A capacitor is connected to a 210 V, 50 Hz source as shown in figure. The peak current in the circuit is nearly ( = 3.14) [2024]

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Solution

Q.
A $10$ $μ F$ 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) [2024]

1 0.58 A

2 0.93 A

3 1.20 A

4 0.35 A

Ans.
(2)

$V_{rms} = 210$ $V, f = 50$ $Hz, C = 10$ $μ F$

$X_{C} = \frac{1}{2 π f C} = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{- 6}} = 318.47$ $Ω$

$I_{rms} = \frac{V_{rms}}{X_{C}} = \frac{210}{318.47} = 0.6594$ $A$

$Peak current, I_{0} = \sqrt{2} I_{rms} = 1.414 \times 0.6594 = 0.93$ $A$

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