PHYS 101 - ELEMENTS OF PHYSICS Ohm's Law Question Bank Set 4

PHYS 101 - ELEMENTS OF PHYSICS

- Ohm’s Law

Question Bank - Set 4

Liberty University

Question 1

Question

A wire has a resistance of 10 Ω. If a current of 2 A flows through the wire, what

is the voltage drop across the wire?

Solution

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

is equal to the product of the current (I) flowing through it and the resistance

(R) of the resistor. Mathematically, this is expressed as:

V=I×R

Step 2: Substitute the given values into the formula. We have I= 2 A and

R= 10 Ω.

V= 2 A ×10 Ω

Step 3: Calculate the voltage drop.

V= 20 V

Step 4: Therefore, the voltage drop across the wire is 20 V.

Question 2

Question

A resistor with resistance R= 100 Ω is connected to a battery with voltage

V= 12 V. Calculate the current passing through the resistor.

Solution

Ohm’s Law states that the current passing through a resistor is given by I=V

where Iis the current, Vis the voltage, and Ris the resistance of the resistor.

Step 1: Substitute the given values into Ohm’s Law to find the current

passing through the resistor.

I=V

R=12 V

100 Ω

Step 2: Calculate the current passing through the resistor.

I=12

100 =3

25 = 0.12 A

Therefore, the current passing through the resistor is 0.12 A.

Question 3

Question

A resistor of resistance R= 30 Ω is connected to a battery with voltage V=

12 V. Calculate the current flowing through the resistor.

Solution

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

resistor is given by 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=12 V

30 Ω .

Step 3: Calculate the current: I=12

30 = 0.4 A.

Step 4: Therefore, the current flowing through the resistor is 0.4 A.

Question 4

Question

A resistor has a voltage drop of 10 V across it when a current of 2 A flows

through it. Calculate the resistance of the resistor.

Solution

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

equal to the current (I) flowing through the resistor multiplied by the resistance

(R) of the resistor. Mathematically, this is expressed as:

V=IR

Step 2: Given that the voltage drop across the resistor is 10 V and the

current flowing through it is 2 A, we can substitute these values into Ohm’s

Law:

10 = 2R

Step 3: Solve for the resistance Rby dividing both sides of the equation by

R=10

R= 5 ohms

Step 4: Therefore, the resistance of the resistor is 5 ohms.

Question 5

Question

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

voltage V. If the current passing through the resistor is given by I=V2

R, find

the power dissipated in the resistor in terms of Vand R.

Solution

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

the formula P=I2R.

Step 2: Substitute the given expression for current Iinto the formula for

power:

P=V2

R2

·R

Step 3: Simplify the expression:

P=V4

R·R=V4

Step 4: Therefore, the power dissipated in the resistor is V4.

Question 6

Question

A circuit consists of a resistor with resistance R= 20Ω connected to a battery

with voltage V= 100 V. Calculate the current passing through the circuit.

Solution

Ohm’s Law states that the current passing through a circuit is proportional to

the voltage and inversely proportional to the resistance. Mathematically, it can

be expressed as I=V

R, where: - Iis the current passing through the circuit, -

Vis the voltage supplied by the battery, and - Ris the resistance in the circuit.

Step 1: Substitute the given values into Ohm’s Law equation.

I=100 V

20Ω

Step 2: Calculate the current passing through the circuit.

I=100

20 = 5 A

Therefore, the current passing through the circuit is 5 A.

Question 7

Question

A resistor has a resistance of 8 Ω and a current of 2 A flowing through it.

Determine the voltage drop across the resistor.

Solution

Ohm’s Law states that the voltage drop (V) across a resistor is equal to the

product of the current (I) flowing through it and the resistance (R) of the

resistor. Mathematically, this can be represented as:

V=I×R

Step 1: Given values are: Resistance, R= 8Ω Current, I= 2 A

Step 2: Substitute the given values into Ohm’s Law equation:

V= 2 A ×8 Ω

Step 3: Calculate the voltage drop:

V= 16 V

Therefore, the voltage drop across the resistor is 16 V.

Question 8

Question

A circuit consists of a resistor with a resistance of 150 Ω connected to a voltage

source with a voltage of 12 V. Find the current flowing through the circuit.

Solution

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

states that the current (I) flowing through a resistor is equal to the voltage (V)

across the resistor divided by the resistance (R) of the resistor. Mathematically,

Ohm’s Law is given by the equation:

I=V

Step 1: Identify the given values: The resistance (R) is 150 Ω and the

voltage (V) is 12 V.

Step 2: Substitute the known values into the formula for Ohm’s Law:

I=12 V

150 Ω

Step 3: Calculate the current flowing through the circuit:

I=12

150 = 0.08 A

Therefore, the current flowing through the circuit is 0.08 A or 80 mA.

Question 9

Question

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

The resistor has a resistance of 50 Ω, the inductor has an inductance of 0.2 H,

and the capacitor has a capacitance of 2 µF. If the frequency of the AC power

source is 100 Hz, determine the total impedance of the circuit.

Solution

Step 1: Calculate the reactance of the inductor. The reactance of an inductor

is given by XL= 2πf L, where fis the frequency and Lis the inductance.

Substituting f= 100 Hz and L= 0.2 H into the formula, we get:

XL= 2π×100 ×0.2 = 40πΩ

Step 2: Calculate the reactance of the capacitor. The reactance of a capacitor

is given by XC=1

2πf C , where fis the frequency and Cis the capacitance.

Substituting f= 100 Hz and C= 2 ×10−6F into the formula, we get:

XC=1

2π×100 ×2×10−6=1

0.0004π=2500

πΩ

Step 3: Calculate the total impedance of the circuit. The total impedance

Zof the circuit is given by Z=R+XL+XC, where Ris the resistance, XL

is the inductive reactance, and XCis the capacitive reactance. Substituting

R= 50 Ω, XL= 40πΩ, and XC=2500

πΩ into the formula, we get:

Z= 50 + 40π+2500

πΩ = 50 + 40π+2500

πΩ

Therefore, the total impedance of the circuit is 50 + 40π+2500

πΩ.

