Alternating Current: NEET and JEE questions with solutions

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

Answer: (B) $7.8$ A and $45°$

$$Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{20^2 + (45 - 25)^2} = 20\sqrt{2} \approx 28.3\ \Omega$$

$$I = \frac{220}{28.3} \approx 7.8\ \text{A}$$

$$\tan\phi = \frac{X_L - X_C}{R} = \frac{20}{20} = 1 \;\Rightarrow\; \phi = 45°$$

Solution by Sreeraj P, M.Sc Physics