =V

Rtotal .

Itotal =10 V

10 Ω = 1 A

Step 5: Since the resistors are in series, the total current passing through

the circuit is the same as the current passing through each resistor.

Step 6: Substitute the total current value into the equation for each resistor

to find the current passing through each: - For the first resistor with 4 Ω:

I1=10 V

4 Ω = 2.5A

- For the second resistor with 6 Ω:

I2=10 V

6 Ω ≈1.67 A

Step 7: Therefore, the current passing through the 4 Ωresistor is 2.5 A and

the current passing through the 6 Ωresistor is approximately 1.67 A.

Question 22

Question

A circuit consists of a battery with emf E= 12 V and internal resistance r= 2

Ω, connected in series with a resistor of resistance R= 8 Ω. Determine the

current in the circuit and the power dissipated in the resistor.

Solution

Step 1: Calculate the total resistance in the circuit. The total resistance Rtotal

in the circuit is the sum of the internal resistance rand the resistance Rof the

resistor:

Rtotal =r+R= 2 Ω + 8 Ω = 10 Ω

Step 2: Calculate the current in the circuit using Ohm’s Law, V=IR. The

total voltage Vacross the circuit is equal to the emf of the battery, E.

E=IRtotal

I=E

Rtotal

=12 V

10 Ω = 1.2A

Step 3: Calculate the power dissipated in the resistor using the formula

P=I2R. The power Pdissipated in the resistor is given by:

P=I2R= (1.2A)2·8 Ω = 11.52 W

Therefore, the current in the circuit is 1.2 A and the power dissipated in the

resistor is 11.52 W.

Question 23

Question

A circuit consists of a 12 V battery connected in series with three resistors:

R1= 4 Ω,R2= 6 Ω, and R3= 8 Ω. Calculate the current flowing through each

resistor in the circuit.

Solution

Step 1: Calculate the total resistance of the circuit. The total resistance (Rtotal)

of resistors in series is the sum of the individual resistances:

Rtotal =R1+R2+R3= 4 Ω + 6 Ω + 8 Ω = 18 Ω

Step 2: Calculate the total current flowing in the circuit using Ohm’s Law

V=IR. The total current flowing through the circuit can be found by:

I=V

Rtotal

I=12 V

18 Ω = 0.67 A

Step 3: Calculate the current through each resistor using Ohm’s Law I=

V/R. For R1= 4 Ω:

I1=V

=12 V

4 Ω = 3 A

For R2= 6 Ω:

I2=V

=12 V

6 Ω = 2 A

For R3= 8 Ω:

I3=V

=12 V

8 Ω = 1.5A

Therefore, the current flowing through R1is 3A, through R2is 2A, and

through R3is 1.5A.

Question 24

Question

A circuit consists of a resistor with resistance Rconnected to a battery with

emf E. When a current Iflows through the circuit, the power dissipated in the

resistor is given by P=I2R. Determine the expression for the power dissipated

in the resistor in terms of the emf E, resistance R, and current I.

Solution

Step 1: Recall the relationship between voltage, current, and resistance in a

circuit according to Ohm’s Law: V=IR, where Vis the voltage drop across

the resistor. Step 2: Since the emf Eis equal to the sum of the voltage drop

across the resistor and the external voltage, we have E=V+IR. Step 3:

Rearrange the equation to solve for V:V=E − IR. Step 4: Substitute the

expression for Vinto the power dissipation formula P=I2R:P=I2(E − IR).

Step 5: Expand the equation: P=I2E−I3R. Step 6: Therefore, the expression

for the power dissipated in the resistor in terms of the emf E, resistance R, and

current Iis P=I2E − I3R.

Question 25

Question

A 10 Ωresistor is connected in series with a 20 Ωresistor and a 5 V battery.

Calculate the current in the circuit and the power dissipated by each resistor.

