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In an ac circuit an alternating voltage $e = 200\sqrt{2}\sin100t$ volts is connected to a capacitor of capacity 1 $\mu$F. The r.m.s. value of the current in the circuit is
A
20 mA
B
10 mA
C
100 mA
D
200 mA
Detailed Solution
Comparing with $e = e_0\sin\omega t$: $e_0 = 200\sqrt{2}$ V and $\omega = 100$ rad/s.
$e_{rms} = \frac{e_0}{\sqrt{2}} = \frac{200\sqrt{2}}{\sqrt{2}} = 200$ V
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100\times1\times10^{-6}} = 10^4\ \Omega$
$I_{rms} = \frac{e_{rms}}{X_C} = \frac{200}{10^4} = 2\times10^{-2}$ A
$I_{rms}$ = 20 mA
$e_{rms} = \frac{e_0}{\sqrt{2}} = \frac{200\sqrt{2}}{\sqrt{2}} = 200$ V
Capacitive reactance $X_C = \frac{1}{\omega C} = \frac{1}{100\times1\times10^{-6}} = 10^4\ \Omega$
$I_{rms} = \frac{e_{rms}}{X_C} = \frac{200}{10^4} = 2\times10^{-2}$ A
$I_{rms}$ = 20 mA