Question 10

Question

A wire has a resistance of 8 Ω and a current of 3 Apassing through it. What is

the voltage across the wire?

Solution

To find the voltage across the wire, we can use Ohm’s Law, which states that

V=I·R, where Vis the voltage (in volts), Iis the current (in amperes), and

Ris the resistance (in ohms).

Step 1: Substitute the given values into Ohm’s Law formula to find the

voltage.

V=I·R

V= 3 A ·8 Ω

V= 24 V

Step 2: Therefore, the voltage across the wire is 24 volts.

Question 11

Question

A circuit consists of a 15 Ω resistor in series with an unknown resistor connected

to a 12 Vbattery. If a current of 1.5Aflows through the circuit, what is the

resistance of the unknown resistor?

Solution

Let Rbe the resistance of the unknown resistor.

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

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

Rtotal = 15 Ω + R

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

V=I·R, where Vis the voltage, Iis the current, and Ris the resistance.

Since the battery provides a voltage of 12 Vand the current flowing through

the circuit is 1.5A, we have:

12 V= 1.5A·(15 Ω + R)

Step 3: Solve for the unknown resistance. Solving the above equation for

R, we get:

R=12 V

1.5A−15 Ω = 8 Ω

Therefore, the resistance of the unknown resistor is 8 Ω.

Question 12

Question

A resistor with resistance R= 150 Ω is connected to a voltage source that

provides a voltage of V= 12 V across the resistor. Calculate the current flowing

through the resistor.

Solution

To find the current flowing through the resistor, we can use Ohm’s Law, which

states that the current through a conductor between two points is directly pro-

portional to the voltage across the two points and inversely proportional to the

resistance. Mathematically, Ohm’s Law is represented as:

V=IR

where: V= voltage across the resistor (12 V), I= current flowing through the

resistor (to be determined), R= resistance of the resistor (150 Ohms).

Step 1: Substitute the given values into Ohm’s Law equation:

12 = I×150

Step 2: Solve for I:

I=12

150

I= 0.08 A

Therefore, the current flowing through the resistor is 0.08 A.

Question 13

Question

A 15 Ω resistor is connected to a battery that produces 12 V. How much current

is flowing through the resistor?

Solution

To find the current flowing 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: Identify the given values: The voltage across the resistor, V, is 12

V. The resistance of the resistor, R, is 15 Ω.

Step 2: Apply Ohm’s Law to find the current:

V=IR

I=V

Substitute the given values:

I=12 V

15 Ω

I= 0.8 A

Therefore, the current flowing through the resistor is 0.8 A.

Question 14

Question

A student is conducting an experiment to investigate Ohm’s Law. They connect

a resistor of resistance R= 10 Ω to a battery with an emf of V= 12 V. When

measuring the current in the circuit, the student notices that the ammeter reads

I= 1.2A.

Is the resistor behaving ohmically? Justify your answer.

Solution

To determine if the resistor is behaving ohmically, we need to check if the ratio

of the voltage to the current remains constant for different values of voltage and

current.

Step 1: Calculate the voltage across the resistor. The voltage across

the resistor can be given by Ohm’s Law: V=IR, where Iis the current flowing

through the resistor and Ris the resistance of the resistor.

V=I·R= 1.2A·10 Ω = 12 V

Step 2: Calculate the ratio of voltage to current. The ratio of voltage

to current for this circuit is:

I=12 V

1.2A= 10 Ω

Step 3: Analyze if the resistor is behaving ohmically. Since the ratio

of voltage to current remains constant at 10 Ω for this resistor, it is behaving

ohmically. Ohm’s Law states that for an ohmic resistor, the ratio of voltage to

current remains constant at a specific resistance. In this case, the resistor is

behaving ohmically.

Question 15

Question

A resistor with resistance Ris connected to a battery with voltage V. If the

current passing through the resistor is I, prove Ohm’s Law.

Solution

To prove Ohm’s Law, we need to show that the voltage across the resistor is

proportional to the current passing through it.

Step 1: Recall Ohm’s Law, which states V=I·Rwhere Vis the voltage

across the resistor, Iis the current passing through the resistor, and Ris the

resistance of the resistor.

Step 2: From the given information, we have that V=I·R. Rearranging

this equation gives us I=V

Step 3: Suppose the voltage Vis kept constant and the resistance Ris

varied. We can express Ohm’s Law as I=k·Rwhere kis a constant.

Step 4: If Vis held constant and Ris increased, the current Ipassing

through the resistor would decrease proportionally because I=V

Step 5: Therefore, by varying the resistance while keeping the voltage con-

stant, we observe that the current passing through the resistor is inversely pro-

portional to the resistance. This confirms Ohm’s Law, V=I·R.

Question 16

Question

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

The resistor has a resistance of 10 Ω, the inductor has an inductance of 2 H,

and the capacitor has a capacitance of 0.002 F. If the circuit is connected to a

voltage source of 20 V at a frequency of 50 Hz, determine the current flowing

through the circuit.

Solution

Step 1: Calculate the total impedance of the circuit. The impedance (Z) of the

circuit is given by:

Z=pR2+ (XL−XC)2

where: - Ris the resistance, - XLis the inductive reactance, given by XL=

2πf L, where fis the frequency and Lis the inductance, - XCis the capacitive

reactance, given by XC=1

2πfC , where Cis the capacitance.

Substitute the given values:

R= 10 Ω, L = 2 H, C = 0.002 F, f = 50 Hz

XL= 2π(50)(2) = 628.32 Ω

XC=1

2π(50)(0.002) = 1591.55 Ω

Z=p102+ (628.32 −1591.55)2

Z=p100 + (−963.23)2

Z=√100 + 927282.4929

Z=√927382.4929

Z≈963.13 Ω

Step 2: Calculate the current flowing through the circuit using Ohm’s Law.

Ohm’s Law states that the current (I) flowing through a circuit is given by:

I=V

where: - Vis the voltage source, which is 20 V in this case, - Zis the impedance

calculated in Step 1.

Substitute the values into the equation:

I=20

963.13

I≈0.021 A

Therefore, the current flowing through the circuit is approximately 0.021 A.

