The quality factor Q of a series LCR circuit is given by

The correct answer is Option (3) → \( \displaystyle \frac{1}{R}\sqrt{\frac{L}{C}} \)

For a series LCR circuit, the quality factor is \( Q=\omega_0L/R \). Here \( \omega_0 \) is the resonant angular frequency, \( \omega_0=1/\sqrt{LC} \). Putting this value of \( \omega_0 \) in the formula gives \( Q \) in terms of \( L \), \( C \) and \( R \).

    \( \begin{aligned} Q &= \frac{\omega_0L}{R} \\ &= \frac{1}{R}\times\frac{L}{\sqrt{LC}} \\ &= \frac{1}{R}\sqrt{\frac{L}{C}} \end{aligned} \)

From the above, \( Q=\frac{1}{R}\sqrt{\frac{L}{C}} \). Hence, a larger inductance or a smaller resistance gives a sharper resonance.

NCERT (2026-27):
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