Question 201

Circuit Diagram:

Calculate the voltage across the 5 Ω resistor if the current source is 17/3 A.

Options:

A) 2.32 V
B) 5.21 V
C) 6.67 V
D) 8.96 V

Answer: B) 5.21 V

Step-by-Step Solution:

Assume the clockwise mesh currents are:

I₁ for the left loop
I₂ for the right loop

The current source supplies:

Is = 17/3 A

Step 1: Apply KVL to Loop 1

The KVL equation for the left loop is:

4 + 2I₁ + 3(I₁ − 17/3) + 4(I₁ − I₂) + 5 = 0

Simplifying,

4 + 2I₁ + 3I₁ − 17 + 4I₁ − 4I₂ + 5 = 0

9I₁ − 4I₂ = 8

Step 2: Apply KVL to Loop 2

For the right loop,

The total resistance is:

4 Ω + 1 Ω + 5 Ω = 10 Ω

Applying KVL,

−4I₁ + 10I₂ = 5

Step 3: Solve the simultaneous equations

From

9I₁ − 4I₂ = 8

and

−4I₁ + 10I₂ = 5

Solving,

I₁ = 1.352 A

I₂ = 1.041 A

Step 4: Calculate the voltage across the 5 Ω resistor

Using Ohm's Law,

V = IR

V = 1.041 × 5

V = 5.205 V

V ≈ 5.21 V

Therefore, the voltage across the 5 Ω resistor is:

5.21 V

Additional Note:

This problem is solved using Mesh Analysis, which is based on Kirchhoff's Voltage Law (KVL).

Procedure:

Assume mesh currents.
Apply KVL to each independent loop.
Solve the simultaneous equations.
Use Ohm's Law (V = IR) to determine the required voltage.

✔ Answer: B) 5.21 V

Question 202

Circuit Diagram:

Calculate $V_{AB}$ .

Options:

A) 3.5 V
B) 12 V
C) 9.5 V
D) 6.5 V

Answer: A) 3.5 V

Step-by-Step Solution:

Both branches are connected in parallel across the 20 V supply.

Step 1: Find the voltage at point A

The left branch consists of:

Upper resistor = 25 Ω
Lower resistor = 15 Ω

Using the Voltage Divider Rule,

VA = 20 × (15 / (25 + 15))

VA = 20 × (15 / 40)

VA = 7.5 V

Step 2: Find the voltage at point B

The right branch consists of:

Upper resistor = 40 Ω
Lower resistor = 10 Ω

Using the Voltage Divider Rule,

VB = 20 × (10 / (40 + 10))

VB = 20 × (10 / 50)

VB = 4 V

Step 3: Calculate $V_{AB}$

By definition,

VAB = VA − VB

Substitute the values,

VAB = 7.5 − 4

VAB = 3.5 V

Verification using KVL

Applying Kirchhoff's Voltage Law around loop A → B → C → A:

VAB + VBC + VCA = 0

Where,

VBC = 4 V
VCA = −7.5 V

Therefore,

VAB + 4 − 7.5 = 0

VAB = 3.5 V

The result is verified.

Additional Note:

The Voltage Divider Rule states:

Voltage across a resistor = Supply Voltage × (Resistor / Total Series Resistance)

For the left branch:

VA = 7.5 V

For the right branch:

VB = 4 V

Therefore,

VAB = VA − VB = 3.5 V

✔ Answer: A) 3.5 V

Question 203

KVL is primarily applied in:

Options:

A) Mesh Analysis
B) Nodal Analysis
C) Both Mesh and Nodal Analysis
D) Neither Mesh nor Nodal Analysis

Answer: A) Mesh Analysis

Step-by-Step Solution:

Kirchhoff's Voltage Law (KVL) is used to write equations around closed loops (meshes).

In mesh analysis, the unknown mesh currents are determined by applying KVL to each independent loop.

Nodal analysis, on the other hand, uses Kirchhoff's Current Law (KCL).

Therefore, KVL is primarily applied in mesh analysis.

Additional Note:

Method	Fundamental Law Used
Mesh Analysis	Kirchhoff's Voltage Law (KVL)
Nodal Analysis	Kirchhoff's Current Law (KCL)

These two techniques are the most common methods used for analyzing electrical circuits.

✔ Answer: A) Mesh Analysis

Question 204

Which of the following is NOT a valid expression for electrical power?

Options:

A) P = VI
B) P = I²R
C) P = V²/R
D) P = I/R

Answer: D) P = I/R

Step-by-Step Solution:

The standard expressions for electrical power are:

P = VI
P = I²R
P = V²/R

These are obtained using Ohm's Law:

V = IR

Substituting V = IR into P = VI,

P = I(IR) = I²R

Similarly,

Substituting I = V/R into P = VI,

P = V(V/R) = V²/R

However,

P = I/R is not a valid power equation.

Additional Note:

Common power formulas:

P = VI
P = I²R
P = V²/R

✔ Answer: D) P = I/R

Question 205

Which of the following statements is true?

Options:

A) Power is proportional only to voltage.
B) Power is proportional only to current.
C) Power is neither proportional to voltage nor current.
D) Power is proportional to both voltage and current.

Answer: D) Power is proportional to both voltage and current

Step-by-Step Solution:

Electrical power is given by:

P = VI

where,

P = Power (W)
V = Voltage (V)
I = Current (A)

From this equation,

Power increases when either voltage or current increases (keeping the other constant).

Therefore, power depends on both voltage and current.

Additional Note:

Power is measured in Watts (W).

1 Watt = 1 Volt × 1 Ampere

✔ Answer: D) Power is proportional to both voltage and current

Question 206

A 250 V lamp draws a current of 0.3 A. Calculate the power consumed by the lamp.

Options:

A) 75 W
B) 50 W
C) 25 W
D) 90 W

Answer: A) 75 W

Step-by-Step Solution:

Given:

Voltage (V) = 250 V
Current (I) = 0.3 A

Using the power formula,

P = VI

Substitute the values,

P = 250 × 0.3

P = 75 W

Therefore, the power consumed by the lamp is:

75 W

Additional Note:

Power can also be calculated using:

P = VI
P = I²R
P = V²/R

Choose the appropriate formula based on the given data.