Question 17

Question

A resistor with resistance R= 50 Ω is connected to a voltage source with V=

120 V. What is 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.

V=IR

where: - Vis the voltage across the resistor (120 V), - Iis the current passing

through the resistor (to be found), - Ris the resistance of the resistor (50 Ω).

Step 2: Rearrange Ohm’s Law to solve for current I.

I=V

Step 3: Substitute the given values into the equation.

I=120 V

50 Ω

I= 2.4A

Step 4: The current passing through the resistor is 2.4A.

Question 18

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 alternating

current (AC) voltage source V(t) = Vmsin(ωt). Show that the total current in

the circuit can be expressed as I(t) = Imsin(ωt −ϕ), where Im,ϕ, and Rare

related by Ohm’s Law.

Solution

Step 1: Calculate the impedance (Z) of the circuit. The impedance Zin a

circuit with a resistor, capacitor, and inductor in series is given by

Z=R+jωL −1

ωC 

where jis the imaginary unit. Substituting Zinto Ohm’s Law V=IZ, we get

I(t) = Vmsin(ωt)

Z=Vmsin(ωt)

R+jωL −1

ωC 

Step 2: Express I(t) in polar form. To simplify the expression, we first

convert Zto polar form:

Z=sR2+ωL −1

ωC 2

cis θ

where θ= arctan ωL−

ωC

R. Substituting this into I(t) gives us

I(t) = Vmsin(ωt)

qR2+ωL −1

ωC 2cis (ωt −θ)

Step 3: Apply Euler’s Formula to simplify I(t). Using Euler’s formula ejθ =

cos(θ) + jsin(θ), we can write the current as

I(t) = Vmsin(ωt)

qR2+ωL −1

ωC 2(cos θ+jsin θ)

=Imsin(ωt −ϕ)

where Im=Vm

qR2+(ωL−

ωC )2and ϕ= arctan ωL−

ωC

R. Hence, we have shown

that the total current in the circuit can be expressed as I(t) = Imsin(ωt −ϕ),

in accordance with Ohm’s Law.

Question 19

Question

A resistor with a resistance of 50 Ω is connected to a 12 V battery. Calculate

the current flowing through the resistor.

Solution

To calculate the current flowing through the resistor, we can use Ohm’s Law,

which states that V=IR, where Vis the voltage, Iis the current, and Ris

the resistance.

Step 1: Write Ohm’s Law equation:

V=IR

Step 2: Rearrange the equation to solve for I:

I=V

Step 3: Substitute in the values given in the question:

I=12 V

50 Ω

Step 4: Calculate the current:

I=12

50 = 0.24 A

Step 5: Therefore, the current flowing through the resistor is 0.24 A.

Question 20

Question

A circuit consists of a resistor R, an inductor L, and a capacitor Cconnected

in series to an AC voltage source. The impedance of the circuit is given by

Z=pR2+ (XL−XC)2, where XL=ωL is the inductive reactance, XC=1

ωC

is the capacitive reactance, and ωis the angular frequency of the AC source.

If the impedance is Z= 10 Ω, the resistance is R= 5 Ω, the inductance is

L= 0.1 H, and the capacitance is C= 10 µF, calculate the angular frequency ω

of the AC source.

Solution

Step 1: Substitute the given values into the impedance formula:

10 = s52+ω·0.1−1

ω·10 ×10−62

Step 2: Simplify the equation by squaring both sides:

100 = 52+ω·0.1−1

ω·10 ×10−62

Step 3: Expand the squared term:

100 = 25 + (ω·0.1)2−2·ω·0.1·1

ω·10 ×10−6+1

ω·10 ×10−62

Step 4: Simplify further:

100 = 25 + ω2·0.01 −1

ω2·102×10−12 +1

ω2·102×10−12

Step 5: Combine like terms and solve for ω:

100 = 25 + ω2·0.01

75 = ω2·0.01

ω2=75

0.01

Step 6: Calculate the angular frequency ω:

ω=r75

0.01 =√7500 ≈86.60 rad/s

Therefore, the angular frequency of the AC source is approximately 86.60 rad/s.

Question 21

Question

A resistor with resistance Ris connected to a battery with voltage V. If the

current flowing through the resistor is inversely proportional to the resistance,

and directly proportional to the voltage, express the current Iin terms of R

and V.

Solution

Step 1: Let’s express the relationships given in the problem statement mathe-

matically:



The current Iis inversely proportional to the resistance R, so we have

I∝1



The current Iis directly proportional to the voltage V, so we have I∝V.

Step 2: Combining the above two relationships, we have:

I=k·V

where kis a constant of proportionality.

Step 3: To find the value of k, we can use Ohm’s Law, which states that

V=I·R. Substituting I=k·V

Rinto Ohm’s Law gives:

V=k·V

R·R

Step 4: Simplifying the above equation, we get:

V=k·V

k= 1

Step 5: Substituting k= 1 back into our expression for current I, we obtain:

I=V

Therefore, the current Iin the circuit is given by I=V

Question 22

Question

A resistor with a resistance of 8 Ω is connected to a battery that supplies a

current of 2.5 A. What is the voltage across the resistor?

Solution

Let’s denote the voltage across the resistor as V.

Step 1: Recall Ohm’s Law, which states that V=I·R, where Vis the

voltage, Iis the current, and Ris the resistance of the resistor.

Step 2: Substitute the given values into the formula V=I·R:

V= 2.5 A ×8 Ω

Step 3: Calculate the voltage:

V= 20 V

Step 4: Therefore, the voltage across the resistor is 20 V.

Question 23

Question

A resistor with a resistance of 15 Ω is connected to a 12V battery. What is the

current flowing through the resistor?

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, Ohm’s Law is represented as:

I=V

Step 1: Given values are V= 12 V and R= 15 Ω, we can plug these values

into Ohm’s Law:

I=12 V

15 Ω

Step 2: Now, we can calculate the current flowing through the resistor:

I=12 V

15 Ω = 0.8 A

Step 3: Therefore, the current flowing through the resistor is 0.8 A.

Question 24

Question

A wire has a resistance of 10 Ω. If a current of 5 A flows through the wire, what

is the voltage across the wire?

Solution

Ohm’s Law states that the voltage (V) across a resistor is equal to the current

(I) flowing through the resistor multiplied by the resistance (R) of the resistor.

Mathematically, this can be expressed as:

V=I×R

Step 1: Given data: The resistance Rof the wire is 10 Ω and the current I

flowing through the wire is 5 A.

Step 2: Applying Ohm’s Law: Substitute the given values of Rand Iinto

Ohm’s Law to find the voltage V:

V= 5 A ×10 Ω

Step 3: Calculate the voltage:

V= 5 ×10 = 50 V

Step 4: Conclusion: The voltage across the wire is 50 V when a current

of 5 A flows through it.

Question 25

Question

A circuit consists of a 20 V battery connected to a resistor with a resistance of

10 Ω. What current is flowing through the circuit?

Solution

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

circuit is equal to the voltage (V) across the circuit divided by the resistance

(R) of the circuit. Mathematically, this can be expressed as:

I=V

Step 2: Given that the voltage Vis 20 V and the resistance Ris 10 Ω, we

can substitute these values into Ohm’s Law to find the current I:

I=20

10 = 2 A

Step 3: Therefore, the current flowing through the circuit is 2 A.

Question 26

Question

A circuit consists of a resistor with resistance R= 150 Ω, a capacitor with

capacitance C= 3 µF , and an inductor with inductance L= 1.5mH connected

in series to a voltage source with voltage V= 12 V. Determine the current

flowing through the circuit at t= 0 seconds.

Solution

Step 1: Calculate the total impedance of the circuit using the formula Z=

pR2+ (XL−XC)2, where XL= 2πf L is the inductive reactance, XC=1

2πfC

is the capacitive reactance, and fis the frequency.

Given: R= 150 Ω, C = 3 µF, L = 1.5mH, V = 12 V

Convert Cand Lto farads and henries: C= 3 ×10−6F, L = 1.5×10−3H

f=1

2π√LC =1

2πp(1.5×10−3)(3 ×10−6)≈318.31Hz

XL= 2πfL = 2π(318.31)(1.5×10−3)≈3 Ω, XC=1

2πf C =1

2π(318.31)(3 ×10−6)≈1.67 kΩ

Z=p1502+ (3 −1.67)2≈150.84 Ω

Step 2: Use Ohm’s Law V=IZ to calculate the current flowing through

the circuit at t= 0 seconds.

I=V

Z=12

150.84 ≈0.0796 A

Therefore, the current flowing through the circuit at t= 0 seconds is ap-

proximately 0.0796 Amps.

Question 27

Question

A resistor with resistance R1is connected in series with a resistor with resistance

R2= 3R1. The combination is connected to a battery of voltage V. If the

current through the circuit is I, determine an expression for the total resistance

Rtotal in terms of R1and R2.

Solution

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

is equal to the product of the current (I) flowing through it and the resistance

(R) of the resistor. Mathematically, this can be written as V=IR.

Step 2: For the circuit described in the question, the total resistance Rtotal

can be written as the sum of R1and R2. Therefore, we have Rtotal =R1+R2.

Step 3: Given that R2= 3R1, we can substitute this expression into the

equation for total resistance: Rtotal =R1+ 3R1.

Step 4: Simplifying the expression, we find Rtotal = 4R1.

Therefore, the total resistance Rtotal in the circuit is 4R1.

Question 28

Question

A resistor with resistance Ris connected to a battery with voltage V. The

current flowing through the resistor is given by Ohm’s Law as I=V

R. If the

resistance Ris doubled, by what factor does the current Ichange?

Solution

Let’s denote the original current as I1when the resistance is R, and the new

current as I2when the resistance is 2R.

Step 1: Find the original current I1when the resistance is R.

I1=V

Step 2: Find the new current I2when the resistance is 2R.

I2=V

Step 3: Calculate the change in current.

Change in I=I2−I1

Change in I=

2R−V

Change in I=

2R−V

Change in I=

2R−V

2R−2V

Change in I=

V−2V

Change in I=

−V

=−1

The current Ichanges by a factor of −1

Question 29

Question

A resistor with resistance R= 10 Ω is connected to a battery with voltage

V= 20 V. Calculate the current passing through the resistor.

Solution

Step 1: The formula for Ohm’s Law is V=IR, where Vis the voltage, Iis the

current, and Ris the resistance. We can rearrange this formula to solve for the

current I:

I=V

Step 2: Substitute the given values into the formula:

I=20 V

10 Ω

Step 3: Simplify the expression to find the current passing through the

resistor:

I= 2 A

Step 4: Therefore, the current passing through the resistor is 2 A .

Question 30

Question

A circuit consists of a resistor with a resistance of 30 Ω and a battery with a

voltage of 12 V. Calculate the current flowing through the circuit according to

Ohm’s Law.

Solution

Step 1: Identify the given values. The resistance of the resistor, R, is 30 Ω, and

the voltage of the battery, V, is 12 V.

Step 2: Recall Ohm’s Law. Ohm’s Law states that the current, I, flowing

through a conductor is equal to the voltage across the conductor divided by the

resistance of the conductor.

V=I×R

Step 3: Substitute the given values into Ohm’s Law.

I=V

I=12 V

30 Ω

Step 4: Calculate the current flowing through the circuit.

I=12

30 = 0.4 A

Therefore, the current flowing through the circuit is 0.4 A.

Question 31

Question

A resistor with resistance Ris connected to a battery of voltage V. If the current

passing through the resistor is I, show that Ohm’s Law can be expressed as

V=IR.

Solution

To show that Ohm’s Law can be expressed as V=IR, we need to analyze the

relationship between voltage, current, and resistance in an electrical circuit.

Step 1: Start with the definition of voltage: Voltage (V) is the electrical

potential difference between two points in a circuit. It is measured in volts.

Step 2: Next, recall the definition of current: Current (I) is the rate of flow

of electric charge through a conductor. It is measured in amperes.

Step 3: Finally, consider the definition of resistance: Resistance (R) is

a measure of how much a material opposes the flow of electric current. It is

measured in ohms.

Step 4: According to Ohm’s Law, the voltage across a resistor is directly

proportional to the current flowing through it, with the constant of propor-

tionality being the resistance. In mathematical terms, Ohm’s Law is expressed

as:

V=IR

This equation shows the relationship between voltage (V), current (I), and

resistance (R) in an electrical circuit.

Therefore, we have shown that Ohm’s Law can be expressed as V=IR.

Question 32

Question

A circuit consists of a resistor with resistance R= 500 Ω connected to a battery

with potential difference V= 12 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 potential difference (voltage), Iis the

current, and Ris the resistance.

Step 1: Write down Ohm’s Law formula: V=IR.

Step 2: Rearrange the formula to solve for current I:I=V

Step 3: Substitute the given values V= 12 V and R= 500 Ω into the

formula:

I=12 V

500 Ω

Step 4: Perform the division to find the current:

I=12

500 = 0.024 A.

Therefore, the current flowing through the circuit is 0.024 Amperes.

Question 33

Question

A resistor with a resistance of 120 Ω is connected to a battery with a voltage of

24 V. Calculate the current passing through the resistor.

Solution

To find the current passing through the resistor, we can 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,

Ohm’s Law can be expressed as:

I=V

Step 1: Substitute the given values into Ohm’s Law formula.

I=24 V

120 Ω

Step 2: Calculate the current passing through the resistor.

I=24

120 = 0.2A=0.2 Amperes

Therefore, the current passing through the resistor is 0.2 Amperes.

Question 34

Question

A circuit consists of a resistor with a resistance of 10 Ω. If a current of 2.5 A

flows through the circuit, what is the voltage across the resistor?

Solution

To find the voltage across the resistor, we can use Ohm’s Law, which states that

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

Step 1: Given values are: Resistance, R= 10 Ω Current, I= 2.5 A

Step 2: Substitute the values into Ohm’s Law equation:

V=I·R

V= 2.5 A ×10 Ω

Step 3: Calculate the voltage:

V= 25 V

Therefore, the voltage across the resistor is 25 V.

Question 35

Question

A resistor with a resistance of 30 ohms is connected to a 12-volt battery. What

is the current flowing through the resistor?

Solution

Step 1: Recall 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, this can be represented as:

I=V

where: I= current (in amperes), V= voltage (in volts), and R= resistance

(in ohms).

Solution

Ohm’s Law states that the current passing through a resistor is given by I=V

where Iis the current, Vis the voltage, and Ris the resistance of the resistor.

Step 1: Substitute the given values into Ohm’s Law to find the current

passing through the resistor.

I=V

R=12 V

100 Ω

Step 2: Calculate the current passing through the resistor.

I=12

100 =3

25 = 0.12 A

Therefore, the current passing through the resistor is 0.12 A.

Question 3

Question

A resistor of resistance R= 30 Ω is connected to a battery with voltage V=

12 V. Calculate the current flowing through the resistor.

Solution

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

resistor is given by 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=12 V

30 Ω .

Step 3: Calculate the current: I=12

30 = 0.4 A.

Step 4: Therefore, the current flowing through the resistor is 0.4 A.

Question 4

Question

A resistor has a voltage drop of 10 V across it when a current of 2 A flows

through it. Calculate the resistance of the resistor.

Solution

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

equal to the current (I) flowing through the resistor multiplied by the resistance

(R) of the resistor. Mathematically, this is expressed as:

V=IR

Step 2: Given that the voltage drop across the resistor is 10 V and the

current flowing through it is 2 A, we can substitute these values into Ohm’s

Law:

10 = 2R

Step 3: Solve for the resistance Rby dividing both sides of the equation by

R=10

R= 5 ohms

Step 4: Therefore, the resistance of the resistor is 5 ohms.

Question 5

Question

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

voltage V. If the current passing through the resistor is given by I=V2

R, find

the power dissipated in the resistor in terms of Vand R.

Solution

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

the formula P=I2R.

Step 2: Substitute the given expression for current Iinto the formula for

power:

P=V2

R2

·R

Step 3: Simplify the expression:

P=V4

R·R=V4

Step 4: Therefore, the power dissipated in the resistor is V4.

Question 6

Question

A circuit consists of a resistor with resistance R= 20Ω connected to a battery

with voltage V= 100 V. Calculate the current passing through the circuit.

Solution

Ohm’s Law states that the current passing through a circuit is proportional to

the voltage and inversely proportional to the resistance. Mathematically, it can

be expressed as I=V

R, where: - Iis the current passing through the circuit, -

Vis the voltage supplied by the battery, and - Ris the resistance in the circuit.

Step 1: Substitute the given values into Ohm’s Law equation.

I=100 V

20Ω

Step 2: Calculate the current passing through the circuit.

I=100

20 = 5 A

Therefore, the current passing through the circuit is 5 A.

Question 7

Question

A resistor has a resistance of 8 Ω and a current of 2 A flowing through it.

Determine the voltage drop across the resistor.

Solution

Ohm’s Law states that the voltage drop (V) across a resistor is equal to the

product of the current (I) flowing through it and the resistance (R) of the

resistor. Mathematically, this can be represented as:

V=I×R

Step 1: Given values are: Resistance, R= 8Ω Current, I= 2 A

Step 2: Substitute the given values into Ohm’s Law equation:

V= 2 A ×8 Ω

Step 3: Calculate the voltage drop:

V= 16 V

Therefore, the voltage drop across the resistor is 16 V.

Question 8

Question

A circuit consists of a resistor with a resistance of 150 Ω connected to a voltage

source with a voltage of 12 V. Find the current flowing through the circuit.

Solution

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

states that the current (I) flowing through a resistor is equal to the voltage (V)

across the resistor divided by the resistance (R) of the resistor. Mathematically,

Ohm’s Law is given by the equation:

I=V

Step 1: Identify the given values: The resistance (R) is 150 Ω and the

voltage (V) is 12 V.

Step 2: Substitute the known values into the formula for Ohm’s Law:

I=12 V

150 Ω

Step 3: Calculate the current flowing through the circuit:

I=12

150 = 0.08 A

Therefore, the current flowing through the circuit is 0.08 A or 80 mA.