Solution

Let’s denote the current in the circuit as I. We can use Ohm’s Law, V=IR, to

calculate the current in the circuit. The total resistance (Rtotal ) of the circuit

is the sum of the two resistors in series: Rtotal =R1+R2.

Step 1: Calculate the total resistance of the circuit

Using the formula forRtotal =R1+R2

Rtotal = 10 Ω + 20 Ω

Rtotal = 30 Ω

Step 2: Calculate the current flowing in the circuit

Using Ohm’s Law,V=IR

I=V

Rtotal

I=5V

30 Ω

I= 0.1667 A

So, the current in the circuit is 0.1667 A.

Step 3: Calculate the power dissipated by each resistor

Power, P=I2R

Power for the 10 Ωresistor, P= (0.1667 A)2×10 Ω

P= 0.0278 W

Power for the 20 Ωresistor, P= (0.1667 A)2×20 Ω

P= 0.0556 W

Therefore, the power dissipated by the 10 Ωresistor is 0.0278 W, and the

power dissipated by the 20 Ωresistor is 0.0556 W.

So, we have: dV

dt =RdI

dt +1

To find I(t), we need to solve this differential equation with the initial con-

dition I(t= 0) = 1.2A. This solution involves calculus and beyond the scope of

this problem due to its complexity.

Question 2

Question

A circuit consists of a battery with an EMF of 9V and an internal resistance of

3Ωconnected in series with a resistor of 6Ω. Find the current flowing through

the circuit.

Solution

Step 1: Calculate the total resistance in the circuit. The total resistance Rtotal

in the circuit is the sum of the internal resistance rand the resistance Rof the

external resistor.

Rtotal =r+R= 3Ω + 6Ω = 9Ω

Step 2: Apply Ohm’s Law to find the total current Itotal. Ohm’s Law states

that the current Iflowing through a circuit is given by the ratio of voltage V

applied across the circuit to the total resistance Rtotal in the circuit.

Itotal =V

Rtotal

Itotal =9V

9Ω = 1A

Step 3: Determine the current flowing through the circuit. The current

flowing through the circuit is equal to the total current Itotal calculated in step

I= 1A

Therefore, the current flowing through the circuit is 1A.

Question 3

Question

A circuit consists of a resistor with resistance R= 10 Ω and an electric potential

difference of V= 12 V. Determine the current flowing through the circuit.

Solution

Step 1: Recall Ohm’s Law, which states that the current (I) flowing through

a resistor is directly proportional to the voltage (V) across the resistor and

inversely proportional to the resistance (R) of the resistor. Mathematically,

Ohm’s Law is expressed as: V=IR.

Step 2: Rearrange Ohm’s Law to solve for the current I:I=V

Step 3: Substitute the given values V= 12 Vand R= 10 Ω into the equation

for current: I=12 V

10 Ω .

Step 4: Perform the division to find the current: I= 1.2A.

Therefore, the current flowing through the circuit is 1.2A.

Question 4

Question

A cylindrical wire made of copper has a resistance of 0.4 ohms. If the diameter

of the wire is doubled, what will be the new resistance of the wire? Assume the

resistivity of copper remains constant.

Solution

Step 1: Find the original cross-sectional area of the wire using the formula for

the area of a circle: A=πr2, where ris the radius of the wire. Given that the

wire is cylindrical, the diameter is doubled, so the radius will also be doubled.

Thus, the original radius r=d

2, where dis the original diameter.

Step 2: Calculate the original cross-sectional area.

A=π(d

2)2

=π(1

4d2)=π

4d2

Step 3: Find the new cross-sectional area after doubling the diameter. The

new radius r′= 2r= 2 ×d

2=d.

A′=π(d)2=πd2

Step 4: Use the formula for resistance Rin terms of resistivity ρ, length L,

and cross-sectional area A:

R=ρL

Step 5: Since the resistivity ρand the length Lof the wire remain constant,

the resistance Ris inversely proportional to the cross-sectional area A.

