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JEE Main 2020 (9 Jan, Shift 2) · Series LCR Circuit

Series LCR Circuit question from JEE Main 2020 (9 Jan, Shift 2)

Q54JEE Main 2020 · 9 JAN · SHIFT 2Medium

Series LCR Circuit

In LC circuit the inductance L = 40 mH and

capacitance C = 100 $\mu$ F. If a voltage

V(t) = 10sin(314t) is applied to the circuit, the

current in the circuit is given as:

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Answer: (a)

$X_L = \omega L = 314 \times 0.04 \approx 12.6\ \Omega$
$X_C = \dfrac{1}{\omega C} = \dfrac{1}{314 \times 10^{-4}} \approx 31.8\ \Omega$
Net reactance $X_C - X_L \approx 19.2\ \Omega$ (capacitive), so current leads voltage by $90^\circ$ .
$I_0 = \dfrac{10}{19.2} \approx 0.52\text{ A}$ , so $I = 0.52\sin(314t + \pi/2) = 0.52\cos 314t$ .
So the answer is (a) $0.52\cos 314t$ .

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