Question 9

Question

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

The resistor has a resistance of 50 Ω, the inductor has an inductance of 0.2 H,

and the capacitor has a capacitance of 2 µF. If the frequency of the AC power

source is 100 Hz, determine the total impedance of the circuit.

Solution

Step 1: Calculate the reactance of the inductor. The reactance of an inductor

is given by XL= 2πf L, where fis the frequency and Lis the inductance.

Substituting f= 100 Hz and L= 0.2 H into the formula, we get:

XL= 2π×100 ×0.2 = 40πΩ

Step 2: Calculate the reactance of the capacitor. The reactance of a capacitor

is given by XC=1

2πf C , where fis the frequency and Cis the capacitance.

Substituting f= 100 Hz and C= 2 ×10−6F into the formula, we get:

XC=1

2π×100 ×2×10−6=1

0.0004π=2500

πΩ

Step 3: Calculate the total impedance of the circuit. The total impedance

Zof the circuit is given by Z=R+XL+XC, where Ris the resistance, XL

is the inductive reactance, and XCis the capacitive reactance. Substituting

R= 50 Ω, XL= 40πΩ, and XC=2500

πΩ into the formula, we get:

Z= 50 + 40π+2500

πΩ = 50 + 40π+2500

πΩ

Therefore, the total impedance of the circuit is 50 + 40π+2500

πΩ.

Question 10

Question

A wire has a resistance of 8 Ω and a current of 3 Apassing through it. What is

the voltage across the wire?

Solution

To find the voltage across the wire, we can use Ohm’s Law, which states that

V=I·R, where Vis the voltage (in volts), Iis the current (in amperes), and

Ris the resistance (in ohms).

Step 1: Substitute the given values into Ohm’s Law formula to find the

voltage.

V=I·R

V= 3 A ·8 Ω

V= 24 V

Step 2: Therefore, the voltage across the wire is 24 volts.

Question 11

Question

A circuit consists of a 15 Ω resistor in series with an unknown resistor connected

to a 12 Vbattery. If a current of 1.5Aflows through the circuit, what is the

resistance of the unknown resistor?

Solution

Let Rbe the resistance of the unknown resistor.

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

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

Rtotal = 15 Ω + R

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

V=I·R, where Vis the voltage, Iis the current, and Ris the resistance.

Since the battery provides a voltage of 12 Vand the current flowing through

the circuit is 1.5A, we have:

12 V= 1.5A·(15 Ω + R)

Step 3: Solve for the unknown resistance. Solving the above equation for

R, we get:

R=12 V

1.5A−15 Ω = 8 Ω

Therefore, the resistance of the unknown resistor is 8 Ω.

Question 12

Question

A resistor with resistance R= 150 Ω is connected to a voltage source that

provides a voltage of V= 12 V across the resistor. Calculate the current flowing

through the resistor.

Solution

To find the current flowing through the resistor, we can use Ohm’s Law, which

states that the current through a conductor between two points is directly pro-

portional to the voltage across the two points and inversely proportional to the

resistance. Mathematically, Ohm’s Law is represented as:

V=IR

where: V= voltage across the resistor (12 V), I= current flowing through the

resistor (to be determined), R= resistance of the resistor (150 Ohms).

Step 1: Substitute the given values into Ohm’s Law equation:

12 = I×150

Step 2: Solve for I:

I=12

150

I= 0.08 A

Therefore, the current flowing through the resistor is 0.08 A.

Question 13

Question

A 15 Ω resistor is connected to a battery that produces 12 V. How much current

is flowing through the resistor?

Solution

To find the current flowing 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: Identify the given values: The voltage across the resistor, V, is 12

V. The resistance of the resistor, R, is 15 Ω.

Step 2: Apply Ohm’s Law to find the current:

V=IR

I=V

Substitute the given values:

I=12 V

15 Ω

I= 0.8 A

Therefore, the current flowing through the resistor is 0.8 A.

Question 14

Question

A student is conducting an experiment to investigate Ohm’s Law. They connect

a resistor of resistance R= 10 Ω to a battery with an emf of V= 12 V. When

measuring the current in the circuit, the student notices that the ammeter reads

I= 1.2A.

Is the resistor behaving ohmically? Justify your answer.

Solution

To determine if the resistor is behaving ohmically, we need to check if the ratio

of the voltage to the current remains constant for different values of voltage and

current.

Step 1: Calculate the voltage across the resistor. The voltage across

the resistor can be given by Ohm’s Law: V=IR, where Iis the current flowing

through the resistor and Ris the resistance of the resistor.

V=I·R= 1.2A·10 Ω = 12 V

Step 2: Calculate the ratio of voltage to current. The ratio of voltage

to current for this circuit is:

I=12 V

1.2A= 10 Ω

Step 3: Analyze if the resistor is behaving ohmically. Since the ratio

of voltage to current remains constant at 10 Ω for this resistor, it is behaving

ohmically. Ohm’s Law states that for an ohmic resistor, the ratio of voltage to

current remains constant at a specific resistance. In this case, the resistor is

behaving ohmically.

Question 15

Question

A resistor with resistance Ris connected to a battery with voltage V. If the

current passing through the resistor is I, prove Ohm’s Law.

Solution

To prove Ohm’s Law, we need to show that the voltage across the resistor is

proportional to the current passing through it.

Step 1: Recall Ohm’s Law, which states V=I·Rwhere Vis the voltage

across the resistor, Iis the current passing through the resistor, and Ris the

resistance of the resistor.

Step 2: From the given information, we have that V=I·R. Rearranging

this equation gives us I=V

Step 3: Suppose the voltage Vis kept constant and the resistance Ris

varied. We can express Ohm’s Law as I=k·Rwhere kis a constant.

Step 4: If Vis held constant and Ris increased, the current Ipassing

through the resistor would decrease proportionally because I=V

Step 5: Therefore, by varying the resistance while keeping the voltage con-

stant, we observe that the current passing through the resistor is inversely pro-

portional to the resistance. This confirms Ohm’s Law, V=I·R.

Question 16

Question

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

The resistor has a resistance of 10 Ω, the inductor has an inductance of 2 H,

and the capacitor has a capacitance of 0.002 F. If the circuit is connected to a

voltage source of 20 V at a frequency of 50 Hz, determine the current flowing

through the circuit.

