Q 12-07-001NEETNEET 2026Top questionEasy
An ac voltage $V = 220\sin(2 \times 10^3 t)$ Volt is applied to a series LCR circuit. Then the current amplitude in this circuit is :
(Given : $L = 10$ mH, $C = 25\ \mu$F, $R = 100\ \Omega$)
Answer: (A) $2.2$ A
$\omega = 2 \times 10^3\ \text{rad s}^{-1}$.
$$X_L = \omega L = 2000 \times 10 \times 10^{-3} = 20\ \Omega$$
$$X_C = \frac{1}{\omega C} = \frac{1}{2000 \times 25 \times 10^{-6}} = 20\ \Omega$$
$X_L = X_C$, so the circuit is at resonance and $Z = R = 100\ \Omega$.
$$I_0 = \frac{V_0}{Z} = \frac{220}{100} = 2.2\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics