An inductor of inductance 100mH is connected in series with a resistance, a variable capacitance and an AC source of frequency 2.0 kHz. the value of the capacitance so that maximum current may be drawn into the circuit.

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50 nF

60 nF

63 nF

79 nF

answer is C.

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Detailed Solution

Given $L = 100 m H, f = 2 k H z, C = ?$

$f = \frac{1}{2 π \sqrt{L C}} \Rightarrow c = \frac{1}{4 π^{2} f^{2} L}$

$= \frac{1}{4 π^{2} \times 4 \times 10^{6} \times 100 \times 10^{- 3}}$

$= \frac{1}{16 \times 10^{6}}$

$= 0.0625 \times 10^{- 6}$

$= 62.5 n F \approx 63 n F$

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