Question

# In an a.c. 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:A. 10 mAB. 100 mAC. 200 mAD. 20 mA

Hint: The emf of the supply voltage is given in the question. Like a resistor, a capacitor also creates some voltage drop across. It has some 'impedance', which when multiplied by current gives the voltage drop across it.

Formula Used:
Capacitor's reactance (or resistance) is obtained as:
$X_C = \dfrac{1}{\omega C}$

Given, the supply:
$e=200 \sqrt{2} \sin 100t$ volts
And the value of C = 1$\mu$ F

We know that an alternating current source will have a supply voltage given by $V_0 \sin \omega t$ type of expression. We only compare this with our given supply voltage. The direct comparison gives us:
$V_0 = 200 \sqrt{2}$ volts
and
$\omega = 100$ Hz
The reactance of the capacitor can be found by plugging in the values as:
$X_C = \dfrac{1}{\omega C} = \dfrac{1}{100 \times 10^{-6}} = 10000$ ohms
We know, by ohm's law that the current I in the circuit with resistance R and supply voltage V is just V/R. Similarly, we can write:
$I_0 = \dfrac{V_0}{X_C}$
Thus,
$I_0 = \dfrac{200 \sqrt{2}}{10000}$A
But, we need to find the rms value of the current, so we use the formula:
$I_{rms} = \dfrac{I_0}{\sqrt{2}}$
Therefore, we can get:
$I_{rms} = \dfrac{2}{100}$A
Or, we can say:
$I_{rms} = 20$mA
Thus, the correct option (D). 20 mA. This is the value of rms ac current in the circuit consisting of 1$\mu$F capacitor.

Note:
If more components are present in the circuit, the situation becomes different. If two capacitors were present in series we had to use equivalent capacitance in the reactance formula. If a resistor would have been present we would have done some vector algebra as the current in a capacitor lags behind from a resistor.