Answer

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**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:

$X_C = \dfrac{1}{\omega C}$

**Complete step by step answer:**

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.

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