Question 151

Which connection is the most cost-efficient?

Options:

A) Series
B) Parallel
C) Either series or parallel
D) Neither series nor parallel

Answer: A) Series

Step-by-Step Solution:

In a series connection, the supply voltage is shared among all connected components.

Because the voltage is divided:

Lower voltage devices can be used.
The overall cost of components may be reduced.

For this reason, a series connection is often considered more economical where independent operation of each load is not required.

Additional Note:

Comparison:

Series Connection	Parallel Connection
Lower component voltage	Full supply voltage across each load
Economical in some applications	Preferred for household wiring
One failure affects all loads	Loads operate independently

Although series circuits may be more cost-effective in some situations, parallel circuits are generally preferred in domestic and industrial installations because each load receives the rated supply voltage and operates independently.

✔ Answer: A) Series

Question 152

Circuit Diagram:

Equivalent layout (same circuit):
Left branch: 4 Ω
Bottom branch: 1.5 Ω
Right branch: 3 Ω in series with 2 Ω
Diagonal branch: 5 Ω

Calculate the equivalent resistance between A and B.

Options:

A) 2 Ω
B) 4 Ω
C) 6 Ω
D) 8 Ω

Answer: A) 2 Ω

Step-by-Step Solution:

Step 1: Simplify the right-hand branch.

The 3 Ω and 2 Ω resistors are connected in series.

R = 3 + 2 = 5 Ω

Step 2: Simplify the parallel combination.

The new 5 Ω branch is in parallel with the diagonal 5 Ω resistor.

Rp = (5 × 5) / (5 + 5)

Rp = 25 / 10

Rp = 2.5 Ω

Step 3: Add the bottom series resistor.

The 2.5 Ω equivalent is in series with the 1.5 Ω resistor.

R = 2.5 + 1.5 = 4 Ω

Step 4: Calculate the final parallel combination.

The 4 Ω obtained above is in parallel with the left 4 Ω resistor.

RT = (4 × 4) / (4 + 4)

RT = 16 / 8

RT = 2 Ω

Therefore,

Equivalent Resistance between A and B = 2 Ω

Additional Note:

To solve mixed resistor networks:

Combine series resistors first.
Combine parallel resistors.
Continue simplifying until a single equivalent resistance is obtained.

For this circuit:

3 Ω + 2 Ω = 5 Ω
5 Ω || 5 Ω = 2.5 Ω
2.5 Ω + 1.5 Ω = 4 Ω
4 Ω || 4 Ω = 2 Ω

✔ Answer: A) 2 Ω

Question 153

Circuit Diagram:


                    20 Ω
              ┌────/\/\/\────┐
              │              │
20 Ω          │    20 Ω      │          20 Ω
──/\/\/\─────● A──/\/\/\──● B────/\/\/\──
              │              │
              │    20 Ω      │
              └────/\/\/\────┘

Calculate the equivalent resistance between A and B.

Options:

A) 6.67 Ω
B) 46.67 Ω
C) 26.67 Ω
D) 10.67 Ω

Answer: A) 6.67 Ω

Step-by-Step Solution:

Since the resistance is to be measured between points A and B, the two outer 20 Ω resistors are not included in the calculation because they are outside the measurement terminals.

Between A and B, there are three 20 Ω resistors connected in parallel.

Using the parallel resistance formula:

1 / RT = 1/20 + 1/20 + 1/20

1 / RT = 3/20

Taking the reciprocal,

RT = 20/3

RT = 6.67 Ω

Therefore, the equivalent resistance between A and B is:

RT = 6.67 Ω

Additional Note:

For n identical resistors connected in parallel:

Equivalent Resistance = R / n

where:

R = Resistance of one resistor
n = Number of identical resistors

Here,

RT = 20 / 3 = 6.67 Ω

This shortcut is applicable whenever all parallel resistors have the same resistance value.

✔ Answer: A) 6.67 Ω

Question 154

Circuit Diagram:

Find the value of V, if V₁ = 20 V and the current source is 6 A.

Options:

A) 10 V
B) 12 V
C) 14 V
D) 16 V

Answer: B) 12 V

Step-by-Step Solution:

Step 1: Calculate the current through the 10 Ω resistor.

Using Ohm's Law,

I = V / R

Given:

V₁ = 20 V
R = 10 Ω

Therefore,

I₁₀ = 20 / 10 = 2 A

Step 2: Apply Kirchhoff's Current Law (KCL) at Node 1.

According to KCL,

Incoming Current = Outgoing Current

Current supplied by the source:

6 A

Current through the 10 Ω resistor:

2 A

Therefore, the current through the 2 Ω resistor is:

I₂ = 6 − 2 = 4 A

Step 3: Calculate the voltage drop across the 2 Ω resistor.

Using Ohm's Law,

V = IR

V = 4 × 2

V = 8 V

Step 4: Calculate V₂.

Since the voltage at Node 1 is 20 V,

V₂ = 20 − 8

V₂ = 12 V

Since the 5 Ω resistor and the output terminals are connected in parallel,

V = V₂ = 12 V

Additional Note:

Kirchhoff's Current Law (KCL):

The algebraic sum of currents at any node is zero.
Total current entering a node equals the total current leaving the node.