R′

R=A

A′=

4d2

πd2=1

Step 6: Find the new resistance of the wire.

R′=1

4R=1

4×0.4 Ω = 0.1 Ω

Therefore, the new resistance of the wire after doubling the diameter is 0.1

ohms.

Question 5

Question

A circuit consists of three resistors connected in series. The resistances of the

three resistors are R1= 5 Ω,R2= 7 Ω, and R3= 9 Ω. A battery of emf

E= 24 Vis connected to the circuit. Calculate: (a) The total resistance of the

circuit. (b) The current flowing through the circuit. (c) The potential difference

across each resistor.

Solution

(a) To find the total resistance of the circuit, we sum the individual resistances

of the resistors in series:

Rtotal =R1+R2+R3

Rtotal = 5 Ω + 7 Ω + 9 Ω

Rtotal = 21 Ω

Step 1: The total resistance of the circuit is 21 Ω.

(b) Using Ohm’s Law V=IR, we can find the current flowing through the

circuit:

I=E

Rtotal

=24 V

21 Ω

I≈1.14 A

Step 2: The current flowing through the circuit is approximately 1.14 A.

V=IR for each resistor:

For R1:

V1=I·R1= 1.14 A·5 Ω

V1≈5.71 V

For R2:

V2=I·R2= 1.14 A·7 Ω

V2≈7.98 V

For R3:

V3=I·R3= 1.14 A·9 Ω

V3≈10.26 V

Step 3: The potential difference across R1,R2, and R3are approximately

5.71 V,7.98 V, and 10.26 V, respectively.

Question 6

Question

A copper wire with a resistance of 10 Ω is connected in series with a carbon

resistor whose resistance varies with temperature as R(T) = R0(1 + αT ). If the

temperature coeﬀicient of resistance for the carbon resistor is α= 2 ×10−3K−1

and the initial resistance R0= 5 Ω, find the total resistance of the circuit at a

temperature of 100 ◦C.

Solution

Step 1: The total resistance of the circuit can be found by summing the in-

dividual resistances in series. Step 2: The resistance of the copper wire is

given as 10 Ω. Step 3: The resistance of the carbon resistor at 100 ◦C is

given by R(100) = 5(1 + 2 ×10−3×100). Step 4: Simplifying, we have

R(100) = 5(1 + 0.2) = 6 Ω. Step 5: Therefore, the total resistance of the

circuit at 100 ◦C is 10 Ω + 6 Ω = 16 Ω.

Question 7

Question

A circuit consists of a 12 V battery connected to two resistors in series. The

first resistor has a resistance of 4 Ωand the second resistor has a resistance of

6Ω. Find:

1. The current flowing through the circuit.

2. The voltage drop across each resistor.

Solution

1. To find the current flowing through the circuit, we can use Ohm’s Law

V=IR, where Vis the voltage, Iis the current, and Ris the total resistance

of the circuit. The total resistance Rtotal in a series circuit is the sum of the

individual resistances. Thus, Rtotal = 4 Ω + 6 Ω = 10 Ω. Given that V= 12 V,

we can rearrange Ohm’s Law to solve for the current I:

I=V

Rtotal

=12 V

10 Ω = 1.2A

2. To find the voltage drop across each resistor, we can use Ohm’s Law

V=IR.

• For the first resistor with R1= 4 Ω:

V1=I·R1= 1.2A·4 Ω = 4.8V

• For the second resistor with R2= 6 Ω:

V2=I·R2= 1.2A·6 Ω = 7.2V

Therefore,

1. The current flowing through the circuit is 1.2A.

2. The voltage drop across the 4 Ωresistor is 4.8V and the voltage drop

across the 6 Ωresistor is 7.2V.

Question 8

Question

A 10 Ωresistor is connected in series with a 5 Ωresistor across a 15 V battery.

Calculate the current flowing through each resistor.

Solution

Step 1: Calculate the total resistance in the circuit.

