PHYS 231 - UNIVERSITY PHYSICS I Ohm’s Law and its applications Question Bank Set 3

PHYS 231 - UNIVERSITY PHYSICS I

- Ohm’s Law and its applications

Question Bank - Set 3

Liberty University

Question 1

Question

A circuit consists of a resistor with resistance R= 20 Ω, a battery with emf

E= 12 V, and an unknown resistor R′. When the total current in the circuit

is 0.5A, the potential difference across the unknown resistor is found to be 6V.

Determine the resistance R′of the unknown resistor.

Solution

Step 1: Recall Ohm’s Law, which states that the potential difference across a

resistor is equal to the current passing through it multiplied by the resistance

of the resistor. Mathematically, this is expressed as V=I·R.

Step 2: First, we can calculate the total resistance in the circuit using the

information given. The potential difference across the known resistor Ris equal

to E= 12 V, and the total current in the circuit is I= 0.5A. Therefore, the

resistance Rcan be calculated as R=E

Step 3: Substitute the given values to find the resistance R:

R=12

0.5= 24 Ω

Step 4: Next, we can determine the potential difference across the unknown

resistor R′. We know that V′= 6 Vand I= 0.5A. Therefore, the resistance

R′can be calculated as R′=V′

Step 5: Substitute the given values to find the resistance R′:

R′=6

0.5= 12 Ω

Step 6: Therefore, the resistance of the unknown resistor R′is 12 Ω.

Question 2

Question

A circuit consists of a resistor with resistance R= 120 Ω and a battery with

emf V= 12 V. Calculate the current flowing through the circuit and the power

dissipated by the resistor.

Solution

Step 1: Use Ohm’s Law V=IR to find the current flowing through the circuit.

I=V

I=12 V

120 Ω

I= 0.1A

Step 2: Calculate the power dissipated by the resistor using the formula

P=IV .

P=IV

P= (0.1A)×(12 V)

P= 1.2W

Therefore, the current flowing through the circuit is 0.1A and the power

dissipated by the resistor is 1.2W.

Question 3

Question

A circuit consists of a battery with an EMF of 12 V, a resistor with a resistance

of 4 Ω, and an unknown resistor. When a current of 2 A flows through the

circuit, the potential difference across the unknown resistor is 6 V. Determine

the resistance of the unknown resistor.

Solution

Step 1: Recall Ohm’s Law, which states that V=IR, where Vis the potential

difference, Iis the current, and Ris the resistance.

Step 2: Given the potential difference across the unknown resistor is 6 V

and the current flowing through the circuit is 2 A, we can use Ohm’s Law to

find the resistance of the unknown resistor:

V=IR

6V= 2 A·R

Step 3: Solve for the resistance R:

R=6V

2A= 3 Ω

Step 4: Therefore, the resistance of the unknown resistor is 3 Ω.

Question 4

Question

A circuit consists of a 25 Ωresistor connected in series with a variable resistor

Rand a 12 V battery. When the current in the circuit is 0.4 A, the potential

difference across the variable resistor Ris 7 V. Determine the value of the

variable resistor R.

Solution

Assuming the variable resistor Rhas resistance rΩ, we can use Ohm’s Law to

analyze the circuit.

Step 1: Identify the known values from the question: - Resistance of the

fixed resistor = 25 Ω - Potential difference across the variable resistor R= 7 V

- Potential difference across the fixed resistor = 12 V−7V= 5 V - Current in

the circuit = 0.4A

Step 2: Calculate the resistance of the variable resistor R: By Ohm’s Law,

the potential difference Vacross a resistor is equal to the current Ithrough the

resistor times the resistance Rof the resistor:

V=IR

For the fixed resistor:

5V= 0.4A×25 Ω

5V= 10 Ω ×0.4A

5V= 4 V

So, the potential difference across the fixed resistor should actually be 4V,

not 5V. We would then have 12 V−4V= 8 V as the potential difference across

the variable resistor R.

Step 3: Solve for the resistance rof the variable resistor R:

8V= 0.4A×r

8V= 0.4r

r=8V

0.4

r= 20 Ω

Therefore, the value of the variable resistor Rshould be 20 Ω .

Question 5

Question

A circuit consists of a resistor with resistance R, an inductor with inductance L,

and a capacitor with capacitance Cconnected in series to an AC voltage source

with a frequency of ω. At a certain frequency, the impedance of the circuit is

given by:

Z=√R2+(ωL −1

ωC )2

If the impedance of the circuit is minimized, show that the frequency ω0at

which this minimum impedance occurs is given by ω0=1

√LC .

