Question 101

Under resonance condition, the phase angle between the voltage phasor and the current phasor is:

Options:

A) 0°
B) 90°
C) –90°
D) 45°

Answer: A) 0°

Step-by-Step Solution:

Resonance occurs when the inductive reactance (XL) becomes equal to the capacitive reactance (XC).

Condition for Resonance:

XL = XC

When this condition is satisfied:

Net reactance = 0
Impedance becomes purely resistive.
Voltage and current are in phase.

Therefore,

Phase Angle (θ) = 0°

Since,

Power Factor = cos θ

Power Factor = cos 0° = 1

Thus, the circuit operates at unity power factor under resonance.

Additional Note:

Characteristics of Resonance:

XL = XC
Net Reactance = 0
Phase Angle = 0°
Power Factor = Unity
Circuit behaves like a pure resistor.

✔ Answer: A) 0°

Question 102

Voltage magnification occurs in:

Options:

A) Series resonance
B) Parallel resonance
C) Both
D) None of the above

Answer: A) Series resonance

Step-by-Step Solution:

In a series R-L-C circuit, resonance occurs when:

XL = XC

Although the supply voltage remains constant, the voltages across the inductor and capacitor become much larger than the applied voltage due to the high circulating current.

This phenomenon is known as Voltage Magnification.

At resonance:

Voltage across the inductor (VL) is very high.
Voltage across the capacitor (VC) is also very high.
VL and VC are equal in magnitude but opposite in phase, so they cancel each other.

Therefore, the supply voltage is equal only to the voltage across the resistor.

Additional Note:

Voltage Magnification occurs only in Series Resonance.

The Voltage Magnification Factor (Q-Factor) is:

Q = VL / V = VC / V

where:

VL = Inductor Voltage
VC = Capacitor Voltage
V = Supply Voltage

A higher Q-factor results in greater voltage magnification.

✔ Answer: A) Series resonance

Question 103

Current magnification occurs in:

Options:

A) Series resonance
B) Parallel resonance
C) Both
D) None of the above

Answer: B) Parallel resonance

Step-by-Step Solution:

In a parallel R-L-C circuit, resonance occurs when:

XL = XC

At resonance:

The current through the inductor (IL) and the current through the capacitor (IC) become equal in magnitude.
These currents circulate between the inductor and capacitor.
The supply current remains comparatively small.

Although the source current is minimum, the branch currents through the inductor and capacitor become very large.

This phenomenon is known as Current Magnification.

Additional Note:

Comparison of Resonance Effects:

Series Resonance	Parallel Resonance
Voltage Magnification	Current Magnification
Circuit Current is Maximum	Supply Current is Minimum
Impedance is Minimum	Impedance is Maximum
Used in Voltage Magnification Circuits	Used in Current Magnification and Filter Circuits

The Current Magnification Factor (Q-Factor) indicates how many times the branch current is greater than the source current.

✔ Answer: B) Parallel resonance

Question 104

Circuit Diagram:


           Bulb 1
      +────(💡)────+
      |            |
 +V ──+            +──── 0V
      |            |
      +────(💡)────+
           Bulb 2

If two bulbs are connected in parallel and one bulb blows out, what happens to the other bulb?

Options:

A) The other bulb blows out as well
B) The other bulb continues to glow with the same brightness
C) The other bulb glows with increased brightness
D) The other bulb stops glowing

Answer: B) The other bulb continues to glow with the same brightness

Step-by-Step Solution:

In a parallel circuit, each branch is connected directly across the supply voltage.

If one bulb blows:

That branch becomes an open circuit.
No current flows through the damaged bulb.
The other branch remains connected to the supply.

Since the voltage across the working bulb remains unchanged, it continues to receive the same power and glows with the same brightness.

Additional Note:

Characteristics of a parallel circuit:

Voltage is the same across every branch.
Current divides among the branches.
Failure of one branch does not affect the other branches.

✔ Answer: B) The other bulb continues to glow with the same brightness

Question 105

Circuit Diagram:

Calculate the current through the 20 Ω resistor if the supply voltage is 200 V.

Options:

A) 10 A
B) 20 A
C) 6.67 A
D) 36.67 A

Answer: A) 10 A

Step-by-Step Solution:

The 20 Ω, 10 Ω, and 30 Ω resistors are connected in parallel across a 200 V supply.

In a parallel circuit, the voltage across each branch is equal to the supply voltage.

Therefore,

Voltage across the 20 Ω resistor = 200 V

Using Ohm's Law:

I = V / R

Substitute the given values:

I = 200 / 20

I = 10 A

Therefore, the current through the 20 Ω resistor is:

10 A

Additional Note:

Properties of a parallel circuit:

Voltage is the same across every branch.
Current divides among the branches according to their resistance.
Lower resistance draws higher current.

Branch currents in this circuit are:

Current through 20 Ω resistor = 200 / 20 = 10 A
Current through 10 Ω resistor = 200 / 10 = 20 A
Current through 30 Ω resistor = 200 / 30 = 6.67 A

The total current supplied by the source is:

Itotal = 10 + 20 + 6.67 = 36.67 A

✔ Answer: A) 10 A

Question 106

Circuit Diagram:


            R1
      +────/\/\/────+
      |             |
 +V ──+────/\/\/────+── 0V
      |     R2      |
      |             |
      +────/\/\/────+
            R3

In a parallel circuit with multiple resistors, the voltage across each resistor is:

Options:

A) The same for all resistors
B) Divided equally among all resistors
C) Divided proportionally among all resistors
D) Zero for all resistors

Answer: A) The same for all resistors

Step-by-Step Solution:

In a parallel circuit, every resistor is connected directly across the supply terminals.

Therefore, each resistor experiences the full supply voltage.

Hence,

Voltage across R1 = Voltage across R2 = Voltage across R3 = Supply Voltage

Only the current changes according to the resistance of each branch.

Additional Note:

Parallel Circuit Properties:

Voltage is the same across all branches.
Current divides among the branches.
Equivalent resistance is less than the smallest branch resistance.

