An inductor of inductance \( 100 \) mH is connected in series with a resistance \( R \), a variable capacitance and an AC source of frequency \( 4.0 \) kHz. To draw maximum current into the circuit, the value of the capacitance should be
(Take \( \pi^{2} = 10 \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 15.625 \) nF
The current in a series LCR circuit is maximum at resonance, when \( X_{L} = X_{C} \), that is \( \omega L = 1/\omega C \).
\( L = 100\,\mathrm{mH} = 0.1\,\mathrm{H} \)
\( f = 4.0\,\mathrm{kHz} = 4\times10^{3}\,\mathrm{Hz} \)
\( \displaystyle C = \frac{1}{\omega^{2}L} = \frac{1}{4\pi^{2}f^{2}L} \)
\( \displaystyle C = \frac{1}{4\times10\times(4\times10^{3})^{2}\times0.1} \)
\( \displaystyle C = \frac{1}{4\times10\times16\times10^{6}\times0.1} = \frac{1}{6.4\times10^{7}}\,\mathrm{F} \)
\( C = 1.5625\times10^{-8}\,\mathrm{F} = 15.625\times10^{-9}\,\mathrm{F} = 15.625\,\mathrm{nF} \)