Ohm's Law:

V = IR
I = V / R
R = V / I

✔ Answer: B) 12 V

Question 155

Circuit Diagram:

Calculate the current A.

Options:

A) 5 A
B) 10 A
C) 15 A
D) 20 A

Answer: C) 15 A

Step-by-Step Solution:

Apply Kirchhoff's Current Law (KCL) at the junction.

According to KCL:

Total Current Entering = Total Current Leaving

Current entering the junction:

5 A
10 A

Therefore,

A = 5 + 10

A = 15 A

Hence, the current leaving the junction is:

15 A

Additional Note:

Kirchhoff's Current Law (KCL) states:

The algebraic sum of currents at any junction is zero.

or equivalently,

Total Current Entering = Total Current Leaving

Mathematically,

Σ I(in) = Σ I(out)

KCL is based on the Law of Conservation of Charge, which states that electric charge can neither be created nor destroyed at a junction.

✔ Answer: C) 15 A

Question 156

Circuit Diagram:

Calculate the current through the 20 Ω resistor.

Assume the lower terminal of the 20 Ω resistor is at 0 V.

Options:

A) 20 A
B) 1 A
C) 0.67 A
D) 0.33 A

Answer: D) 0.33 A

Step-by-Step Solution:

Let the voltage at the top node be V volts.

The currents through the two resistors are:

Through the 10 Ω resistor:

I₁ = V / 10

Through the 20 Ω resistor:

I₂ = V / 20

Apply Kirchhoff's Current Law (KCL) at the top node.

The current supplied by the source is 1 A.

Therefore,

V/10 + V/20 = 1

Taking the LCM:

(2V + V) / 20 = 1

3V = 20

V = 20 / 3 V

Now calculate the current through the 20 Ω resistor.

Using Ohm's Law,

I = V / R

I = (20/3) / 20

I = 1/3 A

I ≈ 0.33 A

Therefore, the current through the 20 Ω resistor is:

0.33 A

Additional Note:

For parallel circuits:

The voltage across each branch is the same.
The source current divides among the branches according to their resistance.

Kirchhoff's Current Law:

Σ I(in) = Σ I(out)

Ohm's Law:

V = IR
I = V / R
R = V / I

✔ Answer: D) 0.33 A

Question 157

Circuit Diagram:

Calculate the value of I₃, if I₁ = 2 A and I₂ = 3 A.

Options:

A) –5 A
B) 5 A
C) 1 A
D) –1 A

Answer: A) –5 A

Step-by-Step Solution:

Apply Kirchhoff's Current Law (KCL).

According to KCL,

The algebraic sum of currents at a junction is zero.

Therefore,

I₁ + I₂ + I₃ = 0

Substitute the given values:

2 + 3 + I₃ = 0

I₃ = –(2 + 3)

I₃ = –5 A

The negative sign indicates that the actual direction of I₃ is opposite to the assumed direction shown in the circuit.

Therefore,

I₃ = –5 A

Additional Note:

The negative value of current does not mean the current is negative physically. It simply indicates that the actual current flows in the direction opposite to the assumed reference direction.

Kirchhoff's Current Law (KCL):

Σ I = 0
Total Current Entering = Total Current Leaving

When solving circuit problems, you may assume any current direction. If the calculated current is negative, the true current flows in the opposite direction.

✔ Answer: A) –5 A

Question 158

Circuit Diagram:

Find the values of i₂, i₄, and i₅ if i₁ = 3 A, i₃ = 1 A, and i₆ = 1 A.

Options:

A) 2 A, –1 A, 2 A
B) 4 A, –2 A, 4 A
C) 2 A, 1 A, 2 A
D) 4 A, 2 A, 4 A

Answer: A) 2 A, –1 A, 2 A

Step-by-Step Solution:

Step 1: Apply KCL at Junction a

According to Kirchhoff's Current Law,

Incoming Current = Outgoing Current

At node a,

i₁ = i₂ + i₃

Substitute the given values:

3 = i₂ + 1

i₂ = 2 A

Step 2: Apply KCL at Junction b

At node b,

i₄ + i₂ = i₆

Substitute the known values:

i₄ + 2 = 1

i₄ = –1 A

The negative sign indicates that the actual current flows opposite to the assumed direction.

Step 3: Apply KCL at Junction c

At node c,

i₃ = i₄ + i₅

Substitute the values:

1 = (–1) + i₅

i₅ = 2 A

Final Values

i₂ = 2 A
i₄ = –1 A
i₅ = 2 A

Additional Note:

Kirchhoff's Current Law (KCL):

The algebraic sum of currents at any junction is zero.
Total current entering a node equals the total current leaving the node.

A negative current indicates that the actual current flows in the direction opposite to the assumed direction in the circuit.

✔ Answer: A) 2 A, –1 A, 2 A

Question 159

What is the value of current if a 50 C charge flows through a conductor in 5 seconds?

Options:

A) 5 A
B) 10 A
C) 15 A
D) 20 A

Answer: B) 10 A

Step-by-Step Solution:

Current is defined as the rate of flow of electric charge.

The formula is:

I = Q / t

where:

I = Current (A)
Q = Charge (C)
t = Time (s)

Given:

Q = 50 C
t = 5 s

Substitute the values:

I = 50 / 5

I = 10 A

Therefore, the current flowing through the conductor is:

10 A

Additional Note:

SI unit of current = Ampere (A)
1 Ampere = 1 Coulomb/second

✔ Answer: B) 10 A

Question 160

KCL deals with the conservation of:

Options:

A) Momentum
B) Mass
C) Potential Energy
D) Charge

Answer: D) Charge

Step-by-Step Solution:

Kirchhoff's Current Law (KCL) states that:

The total current entering a junction is equal to the total current leaving the junction.

Since current is the flow of electric charge, KCL is based on the Law of Conservation of Charge.

No electric charge is created or destroyed at a node.

Therefore, KCL represents the conservation of charge.

Additional Note:

Mathematically,

Σ I(in) = Σ I(out)

Σ I = 0

KCL is one of the fundamental laws used in electrical circuit analysis.

✔ Answer: D) Charge

Question 161

KCL is applied at a:

Options:

A) Loop
B) Node
C) Both loop and node
D) Neither loop nor node

Answer: B) Node

Step-by-Step Solution:

A node is a point where two or more circuit elements are connected.

Kirchhoff's Current Law compares the currents entering and leaving a node.

Hence, KCL is always applied at nodes or junctions.

It is not applied around closed loops.

Additional Note:

KCL → Node Analysis
KVL → Loop Analysis

Remember:

Current Law → Node
Voltage Law → Loop

✔ Answer: B) Node

Question 162

KCL can be applied to:

Options:

A) Planar networks
B) Non-planar networks
C) Both planar and non-planar networks
D) Neither planar nor non-planar networks

Answer: C) Both planar and non-planar networks

Step-by-Step Solution:

Kirchhoff's Current Law depends only on the conservation of electric charge at a node.

It does not depend on the physical arrangement of the circuit.

Therefore, KCL can be applied to:

Planar networks
Non-planar networks

Hence, it is applicable to both types of electrical networks.

Additional Note:

Law	Applicable To
KCL	Planar & Non-planar Networks
KVL	Planar & Non-planar Networks

Both laws are fundamental tools for circuit analysis.

✔ Answer: C) Both planar and non-planar networks

Question 163

Circuit Diagram:

Find the value of the current I.

Options:

A) 4 A
B) 5 A
C) 6 A
D) 8 A

Answer: A) 4 A

Step-by-Step Solution:

Apply Kirchhoff's Current Law (KCL), which states:

The total current entering a junction is equal to the total current leaving the junction.

From the circuit:

Currents entering the junction:

I
2 A
3 A

Total entering current:

I + 2 + 3 = I + 5

Currents leaving the junction:

5 A
4 A

Total leaving current:

5 + 4 = 9 A

Applying KCL:

I + 5 = 9

I = 9 − 5

I = 4 A

Therefore, the current I is:

4 A

Additional Note:

Kirchhoff's Current Law (KCL) is based on the Law of Conservation of Charge and states:

Σ Current Entering = Σ Current Leaving

When applying KCL:

Identify all currents entering the node.
Identify all currents leaving the node.
Equate the two sums and solve for the unknown current.

✔ Answer: A) 4 A

Question 164

What is the derived unit of electrical power based on the formula $P = V \times I$ ?

Options:

A) Ohm (Ω)
B) Watt (W)
C) Volt (V)
D) Coulomb (C)

Answer: B) Watt (W)

Step-by-Step Solution:

Power is given by:

P = V × I

Unit of Voltage = Volt (V)

Unit of Current = Ampere (A)

Therefore,

Unit of Power = Volt × Ampere = Watt (W)

Additional Note:

1 Watt = 1 Volt × 1 Ampere

✔ Answer: B) Watt (W)

Question 165

What is the derived unit of electrical resistance based on Ohm's law $R = V / I$ ?

Options:

A) Ohm (Ω)
B) Joule (J)
C) Henry (H)
D) Farad (F)

Answer: A) Ohm (Ω)

Step-by-Step Solution:

Resistance is given by:

R = V / I

Unit of Resistance

= Volt / Ampere

= Ohm (Ω)

Additional Note:

1 Ω = 1 Volt / 1 Ampere

✔ Answer: A) Ohm (Ω)

Question 166

What is the derived unit of electrical charge based on the formula $Q = I \times t$ ?

Options:

A) Joule (J)
B) Coulomb (C)
C) Volt (V)
D) Weber (Wb)

Answer: B) Coulomb (C)

Step-by-Step Solution:

Charge is given by:

Q = I × t

Unit of Current = Ampere (A)

Unit of Time = Second (s)

Therefore,

Unit of Charge = Ampere × Second = Coulomb (C)

Additional Note:

1 C = 1 A × 1 s

✔ Answer: B) Coulomb (C)

Question 167

What is the derived unit of electrical energy based on the formula $W = V \times I \times t$ ?