Rtotal =R1+R2

Rtotal = 10Ω + 5Ω

Rtotal = 15Ω

Step 2: Use Ohm’s Law to find the total current flowing in the circuit.

V=Itotal ·Rtotal

15 V=Itotal ·15 Ω

Itotal =15 V

15 Ω

Itotal = 1 A

Step 3: Calculate the current flowing through each resistor using the total

current.

I1=V

I1=15 V

10 Ω

I1= 1.5A

Step 4: Calculate the current flowing through the second resistor.

I2=V

I2=15 V

5 Ω

I2= 3 A

Therefore, the current flowing through the 10 Ωresistor is 1.5 A and the

current flowing through the 5 Ωresistor is 3 A.

Question 9

Question

A circuit consists of a resistor with resistance R, a capacitor with capacitance

C, and an AC voltage source of amplitude V0and frequency ω. The current

in the circuit is given by the expression I(t) = I0sin(ωt +ϕ), where I0is the

amplitude of the current and ϕis the phase angle. Suppose the phase angle

is such that tan(ϕ) = ωRC

1−ω2R2C2. Determine the impedance Zof the circuit in

terms of Rand C.

Solution

Step 1: By Ohm’s Law, the impedance Zof the circuit is given by the ratio

of the amplitude of the voltage across the circuit (V0) to the amplitude of the

current (I0):

Z=V0

Step 2: Since V0=I0Z, we need to find the expression for I0.

Step 3: The current is given by I(t) = I0sin(ωt +ϕ). This can also be

written in the form I(t) = I0sin(ϕ) cos(ωt) + I0cos(ϕ) sin(ωt).

Step 4: Comparing this expression to the general form I(t) = I0sin(ωt +ϕ),

we have sin(ϕ) = I0and cos(ϕ) = I0sin(ϕ).

Step 5: We can now express sin(ϕ)and cos(ϕ)in terms of ϕand find I0:

I0= tan(ϕ)

I0=sin(ϕ)

cos(ϕ)=ωRC

1−ω2R2C2

Step 6: Finally, we can find the impedance Z:

Z=V0

=V0

ωRC

1−ω2R2C2

=(V0)(1 −ω2R2C2)

ωRC

Therefore, the impedance Zof the circuit in terms of Rand Cis (V0)(1 −ω2R2C2)

ωRC .

Question 10

Question

A circuit consists of a resistor, an inductor, and a capacitor connected in series

with an AC voltage source. The impedance of the circuit is given by the equation

Z=R+j(XL−XC), where R= 10 Ω,XL= 20 Ω, and XC= 15 Ω. Calculate

the amplitude of the current flowing through the circuit if the AC voltage source

has an amplitude of 5V and a frequency of 50 Hz.

Solution

Step 1: Calculate the total impedance of the circuit using the given values.

Z=R+j(XL−XC)

Z= 10 + j(20 −15)

Z= 10 + j5

Z=√102+ 52∠arctan (5

10)

Z=√125∠arctan(0.5)

Z≈11.18 Ω∠26.57◦

Step 2: Calculate the amplitude of the current using Ohm’s Law.

V=I·Z

I=V

Given V= 5 V and Z= 11.18∠26.57◦Ω,

I=5

11.18∠26.57◦

I=5

11.18∠−26.57◦

I≈0.447 A∠−26.57◦

Therefore, the amplitude of the current flowing through the circuit is 0.447 A.

Question 11

Question

A circuit consists of a resistor with resistance R, a capacitor with capacitance C,

and an inductor with inductance Lconnected in series to an AC voltage source

with angular frequency ω. The voltage across the resistor leads the current by

an angle ϕ, the voltage across the capacitor lags the current by an angle θ, and

the voltage across the inductor leads the current by an angle λ. Show that the

impedances of the resistor, capacitor, and inductor are given respectively by:

ZR=R, ZC=1

iωC , ZL=iωL

and derive an expression for the phase angle ϕin terms of θand λ.

