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
$$X_C = \frac{1}{2\pi fC} = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}} \approx 318.5\ \Omega$$
The $210$ V is the rms value:
$$I_{rms} = \frac{210}{318.5} \approx 0.66\ \text{A}$$
$$I_0 = \sqrt{2}\,I_{rms} \approx 1.414 \times 0.66 \approx 0.93\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics