Electrical Resonance MCQ Quiz in मल्याळम - Objective Question with Answer for Electrical Resonance - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Mar 12, 2025

നേടുക Electrical Resonance ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Electrical Resonance MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

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?

  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

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

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.

Electrical Resonance Question 2:

Which of the following statements is incorrect?

  1.  During parallel resonance impedance at resonance is minimum
  2.  During parallel resonance current is magnified
  3. During series resonance current at resonance is maximum
  4. During series as well as parallel resonance, the power factor is unity

Answer (Detailed Solution Below)

Option 1 :  During parallel resonance impedance at resonance is minimum

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:

  1. current is maximum
  2. circuit is purely inductive
  3. current is minimum
  4. impedance is maximum

Answer (Detailed Solution Below)

Option 1 : current is maximum

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:

F1 U.B. Nita 29.10.2019 D 1

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.

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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.

  1. half resonant
  2. half power
  3. resonant
  4. side band

Answer (Detailed Solution Below)

Option 2 : half power

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, \({f_0} = \sqrt {{f_1}{f_2}}\)

F2 U.B 23.3.20 Pallavi D4

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

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

Answer (Detailed Solution Below)

Option 1 : reduced to one quarter

Electrical Resonance Question 5 Detailed Solution

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

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. 

  1. 0.42 H
  2. 0.28 H
  3. 0.56 H
  4. 0.14 H

Answer (Detailed Solution Below)

Option 2 : 0.28 H

Electrical Resonance Question 6 Detailed Solution

Concept:
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 

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

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

ω= angular resonance frequency

Calculation:

Given that 

fr = 5000/2π hz

Impedance at resonance (Z) = resistance (R)= 56 Ω

ω0 = 2π f= 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

quesOptionImage4401

  1. 100 Hz
  2. 25 Hz
  3. 50 Hz
  4. 16 Hz

Answer (Detailed Solution Below)

Option 4 : 16 Hz

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

  1. 0.707 lagging
  2. 0.5 leading
  3. unity
  4. 0.707 leading

Answer (Detailed Solution Below)

Option 4 : 0.707 leading

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

L 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, \({f_0} = \sqrt {{f_1}{f_2}}\)

F2 U.B 23.3.20 Pallavi D4

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 ______

  1. Figure of efficiency
  2. Figure of merit
  3. Figure of excellence
  4. None of these

Answer (Detailed Solution Below)

Option 2 : Figure of merit

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.

  1. 1/RLC
  2. 1/LC
  3. \(\frac{1}{{2\pi \sqrt {LC} }}\)
  4. 1/RC

Answer (Detailed Solution Below)

Option 3 : \(\frac{1}{{2\pi \sqrt {LC} }}\)

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

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

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