Question

A 10 µF capacitor is connected to a 210 V, 50 Hz source as shown in figure. The peak current in the circuit is nearly (π = 3.14) :
:

• A. 0.58 A
• B. 0.93 A
• C. 1.20 A
• D. 0.35 A

Answer 1

Answer: (B)

0.93 A

Answer 2

The capacitive reactance is given by:
$X_c = \frac{1}{\omega C} = \frac{1}{2\pi fC}$.
Substituting values:
$X_c = \frac{1}{2 \cdot 3.14 \cdot 50 \cdot 10 \cdot 10^{-6}} = 318.31 \, \Omega$.
The peak current:
$I_peak = \frac{V_peak}{X_c} = \frac{210\sqrt{2}}{318.31} \approx 0.93 \, A$.