Calculate the quality factor Q of a series LCR circuit with \( L=2.0\,\mathrm{H} \), \( C=50\,\mu\mathrm{F} \) and \( R=10\,\Omega \).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 20
The quality factor of a series LCR circuit is \( Q=\omega_0L/R \), where \( \omega_0=1/\sqrt{LC} \) is the resonant angular frequency.
The resonant angular frequency is:
\( \begin{aligned} \omega_0 &= \frac{1}{\sqrt{LC}} \\ &= \frac{1}{\sqrt{2.0\,\mathrm{H}\times50\times10^{-6}\,\mathrm{F}}} \\ &= \frac{1}{\sqrt{10^{-4}\,\mathrm{s^{2}}}} \\ &= 100\,\mathrm{rad\,s^{-1}} \end{aligned} \)
The quality factor is then:
\( \begin{aligned} Q &= \frac{\omega_0L}{R} \\ &= \frac{100\,\mathrm{rad\,s^{-1}}\times2.0\,\mathrm{H}}{10\,\Omega} \\ &= 20 \end{aligned} \)
NCERT (2026-27):
Summary
