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

Last updated on Jun 10, 2025

Latest Series Resonance MCQ Objective Questions

Series Resonance Question 1:

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

Series Resonance Question 1 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.

Series Resonance Question 2:

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

Series Resonance Question 2 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

Series Resonance Question 3:

If the quality factor (Q) of a resonant series circuit increases, then what happens to the voltage magnification?

  1. It remains the same because supply voltage is constant.
  2. It increases because energy losses decrease. 
  3. It becomes zero because resonance cancels the reactance
  4. It decreases because resistance increases.

Answer (Detailed Solution Below)

Option 2 : It increases because energy losses decrease. 

Series Resonance Question 3 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:

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

Voltage magnification in such a circuit is directly proportional to the quality factor Q Q" id="MathJax-Element-3-Frame" role="presentation" style="position: relative;" tabindex="0">Q Q . When Q increases, it indicates lower energy loss (primarily due to reduced resistance), and as a result, the voltage across the inductor or capacitor at resonance becomes significantly higher than the supply voltage, showing greater voltage magnification.

Series Resonance Question 4:

Q factor is defined as the ratio of

  1. Resistance /inductance of reactive element
  2. Resistance/capacitance of reactive element  
  3. Resistance to reactance of reactive element 
  4. Resistance to susceptance of reactive element 

Answer (Detailed Solution Below)

Option 1 : Resistance /inductance of reactive element

Series Resonance Question 4 Detailed Solution

Explanation:

Q Factor (Quality Factor)

Definition: The Q factor, or Quality Factor, is a dimensionless parameter that describes the damping or energy loss of an oscillatory system. In the context of electrical circuits, it quantifies the sharpness of the resonance of a circuit and is defined as the ratio of the energy stored in the reactive components (inductors or capacitors) to the energy dissipated per cycle in the resistive elements.

Correct Option Analysis:

The correct option is:

Option 1: Resistance / inductance of reactive element.

The Q factor is defined as:

Q = (Energy stored in the reactive element) / (Energy dissipated per cycle)

For a series RLC circuit, the Q factor can be expressed in terms of the inductance (L) and resistance (R). The formula is:

Q = ωL / R

Where:

  • ω = 2πf, the angular frequency of the circuit.
  • L = Inductance of the inductor.
  • R = Resistance of the circuit.

This equation shows that the Q factor is directly proportional to the inductance and inversely proportional to the resistance of the circuit. A higher Q factor indicates lower energy loss and sharper resonance, which is desirable in many practical applications such as communication systems and filters.

Therefore, the Q factor is essentially the ratio of resistance (R) to the inductance (L) of the reactive element when we consider the relationship between energy dissipation and energy storage in the circuit.

Additional Information

Analysis of Other Options:

Option 2: Resistance / capacitance of reactive element.

This option is incorrect because the Q factor is not defined in terms of capacitance (C) alone. While capacitance is a reactive element, the Q factor depends on the ratio of stored energy to dissipated energy, which is more commonly expressed in terms of inductance (L) and resistance (R) in a series RLC circuit. In circuits where capacitance is the primary reactive element, the Q factor is defined using the reciprocal of capacitance (1/C), not capacitance directly.

Option 3: Resistance to reactance of reactive element.

This option is also incorrect because the Q factor is not directly defined as the ratio of resistance (R) to reactance (X). Instead, the Q factor is related to the ratio of reactance (X) to resistance (R) in a series circuit, as reactance determines the energy storage in the reactive elements. Specifically, for an inductor:

Q = XL / R = ωL / R

For a capacitor:

Q = XC / R = 1 / (ωCR)

Thus, the Q factor is inversely proportional to resistance and directly proportional to the reactance.

Option 4: Resistance to susceptance of reactive element.

This option is incorrect because susceptance (B) is the reciprocal of reactance (1/X), which is used in the analysis of parallel circuits. The Q factor is not defined in terms of resistance (R) to susceptance (B). The relationship between susceptance and Q factor is more relevant for parallel RLC circuits, but even in those cases, the Q factor is not expressed as the ratio of resistance to susceptance.

