A 14.0 uF capacitor is connected to a 220 V, 50 Hz source. The value of peak current in the circuit is

(Given: $π = 22/7$ and $\sqrt{2}=1.414$)

1.37 A

The correct answer is Option (3) → 1.37 A

Given:

Capacitance, $C = 14.0\ \mu\text{F} = 14 \times 10^{-6}\ \text{F}$

Voltage, $V_{\text{rms}} = 220\ \text{V}$

Frequency, $f = 50\ \text{Hz}$

Capacitive reactance: $X_C = \frac{1}{2 \pi f C}$

$X_C = \frac{1}{2 \pi \cdot 50 \cdot 14 \times 10^{-6}}$

$X_C = \frac{1}{0.004398} \approx 227.4\ \Omega$

Peak voltage: $V_0 = \sqrt{2} V_{\text{rms}} = \sqrt{2} \cdot 220 \approx 311\ \text{V}$

Peak current: $I_0 = \frac{V_0}{X_C} = \frac{311}{227.4} \approx 1.37\ \text{A}$

∴ Peak current = 1.37 A