In a series LCR circuit \( C = 5\,\mu\mathrm{F} \) and \( \omega = 1000\,\mathrm{rad\,s^{-1}} \). The value of inductance L for which the current is maximum in this circuit will be
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 200 mH
In a series LCR circuit the current is maximum at resonance, when \( X_L = X_C \) and the impedance is smallest (equal to \( R \)).
\( \displaystyle \omega L = \frac{1}{\omega C} \;\Rightarrow\; L = \frac{1}{\omega^{2} C} \)
\( \displaystyle L = \frac{1}{(1000)^{2} \times 5\times10^{-6}} = \frac{1}{5}\,\mathrm{H} = 0.2\,\mathrm{H} = 200\,\mathrm{mH} \)