Q 12-07-134JEE MainJEE Main 2020 (9 Jan, Shift 2)Medium
In an $LC$ circuit the inductance $L = 40$ mH and capacitance $C = 100\ \mu$F. If a voltage $V(t) = 10\sin(314t)$ is applied to the circuit, the current in the circuit is given as:
Answer: (A) $0.52\cos(314t)$
$X_L = \omega L = 314\times0.04 = 12.56\ \Omega$ and $X_C = \dfrac{1}{\omega C} = \dfrac{1}{314\times10^{-4}} = 31.85\ \Omega$.
Net reactance $X_C - X_L = 19.3\ \Omega$ (capacitive), so the peak current is $\dfrac{10}{19.3} \approx 0.52$ A and the current leads the voltage by $90^\circ$:
$$i = 0.52\sin(314t + \pi/2) = 0.52\cos(314t)$$
Solution by Sreeraj P, M.Sc Physics