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An alternating voltage V = 300 $\sqrt{2} \sin (60 t)$ (in volts) is connected across a 4 $μ$ F capacitor through an a.c. ammeter. The reading of ammeter will be

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72 mA

$40 \sqrt{2} m A$

$30 \sqrt{2} m A$

40 mA

answer is C.

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Detailed Solution

$V_{0} = 300 \sqrt{2} ∴ V_{r m s} = \frac{V_{0}}{\sqrt{2}} = 300 V$

$I = \frac{V_{r m s}}{X_{C}} = V_{r m s} \times ω C = 300 \times 60 \times 4 \times 10^{- 6} A$

$I = 72 \times 10^{- 3} A; I = 72 m A$

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