Solution

Step 1: To find the frequency ω0at which the impedance is minimized, we need

to determine when the derivative of the impedance Zwith respect to ωis equal

to zero:

dω =1

2√R2+(ωL −1

ωC )2·2(ωL −1

ωC )(L+1

ω2C)

Step 2: Setting dZ

dω = 0:

0 = (ωL −1

ωC )(L+1

ω2C)

Step 3: This gives two possibilities:

ωL −1

ωC = 0 or L+1

ω2C= 0

Step 4: Solving ωL =1

ωC for ωgives ω0=1

√LC . This is the frequency at

which the impedance is minimized in the circuit.

Question 6

Question

A circuit consists of a resistor with resistance Rconnected in series to a capacitor

with capacitance C. When a voltage V(t) = V0sin(ωt)is applied across the

circuit, the current I(t)through the circuit is given by:

I(t) = V0ωC

√1+(ωRC)2cos(ωt +ϕ)

where ϕ= arctan(ωRC).

Find an expression for the phase difference ϕand determine under what

conditions the current I(t)is in phase with the voltage V(t).

Solution

Step 1: To find the phase difference ϕ, we first need to express ϕin terms of the

given variables ωand RC. Using the given expression for ϕ, we have:

ϕ= arctan(ωRC)

Step 2: To determine under what conditions the current I(t)is in phase

with the voltage V(t), we need to examine the expression cos(ωt +ϕ)where

ϕ= arctan(ωRC). For the current and voltage to be in phase, the phase

difference ϕshould be zero. This occurs when arctan(ωRC) = 0, which implies

ωRC = 0.

Step 3: Thus, the conditions under which the current I(t)is in phase with

the voltage V(t)are when the product of the angular frequency ωand the time

constant RC equals zero.

Question 7

Question

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

a battery. The resistor has a resistance of 10 Ω, the inductor has an inductance

of 0.5H, and the capacitor has a capacitance of 0.02 F. The battery provides an

emf of 12 Vwith a frequency of 60 Hz. Calculate the impedance of the circuit

and the current flowing through it.

Solution

Step 1: Calculate the total impedance of the circuit. The impedance of the

resistor (ZR) is equal to its resistance (R): ZR= 10 Ω.

The impedance of the inductor (ZL) is given by ZL=jωL, where Lis the

inductance and ω= 2πf is the angular frequency. Substituting the given values,

we get:

ZL=j(2π×60 Hz)(0.5H) = j60πΩ

The impedance of the capacitor (ZC) is given by ZC=1

jωC , where Cis the

capacitance. Substituting the given values:

ZC=1

j(2π×60 Hz)(0.02 F)=1

j2.4πΩ

The total impedance Ztotal of the circuit in series is the sum of the impedances

of the resistor, inductor, and capacitor:

Ztotal =ZR+ZL+ZC= 10 + j60π+1

j2.4πΩ

Step 2: Calculate the current flowing through the circuit. The current (I)

through the circuit is given by Ohm’s law I=V

Ztotal , where Vis the emf of the

battery. Substituting the given values:

I=12

10 + j60π+1

j2.4π

Therefore, the impedance of the circuit is 10+j60π+1

j2.4πΩ, and the current

flowing through it is 12

10+j60π+1

j2.4π

Question 8

Question

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

resistance Rand a 4 Ω resistor. If the current flowing through the circuit is

0.5A, what is the resistance Rof the unknown resistor?

Solution

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

resistor is directly proportional to the voltage across the resistor and inversely

proportional to the resistance of the resistor. Mathematically, this relationship

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 2: In this circuit, the total voltage provided by the battery is 12 V and

the total current flowing through the circuit is 0.5A. The voltage across the 4 Ω

resistor can be determined using Ohm’s Law as V=IR = 0.5A×4 Ω = 2 V.

Step 3: The remaining voltage (12 V−2V= 10 V) will be across the unknown

resistor with resistance R.

Step 4: Using Ohm’s Law for the unknown resistor, we have 10 V= 0.5A×R.

Solving for R, we find R=10 V

0.5A= 20 Ω.

Step 5: Therefore, the resistance of the unknown resistor Ris 20 Ω.