Solution

Step 1: Impedance of the Resistor The impedance of a resistor is simply the

resistance itself, so:

ZR=R

Step 2: Impedance of the Capacitor The impedance of a capacitor is given

by:

ZC=1

iωC

Step 3: Impedance of the Inductor The impedance of an inductor is given

by:

ZL=iωL

Step 4: Deriving the Expression for Phase Angle ϕFrom the given informa-

tion, we have:

Voltage across resistor: VR=IZR=IReiϕ

Voltage across capacitor: VC=IZC=I1

iωC eiθ

Voltage across inductor: VL=IZL=I(iωL)eiλ

The total voltage across the circuit is the sum of the voltages across the

resistor, capacitor, and inductor:

Vtotal =VR+VC+VL=IReiϕ +I1

iωC eiθ +I(iωL)eiλ

Comparing the total voltage to the current Imultiplied by the total impedance

Ztotal, we can write:

Vtotal =IZtotaleiϕ

Equating the expressions for Vtotal gives:

Ztotal =R+1

iωC +iωL =ZR+ZC+ZL

Therefore,

Ztotal =R+1

iωC +iωL

Comparing the real and imaginary parts of Ztotal to the real and imaginary

parts of ZR,ZC, and ZL, we can find the condition for the phase angle ϕin

terms of θand λ.

Question 12

Question

A circuit consists of a 12 V battery connected in series with a resistor of resis-

tance 4 Ω. Calculate the current flowing through the circuit.

Solution

Step 1: Recall Ohm’s Law, which states that the current Iflowing through

a resistor is given by the equation I=V

R, where Vis the voltage across the

resistor and Ris the resistance of the resistor.

Step 2: We are given that the voltage Vis 12 V and the resistance Ris 4 Ω.

Substituting these values into Ohm’s Law, we get:

I=12 V

4 Ω

Step 3: Simplifying the expression, we find:

I= 3 A

Step 4: Therefore, the current flowing through the circuit is 3 A.

Question 13

Question

A circuit consists of a resistor with resistance R, a capacitor with capacitance

C, and an inductor with inductance L, all connected in series to an AC voltage

source. The impedance of the circuit is given by Z=R+j(ωL −1

ωC ), where

ωis the angular frequency of the AC source. If the impedance of the circuit is

Z= 8 + j6 Ω, determine the values of R,L, and C.

Solution

Step 1: We are given that the impedance of the circuit is Z= 8 + j6 Ω. Com-

paring this with the general form Z=R+j(ωL −1

ωC ), we can write:

R= 8 Ω

ωL −1

ωC = 6

Step 2: We know that for an AC circuit with angular frequency ω, the

reactance of the inductor is ωL and the reactance of the capacitor is 1

ωC . From

the given impedance, we can identify the values of R,L, and Cas follows:

R= 8 Ω

ωL = 6

ωC = 0

Step 3: Since the reactance of the capacitor is 0, this implies that the term

ωC is 0. Therefore, ωC =∞or C= 0. However, physically it is not possible

for the capacitance to be 0. Therefore, the correct interpretation is that the

impedance of the capacitor is much larger than the impedance of the inductor:

ωC =∞=⇒C= 0

Step 4: Now, we can substitute R= 8 and C= 0 into the equation ωL = 6

to solve for L:

ωL = 6 =⇒ω·L= 6 =⇒L=6

Step 5: Since we are not given a specific value for ω, the values of R,L, and

Ccan only be determined relative to ω. Therefore, the values of R,L, and C

in terms of ωare:

R= 8 Ω

L=6

ωH

C= 0 F

Question 14

Question

A circuit consists of a resistor with resistance R, a capacitor with capacitance

C, and an inductor with inductance L, all connected in series to an AC voltage

source. The circuit operates at a frequency of ω= 100 Hz. The peak voltage

of the AC source is V0= 10 V. Given that the impedance of the circuit is

Z=√R2+ (ωL −1

ωC )2, calculate the impedance of the circuit in ohms.