✔ Answer: A) 75 W

Question 207

Kilowatt-hour (kWh) is a unit of:

Options:

A) Current
B) Power
C) Energy
D) Resistance

Answer: C) Energy

Step-by-Step Solution:

Energy is calculated by:

Energy = Power × Time

If,

Power is measured in kilowatts (kW)
Time is measured in hours (h)

Then,

Energy = kW × h = kWh

Thus, kilowatt-hour (kWh) is a unit of energy, not power.

Additional Note:

Common energy units:

Joule (J) (SI Unit)
Watt-hour (Wh)
Kilowatt-hour (kWh)

Also,

1 kWh = 1000 Wh = 3.6 × 10⁶ J

✔ Answer: C) Energy

Question 208

Circuit Diagram:

Calculate the power dissipated in the 20 Ω resistor.

Options:

A) 2000 kW
B) 2 kW
C) 200 kW
D) 2 W

Answer: B) 2 kW

Step-by-Step Solution:

Given:

Voltage (V) = 200 V
Resistance (R) = 20 Ω

Using the power formula,

P = V²/R

Substitute the given values,

P = (200)² / 20

P = 40000 / 20

P = 2000 W

Convert watts into kilowatts,

2000 W = 2 kW

Therefore, the power dissipated in the 20 Ω resistor is:

2 kW

Verification using Ohm's Law

First, calculate the current:

I = V/R

I = 200/20

I = 10 A

Now,

P = VI

P = 200 × 10

P = 2000 W = 2 kW

The answer is verified.

Additional Note:

Electrical power can be calculated using any of the following formulas:

P = VI
P = I²R
P = V²/R

Choose the appropriate formula based on the quantities given.

✔ Answer: B) 2 kW

Question 209

A current of 5 A flows through a resistor of 2 Ω. Calculate the energy dissipated in 300 seconds.

Options:

A) 15 kJ
B) 15000 kJ
C) 1500 J
D) 15 J

Answer: A) 15 kJ

Step-by-Step Solution:

Given:

Current (I) = 5 A
Resistance (R) = 2 Ω
Time (t) = 300 s

Step 1: Calculate the power.

Using,

P = I²R

P = (5)² × 2

P = 25 × 2

P = 50 W

Step 2: Calculate the energy.

Using,

E = Pt

E = 50 × 300

E = 15000 J

Convert Joules into kilojoules:

15000 J = 15 kJ

Therefore,

Energy dissipated = 15 kJ

Additional Note:

Energy can be calculated using:

E = Pt
E = VIt
E = I²Rt
E = (V²/R)t

✔ Answer: A) 15 kJ

Question 210

Circuit Diagram:

Calculate the power dissipated in each 20 Ω resistor.

Options:

A) 1000 W, 1000 W
B) 500 W, 500 W
C) 1000 kW, 1000 kW
D) 500 kW, 500 kW

Answer: B) 500 W, 500 W

Step-by-Step Solution:

Since the two 20 Ω resistors are connected in series, the same current flows through both resistors.

Step 1: Calculate the total resistance

RT = 20 + 20

RT = 40 Ω

Step 2: Calculate the circuit current

Using Ohm's Law,

I = V/R

Substitute the given values,

I = 200/40

I = 5 A

Step 3: Calculate the power dissipated in each resistor

Using the power formula,

P = I²R

Substitute the values,

P = (5)² × 20

P = 25 × 20

P = 500 W

Since both resistors have the same resistance and carry the same current,

Power dissipated in each 20 Ω resistor = 500 W

Verification

Voltage across each resistor:

V = IR

V = 5 × 20

V = 100 V

Now,

P = VI

P = 100 × 5

P = 500 W

Hence, the answer is verified.

Additional Note:

For a series circuit:

Total Resistance: RT = R₁ + R₂
Current is the same through all resistors.
Power dissipated in each resistor is calculated using:
- P = I²R
- P = VI
- P = V²/R

✔ Answer: B) 500 W, 500 W

Question 211

Circuit Diagram:

Calculate the power dissipated in each 10 Ω resistor.

Options:

A) 1000 kW, 1000 kW
B) 1 kW, 1 kW
C) 100 W, 100 W
D) 100 kW, 100 kW

Answer: B) 1 kW, 1 kW

Step-by-Step Solution:

The two 10 Ω resistors are connected in parallel across a 100 V supply.

In a parallel circuit,

Voltage across each branch is equal to the supply voltage.

Therefore,

Voltage across each resistor = 100 V

Step 1: Calculate the power in each resistor

Using the power formula,

P = V²/R

Substitute the given values,

P = (100)² / 10

P = 10000 / 10

P = 1000 W

Convert watts into kilowatts,

1000 W = 1 kW

Hence,

Power dissipated in each 10 Ω resistor = 1 kW

Verification using Ohm's Law

Current through each resistor,

I = V/R

I = 100/10

I = 10 A

Now,

P = VI

P = 100 × 10

P = 1000 W = 1 kW

The answer is verified.

Additional Note:

For a parallel circuit:

Voltage across each branch is the same.
Current divides according to the resistance.
Power can be calculated using:
- P = V²/R
- P = VI
- P = I²R

Since both resistors have equal resistance and receive the same voltage, they dissipate equal power.

✔ Answer: B) 1 kW, 1 kW

Question 212

Calculate the work done in a 20 Ω resistor carrying a current of 5 A for 3 hours.

Options:

A) 1.5 J
B) 15 J
C) 1.5 kWh
D) 15 kWh

Answer: C) 1.5 kWh

Step-by-Step Solution:

Given:

Current (I) = 5 A
Resistance (R) = 20 Ω
Time (t) = 3 hours

Step 1: Calculate the power.

Using,

P = I²R

P = (5)² × 20

P = 25 × 20

P = 500 W

Convert into kilowatts:

500 W = 0.5 kW

Step 2: Calculate the work done (energy).

Using,

W = Pt

W = 0.5 × 3

W = 1.5 kWh

Therefore,

Work done = 1.5 kWh

Additional Note:

Electrical energy can be expressed as:

Joule (J) → SI unit
Watt-hour (Wh)
Kilowatt-hour (kWh)

1 kWh = 3.6 × 10⁶ J

✔ Answer: C) 1.5 kWh

Question 213

What is the SI unit of electrical power?

Options:

A) kW (kilowatt)
B) J/s (joule per second)
C) W·s (watt-second)
D) J/h (joule per hour)

Answer: B) J/s (joule per second)

Step-by-Step Solution:

Power is defined as:

Power = Energy / Time

Therefore,

SI Unit of Power = Joule / Second

This unit is known as the Watt (W).