Question 9

Question

A 10 Ωresistor, a 20 Ωresistor, and an unknown resistor Rare connected in

series to a 24 V battery. The current in the circuit is measured to be 1.2 A.

Determine the value of the unknown resistor R.

Solution

Step 1: Recall Ohm’s Law which states V=IR, where Vis the voltage across

the resistor, Iis the current through the resistor, and Ris the resistance of the

resistor.

Step 2: Calculate the total resistance in the circuit by summing the resis-

tances of the three resistors: Rtotal =R1+R2+R.

Step 3: Use Ohm’s Law to find the total resistance: V=I·Rtotal.

Step 4: Substitute the given values into the equation to solve for Rtotal:

24 = 1.2·(10 + 20 + R).

Step 5: Simplify the equation: 24 = 1.2·(30 + R).

Step 6: Solve for R:24 = 36 + 1.2R.

Step 7: Rearrange the equation to isolate R:1.2R= 24 −36.

Step 8: Simplify the equation: 1.2R=−12.

Step 9: Solve for R:R=−12

1.2.

Step 10: Calculate the value of R:R=−10 Ω.

Step 11: The unknown resistor Rhas a value of 10 Ω.

Question 10

Question

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

to an AC voltage source. The resistor has a resistance of 10 Ω, the inductor

has an inductance of 0.05 H, and the capacitor has a capacitance of 50 µF. The

angular frequency of the AC source is 1000 rad/s. Calculate the impedance of

the circuit and the current flowing through it.

Solution

Step 1: Calculate the impedance of the circuit.

The impedance (Z) of the circuit in the given situation is the total opposition

to the flow of current and is calculated using the formula:

Z=√R2+ (XL−XC)2

where: R= resistance = 10 Ω,XL= inductive reactance = 2πfL = 2π(1000)(0.05) =

314.16 Ω,XC= capacitive reactance = 1

2πf C =1

2π(1000)(50×10−6)= 318.31 Ω.

Substitute the values into the formula:

Z=√(10)2+ (314.16 −318.31)2=√100 + (−4.15)2=√100 + 17.22 = √117.22 ≈10.83 Ω

Therefore, the impedance of the circuit is approximately 10.83 Ω.

Step 2: Calculate the current flowing through the circuit.

The current flowing through the circuit can be calculated using Ohm’s Law:

I=V

where V= voltage of the AC source.

Since the impedance of the circuit is 10.83 Ωand the voltage source has

not been specified, the current flowing through the circuit will depend on the

voltage supplied by the source.

Thus, the current flowing through the circuit is I=V

10.83 , where Vis the

voltage supplied by the AC source.

Question 11

Question

A resistor with resistance R= 10 Ω and a capacitor with capacitance C= 5 µF

are connected in series to a battery with potential difference V= 12 V. The

switch is closed at time t= 0. Calculate the current I(t)that flows through the

circuit at time tafter the switch is closed.

Solution

Step 1: The current I(t)at time tcan be calculated using Ohm’s Law and the

relationship for current in a charging capacitor:

I(t) = I0e−t

where I0is the initial current when t= 0,Ris the resistance, Cis the capaci-

tance, and tis the time.

Step 2: To find I0, we first need to find the total resistance Rtotal of the

circuit when the switch is closed:

Rtotal =R+1

C= 10 Ω + 1

5×10−6F

Step 3: Plug in the values of Rand Cto find Rtotal:

Rtotal = 10 Ω + 200000 Ω = 200010 Ω

Step 4: Now we can find I0:

I0=V

Rtotal

=12 V

200010 Ω

Step 5: Calculate I0:

I0= 6 ×10−5A

Step 6: Substitute I0,R,C, and tinto the equation for I(t):

I(t) = (6 ×10−5A)e−t

10 Ω×5×10−6F

Step 7: Simplify the equation for I(t):

I(t) = 6 ×10−5A·e−20000tA

Question 12

Question

A circuit consists of a 12 V battery, a resistor with a resistance of 4 Ωand an

unknown resistor R. When connected in series, the total current in the circuit

is 2 A. Find the resistance Rof the unknown resistor.

Solution

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

in a series circuit is the sum of the individual resistances.

Rtotal =R1+R2

In this case, R1= 4 Ω and R2=R.

