Question: 38. A 10 \mu \mathrm{~F} capacitor is connected to a 210 \mathrm{~V}, 50 \mathrm{~Hz} source as shown in figure. The peak current in the circuit is nearly ( \pi=3.14 ) :
(1) 0.58 A
(2) 0.93 A
(3) 1.20 A
(4) 0.35 A
Answer: Option (2)
Explanation:
For a capacitor connected to an AC source, the capacitive reactance is given by
X_C = \frac{1}{\omega C}, where \omega = 2\pi f.
Given frequency f = 50 \mathrm{~Hz},
the angular frequency is \omega = 2 \times 3.14 \times 50 = 314 \ \mathrm{rad\,s^{-1}}.
The capacitance is C = 10 \mu \mathrm{F} = 10 \times 10^{-6} \mathrm{~F}.
Thus, the capacitive reactance is
X_C = \frac{1}{314 \times 10 \times 10^{-6}} \approx 318 \ \Omega.
The given voltage 210 \mathrm{~V} is the RMS value. The RMS current is
I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{X_C} = \frac{210}{318} \approx 0.66 \ \mathrm{A}.
The peak current is related to RMS current by I_0 = \sqrt{2} \, I_{\mathrm{rms}}.
So,
I_0 = \sqrt{2} \times 0.66 \approx 0.93 \ \mathrm{A}.
Hence, the peak current is nearly 0.93 \ \mathrm{A},
which corresponds to option (2).