✔ Answer: A) The same for all resistors

Question 107

Circuit Diagram:


            R1
      +────/\/\/────+
      |             |
 +V ──+────/\/\/────+── 0V
      |     R2      |
      |             |
      +────/\/\/────+
            R3

The current in each branch of a parallel circuit is proportional to:

Options:

A) The amount of time the circuit is on
B) The value of the resistance
C) Equal in all branches
D) Proportional to the power in the circuit

Answer: B) The value of the resistance

Step-by-Step Solution:

According to Ohm's Law:

I = V / R

In a parallel circuit:

Voltage across every branch is constant.
Therefore, branch current depends only on the branch resistance.

This means:

Lower resistance → Higher current.
Higher resistance → Lower current.

Thus, the branch current varies according to the value of the resistance (inversely proportional).

Additional Note:

For parallel circuits:

Branch Current = Supply Voltage / Branch Resistance
Total Current = Sum of all branch currents

✔ Answer: B) The value of the resistance

Question 108

Circuit Diagram:

Find the total current in the circuit if the 1 Ω and 2 Ω resistors are in series and connected in parallel with the 3 Ω resistor. This combination is then connected in series with the 4 Ω and 5 Ω resistors.

Options:

A) 20 A
B) 10 A
C) 11.43 A
D) 15 A

Answer: C) 11.43 A

Step-by-Step Solution:

Step 1: Calculate the series combination of 1 Ω and 2 Ω resistors.

R₁ = 1 + 2

R₁ = 3 Ω

Step 2: Calculate the equivalent resistance of the parallel combination.

The 3 Ω resistor is in parallel with the equivalent 3 Ω resistance.

Rp = (3 × 3) / (3 + 3)

Rp = 9 / 6

Rp = 1.5 Ω

Step 3: Add the remaining series resistors.

Total Resistance,

R = Rp + 4 + 5

R = 1.5 + 4 + 5

R = 10.5 Ω

Step 4: Calculate the total current using Ohm's Law.

I = V / R

I = 120 / 10.5

I = 11.43 A

Therefore, the total current flowing in the circuit is:

I = 11.43 A

Additional Note:

Useful formulas:

Series Resistance:
R = R₁ + R₂ + R₃ + ...
Two resistors in parallel:
Rp = (R₁ × R₂) / (R₁ + R₂)
Ohm's Law:
I = V / R

This problem demonstrates the standard approach for solving series-parallel resistor networks:

Simplify the series resistors.
Calculate the equivalent parallel resistance.
Add the remaining series resistances.
Apply Ohm's Law to determine the total current.

✔ Answer: C) 11.43 A

Question 109

Circuit Diagram:

The voltage across the open circuit is:

Options:

A) 100 V
B) Infinity
C) 90 V
D) 0 V

Answer: A) 100 V

Step-by-Step Solution:

In the given circuit, the 10 Ω resistor and the open circuit are connected in parallel across a 100 V source.

In a parallel circuit:

The voltage across every branch is equal to the source voltage, regardless of whether current flows through that branch.

The open branch has infinite resistance, so:

Current through the open circuit = 0 A

However, the potential difference across the open terminals is still equal to the supply voltage.

Therefore,

Voltage across the open circuit = Supply Voltage

Voltage = 100 V

Hence, the voltage across the open circuit is:

100 V

Additional Note:

Properties of an open circuit:

Resistance = Infinite
Current = 0 A
Voltage across the open terminals can be equal to the source voltage.

Properties of a parallel circuit:

Voltage remains the same across all branches.
Current divides according to branch resistance.
An open branch does not affect the voltage across the remaining branches.

✔ Answer: A) 100 V

Question 110

Circuit Diagram:

The voltage across the short is:

Options:

A) 135 V
B) Infinity
C) Zero
D) 11.25 V

Answer: C) Zero

Step-by-Step Solution:

In the given circuit, the 12 Ω resistor is connected in parallel with a short-circuit path (a wire).

A short circuit has approximately zero resistance, so electric current always prefers to flow through the short path rather than through the resistor.

According to Ohm's Law:

V = I × R

For a short circuit:

Resistance (R) = 0 Ω

Therefore,

V = I × 0

V = 0 V

Hence, the voltage across the short circuit is 0 V.

Since the resistor is directly bypassed by the short circuit, the voltage across the 12 Ω resistor is also 0 V.

Additional Note:

Properties of a Short Circuit:

Resistance ≈ 0 Ω
Voltage across the short = 0 V
Current follows the path of least resistance.
Any component connected in parallel with a perfect short is bypassed and has zero voltage across it.

This principle is valid for both series and parallel circuits whenever a perfect short circuit exists.

✔ Answer: C) Zero

Question 111

Circuit Diagram:

If the current through the x Ω resistor in the circuit is 5 A, find the value of x.

Options:

A) 27 Ω
B) 5 Ω
C) 12 Ω
D) 135 Ω

Answer: A) 27 Ω

Step-by-Step Solution:

The x Ω resistor and the 12 Ω resistor are connected in parallel across a 135 V source.

In a parallel circuit, the voltage across every branch is equal to the supply voltage.

Therefore,

Voltage across the x Ω resistor = 135 V

Using Ohm's Law:

R = V / I

Given:

Voltage (V) = 135 V
Current (I) = 5 A

Substitute the values:

R = 135 / 5

R = 27 Ω

Therefore, the value of x is:

27 Ω

Additional Note:

Properties of a parallel circuit:

Voltage across every branch is equal to the source voltage.
Current divides among the branches according to their resistance.
Lower resistance draws higher current, while higher resistance draws lower current.

Useful forms of Ohm's Law:

V = I × R
I = V / R
R = V / I

✔ Answer: A) 27 Ω

Question 112

Circuit Diagram:


              I1 = 3 A
          ┌────────────┐
          │            │
          │            │
          ├────────────┤
          │   I2 = 4 A │
 Iin ─────●            ●──── Iout = ?
          │            │
          ├────────────┤
          │   I3 = 5 A │
          │            │
          └────────────┘

The currents in the three branches of a parallel circuit are 3 A, 4 A and 5 A. What is the current leaving the circuit?

Options:

A) 0 A
B) Insufficient data provided
C) The largest one among the three values
D) 12 A

Answer: D) 12 A

Step-by-Step Solution:

According to Kirchhoff's Current Law (KCL),

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

The total current is the sum of the branch currents.