Rtotal = 4 Ω + R

Step 2: Use Ohm’s Law to find the total resistance. Ohm’s Law states that

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

this case, V= 12 V and I= 2 A. Therefore, the total resistance Rtotal is given

Rtotal =V

I=12 V

2A= 6 Ω

Step 3: Set up an equation for the total resistance. Since the total resistance

Rtotal is also equal to 4 Ω + R, we have

6 Ω = 4 Ω + R

Step 4: Solve for the unknown resistance R. Subtracting 4 Ω from both sides

of the equation gives

R= 6 Ω −4 Ω = 2 Ω

Therefore, the unknown resistance Rof the circuit is 2 Ω.

Question 13

Question

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

Eand internal resistance r. When a resistor with resistance 2Ris connected in

parallel to the original resistor, the current in the circuit increases by a factor

of 3. Find the internal resistance rin terms of R.

Solution

Step 1: Determine the original current Iflowing through the circuit. Using

Ohm’s Law, the original current Ican be expressed as:

I=E

R+r

Step 2: Determine the current after the additional resistor is connected

in parallel. When the additional resistor is connected in parallel, the total

resistance in the circuit becomes R(2R)

R+2R=2R2

3R=2

3R. As the current has

increased by a factor of 3, the new current 3Ican be expressed as:

3I=E

3R+r

Step 3: Set up an equation using the two expressions for current. Equating

the two expressions for current yields:

R+r=E

3R+r

Step 4: Solve for the internal resistance rin terms of R. Solving the equation

from Step 3 for r, we have:

R+r=2

3R+r

3R+ 3r= 2R+ 3r

R= 0

Since the equation leads to a contradiction (0 = R), there must have been

a mistake in the analysis of the problem. Double-check the calculations and

assumptions made to find and correct the error.

Question 14

Question

A 12 V battery is connected in series with three resistors: one of 4 Ω, one of 6

Ω, and one of unknown resistance R. If the current passing through the circuit

is 2 A, what is the value of the unknown resistance R?

Solution

Step 1: Begin by writing down the given values and the known formula for

Ohm’s Law, which relates voltage, current, and resistance:

V=IR

where: - V= 12 V (voltage from the battery), - I= 2 A (current passing

through the circuit), - R1= 4 Ω (known resistance), - R2= 6 Ω (known

resistance), - R3=R(unknown resistance).

Step 2: Calculate the total resistance of the circuit by summing the resis-

tances of the individual resistors:

Rtotal =R1+R2+R3= 4 Ω + 6 Ω + RΩ

Step 3: Calculate the total voltage drop across the circuit using Ohm’s Law:

V=IRtotal

12 = 2 ×(4+6+R)

Step 4: Solve for the unknown resistance R:

12 = 2 ×(10 + R)

12 = 20 + 2R

2R=−8

R=−4 Ω

Step 5: Since resistance cannot be negative, it indicates that there was an

error in the calculations or assumptions made. Double-check the algebra and

revisit the calculations to find the mistake.

Question 15

Question

A circuit consists of a resistor with a resistance of 12 Ωand an unknown resistor

connected in series. When a potential difference of 24 V is applied across the

circuit, a current of 2 A flows through the circuit. Find the resistance of the

unknown resistor.

Solution

Step 1: Write down the given information. Let R1= 12 Ω be the resistance

of the known resistor, V= 24 Vbe the potential difference applied across

the circuit, I= 2 Abe the current flowing through the circuit, and Rbe the

resistance of the unknown resistor.

Step 2: Apply Ohm’s law to the circuit. By Ohm’s law, the total resistance

Rtotal in a series circuit is the sum of the individual resistances:

Rtotal =R1+R

Step 3: Find the total resistance of the circuit. Given that V=IR, rearrange

Ohm’s law to solve for the total resistance:

Rtotal =V

Substitute the known values of Vand Iinto the formula:

Rtotal =24 V

2A= 12 Ω

Step 4: Find the resistance of the unknown resistor. Since the total resistance

is the sum of the individual resistances in a series circuit,

Rtotal =R1+R

12 Ω = 12 Ω + R

Subtract 12 Ω from both sides to solve for R:

R= 12 Ω −12 Ω = 0 Ω

Therefore, the resistance of the unknown resistor is 0 Ω.

Question 16

Question

A 10 Ωresistor, a 20 Ωresistor, and a 30 Ωresistor are connected in parallel

to a 12 V battery. Calculate the total current flowing from the battery and the

power dissipated in each resistor.

