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
Concept. Current amplitude $I_0 = \dfrac{V_0}{Z}$ with $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
Step 1. Reactances. $\omega = 2000$ rad/s:
$$X_L = 2000 \times 10^{-2} = 20\ \Omega,\qquad X_C = \frac{1}{2000 \times 25 \times 10^{-6}} = 20\ \Omega$$
Step 2. Impedance. $X_L = X_C$ (resonance), so $Z = R = 100\ \Omega$.
Step 3. Current amplitude. $I_0 = \dfrac{220}{100} = 2.2$ A.
Answer. $2.2$ A (Option A).
Solution by Sreeraj P, M.Sc Physics