Electrical Resonance MCQ Quiz - Objective Question with Answer for Electrical Resonance - Download Free PDF

Last updated on May 14, 2025

Latest Electrical Resonance MCQ Objective Questions

Electrical Resonance Question 1:

A parallel RLC circuit has an inductance of 1 H and a capacitance of 1 µF. What is the resonant frequency (f0 )? 

  1. \(\rm \frac{1}{\pi \times 10^{-3}}Hz\)
  2. \(\rm \frac{1}{\pi}Hz\)
  3. \(\rm \frac{1}{2\pi}Hz\)
  4. \(\rm \frac{1}{2\pi \times 10^{-3}}Hz\)

Answer (Detailed Solution Below)

Option 4 : \(\rm \frac{1}{2\pi \times 10^{-3}}Hz\)

Electrical Resonance Question 1 Detailed Solution

Concept

The resonant frequency of a parallel RLC circuit is given by:

\(f_o={1\over 2\pi \sqrt{LC}}\)

where, fo = Resonant frequency

L = Inductance

C = Capacitance

Calculation

Given, L = 1 H

C = 1 μF = 1 × 10-6 F

\(f_o={1\over 2\pi \sqrt{1\times 1\times 10^{-6}}}\)

\(f_o=\rm \frac{1}{2\pi \times 10^{-3}}Hz\)

Electrical Resonance Question 2:

If the capacitance in a series RLC circuit is increased, the Q-factor will _________.

  1. increase
  2. decrease
  3. remain unchanged
  4. depend on frequency 

Answer (Detailed Solution Below)

Option 2 : decrease

Electrical Resonance Question 2 Detailed Solution

Quality factor

In a resonant series circuit, the quality factor (Q) is a measure of how underdamped the system is and how sharp the resonance is.

It is given by:

\(QF={1\over R}\sqrt{L\over C}={\omega_oL\over R}\)

From the above expression, it is observed that the quality factor is inversely proportional to capacitance.

If the capacitance in a series RLC circuit is increased, the Q-factor will decrease.

Electrical Resonance Question 3:

If the resistance (R) in a parallel RLC circuit increases, the quality factor (Q) will _______. 

  1. remain the same
  2. become zero
  3. decrease
  4. increase

Answer (Detailed Solution Below)

Option 4 : increase

Electrical Resonance Question 3 Detailed Solution

Concept

The quality factor in a parallel RLC circuit is given by:

\(QF=R{\sqrt{C\over L}}\)

where, R = Resistance

C = Capacitance

L = Inductance

Explanation

From the above expression, the quality factor is directly proportional to the resistance.

If the resistance (R) in a parallel RLC circuit increases, the quality factor (Q) will increase.

Electrical Resonance Question 4:

In a parallel RLC circuit, if the inductive current IL is greater than the capacitive current IC, then ________.

  1. XC > XL
  2. XC = XL
  3. XC < XL
  4. circuit is at resonance

Answer (Detailed Solution Below)

Option 1 : XC > XL

Electrical Resonance Question 4 Detailed Solution

Parallel RLC circuit

qImage681846378666492ff3c97a64

The inductive current is given by:

\(I_L={V\over X_L}\)

The capacitive current is given by:

\(I_C={V\over X_C}\)

It is given that IL > IC

\({V\over X_L}>{V\over X_C}\)

XC > XL

The correct answer is option 1.

Electrical Resonance Question 5:

The Q-factor of a resonant circuit is 100. If the resonant frequency is 1 MHz, what is the bandwidth? 

  1. 10 MHz 
  2. 100 kHz
  3. 1 kHz
  4. 10 kHz

Answer (Detailed Solution Below)

Option 4 : 10 kHz

Electrical Resonance Question 5 Detailed Solution

Concept

In a resonant series circuit, the quality factor (Q) is a measure of how underdamped the system is and how sharp the resonance is.

It is given by:

\(QF={ω_o\over BW}\)

where, ωo = Resonance frequency

BW = Bandwidth

Calculation

Given, QF = 100

ω= 1 MHz

\(100={10^6\over BW}\)

BW = 1kHz

Top Electrical Resonance MCQ Objective Questions

For the A.C circuit as shown below, if the rms voltage across the resistor is 120 V. What is the value of the inductor?

Basic of Network 2 images Q6

  1. 0.5 H
  2. 0.6 H
  3. 1.0 H
  4. 1.5 H

Answer (Detailed Solution Below)

Option 4 : 1.5 H

Electrical Resonance Question 6 Detailed Solution

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We have RMS value of source

\({{\rm{V}}_{{\rm{RMS}}}} = \frac{{{{\rm{V}}_{\rm{m}}}}}{{\sqrt 2 }} = 150\)

Then, \({{\rm{V}}_{\rm{L}}} = \sqrt {{\rm{V}}_{{\rm{RMS}}}^2 - {\rm{V}}_{\rm{R}}^2} = \sqrt {{{150}^2} - {{120}^2}}\)

\(\Rightarrow {{\rm{V}}_{\rm{L}}} = 90{\rm{V}}\)

Now,  \({\rm{I}} = \frac{{{{\rm{V}}_{\rm{R}}}}}{{\rm{R}}} = \frac{{120}}{{1{\rm{k}}}} = 120 {\rm{mA}}\)

Again \({{\rm{V}}_{\rm{L}}} = {\rm{I}} \times{\rm{\omega L}}\)

\(\begin{array}{l} \Rightarrow 90 = 0.12 \times 500 \times {\rm{L}}\\ \Rightarrow {\rm{L}} = \frac{{90}}{{0.12 \times 500}} = 1.5{\rm{H}} \end{array}\)

The Q factor of a parallel resonant circuit is given by

  1. \(\frac{1}{R}\sqrt {\frac{L}{C}} \)
  2. \({R}\sqrt {\frac{C}{L}} \)
  3. \(\frac{1}{R}\sqrt {\frac{1}{{LC}}} \)
  4. \(\frac{R}{{\sqrt {LC} }}\)

Answer (Detailed Solution Below)

Option 2 : \({R}\sqrt {\frac{C}{L}} \)

Electrical Resonance Question 7 Detailed Solution

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The quality factor is defined as the ratio of the maximum energy stored to maximum energy dissipated in a cycle

\(Q = 2\pi \frac{{Maximum\;energy\;stored}}{{Total\;energy\;lost\;per\;period}}\)

In a series RLC, Quality factor \(Q = \frac{{\omega L}}{R} = \frac{1}{{\omega RC}} = \frac{{{X_L}}}{R} = \frac{{{X_C}}}{R}\)

In a parallel RLC, \(Q = \frac{R}{{{\rm{\omega L}}}} = \omega RC = \frac{R}{{{X_L}}} = \frac{R}{{{X_C}}}\)

It is defined as, resistance to the reactance of reactive element.

The quality factor Q is also defined as the ratio of the resonant frequency to the bandwidth.

\(Q=\frac{{{f}_{r}}}{BW}\)

For parallel RLC Circuit \({{\rm{Q}}_1} = {\rm{R}}\sqrt {\frac{{\rm{C}}}{{\rm{L}}}}\)

For Series RLC Circuit \({{\rm{Q}}_2} = \frac{1}{{\rm{R}}}\sqrt {\frac{{\rm{L}}}{{\rm{C}}}}\)

An R-L-C series resonant circuit has the following parameters: Resonance frequency = 5000 / 2π Hz; impedance at resonance = 56 Ω and Q-factor = 25. Calculate the capacitance of the capacitor.

  1. 143 μF
  2. 0.143 μF
  3. 1.43 μF
  4. 14.3 μF

Answer (Detailed Solution Below)

Option 2 : 0.143 μF

Electrical Resonance Question 8 Detailed Solution

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Concept:

The quality factor is defined as the ratio of the maximum energy stored to maximum energy dissipated in a cycle

\(Q = 2\pi \frac{{Maximum\;energy\;stored}}{{Total\;energy\;lost\;per\;period}}\)

The quality factor in a series RLC circuit is given by:

\(Q = \frac{{{X_L}}}{R} = \frac{{{X_C}}}{R} = \frac{1}{R}\;\sqrt {\frac{L}{C}} \)

At resonance, the circuit acts as pure resistive and the impedance equals the resistance.

Calculation:

Resonance frequency (f) = 5000 / 2π Hz

Resonant frequency (ω) = 2πf = 5000 rad/sec

Impedance at resonance (Z) = R = 56 Ω

Quality factor = 25

\(Q = \frac{{\omega L}}{R}\)

\( \Rightarrow 25 = \frac{{5000 \times L}}{{56}}\)

⇒ L = 0.28 H

\(Q = \frac{1}{R}\sqrt {\frac{L}{C}} \)

\( \Rightarrow 25 = \frac{1}{{56}}\sqrt {\frac{{0.28}}{C}} \)

⇒ C = 0.143 μF

In a parallel resonance circuit, the admittance is:

  1. zero
  2. Maximum
  3. minimum
  4. infinity

Answer (Detailed Solution Below)

Option 3 : minimum

Electrical Resonance Question 9 Detailed Solution

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Parallel Resonance:

F1 Mrunal Engineering 12.11.2022 D33

The input admittance is given by:

\(Y_{in} = {1 \over R}+j(\omega C-{1\over \omega L})\)

\(Y_{in} = G+j(B_C-B_L)\)

where, Yin = Admittance

G = Conductance

BC = Capacitive susceptance

BL = Inductive susceptance

At resonance, BC = BL 

Hence, Yin = G and is minimum.