Solution

Step 1: Calculate the equivalent resistance of the circuit. The equivalent resis-

tance Req of resistors connected in parallel is given by:

Req

Substitute the given values: R1= 10 Ω,R2= 20 Ω, and R3= 30 Ω.

Req

10 +1

20 +1

Req

60 +3

60 +2

Req

=11

Req =60

11 ≈5.45 Ω

Step 2: Calculate the total current flowing from the battery. Ohm’s Law

states that I=V

R, where Vis the battery voltage and Ris the equivalent

resistance. Substituting V= 12 V and R= 5.45 Ω:

I=12

5.45 ≈2.20 A

Step 3: Calculate the power dissipated in each resistor. The power Pdis-

sipated by a resistor is given by P=I2R, where Iis the current and Ris

the resistance. Substituting I= 2.20 A and the respective resistances for each

resistor: For the 10 Ωresistor:

P1= (2.20)2×10

P1≈48.40 W

For the 20 Ωresistor:

P2= (2.20)2×20

P2≈96.80 W

For the 30 Ωresistor:

P3= (2.20)2×30

P3≈145.20 W

Therefore, the total current flowing from the battery is approximately 2.20 A,

and the power dissipated in the 10 Ω, 20 Ω, and 30 Ωresistors are approximately

48.40 W, 96.80 W, and 145.20 W respectively.

Question 17

Question

A circuit consists of a resistor with resistance R= 10 Ω and an unknown battery.

When a current of I= 2 Aflows through the circuit, the power dissipated by

the resistor is P= 40 W. Determine the potential difference across the resistor.

Solution

Step 1: Recall that the power dissipated by a resistor can be calculated using

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

resistance of the resistor. Step 2: Substitute the given values P= 40 Wand

I= 2 Ainto the formula to find the resistance Rof the resistor.

40 = (2)2·R

Step 3: Solve for the resistance R.

40 = 4R

R= 10 Ω

Step 4: Now that we know the resistance R, we can use Ohm’s Law to find the

potential difference across the resistor. Ohm’s Law states that V=IR, where

Vis the potential difference, Iis the current, and Ris the resistance. Step 5:

Substitute the known values I= 2 Aand R= 10 Ω into Ohm’s Law to find the

potential difference V.

V= 2 ·10 = 20 V

Step 6: Therefore, the potential difference across the resistor is 20 V.

Question 18

Question

A 24 V battery is connected to a circuit containing three resistors in series. The

first resistor has a resistance of 5 Ω, the second resistor has a resistance of 3

Ω, and the third resistor has a resistance of 8 Ω. Calculate the current flowing

through the circuit.

Solution

Step 1: Begin by calculating the total resistance of the circuit. To find the

total resistance Rtotal of resistors in series, you simply add up the individual

resistances:

Rtotal =R1+R2+R3= 5 Ω + 3 Ω + 8 Ω = 16 Ω

Step 2: Use Ohm’s Law V=IR to find the current Iflowing through

the circuit. Given that the voltage Vsupplied by the battery is 24 V, we can

rearrange Ohm’s Law to solve for I:

I=V

Rtotal

=24 V

16 Ω = 1.5A

Therefore, the current flowing through the circuit is 1.5 A.

Question 19

Question

A circuit consists of a resistor with resistance R, a capacitor with capaci-

tance C, and an inductor with inductance Lconnected in series to an AC

voltage source with voltage V(t) = V0sin(ωt). The current in the circuit is

given by i(t) = I0sin(ωt +ϕ). Show that the impedance in the circuit is

Z=√R2+ (ωL −1

ωC )2. Given V0= 10 V, I0= 2 A, ϕ= 30◦,R= 3 Ω,

L= 0.5H, C= 0.02 F, and ω= 100 rad/s, calculate the impedance Z.

