An alternating voltage is given as, $E=100\sqrt{2}\sin 100t$
This voltage E is applied to a capacitor of $1\ \mu\text{F}$. The current reading in an a.c. ammeter in mA is

Given alternating voltage is $E = 100\sqrt{2}\sin(100t)$

Comparing with the standard equation $E = E_0\sin(\omega t)$, we get:

Peak voltage, $E_0 = 100\sqrt{2}$ V

Angular frequency, $\omega = 100$ rad/s

The rms voltage is $E_{rms} = \dfrac{E_0}{\sqrt{2}} = \dfrac{100\sqrt{2}}{\sqrt{2}} = 100$ V

Capacitive reactance, $X_C = \dfrac{1}{\omega C} = \dfrac{1}{100 \times 10^{-6}} = 10^4\ \Omega$

The reading of an a.c. ammeter is the rms current, which is given by:

$I_{rms} = \dfrac{E_{rms}}{X_C} = \dfrac{100}{10^4} = 10^{-2}$ A

Converting to milliamperes:

$I_{rms} = 10^{-2} \times 1000$ mA $= 10$ mA

Answer: $10$