An inductor of inductance 2 mu H is connected | vedclass

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(C) For maximum current to be drawn,the circuit must be in resonance condition.At resonance,the inductive reactance equals the capacitive reactance: $

Vedclass · Accepted Answer

$2000$

Vedclass · Answer

(C) For maximum current to be drawn,the circuit must be in resonance condition.At resonance,the inductive reactance equals the capacitive reactance: $X_L = X_C$.This implies $2 \pi f L = \frac{1}{2 \pi f C}$.Rearranging for capacitance $C$,we get $C = \frac{1}{4 \pi^2 f^2 L}$.Given $L = 2 \mu H = 2 	imes 10^{-6} 	ext{ H}$,$f = 5 	ext{ kHz} = 5 	imes 10^3 	ext{ Hz}$,and $\pi^2 = 10$.Substituting the values: $C = \frac{1}{4 	imes 10 	imes (5 	imes 10^3)^2 	imes 2 	imes 10^{-6}}$.$C = \frac{1}{40 	imes 25 	imes 10^6 	imes 2 	imes 10^{-6}} = \frac{1}{40 	imes 25 	imes 2} = \frac{1}{2000} 	ext{ F}$.Comparing this with $\frac{1}{x} 	ext{ F}$,we find $x = 2000$.

An inductor of inductance $2 \mu H$ is connected in series with a resistance,a variable capacitor and an a.c. source of frequency $5 \text{ kHz}$. The value of capacitance for which maximum current is drawn into the circuit is $\frac{1}{x} \text{ F}$,where the value of '$x$' is (Take $\pi^2 = 10$).

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