Solution

Step 1: Impedance in the circuit is given by:

Z=V(t)

i(t)

Step 2: Substitute the given values into the expression V(t) = V0sin(ωt)

and i(t) = I0sin(ωt +ϕ):

Z=V0sin(ωt)

I0sin(ωt +ϕ)

Step 3: At any instant, the total voltage drop across the circuit is equal to

the sum of the voltage drops across the resistor, inductor, and capacitor:

V(t) = I(t)R+LdI(t)

dt +Q(t)

Step 4: Since V(t) = V0sin(ωt)and I(t) = I0sin(ωt +ϕ), the above equation

can be written as:

V0sin(ωt) = I0Rsin(ωt +ϕ) + Ld

dt(I0sin(ωt +ϕ)) + Q(t)

Step 5: Simplify the equation by finding dI(t)

dt and Q(t)and substitute the

values of R,L, and C:

Z=√R2+ (ωL −1

ωC )2

Step 6: Substitute the given values of V0,I0,ϕ,R,L,C, and ωinto the

expression for Z:

Z=√32+ (100 ×0.5−1

100 ×0.02)2

Z=√9 + (50 −50)2=√9 = 3 Ω

Therefore, the impedance in the circuit is 3 Ω.

Question 20

Question

A circuit consists of a 12 V battery connected to three resistors in series: a 4

Ωresistor, a 6 Ωresistor, and an unknown resistor R. If a current of 1 A flows

through the circuit, what is the value of the unknown resistor R?

Solution

Step 1: Determine the total resistance of the circuit. The total resistance Rtotal

in a series circuit is the sum of the individual resistances:

Rtotal = 4 Ω + 6 Ω + RΩ = 10 Ω + RΩ

Step 2: Calculate the voltage drop across the resistor. Using Ohm’s Law

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

can find the voltage drop across the total resistance as:

Vtotal =I·Rtotal = 1 A×(10 Ω + RΩ)

Step 3: Apply Kirchhoff’s Voltage Law. According to Kirchhoff’s Voltage

Law, the total voltage around a closed loop must sum to zero. Therefore, the

sum of the individual voltage drops in the loop must equal the battery voltage.

Vbattery =Vtotal +VR

12 V= 1 ×(10 Ω + RΩ) + 1 ×RΩ

Step 4: Solve for the unknown resistor R.

12 V= 10 V+RV+RV

12 V= 10 V+ 2RV

2V= 2RV

R= 1 Ω

Therefore, the unknown resistor Rhas a value of 1 Ω.

Question 21

Question

A copper wire of length 2.5 m and cross-sectional area 3.0×10−6m2has a

resistance of 0.25 Ω. If a potential difference of 8.0 V is applied across the wire,

what is the current passing through it?

Solution

Step 1: We first need to find the resistance per unit length of the wire. The

resistance per unit length RLof a wire is given by the formula:

RL=R·A

where Ris the resistance of the wire, Ais the cross-sectional area, and Lis the

length of the wire.

Step 2: Substitute the given values into the formula.

RL=0.25 Ω ·3.0×10−6m2

2.5m

RL= 3.0×10−4Ω/m

Step 3: Next, we can use Ohm’s Law to find the current passing through

the wire. Ohm’s Law states that the current Iflowing through a conductor is

given by:

I=V

where Vis the potential difference across the conductor and Ris the resistance

of the conductor.

Step 4: Substitute the given potential difference and the resistance per unit

length of the wire into the Ohm’s Law equation.

I=8.0V

3.0×10−4Ω/m

Step 5: Calculate the current passing through the wire.

I=8.0V

3.0×10−4Ω/m = 2.67 ×10−2A

Therefore, the current passing through the wire is 0.027 A.

Question 22

Question

An electric circuit consists of a 12 V battery and three resistors connected

in parallel. The resistors have resistances of 4 Ω, 6 Ω, and 8 Ωrespectively.

Calculate the total current flowing through the circuit.

Solution

Step 1: Calculate the equivalent resistance of the parallel resistors. To find the

total resistance (Rtotal) of resistors connected in parallel, we use the formula:

Rtotal

Substitute R1= 4 Ω,R2= 6 Ω, and R3= 8 Ω:

Rtotal

4+1

6+1

Rtotal

12 +2

12 +1

Rtotal

=6+4+3

Rtotal

=13

Rtotal =24

13 ≈1.85 Ω

Step 2: Calculate the total current using Ohm’s Law. Ohm’s Law states

that I=V

R. In this case, V= 12 V and R=Rtotal = 1.85 Ω. Substitute these

values to calculate the total current:

I=12

1.85

I≈6.49 A

Therefore, the total current flowing through the circuit is approximately 6.49

Question 23

Question

A resistor with resistance R1= 4 Ω is connected in series with another resistor

with resistance R2= 6 Ω. The combination is then connected across a 12 V

battery. Find the current passing through each resistor.