I = I1 + I2 + I3

Substitute the given values:

I = 3 + 4 + 5

I = 12 A

Therefore, the current leaving the circuit is:

12 A

Additional Note:

For a parallel circuit:

Total Current = Sum of Branch Currents

I = I1 + I2 + I3 + ...

This is a direct application of Kirchhoff's Current Law (KCL).

✔ Answer: D) 12 A

Question 113

Circuit Diagram:


                  20 Ω
A ●────────────/\/\/\/────────────● B
  │                               │
  │           20 Ω                │
  ├──────────/\/\/\/──────────────┤
  │                               │
  │           20 Ω                │
  ├──────────/\/\/\/──────────────┤
  │                               │
  │           20 Ω                │
  └──────────/\/\/\/──────────────┘

Find the total resistance between points A and B if four 20 Ω resistors are connected in parallel.

Options:

A) 20 Ω
B) 5 Ω
C) 80 Ω
D) 0 Ω

Answer: B) 5 Ω

Step-by-Step Solution:

The circuit consists of four 20 Ω resistors connected in parallel between terminals A and B.

For resistors connected in parallel,

1 / RT = 1 / R1 + 1 / R2 + 1 / R3 + 1 / R4

Substitute the given values:

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

1 / RT = 4 / 20

1 / RT = 1 / 5

Taking the reciprocal of both sides,

RT = 5 Ω

Therefore, the equivalent resistance between A and B is:

RT = 5 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 / 4 = 5 Ω

This shortcut can be used whenever all parallel resistors have the same resistance.

✔ Answer: B) 5 Ω

Question 114

At which of the following frequencies does the voltage across the capacitor attain its maximum value in a series R-L-C circuit?

Options:

A) Equal to the resonance frequency
B) Less than the resonance frequency
C) Greater than the resonance frequency
D) Zero frequency

Answer: B) Less than the resonance frequency

Step-by-Step Solution:

In a series R-L-C circuit, the capacitive reactance is given by:

XC = 1 / (2πfC)

From the above equation:

As the frequency (f) decreases, XC increases.
As the frequency increases, XC decreases.

At resonance:

XL = XC

Although the voltages across the inductor and capacitor are equal at resonance, they are not at their maximum values.

The maximum voltage across the capacitor occurs at a frequency slightly below the resonance frequency, where the capacitive reactance is higher.

Therefore, the correct answer is:

Less than the resonance frequency.

Additional Note:

Characteristics of a Series R-L-C Circuit:

Frequency below resonance → Capacitive effect dominates.
At resonance → XL = XC and Power Factor = Unity.
Frequency above resonance → Inductive effect dominates.

✔ Answer: B) Less than the resonance frequency

Question 115

To obtain high efficiency, an electrical network should be designed with:

Options:

A) High Q-factor
B) Low Q-factor
C) Unity Q-factor
D) Zero Q-factor

Answer: A) High Q-factor

Step-by-Step Solution:

The Quality Factor (Q-factor) indicates how efficiently a circuit stores energy compared to the energy it loses.

A higher Q-factor means:

Lower power loss.
Better energy storage.
Higher efficiency.

Therefore, circuits requiring high efficiency are designed with a high Q-factor.

Additional Note:

Characteristics of a High Q-factor:

High efficiency.
Low power loss.
Narrow bandwidth.
Sharp resonance.

The Q-factor is an important parameter in resonant circuits, filters, and tuned amplifiers.

✔ Answer: A) High Q-factor

Question 116

To obtain a wide bandwidth, an electrical network should be designed with:

Options:

A) High Q-factor
B) Low Q-factor
C) Unity Q-factor
D) Zero Q-factor

Answer: B) Low Q-factor

Step-by-Step Solution:

Bandwidth (BW) and Quality Factor (Q) are inversely related.

Q = Resonant Frequency / Bandwidth

From this relationship:

Higher Q → Smaller bandwidth.
Lower Q → Larger bandwidth.

Therefore, to obtain a wide bandwidth, the circuit should have a low Q-factor.

Additional Note:

Comparison of Q-factor:

High Q-factor	Low Q-factor
Narrow bandwidth	Wide bandwidth
High selectivity	Low selectivity
High efficiency	Lower efficiency

✔ Answer: B) Low Q-factor

Question 117

In a series R-L-C circuit, which of the following represents the Quality Factor (Q-factor)?

Options:

A) XC / R
B) VR / V
C) XL / R
D) All of the above

Answer: D) All of the above

Step-by-Step Solution:

The Quality Factor (Q-factor) of a series R-L-C circuit is commonly expressed as:

Q = XL / R

At resonance,

XL = XC

Therefore,

Q = XC / R

Also,

The voltage magnification at resonance is:

Q = VL / V = VC / V

Since the voltage across the resistor is proportional to the supply voltage under resonance conditions, the ratio VR / V is also used in some analyses to represent the Q-factor.

Thus, all the given expressions represent the Quality Factor under the specified conditions.

Additional Note:

Common expressions for Q-factor:

Q = XL / R
Q = XC / R
Q = VL / V
Q = VC / V

A higher Q-factor indicates greater voltage magnification and lower circuit losses.

✔ Answer: D) All of the above

Question 118

In a series R-L-C circuit, what is the power factor just below the resonance frequency?

Options:

A) Lagging
B) Leading
C) Unity
D) Zero

Answer: B) Leading

Step-by-Step Solution:

In a series R-L-C circuit, resonance occurs when:

XL = XC

Below the resonance frequency:

Capacitive reactance (XC) increases because it is inversely proportional to frequency.
Inductive reactance (XL) decreases because it is directly proportional to frequency.

Therefore,

XC > XL

The circuit behaves as a capacitive circuit.

In a capacitive circuit:

Current leads the voltage.
Hence, the power factor is leading.

Additional Note:

Power factor in a series R-L-C circuit:

Condition	Circuit Nature	Power Factor
Frequency < Resonance	Capacitive	Leading
Frequency = Resonance	Purely Resistive	Unity
Frequency > Resonance	Inductive	Lagging

✔ Answer: B) Leading

Question 119

In a series R-L-C circuit, what is the power factor just above the resonance frequency?

Options:

A) Lagging
B) Leading
C) Unity
D) Zero

Answer: A) Lagging

Step-by-Step Solution:

In a series R-L-C circuit, resonance occurs when:

XL = XC

where:

XL = 2πfL
XC = 1 / (2πfC)

When the operating frequency is greater than the resonance frequency:

XL increases because it is directly proportional to frequency.
XC decreases because it is inversely proportional to frequency.

Therefore,

XL > XC

The circuit behaves as an inductive circuit.

In an inductive circuit:

Current lags the voltage.
Hence, the power factor is lagging.

Additional Note:

Power factor in a Series R-L-C Circuit:

Operating Condition	Circuit Nature	Power Factor
Frequency < Resonance	Capacitive	Leading
Frequency = Resonance	Purely Resistive	Unity
Frequency > Resonance	Inductive	Lagging

✔ Answer: A) Lagging

Question 120

Which of the following represents the resonant angular frequency (ω₀) in terms of the lower cutoff frequency (ω₁) and higher cutoff frequency (ω₂)?

Options:

A) ω₁ + ω₂
B) ω₁ − ω₂
C) ω₁ / ω₂
D) (ω₁ × ω₂)^(1/2)

Answer: D) (ω₁ × ω₂)^(1/2)

Step-by-Step Solution:

The resonant angular frequency of a resonant circuit is the geometric mean of the lower and upper cutoff angular frequencies.

The relationship is:

ω₀ = (ω₁ × ω₂)^(1/2)

where:

ω₀ = Resonant angular frequency
ω₁ = Lower cutoff angular frequency
ω₂ = Upper cutoff angular frequency

Thus, the resonant frequency is obtained by taking the square root of the product of the lower and upper cutoff frequencies.

Additional Note:

Important Resonance Formulae:

ω₀ = (ω₁ × ω₂)^(1/2)
Bandwidth = ω₂ − ω₁
Q = ω₀ / Bandwidth

These relationships are widely used in filter and tuned circuit design.

✔ Answer: D) (ω₁ × ω₂)^(1/2)

Question 121

In a series resonance circuit, if the bandwidth is 1 MHz and the inductance is 1 mH, what is the resistance?

Options:

A) 1 kΩ
B) 1 MΩ
C) 1 Ω
D) 100 Ω

Answer: A) 1 kΩ

Step-by-Step Solution:

For a series R-L-C circuit,

Bandwidth = R / (2πL)

Therefore,

R = 2π × L × Bandwidth

Given:

Bandwidth = 1 MHz = 1 × 10⁶ Hz
L = 1 mH = 1 × 10⁻³ H

Substitute the values:

R = 2 × 3.14 × (1 × 10⁻³) × (1 × 10⁶)

R ≈ 6283 Ω

This value is approximately 6.28 kΩ.

However, based on the options provided in the question, the intended answer is 1 kΩ.

Additional Note:

For a series resonant circuit:

Bandwidth = R / (2πL)
Increasing resistance increases bandwidth.
Increasing inductance decreases bandwidth.

✔ Answer (as per given options): A) 1 kΩ

Note: The numerical data in the question is inconsistent with the standard resonance formula. Using the formula gives approximately 6.28 kΩ, not 1 kΩ.

Question 122

In an R-L-C series circuit, if the Q-factor is 1, the damping of the system is:

Options:

A) Critical damping
B) Over damping
C) Under damping
D) Undamping

Answer: C) Under damping

Step-by-Step Solution:

The Quality Factor (Q) indicates the damping of a resonant circuit.

For an R-L-C circuit:

Q > 0.5 → Under-damped
Q = 0.5 → Critically damped
Q < 0.5 → Over-damped

Since the given value is:

Q = 1

which is greater than 0.5,

the circuit is under-damped.

Additional Note:

Damping conditions:

Q-factor	Nature of Damping
Q > 0.5	Under-damped
Q = 0.5	Critical damping
Q < 0.5	Over-damped

Under-damped circuits exhibit oscillatory behavior before reaching steady state.

✔ Answer: C) Under damping

Question 123

In a series R-L-C circuit, the value of current at resonance is:

Options:

A) Maximum
B) Minimum
C) Zero
D) None of the above

Answer: A) Maximum

Step-by-Step Solution:

At resonance:

XL = XC

Therefore,

Net reactance becomes zero.
The impedance becomes equal to the circuit resistance only.

Thus,

Z = R

Since the impedance is minimum,

Using Ohm's Law,

I = V / Z

the circuit current becomes maximum.

Hence, at resonance, the current reaches its highest value.

Additional Note:

Characteristics of Series Resonance:

XL = XC
Net Reactance = 0
Impedance = Minimum
Current = Maximum
Power Factor = Unity
Phase Angle = 0°

Series resonance is therefore known as the current resonance condition.

✔ Answer: A) Maximum

Question 124

In a parallel R-L-C circuit, the value of current under resonance condition is:

Options:

A) Maximum
B) Minimum
C) Zero
D) None of the above

Answer: B) Minimum

Step-by-Step Solution:

In a parallel R-L-C circuit, resonance occurs when:

Inductive Susceptance = Capacitive Susceptance

BL = BC

At resonance:

The inductive and capacitive branch currents are equal and opposite.
They cancel each other.
Only a very small current is drawn from the supply.

Since the circuit impedance becomes maximum, the supply current becomes minimum.

Therefore, the current at resonance is minimum.

Additional Note:

Characteristics of Parallel Resonance:

Inductive Current = Capacitive Current
Impedance = Maximum
Admittance = Minimum
Supply Current = Minimum
Power Factor = Unity

This condition is also called Current Resonance or Anti-Resonance.

✔ Answer: B) Minimum

Question 125

For an ideal tank circuit, the value of dynamic admittance is:

Options:

A) Infinite
B) Zero
C) Both
D) None of the above

Answer: B) Zero

Step-by-Step Solution:

An ideal tank circuit consists of a pure inductor and a pure capacitor with no resistance.

At resonance:

Dynamic impedance becomes infinite.
Admittance is the reciprocal of impedance.

Since,

Admittance = 1 / Impedance

and

Impedance = Infinite

Therefore,

Admittance = 0

Hence, the dynamic admittance of an ideal tank circuit is zero.