Thus,

1 Watt = 1 Joule/Second

Additional Note:

Some commonly used power units are:

1 W = 1 J/s
1 kW = 1000 W
1 MW = 1000 kW

Although Watt (W) is the SI unit, it is equivalent to Joule per second (J/s).

✔ Answer: B) J/s (joule per second)

Question 214

Which of the following is a unit of electrical energy?

Options:

A) Volt (V)
B) Kilowatt-hour (kWh)
C) Ohm (Ω)
D) Coulomb (C)

Answer: B) Kilowatt-hour (kWh)

Step-by-Step Solution:

Electrical energy is calculated by:

Energy = Power × Time

If,

Power is measured in kilowatts (kW)
Time is measured in hours (h)

Then,

Energy = kW × h = kWh

Hence, kilowatt-hour (kWh) is a unit of electrical energy.

Additional Note:

Common energy units:

Joule (J) → SI Unit
Watt-hour (Wh)
Kilowatt-hour (kWh)

Also,

1 kWh = 3.6 × 10⁶ J

✔ Answer: B) Kilowatt-hour (kWh)

Question 215

A bulb has a power rating of 200 W. Calculate the energy dissipated in 5 minutes.

Options:

A) 60 J
B) 1000 J
C) 60 kJ
D) 1 kJ

Answer: C) 60 kJ

Step-by-Step Solution:

Given:

Power (P) = 200 W
Time (t) = 5 minutes = 300 seconds

Using,

Energy = Power × Time

E = Pt

Substitute the values,

E = 200 × 300

E = 60000 J

Convert into kilojoules,

60000 J = 60 kJ

Therefore,

Energy dissipated = 60 kJ

Additional Note:

Energy can also be calculated using:

E = Pt
E = VIt
E = I²Rt
E = (V²/R)t

✔ Answer: C) 60 kJ

Question 216

Which of the following is NOT a source of electrical energy?

Options:

A) Solar Cell
B) Battery
C) Potentiometer
D) Generator

Answer: C) Potentiometer

Step-by-Step Solution:

Let's examine each option:

Solar Cell: Converts solar (light) energy into electrical energy.
Battery: Converts chemical energy into electrical energy.
Generator: Converts mechanical energy into electrical energy using electromagnetic induction.
Potentiometer: It is a variable resistor used to measure or control voltage. It does not generate electrical energy.

Therefore, a potentiometer is not a source of electrical energy.

Additional Note:

Common electrical energy sources include:

Battery
Generator
Solar Cell
Fuel Cell

Measuring devices like potentiometers and voltmeters consume electrical energy instead of generating it.

✔ Answer: C) Potentiometer

Question 217

Circuit Diagram:

Calculate the energy dissipated in the circuit in 50 seconds.

Options:

A) 50 kJ
B) 50 J
C) 100 J
D) 100 kJ

Answer: A) 50 kJ

Step-by-Step Solution:

Given:

Voltage, V = 100 V
Resistance, R = 10 Ω
Time, t = 50 s

Step 1: Calculate the power dissipated

Using the power formula,

P = V²/R

Substitute the given values,

P = (100)² / 10

P = 10000 / 10

P = 1000 W

Step 2: Calculate the energy dissipated

Using,

E = Pt

Substitute the values,

E = 1000 × 50

E = 50000 J

Convert joules into kilojoules,

E = 50000 / 1000

E = 50 kJ

Therefore, the energy dissipated in the resistor is:

50 kJ

Verification

Current in the circuit,

I = V/R

I = 100/10

I = 10 A

Now,

P = VI

P = 100 × 10

P = 1000 W

Again,

E = Pt

E = 1000 × 50 = 50000 J = 50 kJ

The answer is verified.

Additional Note:

Electrical energy can be calculated using any of the following formulas:

E = Pt
E = VIt
E = I²Rt
E = (V²/R)t

where:

E = Energy (J)
P = Power (W)
V = Voltage (V)
I = Current (A)
R = Resistance (Ω)
t = Time (s)

✔ Answer: A) 50 kJ

Question 218

Which of the following is a correct expression for electrical energy?

Options:

A) V²It
B) V²Rt
C) V²t/R
D) V²t²/R

Answer: C) V²t/R

Step-by-Step Solution:

Electrical energy is given by:

E = Pt

Also,

P = V²/R

Substitute the value of power into the energy equation:

E = (V²/R) × t

Therefore,

E = V²t/R

Hence, the correct expression for electrical energy is:

V²t/R

Additional Note:

Common energy formulas are:

E = Pt
E = VIt
E = I²Rt
E = V²t/R

These formulas are derived using Ohm's Law and the electrical power equations.

✔ Answer: C) V²t/R

Question 219

Circuit Diagram:

Calculate the energy dissipated in the 10 Ω resistor in 10 seconds.

Options:

A) 400 J
B) 40 kJ
C) 4000 J
D) 4 kJ

Answer: B) 40 kJ

Step-by-Step Solution:

The 20 Ω and 10 Ω resistors are connected in parallel across a 200 V supply.

Therefore, the voltage across the 10 Ω resistor is:

V = 200 V

Step 1: Calculate the power dissipated in the 10 Ω resistor

Using the power formula,

P = V²/R

Substitute the given values,

P = (200)² / 10

P = 40000 / 10

P = 4000 W

Step 2: Calculate the energy dissipated in 10 seconds

Using,

E = Pt

Substitute the values,

E = 4000 × 10

E = 40000 J

Convert joules into kilojoules,

E = 40000 / 1000

E = 40 kJ

Therefore, the energy dissipated in the 10 Ω resistor in 10 seconds is:

40 kJ

Verification

Current through the 10 Ω resistor,

I = V/R

I = 200/10

I = 20 A

Now,

P = I²R

P = (20)² × 10

P = 400 × 10

P = 4000 W

Again,

E = Pt

E = 4000 × 10

E = 40000 J = 40 kJ

The answer is verified.

Additional Note:

For parallel circuits:

Voltage across every branch is equal to the supply voltage.
Power in each resistor is calculated using P = V²/R.
Energy is calculated using E = Pt.

