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 )?
Answer (Detailed Solution Below)
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 _________.
Answer (Detailed Solution Below)
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 _______.
Answer (Detailed Solution Below)
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 ________.
Answer (Detailed Solution Below)
Electrical Resonance Question 4 Detailed Solution
Parallel RLC circuit
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?
Answer (Detailed Solution Below)
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
ωo = 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?
Answer (Detailed Solution Below)
Electrical Resonance Question 6 Detailed Solution
Download Solution PDFWe 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
Answer (Detailed Solution Below)
Electrical Resonance Question 7 Detailed Solution
Download Solution PDFThe 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.
Answer (Detailed Solution Below)
Electrical Resonance Question 8 Detailed Solution
Download Solution PDFConcept:
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:
Answer (Detailed Solution Below)
Electrical Resonance Question 9 Detailed Solution
Download Solution PDFParallel Resonance:
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?
Answer (Detailed Solution Below)
Electrical Resonance Question 10 Detailed Solution
Download Solution PDFConcept:
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:
The phasor diagram is as shown below:
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:
The impedance - frequency curve is shown below:
The impedance at resonance offered by a parallel resonant circuit is _____.
Answer (Detailed Solution Below)
Electrical Resonance Question 11 Detailed Solution
Download Solution PDFIn 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.
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?
Answer (Detailed Solution Below)
Electrical Resonance Question 12 Detailed Solution
Download Solution PDFParallel RLC circuit:
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.
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?
Answer (Detailed Solution Below)
Electrical Resonance Question 13 Detailed Solution
Download Solution PDFConcept:
RLC series circuit:
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:
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 ______
Answer (Detailed Solution Below)
Electrical Resonance Question 14 Detailed Solution
Download Solution PDFConcept:
The graph between impedance Z and the frequency of the parallel RLC circuit:
Here,
f1 is the lower cutoff frequency
f2 is 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:
Answer (Detailed Solution Below)
Electrical Resonance Question 15 Detailed Solution
Download Solution PDFThe selectivity/sharpness of the resonance amplifier is measured by the quality factor and is explained in the figure shown below:
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