Additional Information

Series Resonance

Parallel Resonance

Impedance is minimum

Admittance is minimum

Current is maximum

Voltage is maximum

The voltage across the inductor and capacitor is greater than the supply voltage

The current across the inductor and capacitor is greater than the supply current

Voltage magnification circuit

Current magnification circuit

Which of the following phasor diagram represents the series LCR circuit at resonance?

5f6d6520d450ada7fd346c29 16520956408911

  1. 5f6d6520d450ada7fd346c29 16520956408952
  2. F4 Madhuri Engineering 09.05.2022 D1 V3
  3. 5f6d6520d450ada7fd346c29 16520956408995
  4. 5f6d6520d450ada7fd346c29 16520956409006

Answer (Detailed Solution Below)

Option 3 : 5f6d6520d450ada7fd346c29 16520956408995

Electrical Resonance Question 10 Detailed Solution

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Concept:

In series LCR circuits, Resonance is a condition in which the inductive reactance and capacitive reactance are equal and lie opposite in phase, so cancel out each other. Only, resistance is left as impedance.

The frequency at which the series LCR circuit goes into resonance is:

\(f = \frac{1}{{2π }}\sqrt {\frac{1}{{LC}}}\)

At resonance, the impedance of the series LCR circuit is minimum and hence the current is maximum.

A series LCR circuit is as shown:

F1 Neha 12.1.20 Pallavi D6

The phasor diagram is as shown below:

F1 Neha 12.1.20 Pallavi D7

At resonance, XL = XC, i.e. the voltage across the inductor and capacitor are equal and opposite. The resultant phasor diagram at resonant frequency will, therefore, be:

SSC JE Electrical 3

26 June 1

The impedance - frequency curve is shown below:

RRB JE EE 26 10Q FT0 Part1 D1

The impedance at resonance offered by a parallel resonant circuit is _____.

  1. maximum given by 1/CR
  2. maximum given by L/CR
  3. minimum given by L/CR
  4. minimum given by 1/CR

Answer (Detailed Solution Below)

Option 2 : maximum given by L/CR

Electrical Resonance Question 11 Detailed Solution

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In parallel resonant circuit, inductive reactance (XL) is equal to capacitive reactance (XC). The circuit becomes pure resistive.

At resonance XL = XC.

At parallel resonance, impedance becomes maximum.

SSC Network D6

At parallel resonance line current Ir = IL cos ϕ

\(\frac{V}{{{Z_r}}} = \frac{V}{{{Z_1}}}X\frac{R}{{{Z_1}}}or\frac{1}{{{Z_r}}} = \frac{R}{{Z_L^2}}\;\)

\(\frac{1}{{{Z_r}}} = \frac{{\frac{R}{L}}}{C} = \frac{{CR}}{L}\left( {as\;Z_L^2 = \frac{L}{C}} \right)\)

Therefore, the circuit impedance will be given as

\({Z_r} = \frac{L}{{CR}}\)

Consider the following statements with respect to a parallel R-L-C circuit:

1. The bandwidth of the circuit decreases if R is increased.

2. The bandwidth of the circuit remains the same if L is increased.

3. At resonance, the input impedance is a real quantity.

4. At resonance, the magnitude of the input impedance attains its minimum value.

Which of the above statements are correct?

  1. 1, 2 and 4
  2. 1, 3 and 4
  3. 2, 3 and 4
  4. 1, 2 and 3

Answer (Detailed Solution Below)

Option 4 : 1, 2 and 3

Electrical Resonance Question 12 Detailed Solution

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Parallel RLC circuit:

10-May-2019 Rishi D1

The characteristic equation is given as,

\({s^2} + \frac{1}{{RC}}s + \frac{1}{{LC}} = 0\)

The bandwidth of a parallel RLC network is given as:

\({\rm{BW}} = \frac{{{{\rm{\omega}}_{\rm{r}}}}}{{\rm{Q}}}=\frac{1}{RC}\)

Observations:

  • The bandwidth is inversely proportional to the resistance, i.e. as the resistance increases, the bandwidth decreases.
  • The bandwidth is independent of inductance. Hence the bandwidth of the circuit remains the same if L is increased.
  • At resonance, the imaginary part of the net impedance becomes zero, which makes the input impedance a real quantity.
  • Also, at resonance, the input impedance for a parallel resonant circuit attains a maximum value. This is opposite to a series resonance where at resonance, the impedance attains a minimum value.

26 June 1

Specifications

 Series resonance circuit 

 Parallel resonance circuit 

Impedance at resonance

Minimum

Maximum

Current at resonance

Maximum

Minimum

Effective impedance

R

L/CR

It magnifies

Voltage

Current

It is known as

Acceptor circuit

Rejector circuit

Power factor

Unity

Unity

In the circuit shown below, readings of the voltmeter are V1 = 100 V, V2 = 50 V, V3 = 50 V. What will be the source voltage?

F29 Shubham B 19-4-2021 Swati D2

  1. 100 V
  2. 200√2 V
  3. 100√2 V
  4. 200 V

Answer (Detailed Solution Below)

Option 1 : 100 V

Electrical Resonance Question 13 Detailed Solution

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Concept:

RLC series circuit:

quesImage7494

For a series RLC circuit, the net impedance is given by:

Z = R + j (XL - XC)

XL = Inductive Reactance given by:

XL = ωL

XC = Capacitive Reactance given by:

XL = 1/ωC

ω = 2 π f

ω = angular frequency

f = linear frequency

The magnitude of the impedance is given by:

\(|Z|=\sqrt{R^2+(X_L-X_C)^2}\)

\(V = \sqrt {V_1^2 + {{\left( {{V_2} - {V_3}} \right)}^2}} \)

Where, V1 = voltage across the resistor

V2 = voltage across the inductor

V3 = voltage across the capacitor

V = resultant voltage

The current flowing across the series RLC circuit will be:

\(I=\frac{V}{|Z|}\)

At resonance, XL = XC, resulting in the net impedance to be minimum. This eventually results in the flow of the maximum current of:

\(I=\frac{V}{R}\)

Calculation:

quesImage7494

V1 = 100 V, V2 = 50 V, V3 = 50 V 

V2 = V3 (As data provided in the question)

So, the circuit is in series resonance.

V = V1 = 100 V

The source voltage is 100 V

The resonant frequency of a parallel resonant bandpass filter is 20 kHz and its bandwith is 2 kHz. Its upper cutoff frequency is ______

  1. 19 kHz
  2. 22 kHz
  3. 18 kHz
  4. 21 kHz

Answer (Detailed Solution Below)

Option 4 : 21 kHz

Electrical Resonance Question 14 Detailed Solution

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Concept:

The graph between impedance Z and the frequency of the parallel RLC circuit:

F1 Jai 21.11.20 Pallavi D2

Here,

f1 is the lower cutoff frequency

fis the upper cutoff frequency

fr is the resonant frequency

BW is the bandwidth

Formula:

BW = f2 – f1

\({f_1} = {f_r} - \left( {\frac{{BW}}{2}} \right)\)

\({f_2} = {f_r} + \left( {\frac{{BW}}{2}} \right)\)

Calculation:

Given

Resonant frequency fr = 20 kHz

Bandwidth = 2 kHz

The upper cutoff frequency is given as:

\({f_2} = {f_r} + \left( {\frac{{BW}}{2}} \right)\)

\({f_2} = {20kHz} + \left( {\frac{{2kHz}}{2}} \right)\)

f2 = 21 kHz

For a series RLC circuit, selectivity is defined as the ratio of:

  1. bandwidth to the resonant frequency
  2. upper cutoff frequency to lower cutoff frequency
  3. resonant frequency to the bandwidth
  4. lower cutoff frequency to upper cutoff frequency

Answer (Detailed Solution Below)

Option 3 : resonant frequency to the bandwidth

Electrical Resonance Question 15 Detailed Solution

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The selectivity/sharpness of the resonance amplifier is measured by the quality factor and is explained in the figure shown below:

F1 U.B. Nita 29.10.2019 D 3

Selectivity of a resonant circuit is defined as the ratio of resonant frequency fr to the half power bandwidth.

Selectivity = resonance frequency/3-dB Bandwidth
\( = \frac{{{f_r}}}{{{f_2} - {f_1}}}\)

Important Points

The quality factor Q is defined as the ratio of the resonant frequency to the bandwidth, i.e.

\(Q=\frac{{ω}_{r}}{BW}\)

ωr = Resonant frequency

B.W. = Bandwith of the amplifier

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