✔ Answer: B) 40 kJ

Question 220

A battery converts:

Options:

A) Electrical energy into chemical energy
B) Chemical energy into electrical energy
C) Mechanical energy into electrical energy
D) Chemical energy into mechanical energy

Answer: B) Chemical energy into electrical energy

Step-by-Step Solution:

A battery stores chemical energy in the form of electrochemical reactants.

When connected to an electrical circuit, chemical reactions occur inside the battery, producing an electric current.

Thus, a battery converts:

Chemical Energy → Electrical Energy

Additional Note:

Examples of energy conversion:

Battery: Chemical → Electrical
Generator: Mechanical → Electrical
Solar Cell: Light → Electrical
Electric Motor: Electrical → Mechanical

✔ Answer: B) Chemical energy into electrical energy

Question 221

A current of 2 A flows through a wire of resistance 10 Ω for 0.5 hour. Calculate the energy dissipated by the wire.

Options:

A) 72 Wh
B) 72 kJ
C) 7200 J
D) 72 kJh

Answer: B) 72 kJ

Step-by-Step Solution:

Given:

Current (I) = 2 A
Resistance (R) = 10 Ω
Time (t) = 0.5 hour = 1800 s

Step 1: Calculate the power

Using,

P = I²R

P = (2)² × 10

P = 4 × 10

P = 40 W

Step 2: Calculate the energy

Using,

E = Pt

E = 40 × 1800

E = 72000 J

Convert into kilojoules,

E = 72 kJ

Therefore, the energy dissipated is:

72 kJ

Additional Note:

Always convert hours into seconds when calculating energy in joules.

0.5 hour = 30 minutes = 1800 seconds

✔ Answer: B) 72 kJ

Question 222

Circuit Diagram:

Calculate the energy dissipated in the 5 Ω resistor in 20 seconds.

Options:

A) 21.5 kJ
B) 2.15 kJ
C) 2.15 J
D) 21.5 J

Answer: A) 21.5 kJ

Step-by-Step Solution:

The 5 Ω and 10 Ω resistors are connected in series, so the same current flows through both resistors.

Given:

Supply Voltage, V = 220 V
Resistor 1, R₁ = 5 Ω
Resistor 2, R₂ = 10 Ω
Time, t = 20 s

Step 1: Calculate the total resistance

Since the resistors are connected in series,

RT = R₁ + R₂

RT = 5 + 10

RT = 15 Ω

Step 2: Calculate the circuit current

Using Ohm's Law,

I = V / RT

I = 220 / 15

I = 14.67 A

Step 3: Calculate the power dissipated in the 5 Ω resistor

Using the power formula,

P = I²R

Substitute the values,

P = (14.67)² × 5

P = 215.1 × 5

P ≈ 1075.6 W

Step 4: Calculate the energy dissipated

Using,

E = Pt

Substitute the values,

E = 1075.6 × 20

E ≈ 21512 J

Convert joules into kilojoules,

E ≈ 21.5 kJ

Therefore, the energy dissipated in the 5 Ω resistor in 20 seconds is:

21.5 kJ

Verification

Voltage across the 5 Ω resistor,

V₅ = IR

V₅ = 14.67 × 5

V₅ ≈ 73.35 V

Now,

P = VI

P = 73.35 × 14.67

P ≈ 1075.6 W

Again,

E = Pt

E = 1075.6 × 20

E ≈ 21512 J = 21.5 kJ

The answer is verified.

Additional Note:

For a series circuit:

Current is the same through all resistors.
Total resistance is the sum of individual resistances.
Energy can be calculated using:
- E = Pt
- E = I²Rt
- E = VIt
- E = (V²/R)t

✔ Answer: A) 21.5 kJ

Question 223

Practically, if 10 kJ of energy is supplied to a device, how much energy will the device deliver back?

Options:

A) Exactly 10 kJ
B) Less than 10 kJ
C) More than 10 kJ
D) Zero

Answer: B) Less than 10 kJ

Step-by-Step Solution:

According to the Law of Conservation of Energy, energy cannot be created or destroyed.

However, in practical devices, a portion of the supplied energy is lost due to:

Heat
Sound
Friction
Magnetic losses
Mechanical losses

Therefore, the useful output energy is always less than the supplied input energy.

For example:

If the efficiency of a device is 90%,

Input Energy = 10 kJ

Output Energy = 0.9 × 10 = 9 kJ

Thus, the output energy is less than the input energy.

Additional Note:

Efficiency is defined as:

Efficiency = (Output Energy / Input Energy) × 100%

Since no practical electrical device is 100% efficient, some energy is always lost.

✔ Answer: B) Less than 10 kJ

Question 224

Materials that easily allow the flow of electric current are called:

Options:

A) Insulators
B) Conductors
C) Dielectrics
D) Semiconductors

Answer: B) Conductors

Step-by-Step Solution:

A conductor is a material that allows electric current to pass through it easily because it contains a large number of free electrons.

Examples of conductors include:

Copper
Aluminium
Silver
Gold

Since electric current is the flow of electrons, materials with many free electrons have low resistance and conduct electricity well.

Additional Note:

Conductors: Low resistance, high conductivity.
Insulators: High resistance, very poor conductivity.
Semiconductors: Conductivity lies between conductors and insulators.
Dielectrics: Excellent insulating materials used in capacitors.

✔ Answer: B) Conductors

Question 225

Two wires are made of the same material and have the same cross-sectional area. One wire is 2 m long and the other is 5 m long. Which wire has the higher resistance?

Options:

A) Both have equal resistance
B) The 2 m wire has higher resistance
C) The 5 m wire has higher resistance
D) Cannot be determined

Answer: C) The 5 m wire has higher resistance

Step-by-Step Solution:

Resistance is given by:

R = ρL/A

where,

R = Resistance
ρ = Resistivity
L = Length
A = Cross-sectional area

Since both wires have:

Same material (ρ is constant)
Same cross-sectional area (A is constant)

Resistance depends only on the length.

As the length increases, resistance also increases.

Therefore,

5 m wire has greater resistance than the 2 m wire.

Additional Note:

Resistance is:

Directly proportional to Length
Inversely proportional to Area

✔ Answer: C) The 5 m wire has higher resistance

Question 226

Two wires have the same length and are made of the same material. One has a cross-sectional area of 10 m² and the other 15 m². Which wire has the higher resistance?

Options:

A) Both have equal resistance
B) The 10 m² wire has higher resistance
C) The 15 m² wire has higher resistance
D) Cannot be determined

Answer: B) The 10 m² wire has higher resistance

Step-by-Step Solution:

Resistance is given by:

R = ρL/A

Since,

Same material
Same length

Resistance depends only on the cross-sectional area.

Smaller area means larger resistance.

Therefore,

The 10 m² wire has more resistance.

Additional Note:

Increasing the conductor's cross-sectional area decreases its resistance because more electrons can flow through the conductor.

✔ Answer: B) The 10 m² wire has higher resistance

Question 227

Which of the following statements is true regarding electrical resistance?

Options:

A) Resistance is directly proportional to the length of the wire.
B) Resistance is directly proportional to the cross-sectional area.
C) Resistance is inversely proportional to the length.
D) Resistance is inversely proportional to the resistivity.

Answer: A) Resistance is directly proportional to the length of the wire

Step-by-Step Solution:

The resistance of a conductor is given by:

R = ρL/A

From this equation,

Resistance increases with length.
Resistance decreases with cross-sectional area.
Resistance increases with resistivity.

Hence,

Only Option A is correct.

Additional Note:

Resistance is:

Directly proportional to Length (L)
Directly proportional to Resistivity (ρ)
Inversely proportional to Area (A)

✔ Answer: A) Resistance is directly proportional to the length of the wire

Question 228

Circuit Diagram:

A wire has the same resistance as shown in the above circuit. Calculate its resistivity if the wire length is 10 m and its cross-sectional area is 2 m².

Options:

A) 16 Ω·m
B) 8 Ω·m
C) 16 kΩ·m
D) 8 kΩ·m

Answer: B) 8 Ω·m

Step-by-Step Solution:

Given:

Voltage, V = 200 V
Current, I = 5 A
Length of wire, L = 10 m
Cross-sectional area, A = 2 m²

Step 1: Calculate the resistance

Using Ohm's Law,

R = V/I

Substitute the values,

R = 200/5

R = 40 Ω

Step 2: Calculate the resistivity

Using the resistivity formula,

ρ = RA/L

Substitute the values,

ρ = (40 × 2)/10

ρ = 80/10

ρ = 8 Ω·m

Therefore,

Resistivity = 8 Ω·m

Verification

Using,

R = ρL/A

Substitute the values,

R = (8 × 10)/2

R = 80/2

R = 40 Ω

The calculated resistance matches the given circuit.

Additional Note:

The resistance of a conductor is related to its dimensions by:

R = ρL/A

where:

R = Resistance (Ω)
ρ = Resistivity (Ω·m)
L = Length (m)
A = Cross-sectional area (m²)

A material with higher resistivity offers greater opposition to the flow of electric current.

✔ Answer: B) 8 Ω·m

Question 229

Which of the following is the SI unit of resistivity?

Options:

A) Ω/m
B) Ω/m²
C) Ω·m
D) Ω·m²

Answer: C) Ω·m

Step-by-Step Solution:

The resistivity of a material is given by:

ρ = RA/L

where:

ρ = Resistivity
R = Resistance (Ω)
A = Cross-sectional area (m²)
L = Length (m)

Therefore, the SI unit is:

Ω × m² / m = Ω·m

Hence, the SI unit of resistivity is ohm-metre (Ω·m).

Additional Note:

The SI unit of resistivity is:

Ω·m (Ohm-metre)

It represents the resistance of a conductor having:

Length = 1 m
Cross-sectional area = 1 m²

✔ Answer: C) Ω·m

Question 230

What is the resistivity of copper at 20°C?

Options:

A) 1.68 × 10⁻⁸ Ω·m
B) 2.7 × 10⁻⁸ Ω·m
C) 7.3 × 10⁻⁸ Ω·m
D) 5.35 × 10⁻⁸ Ω·m

Answer: A) 1.68 × 10⁻⁸ Ω·m

Step-by-Step Solution:

Resistivity is an intrinsic property of a material.

At 20°C, the resistivity of copper is approximately:

ρ = 1.68 × 10⁻⁸ Ω·m

Copper has a very low resistivity, making it one of the best conductors of electricity.

Additional Note:

Approximate resistivity values at 20°C:

Material	Resistivity (Ω·m)
Silver	1.59 × 10⁻⁸
Copper	1.68 × 10⁻⁸
Gold	2.44 × 10⁻⁸
Aluminium	2.82 × 10⁻⁸

✔ Answer: A) 1.68 × 10⁻⁸ Ω·m

Question 231

Calculate the ratio of the resistivities of two wires having the same length and the same resistance, with cross-sectional areas of 2 m² and 5 m² respectively.

Options:

A) 5 : 7
B) 2 : 7
C) 2 : 5
D) 7 : 5

Answer: C) 2 : 5

Step-by-Step Solution:

The resistivity is given by:

ρ = RA/L

For both wires:

Resistance (R) is the same.
Length (L) is the same.

Therefore,

ρ ∝ A

Hence,

ρ₁/ρ₂ = A₁/A₂

Substitute the given areas,

ρ₁/ρ₂ = 2/5

Therefore,

Ratio of resistivities = 2 : 5

Additional Note:

If R and L are constant,

Resistivity is directly proportional to the cross-sectional area.

✔ Answer: C) 2 : 5

Question 232

Which of the following statements is true regarding resistivity?

Options:

A) Resistivity depends on temperature.
B) Resistivity does not depend on temperature.
C) Resistivity depends on the length of the conductor.
D) Resistivity depends on the cross-sectional area.

Answer: A) Resistivity depends on temperature.

Step-by-Step Solution:

Resistivity is a property of the material.

It is independent of:

Length
Cross-sectional area

However, it changes with temperature.

For metallic conductors:

As temperature increases,
Resistivity also increases.

Hence, the correct statement is:

Resistivity depends on temperature.

Additional Note:

Resistivity depends only on:

Material
Temperature

It does not depend on the dimensions of the conductor.

✔ Answer: A) Resistivity depends on temperature.

Question 233

The reciprocal of resistivity is called:

Options:

A) Conductance
B) Resistance
C) Conductivity
D) Impedance

Answer: C) Conductivity

Step-by-Step Solution:

Conductivity is defined as the reciprocal of resistivity.

Mathematically,

σ = 1/ρ

where:

σ = Conductivity
ρ = Resistivity

A material having high conductivity has low resistivity and conducts electricity easily.

Additional Note:

Relationship between electrical quantities:

Resistance (R) ↔ Conductance (G)

G = 1/R
Resistivity (ρ) ↔ Conductivity (σ)

σ = 1/ρ

Conductivity is measured in:

Siemens per metre (S/m)

✔ Answer: C) Conductivity

Question 234

The resistance of pure metals __________ as temperature increases.

Options:

A) Increases
B) Decreases
C) Remains the same
D) Becomes zero

Answer: A) Increases

Step-by-Step Solution:

In pure metals, there are a large number of free electrons available for conduction.

When the temperature increases:

The atoms of the metal vibrate more vigorously.
These vibrations obstruct the movement of free electrons.
As a result, electron flow becomes more difficult.

Therefore, the electrical resistance of pure metals increases with an increase in temperature.

Additional Note:

Metals have a positive temperature coefficient of resistance, meaning:

Temperature ↑ → Resistance ↑

Examples:

Copper
Aluminium
Silver

✔ Answer: A) Increases

Question 235

The resistance of insulators __________ as temperature increases.

Options:

A) Increases
B) Decreases
C) Remains the same
D) Becomes zero

Answer: B) Decreases

Step-by-Step Solution:

In insulators, very few free electrons are available at normal temperatures.

When the temperature increases:

Some electrons gain enough energy to move into the conduction band.
The number of charge carriers increases.
Conductivity increases.

Since,

Resistance = 1 / Conductance

the resistance decreases.

Additional Note:

Insulators have a negative temperature coefficient.

Examples:

Glass
Rubber
Plastic
Mica

✔ Answer: B) Decreases

Question 236

Which of the following statements is true about metals?

Options:

A) Metals have a positive temperature coefficient.
B) Metals have a negative temperature coefficient.
C) Metals have zero temperature coefficient.
D) Metals have an infinite temperature coefficient.

Answer: A) Metals have a positive temperature coefficient

Step-by-Step Solution:

For metals,

As temperature increases,
Resistance also increases.

Therefore,

The temperature coefficient of resistance is positive.

Mathematically,

α > 0

where α is the temperature coefficient of resistance.

Additional Note:

Examples of materials with positive temperature coefficient:

Copper
Aluminium
Silver
Iron

✔ Answer: A) Metals have a positive temperature coefficient

Question 237

Which of the following statements is true about insulators?

Options:

A) Insulators have a positive temperature coefficient.
B) Insulators have a negative temperature coefficient.
C) Insulators have a zero temperature coefficient.
D) Insulators have an infinite temperature coefficient.

Answer: B) Insulators have a negative temperature coefficient

Step-by-Step Solution:

In insulators,

Increasing temperature creates more free charge carriers.
Conductivity increases.
Resistance decreases.

Therefore,

The temperature coefficient is negative.

Mathematically,

α < 0

Additional Note:

Materials having a negative temperature coefficient include:

Glass
Rubber
Silicon
Germanium (Semiconductors)

✔ Answer: B) Insulators have a negative temperature coefficient

Question 238

What is the SI unit of the temperature coefficient of resistance?

Options:

A) Ω/°C
B) Ω·°C
C) °C⁻¹
D) °C

Answer: C) °C⁻¹

Step-by-Step Solution:

The resistance at any temperature is given by:

R = R₀ [1 + α(T − T₀)]

where:

R = Resistance at temperature T
R₀ = Resistance at reference temperature T₀
α = Temperature coefficient of resistance

Rearranging,

α = (R/R₀ − 1)/(T − T₀)

Since the numerator is dimensionless, the unit of α is the reciprocal of temperature.

Therefore,

Unit of α = 1/°C = °C⁻¹

Additional Note:

The temperature coefficient indicates how much the resistance changes per degree change in temperature.

Typical values:

Copper: 0.0039 °C⁻¹
Aluminium: 0.0043 °C⁻¹

A positive value indicates that resistance increases with temperature.

✔ Answer: C) °C⁻¹

Question 239

A copper coil has a resistance of 200 Ω at 0°C. Calculate its resistance at 80°C if the temperature coefficient of resistance is 0.004041 °C⁻¹.

Options:

A) 264.65 Ω
B) 264.65 kΩ
C) 286.65 Ω
D) 286.65 kΩ

Answer: A) 264.65 Ω

Step-by-Step Solution:

Given:

Initial resistance, R₀ = 200 Ω
Initial temperature, T₀ = 0°C
Final temperature, T = 80°C
Temperature coefficient, α = 0.004041 °C⁻¹

The resistance at temperature T is calculated using:

R = R₀ [1 + α(T − T₀)]

Substitute the given values,

R = 200 [1 + (0.004041 × 80)]

R = 200 [1 + 0.32328]

R = 200 × 1.32328

R = 264.656 Ω

R ≈ 264.65 Ω

Therefore, the resistance of the copper coil at 80°C is:

264.65 Ω

Additional Note:

For conductors,

Temperature ↑ ⇒ Resistance ↑

General formula:

R = R₀ [1 + αΔT]

where,

ΔT = T − T₀

✔ Answer: A) 264.65 Ω

Question 240

The temperature of a coil cannot be measured by which of the following methods?

Options:

A) Thermometer
B) Increase in the resistance of the coil
C) Thermocouple (Thermo-junction) embedded in the coil
D) Calorimeter

Answer: D) Calorimeter

Step-by-Step Solution:

A calorimeter is used to measure the quantity of heat, not the temperature of an object.

The temperature of a coil can be measured by:

Thermometer
Measuring the increase in resistance
Embedded thermocouples (thermo-junctions)

Therefore, a calorimeter is not used for measuring the coil temperature.

Additional Note:

Common methods of measuring coil temperature:

Thermometer
RTD (Resistance Temperature Detector)
Thermocouple
Resistance method

✔ Answer: D) Calorimeter

Question 241

The rise or fall in resistance with an increase in temperature depends on:

Options:

A) The property of the conductor material
B) The current flowing through the material
C) Both the material property and the current
D) None of the above

Answer: A) The property of the conductor material

Step-by-Step Solution:

The change in resistance with temperature depends on the temperature coefficient of resistance, which is a property of the material.

For example:

Metals → Resistance increases with temperature.
Semiconductors → Resistance decreases with temperature.
Insulators → Resistance generally decreases with temperature.

Thus, the behavior depends on the material, not on the current flowing through it.

Additional Note:

The relationship is given by:

R = R₀ [1 + αΔT]

where α depends only on the material.

✔ Answer: A) The property of the conductor material

Question 242

If the temperature of a semiconductor keeps increasing and its resistance continuously decreases, this phenomenon is called:

Options:

A) Avalanche Breakdown
B) Zener Breakdown
C) Thermal Runaway
D) Avalanche Runway

Answer: C) Thermal Runaway

Step-by-Step Solution:

In a semiconductor:

Temperature increases.
Resistance decreases.
Current increases.
Increased current produces more heat.
More heat further decreases the resistance.

This continuous cycle eventually damages the device.

This phenomenon is called Thermal Runaway.

Additional Note:

Thermal runaway is prevented by:

Heat sinks
Cooling fans
Proper biasing
Temperature compensation circuits

It commonly occurs in:

Transistors
Power semiconductors
Diodes

✔ Answer: C) Thermal Runaway

Question 243

Materials having resistance almost equal to zero are called:

Options:

A) Semiconductors
B) Conductors
C) Superconductors
D) Insulators

Answer: C) Superconductors

Step-by-Step Solution:

A superconductor is a material whose electrical resistance becomes nearly zero when cooled below its critical temperature.

At this condition:

Resistance ≈ 0 Ω
Conductivity becomes extremely high.
No I²R power loss occurs.

Therefore, such materials are called Superconductors.

Additional Note:

Properties of superconductors:

Zero electrical resistance.
Perfect diamagnetism (Meissner Effect).
Extremely high current-carrying capability.

Applications include:

MRI machines
Maglev trains
Particle accelerators
Superconducting power cables

✔ Answer: C) Superconductors

Circuit Diagram:

Find the values of $I_1$ , $I_2$ , and $I_3$ using the matrix method.

Options:

A) −0.566 A, 1.29 A, −1.91 A
B) −1.29 A, −0.566 A, 1.91 A
C) 1.29 A, −0.566 A, −1.91 A
D) 1.91 A, 0.566 A, 1.29 A

Answer: C) 1.29 A, −0.566 A, −1.91 A

Step-by-Step Solution:

Using Mesh Analysis, the three loop equations can be written in matrix form as:


|  3  -2   0 |   | I1 |   |   5  |
| -2   9  -4 | × | I2 | = |   0  |
|  0  -4   9 |   | I3 |   | -15  |

Where:

3 = 1 + 2
9 = 2 + 3 + 4
9 = 4 + 5
Shared resistances appear as negative terms.

Step 1: Solve the matrix

Using matrix inversion or Cramer's Rule,

the mesh currents are obtained as:

I₁ = 1.29 A

I₂ = −0.566 A

I₃ = −1.91 A

Step 2: Interpret the negative values

The assumed mesh currents are clockwise.

The negative signs indicate that:

I₂ actually flows opposite to the assumed direction.
I₃ also flows opposite to the assumed direction.

Final Answer

I₁ = 1.29 A
I₂ = −0.566 A
I₃ = −1.91 A

Additional Note:

For mesh analysis using the matrix method:

Diagonal elements = Sum of resistances in each mesh.
Off-diagonal elements = Negative of the common resistance between two adjacent meshes.
Right-hand side = Algebraic sum of source voltages.

The matrix equation is written as:

[R][I] = [V]

where:

[R] = Resistance matrix
[I] = Mesh current matrix
[V] = Voltage source matrix

✔ Answer: C) 1.29 A, −0.566 A, −1.91 A

Question 245

Circuit Diagram:

Find the value of the source voltage $V$ if $I_3 = 0$ A.

Options:

A) 1.739 V
B) 6.5 V
C) 4.5 V
D) 2.739 V

Answer: A) 1.739 V

Step-by-Step Solution:

Assume clockwise mesh currents I₁, I₂, and I₃.

Since I₃ = 0 A, only the first two mesh currents need to be determined.

Step 1: Apply KVL to Mesh 1

$5-1I_1-2(I_1-I_2)=0$

Simplifying,

$5-3I_1+2I_2=0$

Step 2: Apply KVL to Mesh 2

Since I₃ = 0,

$2(I_2-I_1)+3I_2+4(I_2-0)=0$

Simplifying,

$-2I_1+9I_2=0$

Step 3: Solve the simultaneous equations

From

$9I_2=2I_1$ $I_1=4.5I_2$

Substitute into the first equation,

$5-3(4.5I_2)+2I_2=0$ $5-13.5I_2+2I_2=0$ $5=11.5I_2$ $I_2=0.4348\text{ A}$

Step 4: Calculate the unknown source voltage

Applying KVL to Mesh 3,

$-4(I_3-I_2)-5I_3+V=0$

Since I₃ = 0,

$4I_2=V$

Substitute the value of I₂,

$V=4\times0.4348$ $V=1.739\text{ V}$

Final Answer

V = 1.739 V

Additional Note:

For mesh analysis:

Apply Kirchhoff's Voltage Law (KVL) to each independent mesh.
If a mesh current is specified (such as I₃ = 0 A), substitute it before solving.
Once the mesh currents are known, the unknown source voltage is obtained using the corresponding mesh equation.

✔ Answer: A) 1.739 V

Question 246

Circuit Diagram:

Find the value of resistance $R$ if the power consumed in the circuit is 1000 W.

Options:

A) 10 Ω
B) 9 Ω
C) 8 Ω
D) 7 Ω

Answer: C) 8 Ω

Step-by-Step Solution:

Given:

Supply Voltage, V = 100 V
Total Power, P = 1000 W

Step 1: Calculate the circuit current

Using the power formula,

P = VI

Substitute the values,

1000 = 100 × I

I = 1000 / 100

I = 10 A

Step 2: Calculate the voltage across the 2 Ω resistor

Using Ohm's Law,

V₂ = IR

V₂ = 10 × 2

V₂ = 20 V

Step 3: Calculate the voltage across resistor $R$

Since the resistors are connected in series, the sum of voltage drops equals the supply voltage.

VR = 100 − 20

VR = 80 V

Step 4: Calculate the value of $R$

Using Ohm's Law,

R = V/I

R = 80 / 10

R = 8 Ω

Verification

Total resistance,

RT = 2 + 8 = 10 Ω

Current,

I = V/RT = 100/10 = 10 A

Power,

P = VI = 100 × 10 = 1000 W

The result matches the given power, so the answer is correct.

Additional Note:

For a series circuit:

Total resistance: RT = R₁ + R₂
Current is the same through all resistors.
Voltage divides in proportion to resistance.

Useful formulas:

P = VI
P = I²R
P = V²/R

✔ Answer: C) 8 Ω

Question 247

Circuit Diagram:

Find the current through the 4 Ω resistor.

Options:

A) 5 A
B) 0 A
C) 2.2 A
D) 20 A

Answer: B) 0 A

Step-by-Step Solution:

Observe the circuit carefully.

The 4 Ω resistor is connected in parallel with an ideal short circuit (a wire).

An ideal wire has:

Resistance = 0 Ω
Voltage drop = 0 V

Since the voltage across the short circuit is zero, the voltage across the 4 Ω resistor is also zero.

Using Ohm's Law,

I = V/R

Substitute the values,

I = 0/4

I = 0 A

Therefore, no current flows through the 4 Ω resistor.

Why does this happen?

Electric current always follows the path of least resistance.

Since the resistor is bypassed by a wire of zero resistance, all the current flows through the short-circuit path instead of the 4 Ω resistor.

Thus, the 4 Ω resistor is short-circuited.

Additional Note:

When a resistor is connected in parallel with an ideal short circuit:

Voltage across the resistor = 0 V
Current through the resistor = 0 A
The resistor has no effect on the circuit.

This is a common concept used in basic circuit analysis.

✔ Answer: B) 0 A

Question 248

Nodal analysis is generally used to determine:

Options:

A) Voltage
B) Current
C) Resistance
D) Power

Answer: A) Voltage

Step-by-Step Solution:

Nodal Analysis is based on Kirchhoff's Current Law (KCL).

The method involves:

Selecting a reference (ground) node.
Assigning voltages to the remaining nodes.
Applying KCL at each node.
Solving the equations to determine the unknown node voltages.

Therefore, nodal analysis is primarily used to determine voltages.

Additional Note:

Analysis Method	Primary Quantity	Law Used
Nodal Analysis	Node Voltage	Kirchhoff's Current Law (KCL)
Mesh Analysis	Mesh Current	Kirchhoff's Voltage Law (KVL)

✔ Answer: A) Voltage

Question 249

Mesh analysis is generally used to determine:

Options:

A) Voltage
B) Current
C) Resistance
D) Power

Answer: B) Current

Step-by-Step Solution:

Mesh Analysis is based on Kirchhoff's Voltage Law (KVL).

The procedure is:

Identify all independent meshes.
Assume a mesh current for each loop.
Apply KVL around each mesh.
Solve the simultaneous equations to obtain the mesh currents.

Therefore, mesh analysis is primarily used to determine currents in a circuit.

Additional Note:

Method	Based on	Determines
Mesh Analysis	KVL	Mesh Currents
Nodal Analysis	KCL	Node Voltages

✔ Answer: B) Current

Question 250

Circuit Diagram:

What is the current in the circuit?

Options:

A) 0 A
B) 15 A
C) 5 A
D) 10 A

Answer: A) 0 A

Step-by-Step Solution:

Given:

Resistors: 1 Ω, 2 Ω, and 3 Ω
Voltage Sources: 10 V, 20 V, and 30 V

Step 1: Calculate the total resistance

Since all resistors are connected in series,

R = 1 + 2 + 3 = 6 Ω

Step 2: Apply Kirchhoff's Voltage Law (KVL)

Traversing the loop in the clockwise direction,

−10 − 20 + 30 = 0 V

Hence, the net applied voltage is:

V = 0 V

Step 3: Calculate the circuit current

Using Ohm's Law,

I = V/R

Substituting the values,

I = 0/6

I = 0 A

Therefore, no current flows in the circuit.

Verification

Using KVL,

ΣV = 0

Since the driving voltage is zero,

I = 0 A

The result satisfies both Ohm's Law and Kirchhoff's Voltage Law.

Additional Note:

If the algebraic sum of all voltage sources in a closed loop is zero, there is no net electromotive force (EMF) to drive current through the circuit.

Therefore, regardless of the total resistance, the circuit current is:

I = 0 A

✔ Answer: A) 0 A

2000 Basic Electrical Engineering Fully Solved MCQs-5

Question 201

Answer: B) 5.21 V

Step-by-Step Solution:

Step 1: Apply KVL to Loop 1

Step 2: Apply KVL to Loop 2

Step 3: Solve the simultaneous equations

Step 4: Calculate the voltage across the 5 Ω resistor

Additional Note:

Question 202

Answer: A) 3.5 V

Step-by-Step Solution:

Step 1: Find the voltage at point A

Step 2: Find the voltage at point B

Step 3: Calculate VABV_{AB}VAB​

Verification using KVL

Additional Note:

Question 203

Answer: A) Mesh Analysis

Step-by-Step Solution:

Additional Note:

Question 204

Answer: D) P = I/R

Step-by-Step Solution:

Additional Note:

Question 205

Answer: D) Power is proportional to both voltage and current

Step-by-Step Solution:

Additional Note:

Question 206

Answer: A) 75 W

Step-by-Step Solution:

Additional Note:

Question 207

Answer: C) Energy

Step-by-Step Solution:

Additional Note:

Question 208

Answer: B) 2 kW

Step-by-Step Solution:

Verification using Ohm's Law

Additional Note:

Question 209

Answer: A) 15 kJ

Step-by-Step Solution:

Additional Note:

Question 210

Answer: B) 500 W, 500 W

Step-by-Step Solution:

Step 1: Calculate the total resistance

Step 2: Calculate the circuit current

Step 3: Calculate the power dissipated in each resistor

Verification

Additional Note:

Question 211

Answer: B) 1 kW, 1 kW

Step-by-Step Solution:

Step 1: Calculate the power in each resistor

Verification using Ohm's Law

Additional Note:

Question 212

Answer: C) 1.5 kWh

Step-by-Step Solution:

Additional Note:

Question 213

Answer: B) J/s (joule per second)

Step-by-Step Solution:

Additional Note:

Question 214

Answer: B) Kilowatt-hour (kWh)

Step-by-Step Solution:

Additional Note:

Question 215

Answer: C) 60 kJ

Step-by-Step Solution:

Additional Note:

Question 216

Answer: C) Potentiometer

Step-by-Step Solution:

Additional Note:

Step 3: Calculate $V_{AB}$