8. In an AC circuit an alternating voltage $E = 200\sqrt{2} \sin 100t volts$ is connected to a capacitor of capacity $1 \mu F$. The r.m.s value of the current in the circuit is: (2011 Pre)
Solution:
Peak voltage is $E_0 = 200\sqrt{2} V$, so $E_{rms} = \frac{E_0}{\sqrt{2}} = 200 V$.
Capacitive reactance is $X_C = \frac{1}{\omega C} = \frac{1}{100 \times 10^{-6}} = 10^4 \Omega$.
The rms current is $I_{rms} = \frac{E_{rms}}{X_C} = \frac{200}{10^4} = 20 \times 10^{-3} A = 20 mA$.
Leave a Reply