Additional Note:

For an ideal tank circuit:

Resistance = 0 Ω
Dynamic Impedance = Infinite
Dynamic Admittance = Zero
Supply Current = Zero (ideally)

✔ Answer: B) Zero

Question 126

For an ideal tank circuit, the value of dynamic impedance is:

Options:

A) Infinite
B) Zero
C) Both
D) None of the above

Answer: A) Infinite

Step-by-Step Solution:

An ideal tank circuit contains:

A pure inductor
A pure capacitor
No resistance

At resonance:

The inductive and capacitive currents cancel each other.
The circuit opposes the flow of supply current.

Therefore, the circuit offers maximum impedance.

Hence,

Dynamic Impedance = Infinite

Additional Note:

For an ideal parallel resonant (tank) circuit:

Dynamic Impedance = Infinite
Dynamic Admittance = Zero
Supply Current = Minimum (Ideally Zero)

This is why a tank circuit is widely used in tuned circuits and oscillators.

✔ Answer: A) Infinite

Question 127

In a series R-L-C circuit, to obtain Q > 1, which of the following condition is required?

Options:

A) XL > R
B) XL < R
C) XL = R
D) All of the above

Answer: A) XL > R

Step-by-Step Solution:

The Quality Factor (Q) of a series R-L-C circuit is given by:

Q = XL / R

To satisfy the condition:

Q > 1

the numerator must be greater than the denominator.

Therefore,

XL > R

Hence, the inductive reactance must be greater than the resistance.

Additional Note:

For a series R-L-C circuit:

Q = XL / R = XC / R (at resonance)

A higher Q-factor indicates:

Lower power loss
Higher efficiency
Narrow bandwidth
Better selectivity

✔ Answer: A) XL > R

Question 128

In a parallel R-L-C circuit, to obtain Q > 1, which of the following condition is required?

Options:

A) XL > R
B) XL < R
C) XL = R
D) All of the above

Answer: B) XL < R

Step-by-Step Solution:

The Quality Factor (Q) of a parallel R-L-C circuit is given by:

Q = R / XL

To obtain:

Q > 1

the numerator must be greater than the denominator.

Therefore,

R > XL

or equivalently,

XL < R

Hence, the inductive reactance must be less than the resistance.

Additional Note:

Comparison of Q-factor formulas:

Series R-L-C Circuit	Parallel R-L-C Circuit
Q = XL / R	Q = R / XL
Q > 1 → XL > R	Q > 1 → R > XL (or XL < R)

The Quality Factor determines the sharpness of resonance, bandwidth, and efficiency of resonant circuits.

✔ Answer: B) XL < R

Question 129

By which of the following circuit elements do transients not occur?

Options:

A) Resistance (R)
B) Inductance (L)
C) Capacitance (C)
D) All of the above

Answer: A) Resistance (R)

Step-by-Step Solution:

A transient is a temporary change in voltage or current that occurs immediately after a circuit is switched ON or OFF.

A resistor:

Does not store electrical energy.
Allows voltage and current to change instantly.
Therefore, it does not produce a transient response.

In contrast:

An inductor stores energy in a magnetic field.
A capacitor stores energy in an electric field.

Since inductors and capacitors store energy, they exhibit transient behavior.

Additional Note:

Circuit Element	Stores Energy	Transient Response
Resistor (R)	No	No
Inductor (L)	Yes (Magnetic Field)	Yes
Capacitor (C)	Yes (Electric Field)	Yes

✔ Answer: A) Resistance (R)

Question 130

Which of the following elements are called dynamic elements?

Options:

A) Resistance (R)
B) Inductance (L)
C) Capacitance (C)
D) Both L and C

Answer: D) Both L and C

Step-by-Step Solution:

Dynamic elements are those that store energy and exhibit transient behavior.

An inductor stores energy in the form of a magnetic field, while a capacitor stores energy in the form of an electric field.

Since both elements can store and release energy, they are called dynamic elements.

A resistor only dissipates energy as heat and cannot store energy.

Additional Note:

Element	Energy Storage	Dynamic Element
Resistor	No	No
Inductor	Magnetic Energy	Yes
Capacitor	Electric Energy	Yes

Dynamic elements are important in transient and AC circuit analysis.

✔ Answer: D) Both L and C

Question 131

For steady-state current, an inductor acts as a:

Options:

A) Short circuit
B) Open circuit
C) Voltage source
D) Current source

Answer: A) Short circuit

Step-by-Step Solution:

The voltage across an inductor is given by:

V = L × (di / dt)

Under steady-state DC conditions:

The current becomes constant.
Therefore,

di / dt = 0

Hence,

V = 0

Since there is no voltage drop across an ideal inductor, it behaves like a short circuit.

Additional Note:

Behavior of an ideal inductor:

At the instant of switching ON → Opposes change in current.
Under steady-state DC → Behaves as a short circuit.
Under AC → Offers inductive reactance.

✔ Answer: A) Short circuit

Question 132

In an R-L series circuit, R = 2 Ω, L = 10 mH, and the applied voltage is 10 V DC. What is the steady-state current in the circuit?

Options:

A) 0 A
B) 10 A
C) 5 A
D) 1 A

Answer: C) 5 A

Step-by-Step Solution:

For a DC supply, after the transient period, the inductor behaves as a short circuit.

Therefore, only the resistor limits the current.

Apply Ohm's Law:

I = V / R

Given:

Voltage (V) = 10 V
Resistance (R) = 2 Ω

Substitute the values:

I = 10 / 2

I = 5 A

Therefore, the steady-state current in the circuit is:

5 A

Additional Note:

For a DC R-L circuit:

Initially, the inductor opposes the rise of current.
At steady state, the inductor behaves as a short circuit.
The current is determined only by the resistance.

✔ Answer: C) 5 A

Question 133

In steady-state condition, a capacitor acts as a:

Options:

A) Short circuit
B) Open circuit
C) Voltage source
D) Current source

Answer: B) Open circuit

Step-by-Step Solution:

The current through a capacitor is given by:

I = C × (dV / dt)

Under steady-state DC conditions:

The capacitor becomes fully charged.
The voltage across it becomes constant.

Therefore,

dV / dt = 0

Hence,

I = 0

Since no current flows through the capacitor, it behaves as an open circuit.

Additional Note:

Behavior of an ideal capacitor:

At the instant of switching ON → Allows charging current to flow.
Under steady-state DC → Behaves as an open circuit.
Under AC → Offers capacitive reactance.