Conclusion:

The correct definition of the Q factor involves the ratio of resistance (R) to the inductance (L) of the reactive element, as given in Option 1. The Q factor plays a critical role in determining the performance of resonant circuits, with higher Q values indicating better performance due to lower energy losses. Understanding the Q factor and its mathematical relationship with circuit parameters is essential for designing efficient electrical systems, especially in applications requiring high selectivity and minimal energy dissipation.

Series Resonance Question 5:

An AC circuit has R = 100 Ω, C = 2 μF and L = 80 mH, connected in series. The quality factor of the circuit is

  1. 2
  2. 0.5
  3. 20
  4. 400

Answer (Detailed Solution Below)

Option 1 : 2

Series Resonance Question 5 Detailed Solution

Calculation:
The formula for the quality factor (Q) of an LC circuit is given by:

Q = (1 / R) × √(L / C)

Where:

  • R = resistance = 100 Ω
  • L = inductance = 80 mH = 80 × 10-3 H
  • C = capacitance = 2 μF = 2 × 10-6 F

Substituting the values into the formula:

Q = (1 / 100) × √((80 × 10-3) / (2 × 10-6))

Q = (1 / 100) × √(40 × 103)

Q = (1 / 100) × 200

Q = 2

The quality factor of the circuit is 2.

Top Series Resonance MCQ Objective Questions

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

Series Resonance Question 6 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

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

Series Resonance Question 7 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

A RLC series circuit has a resistance of 1 kΩ. Its half power frequencies are 10 kHz and 90 kHz. Quality factor of the circuit is:

  1. 2.333
  2. 0.375
  3. 0.925
  4. 1.625

Answer (Detailed Solution Below)

Option 2 : 0.375

Series Resonance Question 8 Detailed Solution

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Concept:

In a series resonant circuit,

Bandwidth (BW) \(= \frac{{{f_r}}}{Q} = {f_H} - {f_L} = \frac{R}{L}\)

Where fr is the resonant frequency

Q is the quality factor

fH is the higher half-power frequency

fL is the lower half-power frequency

Resonant frequency (fr) is defined as:

\(f_r= \frac{1}{{2\pi \sqrt {LC} }} = \sqrt {{f_H}{f_L}}\)

Calculation:

Given that,

Higher half-power frequency (fH) = 90 kHz

Lower half-power frequency (fL) = 10 kHz

Bandwidth (BW) = fH – fL = 90 – 10 = 80 kHz

Resonant frequency (fr) will be:

\(= \sqrt {{f_H}{f_L}} = \sqrt {90 \times 10} = 30\;kHz\)

Quality Factor Q = fr / BW = 30 / 80 = 0.375

For the series RLC circuit, the Current- Frequency Curve is shown below. which of the following statement is incorrect

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  1. Magnitude of power factor Angle at points A and B is 45°
  2. ωU - ω=  ωr + ωL
  3. Power Factor is unity at point C.
  4. For ω > ωthe circuit behaves as inductive circuit.

Answer (Detailed Solution Below)

Option 2 : ωU - ω=  ωr + ωL

Series Resonance Question 9 Detailed Solution

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Concept: 

1. Consider the information Shown in the figure below
F1 Vinanti Engineering 10.08.23 D1 V2

2. For the series RLC circuit ωr is geometrical mean of ωL and ωU or \({{\rm{ω}}_{{\rm{U}}}} = \frac{{{\rm{ω}}_r^2}}{{{{\rm{ω}}_{\rm{L}}}}}\;\)

3. At upper and lower cutoff frequencies current through the circuit is 70.7% of I0 because at these frequency Z = √ 2 R

4. At resonant frequency, the circuit behaves as purely resistive circuit and hence voltage and current are in same phase. Also power factor become unity.

Analysis:

1. At point A and B  current through the circuit is 70.7% of I0 because at these frequency Z = √ 2 R 

As we know power factor = cos ϕ = \(\dfrac{R}{Z}\) = 0.707

ϕ = 45° hence option 1 is correct.

2.The curve is symmetrical about resonance frequency 

So ωU - ω=  ωr - ωL

hence option 2 is incorrect.

3. At resonant frequency, the circuit behaves as purely resistive circuit and hence voltage and current are in same phase. Also power factor become unity.

4. for  ω > ω  XL > XC so circuit behave as inductive circuit.

 for  ω < ω  XL < XC so circuit behave as Capacitive circuit.

In a RLC circuit Inductance is 20 mH and capacitance is 200 micro Farad. Find the resonance frequency of the circuit.

  1. 1000 rad/sec
  2. 250 rad/sec
  3. 500 rad/sec
  4. 50 rad/sec

Answer (Detailed Solution Below)

Option 3 : 500 rad/sec

Series Resonance Question 10 Detailed Solution

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RLC series circuit:

F1 P.Y 7.5.20 Pallavi D2

 An RLC circuit is an electrical circuit consisting of Inductor (L)Capacitor (C)Resistor (R) it can be connected either parallel or series.

When the LCR circuit is set to resonate (X= XC), the resonant frequency is expressed as 

 \(f = \dfrac{1}{{2π }}\sqrt {\dfrac{1}{{LC}}}\;Hz\)

\(\omega = \sqrt {\dfrac{1}{{LC}}}\;rad/sec\)

Quality factor:

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

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

Mathematically, for a coil, the quality factor is given by:

 \(Q=\frac{{{\omega }_{0}}L}{R}=\frac{1}{R}\sqrt{\frac{L}{C}}\)

Where,

XL & XC = Impedance of inductor and capacitor respectively

L, R & C = Inductance, resistance, and capacitance respectively

fr = frequency

ω= angular resonance frequency

Calculation:

Given, L = 20 mH = 20 × 10-3 H

C = 200 μF = 200 × 10-6 F

\(\omega = \sqrt {\dfrac{1}{{LC}}} =\sqrt {\dfrac{1}{{20\times 10^{-3}\times 200\times10^{-6}}}}=\;500\;rad/sec\)

How is the transient current in a loss-free R-L-C circuit?

  1. Oscillating
  2. None-oscillating
  3. Square wave
  4. Sinusoidal

Answer (Detailed Solution Below)

Option 4 : Sinusoidal

Series Resonance Question 11 Detailed Solution

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The correct answer is option 4.

Concept: 

The RLC circuit means the series or parallel connection of the resistance(R), inductor(L), and capacitor(C).

A series circuit is given below.

F1 P.Y 7.5.20 Pallavi D2

Since it is mentioned that the circuit is a loss-free circuit the current will behave similarly to that of an LC series circuit i.e. an undamped sinusoidal waveform will be observed.

In a loss-free circuit, the energy keeps shifting between L and C.

 Important PointsIf the case would have been of loss-making RLC series circuit then the current would have a damped sinusoidal equation.

In a series AC circuit, XL = 2350 ohms, C = 0.005 μF, and R = 500Ω. What is the impedance at resonance?

  1. The frequency must be known
  2. 2.1 kΩ
  3. 4200 Ω
  4. 0.5 kΩ

Answer (Detailed Solution Below)

Option 4 : 0.5 kΩ

Series Resonance Question 12 Detailed Solution

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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

The magnitude of the impedance is given by:

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

At resonance, XL = XC,

|Z| = R

Hence, the impedance at resonance is |Z| = R = 0.5 kΩ

26 June 1

Note:

The current flowing across the series RLC circuit will be:

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

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

Hence impedance is minimum and current will be maximum at resonance.

In the series RLC circuit, the current (I) Vs frequency (f) graph for the series resonance circuit is shown below. :-

F1 Jai Prakash Anil 08.12.20 D1

From above we can conclude:-

  • For resonance to occur in any circuit it must have at least one inductor and one capacitor.
  • Resonance is the result of oscillations in a circuit as stored energy is passed from the inductor to the capacitor.
  • Resonance occurs when XL = XC and the imaginary part of the transfer function is zero.
  • At resonance, the impedance of the circuit is equal to the resistance value as Z = R.
  • At low frequencies the series circuit is capacitive as XC > XL, this gives the circuit a leading power factor.
  • At high frequencies the series circuit is inductive as XL > XC, this gives the circuit a lagging power factor.
  • The high value of current at resonance produces very high values of voltage across the inductor and capacitor.
  • Because impedance is minimum and current is maximum, series resonance circuits are also called Acceptor Circuits.

To double the resonant frequency of an LC circuit with a fixed value of L, the capacitance must be

  1. reduced to one quarter
  2. reduced by one-half
  3. doubled
  4. quadrupled

Answer (Detailed Solution Below)

Option 1 : reduced to one quarter

Series Resonance Question 13 Detailed Solution

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Concept:

An LC circuit is represented as:

F3 S.B Madhu 13.03.20 D5

There is a characteristic frequency at which the circuit will oscillate, called the resonant frequency given by:

\(f_{0}=\frac{1}{2\pi\sqrt{LC}}\)

Application:

For a fixed value of L, the resonant frequency is proportional to:

\(f_{0}\propto\frac{1}{\sqrt{C}}\)

To double the resonant frequency, C must be reduced to 1/4 (quarter), i.e. for C' = C/4, f0 becomes:

f'0 = 2 f0

A series RLC circuit resonates at 1.5 kHz and consumes 50 W from a 50 V AC source operating at the resonant frequency. If the bandwidth is 0.75 kHz, then what are the values of the circuit elements R and L?

  1. 25 Ω and 5.31 mH
  2. 50 Ω and 10.6 mH
  3. 50 Ω and 66.6 mH
  4. 2.5 Ω and 1.06 mH

Answer (Detailed Solution Below)

Option 2 : 50 Ω and 10.6 mH

Series Resonance Question 14 Detailed Solution

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Concept:

At resonance, only the resistance will remain (as both the inductive and capacitive reactance will be equal), because of which all the power will be delivered to the resistance only.

So, power will be:

\(P = \frac{{{V^2}}}{R}\;W\)

The bandwidth of a resonant circuit is the difference between the upper cut-off frequency and lower cut-off frequency (with the cut-off frequency being the frequency at which the power is half of the maximum power).

For a series RLC circuit, the bandwidth is calculated as:

\(B = \frac{{{\omega _o}}}{Q}\)

Where Q = quality factor and is defined as:

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

From both these equations, the Bandwidth can be calculated as:

\(B= \frac{R}{L}\).

Calculation:

Given resonant frequency (f0) = 1.5 kHz

Bandwidth (BW) = 0.75 kHz

Voltage (V) = 50 V

Power (P) = 50 W

\(R = \frac{{{V^2}}}{P} = 50\;{\Omega}\)

B.W. in radians = R/L

B.W. in Hertz = \(\frac{R}{2\pi L}\)

\( L = \frac{{50}}{{2\pi \times 0.75 \times {{10}^3}}} = 10.6\;mH\)

At half power points of a resonance curve, the current is

  1. 2 times the maximum current.
  2. \(\frac{1}{2}\) times the maximum current.
  3. 0.707 times the maximum current.
  4. 1.414 times the maximum current.

Answer (Detailed Solution Below)

Option 3 : 0.707 times the maximum current.

Series Resonance Question 15 Detailed Solution

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Key Points

 At half-power points of a resonance curve, the current is 0.707 times the maximum current.

F4 Madhuri Engineering 18.01.2023 D2

Important points regarding half-power points are:

  • The current is Imax/√2.
  • Impedance is √2 R or √2 Zmin.
  • The circuit phase angle → ±45° or π/4 rad

 

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