Q 12-07-004NEETNEET 2024Top questionMedium
A $10\ \mu$F capacitor is connected to a $210$ V, $50$ Hz source as shown in figure. The peak current in the circuit is nearly ($\pi = 3.14$):
Answer: (D) $0.93$ A
Concept. $I_{\text{rms}} = \dfrac{V_{\text{rms}}}{X_C}$ and $I_0 = \sqrt2I_{\text{rms}}$.
Step 1. Reactance.
$$X_C = \frac{1}{2 \times 3.14 \times 50 \times 10^{-5}} \approx 318.5\ \Omega$$
Step 2. rms current. $\dfrac{210}{318.5} \approx 0.66$ A.
Step 3. Peak current. $1.414 \times 0.66 \approx 0.93$ A.
Answer. $0.93$ A (Option D).
Solution by Sreeraj P, M.Sc Physics