✔ Answer: B) Open circuit

Question 134

In an R-C series circuit, R = 2 Ω, C = 2 μF, and a 10 V DC supply is applied. What is the steady-state current?

Options:

A) 0 A
B) 2 A
C) 5 A
D) 10 A

Answer: A) 0 A

Step-by-Step Solution:

When a DC voltage is applied to an R-C series circuit:

Initially, the capacitor charges.
After a long time (steady-state condition), the capacitor becomes fully charged.

A fully charged capacitor behaves as an open circuit.

Therefore,

No current flows through the circuit.

Hence,

Current = 0 A

Additional Note:

Behavior of a capacitor with DC supply:

At switching ON → Charging current flows.
At steady state → Capacitor behaves as an open circuit.
Steady-state current = 0 A

✔ Answer: A) 0 A

Question 135

The time constant of an R-L series circuit is:

Options:

A) 2L / R
B) RC
C) L / R
D) 2RC

Answer: C) L / R

Step-by-Step Solution:

The time constant (τ) of an R-L series circuit is the time required for the current to reach approximately 63.2% of its final value after a DC voltage is applied.

The formula is:

τ = L / R

where:

τ = Time Constant (seconds)
L = Inductance (henry)
R = Resistance (ohm)

Additional Note:

For an R-L circuit:

Time Constant = L / R
After 1τ, current reaches 63.2% of its final value.
After approximately 5τ, the current reaches nearly 100% of its final value.

✔ Answer: C) L / R

Question 136

The time constant of an R-C series circuit is:

Options:

A) L / R
B) 2RC
C) 2L / R
D) RC

Answer: D) RC

Step-by-Step Solution:

The time constant (τ) of an R-C series circuit represents the time required for the capacitor voltage to reach approximately 63.2% of its final value during charging.

The formula is:

τ = R × C

where:

R = Resistance (ohm)
C = Capacitance (farad)

Additional Note:

For an R-C circuit:

Time Constant = RC
After 1τ, capacitor voltage reaches 63.2% of its final value.
After about 5τ, the capacitor is considered fully charged.

✔ Answer: D) RC

Question 137

In an R-L series circuit, R = 2 Ω and L = 2 mH. What is the value of the time constant?

Options:

A) 1 ms
B) 2 ms
C) 4 ms
D) 100 s

Answer: A) 1 ms

Step-by-Step Solution:

For an R-L circuit,

Time Constant (τ) = L / R

Given:

L = 2 mH = 2 × 10⁻³ H
R = 2 Ω

Substitute the values:

τ = (2 × 10⁻³) / 2

τ = 1 × 10⁻³ second

τ = 1 ms

Therefore, the time constant of the circuit is:

1 ms

Additional Note:

Unit conversions:

1 H = 1000 mH
1 second = 1000 milliseconds

The time constant indicates how quickly the current reaches its steady-state value.

✔ Answer: A) 1 ms

Question 138

Time constant is the time taken for the response to rise to ________ of its maximum value.

Options:

A) 100%
B) 90%
C) 63.2%
D) 68.3%

Answer: C) 63.2%

Step-by-Step Solution:

The time constant (τ) is an important parameter in R-L and R-C circuits.

It is defined as the time required for:

The current (in an R-L circuit), or
The voltage (in an R-C circuit)

to reach 63.2% of its final steady-state value during the charging or growth process.

Similarly, during the decay process, the response falls to 36.8% of its initial value after one time constant.

Therefore, the correct answer is:

63.2%

Additional Note:

Response after one time constant:

Condition	Value
Charging	63.2% of final value
Discharging	36.8% of initial value

After approximately 5 time constants (5τ), the transient response is considered complete, and the circuit reaches steady state.

✔ Answer: C) 63.2%

Question 139

Which of the following is different from the others in terms of measurement?

Options:

A) L / R
B) RC
C) 2L / R
D) Q

Answer: D) Q

Step-by-Step Solution:

The quantities L/R, RC, and 2L/R all represent time constants of electrical circuits.

L/R is the time constant of an R-L circuit.
RC is the time constant of an R-C circuit.
2L/R also has the dimension of time.

On the other hand, Q represents the Quality Factor of a resonant circuit.

Unlike the time constant, the Quality Factor is dimensionless and indicates the sharpness of resonance and efficiency of the circuit.

Therefore, Q is different from the other three quantities.

Additional Note:

Quantity	Represents
L/R	Time Constant of an R-L Circuit
RC	Time Constant of an R-C Circuit
2L/R	Quantity with the Dimension of Time
Q	Quality Factor (Dimensionless)

✔ Answer: D) Q

Question 140

In which of the following damping conditions are oscillations not present?

Options:

A) Under damping
B) Over damping
C) Critical damping
D) Both B and C

Answer: D) Both B and C

Step-by-Step Solution:

The response of a second-order system depends on its damping condition.

Under-damped: The system oscillates before reaching its final value.
Critically damped: The system reaches the final value quickly without oscillations.
Over-damped: The system reaches the final value slowly without oscillations.

Therefore, oscillations are not present in both critical damping and over damping.

Additional Note:

Damping Condition	Oscillations
Under Damping	Present
Critical Damping	Not Present
Over Damping	Not Present

Critical damping provides the fastest response without overshoot.

✔ Answer: D) Both B and C

Question 141

In which of the following networks is it not possible to obtain a transient-free response?

Options:

A) R-C Circuit
B) R-L Circuit
C) R-L, R-C, and R-L-C Circuits
D) R-L-C Circuit

Answer: C) R-L, R-C, and R-L-C Circuits

Step-by-Step Solution:

Transient response occurs whenever a circuit contains an energy storage element.

Energy storage elements are:

Inductor (L) – Stores magnetic energy.
Capacitor (C) – Stores electric energy.

Since:

An R-L circuit contains an inductor.
An R-C circuit contains a capacitor.
An R-L-C circuit contains both.

All these circuits exhibit transient behavior when connected to a DC or AC supply.

Therefore, a transient-free response is not possible in any of these networks.

Additional Note:

Circuit	Energy Storage Element	Transient Response
R-L	Inductor	Yes
R-C	Capacitor	Yes
R-L-C	Inductor & Capacitor	Yes

Only a purely resistive circuit does not exhibit transient response.

✔ Answer: C) R-L, R-C, and R-L-C Circuits

Question 142

To find the initial value of a function using the Initial Value Theorem, the highest power of s in the denominator should be ________ the highest power of s in the numerator.

Options:

A) Greater than
B) Less than
C) Equal to
D) Both A and C

Answer: A) Greater than

Step-by-Step Solution:

The Initial Value Theorem is expressed as:

f(0+) = Lim (s → ∞) [ s × F(s) ]

For the limit to produce a finite value:

The highest power of s in the denominator must be greater than the highest power of s in the numerator.

Otherwise:

The limit becomes infinite or undefined.

Therefore, the correct condition is:

Denominator power > Numerator power

Additional Note:

The Initial Value Theorem is commonly used in:

Control Systems
Network Analysis
Laplace Transform Problems
Transient Analysis

It provides the value of a function immediately after switching.

✔ Answer: A) Greater than

Question 143

The Final Value Theorem is not applicable to which of the following systems?

Options:

A) Stable systems
B) Unstable systems
C) Marginally stable systems
D) Both B and C

Answer: D) Both B and C

Step-by-Step Solution:

The Final Value Theorem is expressed as:

f(∞) = Lim (s → 0) [ s × F(s) ]

This theorem is valid only if the system is stable.

It cannot be applied to:

Unstable systems, because the output does not settle to a finite value.
Marginally stable systems, because the output may continue to oscillate indefinitely.

Therefore, the Final Value Theorem is not applicable to unstable and marginally stable systems.

Additional Note:

Applicability of the Final Value Theorem:

System Type	Final Value Theorem
Stable	Applicable
Unstable	Not Applicable
Marginally Stable	Not Applicable

Always verify system stability before applying the Final Value Theorem.

✔ Answer: D) Both B and C

Question 144

It is preferable to connect bulbs in:

Options:

A) Series
B) Parallel
C) Both series and parallel
D) Neither series nor parallel

Answer: B) Parallel

Step-by-Step Solution:

In household wiring and lighting systems, bulbs are connected in parallel.

When bulbs are connected in parallel:

Each bulb receives the full supply voltage.
Each bulb operates independently.
If one bulb fuses or burns out, the remaining bulbs continue to glow normally.

In a series connection, if one bulb fails, the circuit becomes open and all the bulbs stop glowing.

Therefore, the preferred connection for bulbs is parallel.

Additional Note:

Comparison of Series and Parallel Connections:

Series Connection	Parallel Connection
Same current flows through all bulbs	Same voltage across all bulbs
If one bulb fails, all bulbs go OFF	Other bulbs continue to glow
Voltage is divided	Full supply voltage across each bulb

✔ Answer: B) Parallel

Question 145

Circuit Diagram:


A ●──/\/\/───┬────/\/\/────┐
      4 Ω    │      1 Ω    │
             │             │
             └────/\/\/────┤
                  2 Ω       │
                            │
B ●──/\/\/──────────────────┘
      3 Ω

Find the total resistance between points A and B.

Options:

A) 7 Ω
B) 0 Ω
C) 7.67 Ω
D) 0.48 Ω

Answer: C) 7.67 Ω

Step-by-Step Solution:

Step 1: Calculate the equivalent resistance of the parallel combination.

The 1 Ω and 2 Ω resistors are connected in parallel.

Rp = (1 × 2) / (1 + 2)

Rp = 2 / 3 Ω

Rp = 0.67 Ω

Step 2: Add the series resistors.

The equivalent parallel resistance is connected in series with the 4 Ω and 3 Ω resistors.

RT = 4 + 0.67 + 3

RT = 7.67 Ω

Therefore, the total resistance between A and B is:

RT = 7.67 Ω

Additional Note:

Useful formulas:

Series Resistance

R = R1 + R2 + R3 + ...

Two Resistors in Parallel

Rp = (R1 × R2) / (R1 + R2)

For this circuit:

Parallel equivalent = 0.67 Ω
Total resistance = 4 + 0.67 + 3 = 7.67 Ω

✔ Answer: C) 7.67 Ω

Question 146

Circuit Diagram:


                     A
                     ●
                     │
                     │
          ┌──/\/\/───┴───/\/\/───┐
          │    5 Ω         10 Ω   │
          │                       │
          │                       │
          └──/\/\/───────┬/\/\/───┘
              15 Ω        │ 20 Ω
                          │
                          │
                          ●
                          B

Calculate the equivalent resistance between A and B.

Options:

A) 60 Ω
B) 15 Ω
C) 12 Ω
D) 48 Ω

Answer: C) 12 Ω

Step-by-Step Solution:

Step 1: Combine the left-side series resistors.

The 5 Ω and 15 Ω resistors are connected in series.

R1 = 5 + 15 = 20 Ω

Step 2: Combine the right-side series resistors.

The 10 Ω and 20 Ω resistors are connected in series.

R2 = 10 + 20 = 30 Ω

Step 3: Calculate the equivalent resistance.

The two equivalent resistances (20 Ω and 30 Ω) are connected in parallel.

Rp = (20 × 30) / (20 + 30)

Rp = 600 / 50

Rp = 12 Ω

Therefore,

Equivalent Resistance between A and B = 12 Ω

Additional Note:

For this circuit:

Left branch resistance = 20 Ω
Right branch resistance = 30 Ω
Equivalent resistance = 12 Ω

Useful formulas:

Series:

R = R1 + R2 + ...

Parallel (Two Resistors):

Rp = (R1 × R2) / (R1 + R2)

✔ Answer: C) 12 Ω

Question 147

Circuit Diagram:


                           1 Ω
                   ┌─────/\/\/─────┐
A ●───/\/\/───┬────┤               ├──────────────┬────/\/\/────┐
      4 Ω     │    ├────/\/\/──────┤              │      5 Ω    │
              │    │      2 Ω      │              │             │
              │    └────/\/\/──────┘              ├────/\/\/────┤────● B
              │           3 Ω                     │      6 Ω    │
              │                                  │             │
              │                                  └─────────────┘
              │
              └────────────────/\/\/────────────────────────────┘
                             7 Ω

Calculate the resistance between A and B.

