6. 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: (2020)
Solution:
Capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$.
RMS current is given by $I_{rms} = \frac{V_{rms}}{X_C} = V_{rms} \times 2\pi f C$.
Substituting the values: $I_{rms} = 200 \times 2\pi \times 50 \times 40 \times 10^{-6} = 0.8\pi \approx 2.51 A \approx 2.5 A$.
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