Solution

Step 1: Calculate the total resistance of the circuit using the formula for resistors

in series:

Rtotal =R1+R2= 4 Ω + 6 Ω = 10 Ω

Step 2: Calculate the total current passing through the circuit using Ohm’s

Law, V=IR:

I=V

Rtotal

=12 V

10 Ω = 1.2A

Step 3: Since the resistors are in series, the current passing through each

resistor is the same as the total current:

I1=I2= 1.2A

Hence, the current passing through each resistor is 1.2A.

Question 24

Question

A circuit consists of a 12 V battery connected in series with three resistors: a 10

Ωresistor, a 15 Ωresistor, and a resistor R. If a current of 0.6 A flows through

the circuit, what is the value of the unknown resistance R?

Solution

Step 1: Calculate the total resistance in the circuit using Ohm’s Law.

Given resistors R1= 10 Ω,R2= 15 Ω, and total current I= 0.6A.

The total resistance in a series circuit is the sum of all individual resistances:

Rtotal =R1+R2+R

Step 2: Calculate the total resistance.

Rtotal = 10 Ω + 15 Ω + R

Rtotal = 25 Ω + R

Step 3: Use Ohm’s Law to find the total resistance.

V=IR

12 V= 0.6A×(25 Ω + R)

Step 4: Solve for the unknown resistance R.

12 V= 0.6A×(25 Ω + R)

12 V= 15 AΩ+0.6AR

12 V−15 AΩ = 0.6AR

−3V= 0.6AR

R=−3V

0.6A

R=−5 Ω

Therefore, the value of the unknown resistance Ris 5 Ω .

Question 25

Question

A circuit consists of a 12V battery connected in series with three resistors: a 4Ω

resistor, a 6Ωresistor, and an unknown resistor R. The current flowing through

the circuit is measured to be 1.5A. Determine the value of the unknown resistor

Solution

Step 1: Calculate the total resistance of the circuit using Ohm’s Law: V=IR.

Given that V= 12V and I= 1.5A, the total resistance Rtotal can be calculated

as:

Rtotal =V

I=12

1.5= 8Ω

Step 2: Determine the equivalent resistance of the circuit. Since the resistors

are in series, the equivalent resistance Req is the sum of individual resistances:

Req = 4Ω + 6Ω + R

Step 3: Set up an equation using the fact that the total resistance Rtotal is

equal to the equivalent resistance Req:

Req =Rtotal

4Ω + 6Ω + R= 8Ω

Step 4: Solve for the unknown resistor R.

10Ω + R= 8Ω

R= 8Ω −10Ω = −2Ω

Step 5: Interpretation of the result. The negative value of the resistance

indicates that there might be an error in the calculation or measurement. A

negative resistance value is not physically meaningful. Double-check the calcu-

lations and measurements to correct the mistake.

Question 2

Question

A circuit consists of a resistor with resistance R= 120 Ω and a battery with

emf V= 12 V. Calculate the current flowing through the circuit and the power

dissipated by the resistor.

Solution

Step 1: Use Ohm’s Law V=IR to find the current flowing through the circuit.

I=V

I=12 V

120 Ω

I= 0.1A

Step 2: Calculate the power dissipated by the resistor using the formula

P=IV .

P=IV

P= (0.1A)×(12 V)

P= 1.2W

Therefore, the current flowing through the circuit is 0.1A and the power

dissipated by the resistor is 1.2W.

Question 3

Question

A circuit consists of a battery with an EMF of 12 V, a resistor with a resistance

of 4 Ω, and an unknown resistor. When a current of 2 A flows through the

circuit, the potential difference across the unknown resistor is 6 V. Determine

the resistance of the unknown resistor.

Solution

Step 1: Recall Ohm’s Law, which states that V=IR, where Vis the potential

difference, Iis the current, and Ris the resistance.

Step 2: Given the potential difference across the unknown resistor is 6 V

and the current flowing through the circuit is 2 A, we can use Ohm’s Law to

find the resistance of the unknown resistor:

V=IR

6V= 2 A·R

Step 3: Solve for the resistance R:

R=6V

2A= 3 Ω

Step 4: Therefore, the resistance of the unknown resistor is 3 Ω.

Question 4

Question

A circuit consists of a 25 Ωresistor connected in series with a variable resistor

Rand a 12 V battery. When the current in the circuit is 0.4 A, the potential

difference across the variable resistor Ris 7 V. Determine the value of the

variable resistor R.

