Question

An alternating voltage $E$ (in volts)=$200\sqrt{2}\sin \left(100t\right)$ is connected to a $1\mu F$ capacitor through an a.c ammeter. The reading of the ammeter shall be:

• A. $10mA$
• B. $20mA$
• C. $40mA$
• D. $80mA$

Answer 1

The correct option is B $20mA$

Given,

$E=200\sqrt{2}\sin \left(100t\right)V$

$\omega =100s^{-1}$

Now,

As the capacitor is only connected to the source,

$V_{C\left(rms\right)}=\frac{E}{\sqrt{2}}=200V$

$X_C=\frac{1}{\omega C}=\frac{1}{100\times {10}^{-6}}=\frac{1}{{10}^{-4}}=10000\Omega$

$I_{rms}=\frac{V_C\left(rms\right)}{X_C}=\frac{200}{10000}=20mA$

Note: Ammeter reads rms value of current.

Option $B$ is correct answer.