In an LC circuit,the inductance L = 40 ; mH a | vedclass

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(A) Given: $L = 40 	imes 10^{-3} \; H$,$C = 100 	imes 10^{-6} \; F$,$V(t) = 10 \sin (314 t)$.Here,$\omega = 314 \; rad/s$.Inductive reactance $X_{L}

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$0.52 \cos 314 t$

Vedclass · Answer

(A) Given: $L = 40 	imes 10^{-3} \; H$,$C = 100 	imes 10^{-6} \; F$,$V(t) = 10 \sin (314 t)$.Here,$\omega = 314 \; rad/s$.Inductive reactance $X_{L} = \omega L = 314 	imes 40 	imes 10^{-3} = 12.56 \; \Omega$.Capacitive reactance $X_{C} = \frac{1}{\omega C} = \frac{1}{314 	imes 100 	imes 10^{-6}} = \frac{10^{4}}{314} \approx 31.85 \; \Omega$.Since $X_{C} > X_{L}$,the circuit is capacitive.The net reactance $X = X_{C} - X_{L} = 31.85 - 12.56 = 19.29 \; \Omega$.The peak current $I_{m} = \frac{V_{m}}{X} = \frac{10}{19.29} \approx 0.52 \; A$.In a capacitive circuit,the current leads the voltage by $\frac{\pi}{2}$.Thus,$I(t) = I_{m} \sin (\omega t + \frac{\pi}{2}) = 0.52 \sin (314 t + \frac{\pi}{2}) = 0.52 \cos (314 t)$.

In an $LC$ circuit,the inductance $L = 40 \; mH$ and capacitance $C = 100 \; \mu F$. If a voltage $V(t) = 10 \sin (314 t)$ is applied to the circuit,the current in the circuit is given by:

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