Solution

Step 1: Calculate the total impedance of the circuit. The impedance (Z) of the

circuit is given by:

Z=pR2+ (XL−XC)2

where: - Ris the resistance, - XLis the inductive reactance, given by XL=

2πf L, where fis the frequency and Lis the inductance, - XCis the capacitive

reactance, given by XC=1

2πfC , where Cis the capacitance.

Substitute the given values:

R= 10 Ω, L = 2 H, C = 0.002 F, f = 50 Hz

XL= 2π(50)(2) = 628.32 Ω

XC=1

2π(50)(0.002) = 1591.55 Ω

Z=p102+ (628.32 −1591.55)2

Z=p100 + (−963.23)2

Z=√100 + 927282.4929

Z=√927382.4929

Z≈963.13 Ω

Step 2: Calculate the current flowing through the circuit using Ohm’s Law.

Ohm’s Law states that the current (I) flowing through a circuit is given by:

I=V

where: - Vis the voltage source, which is 20 V in this case, - Zis the impedance

calculated in Step 1.

Substitute the values into the equation:

I=20

963.13

I≈0.021 A

Therefore, the current flowing through the circuit is approximately 0.021 A.

Question 17

Question

A resistor with resistance R= 50 Ω is connected to a voltage source with V=

120 V. What is 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.

V=IR

where: - Vis the voltage across the resistor (120 V), - Iis the current passing

through the resistor (to be found), - Ris the resistance of the resistor (50 Ω).

Step 2: Rearrange Ohm’s Law to solve for current I.

I=V

Step 3: Substitute the given values into the equation.

I=120 V

50 Ω

I= 2.4A

Step 4: The current passing through the resistor is 2.4A.

Question 18

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 alternating

current (AC) voltage source V(t) = Vmsin(ωt). Show that the total current in

the circuit can be expressed as I(t) = Imsin(ωt −ϕ), where Im,ϕ, and Rare

related by Ohm’s Law.

Solution

Step 1: Calculate the impedance (Z) of the circuit. The impedance Zin a

circuit with a resistor, capacitor, and inductor in series is given by

Z=R+jωL −1

ωC 

where jis the imaginary unit. Substituting Zinto Ohm’s Law V=IZ, we get

I(t) = Vmsin(ωt)

Z=Vmsin(ωt)

R+jωL −1

ωC 

Step 2: Express I(t) in polar form. To simplify the expression, we first

convert Zto polar form:

Z=sR2+ωL −1

ωC 2

cis θ

where θ= arctan ωL−

ωC

R. Substituting this into I(t) gives us

I(t) = Vmsin(ωt)

qR2+ωL −1

ωC 2cis (ωt −θ)

Step 3: Apply Euler’s Formula to simplify I(t). Using Euler’s formula ejθ =

cos(θ) + jsin(θ), we can write the current as

I(t) = Vmsin(ωt)

qR2+ωL −1

ωC 2(cos θ+jsin θ)

=Imsin(ωt −ϕ)

where Im=Vm

qR2+(ωL−

ωC )2and ϕ= arctan ωL−

ωC

R. Hence, we have shown

that the total current in the circuit can be expressed as I(t) = Imsin(ωt −ϕ),

in accordance with Ohm’s Law.

Question 19

Question

A resistor with a resistance of 50 Ω is connected to a 12 V battery. Calculate

the current flowing through the resistor.

Solution

To calculate the current flowing through the resistor, we can use Ohm’s Law,

which states that V=IR, where Vis the voltage, Iis the current, and Ris

the resistance.

Step 1: Write Ohm’s Law equation:

V=IR

Step 2: Rearrange the equation to solve for I:

I=V

Step 3: Substitute in the values given in the question:

I=12 V

50 Ω

Step 4: Calculate the current:

I=12

50 = 0.24 A

Step 5: Therefore, the current flowing through the resistor is 0.24 A.

Question 20

Question

A circuit consists of a resistor R, an inductor L, and a capacitor Cconnected

in series to an AC voltage source. The impedance of the circuit is given by

Z=pR2+ (XL−XC)2, where XL=ωL is the inductive reactance, XC=1

ωC

is the capacitive reactance, and ωis the angular frequency of the AC source.

If the impedance is Z= 10 Ω, the resistance is R= 5 Ω, the inductance is

L= 0.1 H, and the capacitance is C= 10 µF, calculate the angular frequency ω

of the AC source.

Solution

Step 1: Substitute the given values into the impedance formula:

10 = s52+ω·0.1−1

ω·10 ×10−62

Step 2: Simplify the equation by squaring both sides:

100 = 52+ω·0.1−1

ω·10 ×10−62

Step 3: Expand the squared term:

100 = 25 + (ω·0.1)2−2·ω·0.1·1

ω·10 ×10−6+1

ω·10 ×10−62

Step 4: Simplify further:

100 = 25 + ω2·0.01 −1

ω2·102×10−12 +1

ω2·102×10−12

Step 5: Combine like terms and solve for ω:

100 = 25 + ω2·0.01

75 = ω2·0.01

ω2=75

0.01

Step 6: Calculate the angular frequency ω:

ω=r75

0.01 =√7500 ≈86.60 rad/s

Therefore, the angular frequency of the AC source is approximately 86.60 rad/s.

Question 21

Question

A resistor with resistance Ris connected to a battery with voltage V. If the

current flowing through the resistor is inversely proportional to the resistance,

and directly proportional to the voltage, express the current Iin terms of R

and V.

Solution

Step 1: Let’s express the relationships given in the problem statement mathe-

matically:



The current Iis inversely proportional to the resistance R, so we have

I∝1



The current Iis directly proportional to the voltage V, so we have I∝V.

Step 2: Combining the above two relationships, we have:

I=k·V

where kis a constant of proportionality.

Step 3: To find the value of k, we can use Ohm’s Law, which states that

V=I·R. Substituting I=k·V

Rinto Ohm’s Law gives:

V=k·V

R·R

Step 4: Simplifying the above equation, we get:

V=k·V

k= 1

Step 5: Substituting k= 1 back into our expression for current I, we obtain:

I=V

Therefore, the current Iin the circuit is given by I=V

Question 22

Question

A resistor with a resistance of 8 Ω is connected to a battery that supplies a

current of 2.5 A. What is the voltage across the resistor?

Solution

Let’s denote the voltage across the resistor as V.

Step 1: Recall Ohm’s Law, which states that V=I·R, where Vis the

voltage, Iis the current, and Ris the resistance of the resistor.

Step 2: Substitute the given values into the formula V=I·R:

V= 2.5 A ×8 Ω

Step 3: Calculate the voltage:

V= 20 V

Step 4: Therefore, the voltage across the resistor is 20 V.

Question 23

Question

A resistor with a resistance of 15 Ω is connected to a 12V battery. What is the

current flowing through the resistor?

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, Ohm’s Law is represented as:

I=V

Step 1: Given values are V= 12 V and R= 15 Ω, we can plug these values

into Ohm’s Law:

I=12 V

15 Ω

Step 2: Now, we can calculate the current flowing through the resistor:

I=12 V

15 Ω = 0.8 A

Step 3: Therefore, the current flowing through the resistor is 0.8 A.

Question 24

Question

A wire has a resistance of 10 Ω. If a current of 5 A flows through the wire, what

is the voltage across the wire?

Solution

Ohm’s Law states that the voltage (V) across a resistor is equal to the current

(I) flowing through the resistor multiplied by the resistance (R) of the resistor.

Mathematically, this can be expressed as:

V=I×R

Step 1: Given data: The resistance Rof the wire is 10 Ω and the current I

flowing through the wire is 5 A.

Step 2: Applying Ohm’s Law: Substitute the given values of Rand Iinto

Ohm’s Law to find the voltage V:

V= 5 A ×10 Ω

Step 3: Calculate the voltage:

V= 5 ×10 = 50 V

Step 4: Conclusion: The voltage across the wire is 50 V when a current

of 5 A flows through it.

Question 25

Question

A circuit consists of a 20 V battery connected to a resistor with a resistance of

10 Ω. What current is flowing through the circuit?

Solution

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

circuit is equal to the voltage (V) across the circuit divided by the resistance

(R) of the circuit. Mathematically, this can be expressed as:

I=V

Step 2: Given that the voltage Vis 20 V and the resistance Ris 10 Ω, we

can substitute these values into Ohm’s Law to find the current I:

I=20

10 = 2 A

Step 3: Therefore, the current flowing through the circuit is 2 A.

Question 26

Question

A circuit consists of a resistor with resistance R= 150 Ω, a capacitor with

capacitance C= 3 µF , and an inductor with inductance L= 1.5mH connected

in series to a voltage source with voltage V= 12 V. Determine the current

flowing through the circuit at t= 0 seconds.

Solution

Step 1: Calculate the total impedance of the circuit using the formula Z=

pR2+ (XL−XC)2, where XL= 2πf L is the inductive reactance, XC=1

2πfC

is the capacitive reactance, and fis the frequency.

Given: R= 150 Ω, C = 3 µF, L = 1.5mH, V = 12 V

Convert Cand Lto farads and henries: C= 3 ×10−6F, L = 1.5×10−3H

f=1

2π√LC =1

2πp(1.5×10−3)(3 ×10−6)≈318.31Hz

XL= 2πfL = 2π(318.31)(1.5×10−3)≈3 Ω, XC=1

2πf C =1

2π(318.31)(3 ×10−6)≈1.67 kΩ

Z=p1502+ (3 −1.67)2≈150.84 Ω

Step 2: Use Ohm’s Law V=IZ to calculate the current flowing through

the circuit at t= 0 seconds.

I=V

Z=12

150.84 ≈0.0796 A

Therefore, the current flowing through the circuit at t= 0 seconds is ap-

proximately 0.0796 Amps.

Question 27

Question

A resistor with resistance R1is connected in series with a resistor with resistance

R2= 3R1. The combination is connected to a battery of voltage V. If the

current through the circuit is I, determine an expression for the total resistance

Rtotal in terms of R1and R2.

Solution

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

is equal to the product of the current (I) flowing through it and the resistance

(R) of the resistor. Mathematically, this can be written as V=IR.

Step 2: For the circuit described in the question, the total resistance Rtotal

can be written as the sum of R1and R2. Therefore, we have Rtotal =R1+R2.

Step 3: Given that R2= 3R1, we can substitute this expression into the

equation for total resistance: Rtotal =R1+ 3R1.

Step 4: Simplifying the expression, we find Rtotal = 4R1.

Therefore, the total resistance Rtotal in the circuit is 4R1.

Question 28

Question

A resistor with resistance Ris connected to a battery with voltage V. The

current flowing through the resistor is given by Ohm’s Law as I=V

R. If the

resistance Ris doubled, by what factor does the current Ichange?

Solution

Let’s denote the original current as I1when the resistance is R, and the new

current as I2when the resistance is 2R.

Step 1: Find the original current I1when the resistance is R.

I1=V

Step 2: Find the new current I2when the resistance is 2R.

I2=V

Step 3: Calculate the change in current.

Change in I=I2−I1

Change in I=

2R−V

Change in I=

2R−V

Change in I=

2R−V

2R−2V

Change in I=

V−2V

Change in I=

−V

=−1

The current Ichanges by a factor of −1

Question 29

Question

A resistor with resistance R= 10 Ω is connected to a battery with voltage

V= 20 V. Calculate the current passing through the resistor.

Solution

Step 1: The formula for Ohm’s Law is V=IR, where Vis the voltage, Iis the

current, and Ris the resistance. We can rearrange this formula to solve for the

current I:

I=V

Step 2: Substitute the given values into the formula:

I=20 V

10 Ω

Step 3: Simplify the expression to find the current passing through the

resistor:

I= 2 A

Step 4: Therefore, the current passing through the resistor is 2 A .

Question 30

Question

A circuit consists of a resistor with a resistance of 30 Ω and a battery with a

voltage of 12 V. Calculate the current flowing through the circuit according to

Ohm’s Law.

Solution

Step 1: Identify the given values. The resistance of the resistor, R, is 30 Ω, and

the voltage of the battery, V, is 12 V.