Solution

Assuming the variable resistor Rhas resistance rΩ, we can use Ohm’s Law to

analyze the circuit.

Step 1: Identify the known values from the question: - Resistance of the

fixed resistor = 25 Ω - Potential difference across the variable resistor R= 7 V

- Potential difference across the fixed resistor = 12 V−7V= 5 V - Current in

the circuit = 0.4A

Step 2: Calculate the resistance of the variable resistor R: By Ohm’s Law,

the potential difference Vacross a resistor is equal to the current Ithrough the

resistor times the resistance Rof the resistor:

V=IR

For the fixed resistor:

5V= 0.4A×25 Ω

5V= 10 Ω ×0.4A

5V= 4 V

So, the potential difference across the fixed resistor should actually be 4V,

not 5V. We would then have 12 V−4V= 8 V as the potential difference across

the variable resistor R.

Step 3: Solve for the resistance rof the variable resistor R:

8V= 0.4A×r

8V= 0.4r

r=8V

0.4

r= 20 Ω

Therefore, the value of the variable resistor Rshould be 20 Ω .

Question 5

Question

A circuit consists of a resistor with resistance R, an inductor with inductance L,

and a capacitor with capacitance Cconnected in series to an AC voltage source

with a frequency of ω. At a certain frequency, the impedance of the circuit is

given by:

Z=√R2+(ωL −1

ωC )2

If the impedance of the circuit is minimized, show that the frequency ω0at

which this minimum impedance occurs is given by ω0=1

√LC .

Solution

Step 1: To find the frequency ω0at which the impedance is minimized, we need

to determine when the derivative of the impedance Zwith respect to ωis equal

to zero:

dω =1

2√R2+(ωL −1

ωC )2·2(ωL −1

ωC )(L+1

ω2C)

Step 2: Setting dZ

dω = 0:

0 = (ωL −1

ωC )(L+1

ω2C)

Step 3: This gives two possibilities:

ωL −1

ωC = 0 or L+1

ω2C= 0

Step 4: Solving ωL =1

ωC for ωgives ω0=1

√LC . This is the frequency at

which the impedance is minimized in the circuit.

Question 6

Question

A circuit consists of a resistor with resistance Rconnected in series to a capacitor

with capacitance C. When a voltage V(t) = V0sin(ωt)is applied across the

circuit, the current I(t)through the circuit is given by:

I(t) = V0ωC

√1+(ωRC)2cos(ωt +ϕ)

where ϕ= arctan(ωRC).

Find an expression for the phase difference ϕand determine under what

conditions the current I(t)is in phase with the voltage V(t).

Solution

Step 1: To find the phase difference ϕ, we first need to express ϕin terms of the

given variables ωand RC. Using the given expression for ϕ, we have:

ϕ= arctan(ωRC)

Step 2: To determine under what conditions the current I(t)is in phase

with the voltage V(t), we need to examine the expression cos(ωt +ϕ)where

ϕ= arctan(ωRC). For the current and voltage to be in phase, the phase

difference ϕshould be zero. This occurs when arctan(ωRC) = 0, which implies

ωRC = 0.

Step 3: Thus, the conditions under which the current I(t)is in phase with

the voltage V(t)are when the product of the angular frequency ωand the time

constant RC equals zero.

Question 7

Question

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

a battery. The resistor has a resistance of 10 Ω, the inductor has an inductance

of 0.5H, and the capacitor has a capacitance of 0.02 F. The battery provides an

emf of 12 Vwith a frequency of 60 Hz. Calculate the impedance of the circuit

and the current flowing through it.

Solution

Step 1: Calculate the total impedance of the circuit. The impedance of the

resistor (ZR) is equal to its resistance (R): ZR= 10 Ω.

The impedance of the inductor (ZL) is given by ZL=jωL, where Lis the

inductance and ω= 2πf is the angular frequency. Substituting the given values,

we get:

ZL=j(2π×60 Hz)(0.5H) = j60πΩ

The impedance of the capacitor (ZC) is given by ZC=1

jωC , where Cis the

capacitance. Substituting the given values:

ZC=1

j(2π×60 Hz)(0.02 F)=1

j2.4πΩ

The total impedance Ztotal of the circuit in series is the sum of the impedances

of the resistor, inductor, and capacitor:

Ztotal =ZR+ZL+ZC= 10 + j60π+1

j2.4πΩ