An a.c. e.m.f. of peak value V_0 = 230 V and | vedclass

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(B) For the current in an $LCR$ series circuit to be maximum,the circuit must be in resonance. At resonance,the inductive reactance equals the capacit

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$4 \ \mu F, 20 \ A$

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(B) For the current in an $LCR$ series circuit to be maximum,the circuit must be in resonance. At resonance,the inductive reactance equals the capacitive reactance: $X_L = X_C$,which implies $\omega L = \frac{1}{\omega C}$.Given $f = 50 \ Hz$,the angular frequency is $\omega = 2 \pi f = 2 \pi (50) = 100 \pi \ rad/s$.The resonance condition is $C = \frac{1}{\omega^2 L} = \frac{1}{(100 \pi)^2 	imes 2.5} = \frac{1}{10000 	imes 10 	imes 2.5} = \frac{1}{250000} \ F$.$C = 4 	imes 10^{-6} \ F = 4 \ \mu F$.At resonance,the impedance $Z$ is equal to the resistance $R$,so $Z = R = 11.5 \ \Omega$.The maximum current is $I_{max} = \frac{V_0}{Z} = \frac{230}{11.5} = 20 \ A$.Thus,the values are $4 \ \mu F$ and $20 \ A$.

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