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) :
:
:- 0.58 A
- 0.93 A
- 1.20 A
0.35 A
The Correct Option is B
Solution and Explanation
Capacitive Reactance
$$ X_C = \frac{1}{\omega C} = \frac{1}{2 \pi f C} $$
$$ X_C = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}} = \frac{1000}{3.14} $$
$$ V_{rms} = 210 \ V $$
$$ i_{rms} = \frac{V_{rms}}{X_c} = \frac{210}{X_c} $$
Peak current \( = \sqrt{2} i_{rms} \)
$$ = \sqrt{2} \times \frac{210}{1000} \times 3.14 = 0.932 \simeq 0.93 \ A $$
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