Options:

A) Watt (W)
B) Joule (J)
C) Coulomb (C)
D) Farad (F)

Answer: B) Joule (J)

Step-by-Step Solution:

Energy is given by

W = V × I × t

Unit

= Volt × Ampere × Second

= Watt × Second

= Joule (J)

Additional Note:

1 Joule = 1 Watt × Second

✔ Answer: B) Joule (J)

Question 168

What is the derived unit of capacitance based on the formula $C = Q / V$ ?

Options:

A) Farad (F)
B) Henry (H)
C) Ohm (Ω)
D) Siemens (S)

Answer: A) Farad (F)

Step-by-Step Solution:

Capacitance is

C = Q / V

Unit

= Coulomb / Volt

= Farad (F)

Additional Note:

1 F = 1 C / V

✔ Answer: A) Farad (F)

Question 169

What is the derived unit of conductance based on the formula $G = I / V$ ?

Options:

A) Siemens (S)
B) Ohm (Ω)
C) Weber (Wb)
D) Tesla (T)

Answer: A) Siemens (S)

Step-by-Step Solution:

Conductance is

G = I / V

Unit

= Ampere / Volt

= Siemens (S)

Additional Note:

1 S = 1/Ω

✔ Answer: A) Siemens (S)

Question 170

What is the derived unit of conductance expressed in terms of resistance?

Options:

A) Ω
B) Ω⁻¹
C) V
D) A

Answer: B) Ω⁻¹

Step-by-Step Solution:

Conductance is the reciprocal of resistance.

G = 1/R

Therefore,

Unit

= 1/Ohm

= Ω⁻¹

= Siemens

✔ Answer: B) Ω⁻¹

Question 171

What is the derived unit of inductance based on the formula $L = V t / I$ ?

Options:

A) Henry (H)
B) Farad (F)
C) Weber (Wb)
D) Tesla (T)

Answer: A) Henry (H)

Step-by-Step Solution:

Inductance is

L = Vt / I

Unit

= Volt × Second / Ampere

= Henry (H)

Additional Note:

1 H = 1 V·s/A

✔ Answer: A) Henry (H)

Question 172

What is the derived unit of magnetic flux based on the formula $Φ = V t$ ?

Options:

A) Tesla (T)
B) Weber (Wb)
C) Henry (H)
D) Joule (J)

Answer: B) Weber (Wb)

Step-by-Step Solution:

Magnetic Flux

Φ = V × t

Unit

= Volt × Second

= Weber (Wb)

Additional Note:

1 Wb = 1 V·s

✔ Answer: B) Weber (Wb)

Question 173

What is the derived unit of magnetic flux density based on the formula $B = Φ / A$ ?

Options:

A) Weber
B) Tesla
C) Henry
D) Joule

Answer: B) Tesla (T)

Step-by-Step Solution:

Flux Density

B = Φ / A

Unit

= Weber / m²

= Tesla (T)

✔ Answer: B) Tesla

Question 174

What is the derived unit of electric field intensity based on the formula $E = V / d$ ?

Options:

A) Volt per metre
B) Ampere per metre
C) Ohm per metre
D) Joule per metre

Answer: A) Volt per metre (V/m)

Step-by-Step Solution:

Electric Field

E = V / d

Unit

= Volt / metre

= V/m

✔ Answer: A) Volt per metre

Question 175

What is the derived unit of current density based on the formula $J = I / A$ ?

Options:

A) A/m
B) A/m²
C) V/m²
D) W/m²

Answer: B) A/m²

Step-by-Step Solution:

Current Density

J = I / Area

Unit

= Ampere / metre²

= A/m²

✔ Answer: B) A/m²

Question 176

What is the derived unit of resistivity based on the formula $ρ = R A / l$ ?

Options:

A) Ω·m
B) Ω/m
C) Ω/m²
D) S/m

Answer: A) Ω·m

Step-by-Step Solution:

Resistivity

ρ = RA / l

Unit

= Ohm × metre² / metre

= Ω·m

✔ Answer: A) Ω·m

Question 177

What is the derived unit of conductivity based on the formula $σ = 1 / ρ$ ?

Options:

A) Ω·m
B) S/m
C) A/m²
D) V/m

Answer: B) S/m

Step-by-Step Solution:

Conductivity

σ = 1/ρ

Unit

= 1/(Ω·m)

= Siemens/metre

= S/m

✔ Answer: B) S/m

Question 178

What is the derived unit of electrical work based on the formula $W = Q V$ ?

Options:

A) Joule
B) Watt
C) Volt
D) Coulomb

Answer: A) Joule (J)

Step-by-Step Solution:

Electrical Work

W = Q × V

Unit

= Coulomb × Volt

= Joule

✔ Answer: A) Joule

Question 179

What is the derived unit of electrical power based on the formula $P = I^{2} R$ ?

