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A $40\ \mu F$ capacitor is connected to a 200 V, 50 Hz ac supply. The rms value of the current in the circuit is, nearly:
Detailed Solution
$i_{rms} = \frac{\varepsilon_{rms}}{X_C} = \varepsilon_{rms}\,\omega C$, with $\omega = 2\pi f = 100\pi$
$i_{rms} = 200 \times 2\pi \times 50 \times 40 \times 10^{-6}$
$= 2.5$ A