Solution

Step 1: Recall that the impedance of a circuit in an AC circuit is given by the

formula Z=√R2+ (ωL −1

ωC )2.

Step 2: Substitute the given values into the formula to find the impedance:

Z=√R2+ (ωL −1

ωC )2

Z=√R2+ ((100)(L)−1

(100)(C))2

Step 3: Since we have not been given specific values for R,L, and C, we

cannot calculate the impedance further without additional information.

Question 15

Question

A circuit consists of a resistor with resistance Rconnected across a battery with

voltage V. The current flowing through the resistor is I. If the resistance of

the resistor is doubled while the battery voltage remains constant, how will the

current through the resistor change?

Solution

Let’s denote the original resistance as R, the final resistance as 2R, the original

current as I, and the battery voltage as V.

Step 1: Recall Ohm’s Law, which states that the current flowing through

a resistor is given by I=V

Rwhere Vis the voltage across the resistor and Ris

the resistance of the resistor.

Step 2: For the original circuit, the current Iis given by I=V

Step 3: Now, consider the circuit with the resistance doubled. The current

I′in this circuit is given by I′=V

2R.

Step 4: Comparing Iand I′, we see that I′=V

2R=1

2·V

R=1

2·I.

Step 5: Therefore, if the resistance of the resistor is doubled while the

battery voltage remains constant, the current through the resistor will be halved.

Question 16

Question

A circuit consists of a resistor with resistance R, a capacitor with capacitance

C, and a battery with emf Econnected in series. Initially, the capacitor is

uncharged. At t= 0, the switch is closed. After a long time t, the charge on

the capacitor is Qf. Show that the steady-state current in the circuit is given

by Iss =E

Solution

Step 1: After a long time t, the capacitor is fully charged. The potential dif-

ference across the capacitor terminals is equal to the emf of the battery, which

means VC=E.

Step 2: The current Iin the circuit can be calculated using the loop rule

(Kirchhoff’s voltage law):

E=IR +Q

where Qis the charge on the capacitor at time t. Since the capacitor is fully

charged (Q=Qf), the current in the circuit is given by Iss:

Iss =E

Therefore, the steady-state current in the circuit is Iss =E

Question 17

Question

A resistor with resistance R1= 10 Ω is connected in series with a resistor with

resistance R2= 20 Ω. The combination is connected to a 12 Vbattery. Calcu-

late: (a) the total resistance of the circuit, (b) the current passing through the

circuit, and (c) the voltage across each resistor.

Solution

(a) To find the total resistance of the circuit in series, we simply add the resis-

tances of the two resistors:

Rtotal =R1+R2= 10 Ω + 20 Ω = 30 Ω

(b) The current passing through the circuit can be found using Ohm’s Law,

V=IR. Since the total resistance is 30 Ω and the voltage across the circuit is

12 V:

I=V

Rtotal

=12 V

30 Ω = 0.4A

The voltage across R1is:

VR1=I×R1= 0.4A×10 Ω = 4 V

The voltage across R2is:

VR2=I×R2= 0.4A×20 Ω = 8 V

Question 18

Question

A circuit consists of three resistors connected in series with a 12 V battery. The

resistors have resistance values of 4 Ω, 6 Ω, and 8 Ω. Calculate the total current

in the circuit.

Solution

Step 1: Calculate the total resistance in the circuit using Ohm’s law, which

states that the total resistance in a series circuit is the sum of the individual

resistances.

Rtotal =R1+R2+R3= 4 Ω + 6 Ω + 8 Ω = 18 Ω

Step 2: Use Ohm’s law (V=IR) to find the total current in the circuit.

Given that the battery voltage is 12 V:

I=V

Rtotal

=12 V

18 Ω = 0.67 A

Therefore, the total current in the circuit is 0.67 A.

