What is the quality factor for the given series resonance circuit? 

F1 Vinanti Engineering 08.12.23 D2

This question was previously asked in
SSC JE Electrical 4 Dec 2023 Official Paper II
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  1. 160
  2. 80
  3. 320
  4. 40

Answer (Detailed Solution Below)

Option 2 : 80
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Detailed Solution

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The correct option is 2

Concept:

The quality factor is defined as the ratio of the maximum energy stored to the maximum energy dissipated in a cycle

\(Q = 2\pi \frac{{Maximum\;energy\;stored}}{{Total\;energy\;lost\;per\;period}}\)

In a series RLC, the 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 elements.

Calculation:

Since it is given that it is a series RLC Circuit, hence the quality factor will be:

\(Q = \frac{{\omega L}}{R} = \frac{1}{{\omega RC}} = \frac{{{X_L}}}{R} = \frac{{{X_C}}}{R} =\frac{480}{6} = 80\)

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