An inductor of inductance 1000 mH is connected in series with a resistance, a variable capacitance and an AC source of frequency 2.0 kHz. What value of the capacitance result in drawing the maximum current from the circuit?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 6.3\,\mathrm{nF} \)
The current in a series LCR circuit is maximum at resonance, when \( X_L = X_C \) and the impedance reduces to \( R \).
\( \displaystyle \omega L = \frac{1}{\omega C} \;\Rightarrow\; C = \frac{1}{\omega^{2} L} = \frac{1}{4\pi^{2} f^{2} L} \)
\( L = 1000\,\mathrm{mH} = 1\,\mathrm{H} \)
\( f = 2.0\,\mathrm{kHz} = 2\times10^{3}\,\mathrm{Hz} \)
\( \displaystyle C = \frac{1}{4 \times 9.87 \times (2\times10^{3})^{2} \times 1} = \frac{1}{1.58\times10^{8}}\,\mathrm{F} \)
\( \displaystyle C = 6.3\times10^{-9}\,\mathrm{F} = 6.3\,\mathrm{nF} \)