Question 19

Question

A circuit consists of a 12 V battery connected in series with a resistor and

an unknown device. When the current through the circuit is 2 A, the voltage

across the unknown device is measured to be 8 V. Determine the resistance of

the unknown device.

Solution

Step 1: Recall Ohm’s Law, which states that the voltage (V) across a resistor

is equal to the current (I) flowing through it multiplied by the resistance (R):

V=IR

Step 2: In this circuit, the voltage across the unknown device is 8 V and the

current through the circuit is 2 A. Therefore, we have:

8 = 2R

Step 3: Solve for Rby dividing both sides by 2:

R=8

2= 4 ohms

Step 4: Therefore, the resistance of the unknown device is 4 ohms.

Question 20

Question

A resistor with a resistance of 25 Ω is connected to a battery with an emf of 12

V. If the current in the circuit is 0.4 A, determine the power dissipated by the

resistor.

Solution

Given: Resistance, R= 25 Ω EMF of the battery, ε= 12 VCurrent in the

circuit, I= 0.4A

We can determine the power dissipated by the resistor using the formula

P=I2R, where Pis the power, Iis the current, and Ris the resistance.

Step 1: Find the power dissipated by the resistor.

Power, P=I2R= (0.4A)2×25 Ω

P= 0.16 A2×25 Ω = 4 W

Therefore, the power dissipated by the resistor is 4 W.

Question 21

Question

A 10 V battery is connected in a circuit with two resistors in series. The first

resistor has a resistance of 4 Ωand the second resistor has a resistance of 6 Ω.

Determine the current passing through each 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, and Ris

the resistance.

Step 2: To find the total resistance in the circuit, we sum the resistances in

series. The total resistance, Rtotal, is given by Rtotal =R1+R2.

Step 3: Substitute the given values into the equation to find the total resis-

tance:

Rtotal = 4 Ω + 6 Ω = 10 Ω

Step 4: Now, use Ohm’s Law to find the total current passing through the

circuit. The total current, Itotal, is given by Itotal =V

Rtotal .

Itotal =10 V

10 Ω = 1 A

Step 5: Since the resistors are in series, the total current passing through

the circuit is the same as the current passing through each resistor.

Step 6: Substitute the total current value into the equation for each resistor

to find the current passing through each: - For the first resistor with 4 Ω:

I1=10 V

4 Ω = 2.5A

- For the second resistor with 6 Ω:

I2=10 V

6 Ω ≈1.67 A

Step 7: Therefore, the current passing through the 4 Ωresistor is 2.5 A and

the current passing through the 6 Ωresistor is approximately 1.67 A.

Question 22

Question

A circuit consists of a battery with emf E= 12 V and internal resistance r= 2

Ω, connected in series with a resistor of resistance R= 8 Ω. Determine the

current in the circuit and the power dissipated in the resistor.

Solution

Step 1: Calculate the total resistance in the circuit. The total resistance Rtotal

in the circuit is the sum of the internal resistance rand the resistance Rof the

resistor:

Rtotal =r+R= 2 Ω + 8 Ω = 10 Ω

Step 2: Calculate the current in the circuit using Ohm’s Law, V=IR. The

total voltage Vacross the circuit is equal to the emf of the battery, E.

E=IRtotal

I=E

Rtotal

=12 V

10 Ω = 1.2A

Step 3: Calculate the power dissipated in the resistor using the formula

P=I2R. The power Pdissipated in the resistor is given by:

P=I2R= (1.2A)2·8 Ω = 11.52 W

Therefore, the current in the circuit is 1.2 A and the power dissipated in the

resistor is 11.52 W.

Question 23

Question

A circuit consists of a 12 V battery connected in series with three resistors:

R1= 4 Ω,R2= 6 Ω, and R3= 8 Ω. Calculate the current flowing through each

resistor in the circuit.