A $40 μF$ capacitor is connected to a 200 V , 50 Hz ac supply. The rms value of the current in the circuit is , nearly :

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2.5 A

25.1 A

1.7 A

2.05 A

answer is B.

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Detailed Solution

$\begin{array}{l} Z = X_{C} = \frac{1}{Cω} = \frac{1}{40 \times 10^{- 8} \times 2 π \times 50} \\ I = \frac{V}{Z} = 200 \times 40 \times 10^{- 6} \times 2 π \times 50 = 2 .512 \end{array}$

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