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

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Question

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

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$4 \ \mu F, \quad 20 \ A$

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(A) 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: $X_L = X_C$.Given $L = 2.5 \ H$ and $f = 50 \ Hz$,the inductive reactance is $X_L = 2 \pi f L = 2 \pi (50) (2.5) = 250 \pi \ \Omega$.Using $\pi^2 = 10$,we have $\pi = \sqrt{10} \approx 3.162$,so $X_L = 250 	imes 3.162 = 790.5 \ \Omega$.However,the resonance condition is $X_C = X_L$,so $\frac{1}{2 \pi f C} = 2 \pi f L$.$C = \frac{1}{4 \pi^2 f^2 L} = \frac{1}{4 	imes 10 	imes (50)^2 	imes 2.5} = \frac{1}{40 	imes 2500 	imes 2.5} = \frac{1}{250000} = 4 	imes 10^{-6} \ F = 4 \ \mu F$.At resonance,the impedance $Z = R = 11.5 \ \Omega$.The maximum current $I_{max} = \frac{V_{peak}}{Z} = \frac{230}{11.5} = 20 \ A$.Thus,the values are $4 \ \mu F$ and $20 \ A$.

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