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A 40 $μ$ F capacitor is connected to a 200 V, 40Hz ac supply. The rms value of the current in the circuit is, nearly

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2 A

2.5 A

25 A

1.7 A

answer is A.

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Detailed Solution

$\begin{array}{l} I_{rms} = \frac{ε_{rms}}{X_{C}} = ε_{rms} ωC \\ = 200 \times 2 π \times 40 \times 40 \times 10^{- 6} A = 2 A \end{array}$

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