Electrical Resonance MCQ Quiz in मल्याळम - Objective Question with Answer for Electrical Resonance - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 12, 2025
Latest Electrical Resonance MCQ Objective Questions
Top Electrical Resonance MCQ Objective Questions
Electrical Resonance Question 1:
In a series AC circuit, XL = 2350 ohms, C = 0.005 μF, and R = 500Ω. What is the impedance at resonance?
Answer (Detailed Solution Below)
Electrical Resonance Question 1 Detailed Solution
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
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. :-
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.
Electrical Resonance Question 2:
Which of the following statements is incorrect?
Answer (Detailed Solution Below)
Electrical Resonance Question 2 Detailed Solution
Concept:
- In a series RLC circuit, at resonance condition impedance is purely resistive and it is equal to R. It is the minimum impedance condition. Hence the current at this condition is maximum.
- In parallel RLC circuit, at resonance condition impedance is purely resistive and it is equal to R. It is the maximum impedance condition. Hence the current at this condition is minimum.
Additional Information
- In parallel RLC circuit current through the inductor and current through the capacitor is greater than the source current, this phenomenon is called current magnification.
- The magnitude of current flowing through the inductor and capacitor is equal to Q times the input sinusoidal current.
Important Points
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 |
Electrical Resonance Question 3:
In series RLC circuit, at resonance:
Answer (Detailed Solution Below)
Electrical Resonance Question 3 Detailed Solution
Series Resonance:
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: XC = 1/ωC
The magnitude of the impedance is given by:
\(|Z|=\sqrt{R^2+(X_L-X_C)^2}\)
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 maximum current.
The plot of the frequency response of the series circuit is as shown in the figure:
We observe that at resonant frequency ω0, the current is maximum.
Parallel Resonance:
For a parallel RLC circuit, the net admittance is calculated as:
\(Y=G+j(X_C-\frac{1}{X_L})\)
At resonance, the imaginary part will be zero resulting in a minimum value of admittance, i.e.
Y = G
Since \(Z = \frac{1}{Y}\), the impedance will be maximum at resonance for a parallel RLC circuit.
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 |
Electrical Resonance Question 4:
The lower and upper cut-off frequencies are also called as _______ frequencies.
Answer (Detailed Solution Below)
Electrical Resonance Question 4 Detailed Solution
Explanation:
Half power frequencies:
Half power frequencies are the frequencies at which current is \(\frac{1}{{\sqrt 2 }}\) times of maximum current.
The lower and upper cut-off frequencies are also called as half power frequencies.
The half-power frequencies are shown in the figure below.
f0 is the resonant frequency
f1 is the lower half-power frequency
f2 is the higher half-power frequency
The relation between them is, \({f_0} = \sqrt {{f_1}{f_2}}\)
As the lower half power frequency is less than the resonant frequency, the nature of power factor is leading.
For half power frequency, \(P = \frac{{{P_{max}}}}{2}\)
\(I = \frac{{{I_{max}}}}{{\sqrt 2 }}\)
\(Z = \sqrt 2 {Z_{min}} = \sqrt 2 R\)
At half power frequencies, the impedance is √2 times of the resistance.
Electrical Resonance Question 5:
To double the resonant frequency of an LC circuit with a fixed value of L, the capacitance must be
Answer (Detailed Solution Below)
Electrical Resonance Question 5 Detailed Solution
Concept:
An LC circuit is represented as:
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
Electrical Resonance Question 6:
An RLC series resonance circuit has a resonance frequency of (5000/2π) Hz and impedance at resonance is 56 Ω. Find the value of the inductance if the Q-factor is 25.
Answer (Detailed Solution Below)
Electrical Resonance Question 6 Detailed Solution
Concept:
RLC series circuit:
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 (XL = XC), the resonant frequency is expressed as
\(f = \frac{1}{{2π }}\sqrt {\frac{1}{{LC}}}\)
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
ω0 = angular resonance frequency
Calculation:
Given that
fr = 5000/2π hz
Impedance at resonance (Z) = resistance (R)= 56 Ω
ω0 = 2π fr = 5000 rad/sec
∴ \(Q=\frac{{{\omega }_{0}}L}{R}\)
\(L=\frac{{}25\times 56}{5000}\)
L = 0.28 H
Electrical Resonance Question 7:
In the circuit shown, the frequency of the sinusoidal source is about
Answer (Detailed Solution Below)
Electrical Resonance Question 7 Detailed Solution
Concept:
Series resonance:
- The state of an electrical network at which reactance of the circuit will be zero means at fixed frequency L and C elements will exchange energy freely as a function of time, which results in sinusoidal oscillations across either inductor or capacitor.
- At resonance inductive reactance is equal to capacitive reactance, so the circuit is purely resistive in nature under series resonance.
- As the circuit is purely resistive at resonance condition. The power factor of the series RLC circuit is unity, hence the current flowing through the circuit is in phase with supply voltage.
- At resonance series, the RLC circuit offers minimum impedance which is equal to resistance only. It allows maximum current at resonance condition so it is called an acceptor circuit.
Calculation:
Given that
Resistance = R ohm
Inductance L = 0.1 H
Inductive reactance XL = ωL
Where ω is the angular frequency in rad/sec
Capacitive reactance Xc = 1 / ωC = 10 ohm
Supply voltage V = 100∠300
Current through the circuit is I = 10∠300
From the values of voltage and current we can observe that supply voltage V and current I are in phase. That means the circuit is under resonance.
At resonance XL = XC
⇒ ω L = Xc
⇒ ω × 0.1 = 10
ω = 100 rad/sec
2π f = 100
f = 100/2π = 15.92 Hz
Electrical Resonance Question 8:
For a series RLC circuit, the power factor at the lower half power frequency is
Answer (Detailed Solution Below)
Electrical Resonance Question 8 Detailed Solution
In a series RLC circuit, the impedance is given by
\(Z = \sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}}\)
R is resistance
XL is inductive reactance = ωL
XC is capacitive reactance = -1/ωC
(XL – XC) is net reactance
At resonant frequency, inductive reactance is equal to capacitive reactance i.e. XL = XC
So, at this condition the impedance is minimum, and it is equivalent to R.
Nature of power factor:
- At resonant frequency, the net reactance is zero and the circuit acts as pure resistive circuit. So, the power factor is unity.
- At a frequency less than resonant frequency, the net reactance is negative, and it is capacitive in nature. So, the power factor is leading.
- At a frequency more than resonant frequency, the net reactance is positive, and it is inductive in nature. So, the power factor is lagging.
Half power frequencies:
Half power frequencies are the frequencies at which current is \(\frac{1}{{\sqrt 2 }}\) times of maximum current.
The half power frequencies are shown in the figure below.
f0 is the resonant frequency
f1 is the lower half power frequency
f2 is the higher half power frequency
The relation between them is, \({f_0} = \sqrt {{f_1}{f_2}}\)
As the lower half power frequency is less than the resonant frequency, the nature of power factor is leading.
For half power frequency, \(P = \frac{{{P_{max}}}}{2}\)
\(I = \frac{{{I_{max}}}}{{\sqrt 2 }}\)
\(Z = \sqrt 2 {Z_{min}} = \sqrt 2 R\)
At half power frequencies, the impedance is √2 times of the resistance.
Now, power factor \(\cos \phi = \frac{R}{Z} = \frac{R}{{\sqrt 2 R}} = 0.707\)
Therefore, the power factor at lower half power frequency is 0.707 leading.
Electrical Resonance Question 9:
The quality factor of an RLC circuit is also known as ______
Answer (Detailed Solution Below)
Electrical Resonance Question 9 Detailed Solution
The quality factor of an RLC circuit is also known as figure of merit it represents the efficiency of the resonant circuit to store energy
The Q factor is defined as the ratio of energy stored per cycle to the energy lost per cycle
Q = 2π × Maximum energy stored per cycle/Energy dissipated per cycle
Additional Information
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.
Electrical Resonance Question 10:
At f = _________; the RLC series circuit operates at unity power factor.
Answer (Detailed Solution Below)
Electrical Resonance Question 10 Detailed Solution
- At resonance in the series RLC circuit, the voltage across inductor and capacitor is equal in magnitude and opposite in direction and thereby they cancel each other.
- So, in a series, resonant circuit voltage across the resistor is equal to supply voltage at resonance condition.
- At resonance, both inductive and capacitive reactance cancel each other.
- The total impedance of the circuit is resistive only.
- Therefore, the circuit behaves like a pure resistive circuit and we know that in pure resistive circuit, voltage and the current are in the same phase.
- Therefore, the phase angle between voltage and current is zero and the power factor is unity.
Then the expression of resonant frequency fr for series RLC circuit is
\({f_r} = \frac{1}{{2\pi \sqrt {LC} }}\)