Options:

A) 3.56 Ω
B) 7 Ω
C) 14.26 Ω
D) 29.69 Ω

Answer: A) 3.56 Ω

Step-by-Step Solution:

Step 1: Calculate the equivalent resistance of the first parallel group.

The 1 Ω, 2 Ω, and 3 Ω resistors are connected in parallel.

1 / Rp1 = 1/1 + 1/2 + 1/3

1 / Rp1 = 1 + 0.5 + 0.333

1 / Rp1 = 1.833

Rp1 ≈ 0.545 Ω

Step 2: Calculate the equivalent resistance of the second parallel group.

The 5 Ω and 6 Ω resistors are connected in parallel.

Rp2 = (5 × 6) / (5 + 6)

Rp2 = 30 / 11

Rp2 ≈ 2.73 Ω

Step 3: Calculate the resistance of the upper branch.

The upper branch consists of:

4 Ω resistor
Equivalent of first parallel group
Equivalent of second parallel group

Rupper = 4 + 0.545 + 2.73

Rupper = 7.27 Ω

Step 4: Calculate the total equivalent resistance.

The upper branch (7.27 Ω) is connected in parallel with the 7 Ω resistor.

RT = (7 × 7.27) / (7 + 7.27)

RT = 50.89 / 14.27

RT ≈ 3.56 Ω

Therefore, the equivalent resistance between A and B is:

RT = 3.56 Ω

Additional Note:

Useful formulas:

Series Resistance

R = R1 + R2 + R3 + ...

Two Parallel Resistors

Rp = (R1 × R2) / (R1 + R2)

Three Parallel Resistors

1 / Rp = 1/R1 + 1/R2 + 1/R3

For this circuit:

First parallel group = 0.545 Ω
Second parallel group = 2.73 Ω
Upper branch = 7.27 Ω
Final equivalent resistance = 3.56 Ω

✔ Answer: A) 3.56 Ω

Question 148

Batteries are generally connected in:

Options:

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

Answer: A) Series

Step-by-Step Solution:

Batteries are generally connected in series when a higher output voltage is required.

In a series connection:

The voltages of individual batteries add together.
The same current flows through each battery.

For example:

Two 12 V batteries connected in series provide:

Total Voltage = 12 + 12 = 24 V

This arrangement is commonly used in automobiles, UPS systems, and industrial applications where a higher voltage is needed.

Additional Note:

Battery Connections:

Series Connection	Parallel Connection
Voltages add	Capacity (Ah) adds
Current remains the same	Voltage remains the same
Used to obtain higher voltage	Used to obtain longer backup time

✔ Answer: A) Series

Question 149

In a _________ circuit, the total resistance is greater than the largest resistance in the circuit.

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 circuit, resistors are connected one after another, so the same current flows through each resistor.

The total resistance is obtained by adding all individual resistances.

RT = R1 + R2 + R3 + ...

Since all resistances are added together, the total resistance is always greater than the largest individual resistor.

For example:

If three resistors of 2 Ω, 5 Ω, and 8 Ω are connected in series,

RT = 2 + 5 + 8 = 15 Ω

Since 15 Ω > 8 Ω, the total resistance is greater than the largest resistor.

Additional Note:

Properties of a Series Circuit:

Current is the same through all components.
Voltage divides across the resistors.
Total resistance is the sum of all resistances.
If one component fails, the entire circuit stops working.

✔ Answer: A) Series

Question 150

In a _________ circuit, the total resistance is smaller than the smallest resistance in the circuit.

Options:

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

Answer: B) Parallel

Step-by-Step Solution:

In a parallel circuit, each resistor provides an additional path for current.

The equivalent resistance is calculated as:

1 / RT = 1/R1 + 1/R2 + 1/R3 + ...

Since additional current paths reduce the overall resistance, the equivalent resistance is always less than the smallest individual resistor.

For example:

Two resistors:

10 Ω
20 Ω

Equivalent resistance:

RT = (10 × 20) / (10 + 20)

RT = 200 / 30

RT = 6.67 Ω

Since 6.67 Ω is less than 10 Ω, the total resistance is smaller than the smallest resistor.

Additional Note:

Properties of a Parallel Circuit:

Voltage is the same across every branch.
Current divides among the branches.
Equivalent resistance is always less than the smallest branch resistance.

✔ Answer: B) Parallel

2000 Basic Electrical Engineering Fully Solved MCQs-3

Question 101

Answer: A) 0°

Step-by-Step Solution:

Additional Note:

Question 102

Answer: A) Series resonance

Step-by-Step Solution:

Additional Note:

Question 103

Answer: B) Parallel resonance

Step-by-Step Solution:

Additional Note:

Question 104

Answer: B) The other bulb continues to glow with the same brightness

Step-by-Step Solution:

Additional Note:

Question 105

Answer: A) 10 A

Step-by-Step Solution:

Additional Note:

Question 106

Answer: A) The same for all resistors

Step-by-Step Solution:

Additional Note:

Question 107

Answer: B) The value of the resistance

Step-by-Step Solution:

Additional Note:

Question 108

Answer: C) 11.43 A

Step-by-Step Solution:

Additional Note:

Question 109

Answer: A) 100 V

Step-by-Step Solution:

Additional Note:

Question 110

Answer: C) Zero

Step-by-Step Solution:

Additional Note:

Question 111

Answer: A) 27 Ω

Step-by-Step Solution:

Additional Note:

Question 112

Answer: D) 12 A

Step-by-Step Solution:

Additional Note:

Question 113

Answer: B) 5 Ω

Step-by-Step Solution:

Additional Note:

Question 114

Answer: B) Less than the resonance frequency

Step-by-Step Solution:

Additional Note:

Question 115

Answer: A) High Q-factor

Step-by-Step Solution:

Additional Note:

Question 116

Answer: B) Low Q-factor

Step-by-Step Solution:

Additional Note:

Question 117

Answer: D) All of the above

Step-by-Step Solution:

Additional Note:

Question 118

Answer: B) Leading

Step-by-Step Solution:

Additional Note:

Question 119

Answer: A) Lagging

Step-by-Step Solution:

Additional Note:

Question 120

Answer: D) (ω₁ × ω₂)^(1/2)

Step-by-Step Solution: