Quality Factor In An AC Circuit MCQ - Practice Questions & Answers

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Match List I with List II

List I	List 2
A. AC generator	I. Presence of both Land C
B. Transformer	II. Electromagnetic Induction
C. Resonance phenomenon to occur	III. Quality factor
D. Sharpness of resonance	IV. Mutual Induction

Choose the correct answer from the options given below:

A-IV, B-III, C-I, D-II

A-IV, B-II, C-I, D-III

A-II, B-IV, C-I, D-III

A-II, B-I, C-III, D-IV

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A series combination of resistors of resistance $100\, \Omega$ , inductors of inductance, $1 \mathrm{H}$ and capacitors of capacitance $6.25\, \mu \mathrm{F}$ is connected to an AC source. The quality factor of the circuit will be________

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Concepts Covered - 0

Quality factor

Quality factor-

The quality factor Q is a parameter which is used to describe the sharpness of resonance curve. So it is defined as the ratio of voltage drop across inductor or capacitor at resonance to the applied voltage. So,

$Q = \frac{Voltage \ across \ L \ or \ C \ at \ resonance}{Applied \ voltage}$

$Q = \frac{I_v \omega_o L}{I_v R} = \omega_o \frac{L}{R}$

As we know that, at the resonance -

$\omega_o = \frac{1}{\sqrt{LC}}$

So,

$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$

We can also say that the characteristic of a series resonant circuit is determined by the quality factor (Q - factor) of the circuit. So, if the value of Q-factor is high then the sharpness of the resonant curve is more and vice-versa.

We can also define the Q -factor that is is defined as $2 \pi$ times the ratio of the energy stored in L or C to the average energy loss per period. So,

$Q = 2 \pi[\frac{Maximum\ energy \ stored \ in \ the \ capacitor}{Energy \ loss \ per \ period}] \ \ \ \ \ \ \ \ \ \ \ . . . . . . . (1)$

Now, the maximum energy stored in the inductor =

$U = \frac{1}{2}L(I_o)^2$

Also the energy dissipated per second =

$P_R = I^2_{rms} R = \frac{I_o^2 R}{2}$

Energy dissipated per time period =

$U_R = \frac{I_o^2 R}{2} \times T$

Putting all these in the (1)

$Q = \frac{1}{R} \sqrt{\frac{L}{C}}$

The Q-factor of the circuit varies inversely as R. Thus, at resonance, the voltage drop across inductance or capacitance is Q-times the applied voltage.

From the graph we can see that when the Q-factor tends to infinity, then the current become infinite. And as the Q-factor become very low then the amplitude of the current will become very low.

In an ac circuit, If,

$R=0\ or \;\cos \phi = 0$