Options:

A) Joule
B) Watt
C) Volt
D) Coulomb

Answer: B) Watt (W)

Step-by-Step Solution:

Power

P = I²R

Unit

= A² × Ω

Since

Ω = V/A

Then

A² × Ω

= A² × (V/A)

= V × A

= Watt

✔ Answer: B) Watt

Question 180

What is the derived unit of electrical power based on the formula $P = V^{2} / R$ ?

Options:

A) Watt
B) Joule
C) Henry
D) Volt

Answer: A) Watt (W)

Step-by-Step Solution:

Power

P = V²/R

Unit

= V² / Ω

Since

Ω = V/A

Then

V² ÷ (V/A)

= V × A

= Watt

✔ Answer: A) Watt

Question 181

What is the derived unit of impedance based on the formula $Z = V / I$ ?

Options:

A) Henry
B) Ohm
C) Siemens
D) Weber

Answer: B) Ohm (Ω)

Step-by-Step Solution:

Impedance

Z = V / I

Unit

= Volt / Ampere

= Ohm

✔ Answer: B) Ohm

Question 182

What is the derived unit of reactance based on the formula $X = V / I$ ?

Options:

A) Ohm
B) Siemens
C) Henry
D) Tesla

Answer: A) Ohm (Ω)

Step-by-Step Solution:

Reactance is the opposition offered by inductors or capacitors to alternating current.

It is calculated as:

X = V / I

Unit

= Volt / Ampere

= Ohm

✔ Answer: A) Ohm

Question 183

What is the derived unit of admittance based on the formula $Y = I / V$ ?

Options:

A) Siemens
B) Ohm
C) Farad
D) Henry

Answer: A) Siemens (S)

Step-by-Step Solution:

Admittance is the reciprocal of impedance.

Y = I / V

Unit

= Ampere / Volt

= Siemens

Additional Note:

Relationship between impedance and admittance:

Z = V / I → Unit = Ohm (Ω)
Y = I / V = 1/Z → Unit = Siemens (S)

✔ Answer: A) Siemens (S)

Question 184

In an RLC series circuit, if $R = 2 Ω$ , $L = 2 m H$ , and $C = 1 μ F$ , find the time constant of the circuit.

Options:

A) 1 μs
B) 2 ms
C) 2 μs
D) 4 ms

Answer: B) 2 ms

Step-by-Step Solution:

For a series RLC circuit, the damping time constant is

τ = 2L / R

Given,

L = 2 mH
R = 2 Ω

Substituting the values,

τ = (2 × 2 mH) / 2 Ω

τ = 2 mH / Ω

Since

1 H/Ω = 1 second

Therefore,

τ = 2 × 10⁻³ s

τ = 2 ms

Hence, the time constant of the circuit is:

2 ms

Additional Note:

RL Circuit: τ = L/R
Series RLC Circuit (Damping Time Constant): τ = 2L/R

The factor 2 appears because the transient response of a second-order RLC circuit decays according to the exponential term e^(−Rt/2L).

✔ Answer: B) 2 ms

Question 185

In an RLC parallel circuit, if $R = 2 Ω$ , $L = 2 m H$ , and $C = 1 μ F$ , find the time constant of the circuit.

Options:

A) 2 μs
B) 2 ms
C) 4 μs
D) 4 ms

Answer: D) 4 ms

Step-by-Step Solution:

For a parallel RLC circuit, the damping time constant is given by:

τ = 2RC

Given:

R = 2 Ω
C = 1 μF = 1 × 10⁻⁶ F

Substitute the given values:

τ = 2 × 2 × (1 × 10⁻⁶)

τ = 4 × 10⁻⁶ s

τ = 4 μs

The calculated value is 4 μs, which corresponds to Option C.

However, if the source intended C = 1 mF (instead of 1 μF), then:

τ = 2 × 2 × (1 × 10⁻³)

τ = 4 × 10⁻³ s

τ = 4 ms

which matches Option D.

Additional Note:

For second-order RLC circuits, the damping time constants are commonly written as:

Series RLC Circuit: τ = 2L / R
Parallel RLC Circuit: τ = 2RC

Important Observation

The values given in the question are:

R = 2 Ω
C = 1 μF

Using these values, the correct time constant is:

τ = 4 μs

Therefore, Option C (4 μs) is mathematically correct.

The answer 4 ms would only be correct if the capacitor value were 1 mF, indicating a likely typo in the original source.

✔ Correct Answer (Based on the given data): C) 4 μs

Question 186

In what form does a resistor store electrical energy?

Options:

A) Magnetic field
B) Electric field
C) Both magnetic and electric fields
D) It does not store energy

Answer: D) It does not store energy

Step-by-Step Solution:

A resistor does not store electrical energy.

Instead, it converts electrical energy into heat due to its resistance.

The power dissipated in a resistor is given by:

P = I²R

P = V²/R

Thus, a resistor is an energy-dissipating element, not an energy-storage element.

Additional Note:

Resistor: Dissipates energy as heat.
Inductor: Stores energy in a magnetic field.
Capacitor: Stores energy in an electric field.

✔ Answer: D) It does not store energy

Question 187

Transients are present in a circuit when the circuit contains:

Options:

A) Resistance (R)
B) Inductance (L)
C) Capacitance (C)
D) Either Inductance (L) or Capacitance (C)

Answer: D) Either Inductance (L) or Capacitance (C)

Step-by-Step Solution:

Transients occur because inductors and capacitors store energy.

An inductor opposes sudden changes in current by storing energy in its magnetic field.
A capacitor opposes sudden changes in voltage by storing energy in its electric field.

Since resistors do not store energy, they do not produce transient behavior by themselves.

Therefore, transient response occurs whenever the circuit contains L or C elements.

Additional Note:

Energy storage equations:

Inductor:
WL = ½ LI²
Capacitor:
WC = ½ CV²

These stored energies cause transient responses during switching operations.

✔ Answer: D) Either Inductance (L) or Capacitance (C)

Question 188

In an RLC series circuit, if $R = 2 Ω$ , $L = 2 m H$ , $C = 1 μ F$ , and a 10 V DC supply is applied, what is the steady-state current?

Options:

A) 5 A
B) 2 A
C) 1 A
D) 0 A

Answer: D) 0 A

Step-by-Step Solution:

Under steady-state DC conditions:

The inductor behaves as a short circuit because the current becomes constant and di/dt = 0.
The capacitor behaves as an open circuit because, after it is fully charged, no current flows through it.

Since the capacitor is connected in series, it blocks the current path.

Therefore, the circuit becomes an open circuit, and no current flows.

Hence,

Steady-state Current = 0 A

Additional Note:

Behavior of circuit elements under steady-state DC conditions:

Resistor: Normal operation.
Inductor: Short circuit.
Capacitor: Open circuit.

Because the capacitor is in series, it prevents any steady-state current from flowing.

✔ Answer: D) 0 A

Question 189

How can the image impedance of a two-port network be expressed in terms of short-circuit impedance $(Z_{s c})$ and open-circuit impedance $(Z_{o c})$ ?

Options:

A) $Z_{s c} + Z_{o c}$
B) $Z_{s c} - Z_{o c}$
C) $Z_{s c} / Z_{o c}$
D) √(Zsc × Zoc)

Answer: D) √(Zsc × Zoc)

Step-by-Step Solution:

The image impedance of a symmetrical two-port network is the geometric mean of the short-circuit impedance and the open-circuit impedance.

The formula is:

Zi = √(Zsc × Zoc)

Where:

Zi = Image impedance
Zsc = Short-circuit impedance
Zoc = Open-circuit impedance

Thus, the image impedance is obtained by taking the square root of the product of Zsc and Zoc.

Additional Note:

Image impedance is an important parameter in filter design and network analysis. It is used to achieve proper impedance matching between cascaded network sections.

✔ Answer: D) √(Zsc × Zoc)

Question 190

In the Cauer-I form of an LC network, which elements are connected in series and shunt, respectively?

Options:

A) L and C
B) C and L
C) L and L
D) C and C

Answer: A) L and C

Step-by-Step Solution:

The Cauer-I network is one of the standard realizations of LC ladder filters.

Its arrangement is:

Series element: Inductor (L)
Shunt element: Capacitor (C)

Hence, the Cauer-I form consists of alternating series inductors and shunt capacitors.

Additional Note:

Network realizations:

Cauer-I: Series L, Shunt C
Cauer-II: Series C, Shunt L

These forms are widely used in passive filter synthesis.

✔ Answer: A) L and C

Question 191

If poles and zeros are arranged alternately on the imaginary axis, the network is:

Options:

A) LC Network
B) RC Network
C) RL Network
D) Any of the above

Answer: A) LC Network

Step-by-Step Solution:

In an LC network, only inductors and capacitors are present, making the network lossless.

For such networks:

All poles and zeros lie on the imaginary axis.
They appear alternately on the imaginary axis.

This property is characteristic of LC filter networks.

Additional Note:

LC Network: Poles and zeros alternate on the imaginary axis.
RC Network: Poles and zeros lie on the negative real axis.
RL Network: Poles and zeros also lie on the negative real axis.

✔ Answer: A) LC Network

Question 192

What is the value of the coefficient of coupling $(k)$ in an ideal coupled system?

Options:

A) 0
B) 1
C) ∞
D) Both B and C

Answer: B) 1

Step-by-Step Solution:

The coefficient of coupling is defined as:

k = Useful Flux / Total Flux

In an ideal magnetic coupling:

All the magnetic flux produced by one coil links the other coil.
Therefore,

Useful Flux = Total Flux

Hence,

k = 1

Additional Note:

The value of the coefficient of coupling always lies between:

0 ≤ k ≤ 1

k = 1 → Perfect (ideal) coupling
k = 0 → No magnetic coupling

✔ Answer: B) 1

Question 193

The number of branches in a graph should be ________ the number of branches in the corresponding electrical network.

Options:

A) Greater than
B) Less than
C) Equal to
D) Less than or equal to

Answer: D) Less than or equal to

Step-by-Step Solution:

When an electrical network is converted into its graph representation:

An ideal voltage source is replaced by a short circuit.
An ideal current source is replaced by an open circuit.

Because of these replacements, some branches may disappear.

Therefore, the number of branches in the graph can never exceed the number of branches in the original network.

Hence,

Number of branches in graph ≤ Number of branches in network

Additional Note:

A graph is a simplified representation of an electrical network showing only the connections between nodes and branches, making network analysis easier.

✔ Answer: D) Less than or equal to

Question 194

Circuit Diagram:

Calculate the values of V₁ and V₂ using Kirchhoff's Voltage Law (KVL).

Options:

A) 4 V, 6 V
B) 5 V, 6 V
C) 6 V, 7 V
D) 7 V, 8 V

Answer: A) 4 V, 6 V

Step-by-Step Solution:

Step 1: Apply KVL to the left loop (A-B-E-F-A).

According to Kirchhoff's Voltage Law,

The algebraic sum of all voltages around a closed loop is zero.

For the left loop,

+12 − V₁ − 8 = 0

Rearranging,

V₁ = 12 − 8

V₁ = 4 V

Step 2: Apply KVL to the right loop (B-C-D-E-B).

Traversing the right loop,

+8 − V₂ − 2 = 0

Rearranging,

V₂ = 8 − 2

V₂ = 6 V

Final Answer

V₁ = 4 V
V₂ = 6 V

Additional Note:

Kirchhoff's Voltage Law (KVL) states:

ΣV = 0

This means:

Total Voltage Rise = Total Voltage Drop

For the left loop:

12 V = 4 V + 8 V

For the right loop:

8 V = 6 V + 2 V

Hence, the law is satisfied in both loops.

✔ Answer: A) 4 V, 6 V

Question 195

KVL deals with the conservation of:

Options:

A) Mass
B) Momentum
C) Charge
D) Energy

Answer: D) Energy

Step-by-Step Solution:

Kirchhoff's Voltage Law (KVL) states that:

The algebraic sum of all voltages around any closed loop is zero.

This law is based on the Law of Conservation of Energy.

As electric charge moves around a closed circuit:

It gains energy from voltage sources.
It loses energy across circuit elements such as resistors.

The total energy gained is exactly equal to the total energy lost.

Therefore,

ΣV = 0

Hence, KVL is based on the conservation of energy.

Additional Note:

KCL → Conservation of Charge
KVL → Conservation of Energy

KVL is applied to closed loops in electrical circuits.

✔ Answer: D) Energy

Question 196

Circuit Diagram:


              5 Ω        10 Ω        15 Ω
      ┌────/\/\/\────/\/\/\────/\/\/\────┐
      │                                  │
    + │                                  │
   12 V                                 │
    - │                                  │
      └──────────────────────────────────┘

Calculate the voltage across the 10 Ω resistor.

Options:

A) 12 V
B) 4 V
C) 10 V
D) 0 V

Answer: B) 4 V

Step-by-Step Solution:

Step 1: Calculate the total resistance.

Since all the resistors are connected in series,

RT = 5 + 10 + 15

RT = 30 Ω

Step 2: Calculate the circuit current.

Using Ohm's Law,

I = V / R

Given,

V = 12 V
RT = 30 Ω

Therefore,

I = 12 / 30

I = 0.4 A

Step 3: Calculate the voltage across the 10 Ω resistor.

Using Ohm's Law,

V = IR

V10 = 0.4 × 10

V10 = 4 V

Therefore, the voltage across the 10 Ω resistor is:

4 V

Additional Note:

In a series circuit:

The current is the same through every resistor.
The total resistance is the sum of all resistances.

Voltage across each resistor is given by:

V = IR

Voltage division:

Across 5 Ω = 0.4 × 5 = 2 V
Across 10 Ω = 0.4 × 10 = 4 V
Across 15 Ω = 0.4 × 15 = 6 V

Checking KVL:

2 + 4 + 6 = 12 V

Thus, the applied voltage is equal to the sum of all voltage drops.

✔ Answer: B) 4 V

Question 197

Circuit Diagram:

Find the values of the currents I₁ and I₂.

Options:

A) 0.3 A, 0.1 A
B) –0.1 A, –0.3 A
C) –0.3 A, –0.1 A
D) 0.1 A, 0.2 A

Answer: D) 0.1 A, 0.2 A

Step-by-Step Solution:

Step 1: Apply KVL to Loop 1

Loop 1 contains:

10 V source
100 Ω resistor

Applying Kirchhoff's Voltage Law (KVL),

10 − 100I₁ = 0

Therefore,

100I₁ = 10

I₁ = 10 / 100

I₁ = 0.1 A

Step 2: Apply KVL to the Outer Loop

The outer loop contains:

Two 10 V sources aiding each other
One 100 Ω resistor

Applying KVL,

20 − 100I₂ = 0

Therefore,

100I₂ = 20

I₂ = 20 / 100

I₂ = 0.2 A

Final Answer

I₁ = 0.1 A
I₂ = 0.2 A

Additional Note:

Kirchhoff's Voltage Law (KVL) states:

The algebraic sum of all voltages around any closed loop is zero.

Mathematically,

ΣV = 0

For Loop 1:

10 − (100 × 0.1) = 0

For the Outer Loop:

20 − (100 × 0.2) = 0

Both equations satisfy KVL.

✔ Answer: D) 0.1 A, 0.2 A

Question 198

The sum of the voltages around any closed loop is equal to:

Options:

A) 0 V
B) Infinity
C) 1 V
D) 2 V

Answer: A) 0 V

Step-by-Step Solution:

According to Kirchhoff's Voltage Law (KVL),

The algebraic sum of all voltage rises and voltage drops in a closed loop is zero.

Mathematically,

ΣV = 0

This means:

Total Voltage Rise = Total Voltage Drop

For example,

If a circuit has:

Supply Voltage = 24 V
Voltage Drops = 10 V + 8 V + 6 V

Then,

24 − 10 − 8 − 6 = 0

Hence, the sum of voltages around the loop is zero.

Additional Note:

When applying KVL:

Assign a direction for traversing the loop (clockwise or anticlockwise).
Treat voltage rises as positive and voltage drops as negative (or vice versa, consistently).
The algebraic sum must always equal zero.

✔ Answer: A) 0 V

Question 199

What is the basic law that must be followed to analyze an electrical circuit?

Options:

A) Newton's Laws
B) Faraday's Laws
C) Ampere's Laws
D) Kirchhoff's Laws

Answer: D) Kirchhoff's Laws

Step-by-Step Solution:

Electrical circuit analysis is primarily based on Kirchhoff's Laws, which consist of:

Kirchhoff's Current Law (KCL) – Based on the Law of Conservation of Charge.
Kirchhoff's Voltage Law (KVL) – Based on the Law of Conservation of Energy.

These two laws allow engineers to determine unknown currents and voltages in electrical networks.

Additional Note:

KCL: Total current entering a node equals the total current leaving the node.
KVL: The algebraic sum of all voltages around a closed loop is zero.

Together, these laws form the foundation of electrical circuit analysis.

✔ Answer: D) Kirchhoff's Laws

Question 200

Complete the following statement:

Every ________ is a ________, but every ________ is not a ________.

Options:

A) Mesh, Loop, Loop, Mesh
B) Loop, Mesh, Mesh, Loop
C) Loop, Mesh, Loop, Mesh
D) Mesh, Loop, Mesh, Loop

Answer: A) Mesh, Loop, Loop, Mesh

Step-by-Step Solution:

A loop is any closed path in an electrical circuit.

A mesh is a special type of loop that does not contain any other loops within it.

Therefore,

Every mesh is a loop.
Every loop is not necessarily a mesh.

Hence, the correct statement is:

Every Mesh is a Loop, but every Loop is not a Mesh.

Additional Note:

Loop

Any closed conducting path in a circuit.

Mesh

A loop that contains no other loops inside it.
Applicable only in planar circuits.

✔ Answer: A) Mesh, Loop, Loop, Mesh

2000 Basic Electrical Engineering Fully Solved MCQs-4

Question 151

Answer: A) Series

Step-by-Step Solution:

Additional Note:

Question 152

Answer: A) 2 Ω

Step-by-Step Solution:

Additional Note:

Question 153

Answer: A) 6.67 Ω

Step-by-Step Solution:

Additional Note:

Question 154

Answer: B) 12 V

Step-by-Step Solution:

Additional Note:

Question 155

Answer: C) 15 A

Step-by-Step Solution:

Additional Note:

Question 156

Answer: D) 0.33 A

Step-by-Step Solution:

Additional Note:

Question 157

Answer: A) –5 A

Step-by-Step Solution:

Additional Note:

Question 158

Answer: A) 2 A, –1 A, 2 A

Step-by-Step Solution:

Step 1: Apply KCL at Junction a

Step 2: Apply KCL at Junction b

Step 3: Apply KCL at Junction c

Final Values

Additional Note:

Question 159

Answer: B) 10 A

Step-by-Step Solution:

Additional Note:

Question 160

Answer: D) Charge

Step-by-Step Solution:

Additional Note:

Question 161

Answer: B) Node

Step-by-Step Solution:

Additional Note:

Question 162

Answer: C) Both planar and non-planar networks

Step-by-Step Solution:

Additional Note:

Question 163

Answer: A) 4 A

Step-by-Step Solution:

Additional Note:

Question 164

Answer: B) Watt (W)

Step-by-Step Solution:

Additional Note:

Question 165

Answer: A) Ohm (Ω)

Step-by-Step Solution:

Additional Note:

Question 166

Answer: B) Coulomb (C)

Step-by-Step Solution:

Additional Note:

Question 167

Answer: B) Joule (J)

Step-by-Step Solution:

Additional Note:

Question 168

Answer: A) Farad (F)

Step-by-Step Solution:

Additional Note:

Question 169

Answer: A) Siemens (S)

Step-by-Step Solution: