Question

An alternating voltage (in volts) given by $V=200\sqrt{2}sin\left(100\right)t$ is connected to $1\mu F$ capacitor through an ideal ac ammeter in series. The reading of the ammeter and the average power consumed in the circuit shall be

• A. $20mA,4W$
• B. $20mA,0$
• C. $20\sqrt{2}mA,8W$
• D. $20\sqrt{2}mA,4\sqrt{2}W$

Answer 1

The correct option is B $20mA,0$

On comparing $V=200\sqrt{2}sin\left(100t\right)$ with
$V=V_{0}sin\omega t,$ we get
$V_0=200\sqrt{2}V,\omega =100rad{s}^{-1}$

$\therefore V_{rms}=\frac{V_0}{\sqrt{2}}=\frac{200\sqrt{2}V}{\sqrt{2}}=200V$

The capacitive reactance is
$X_C=\frac{1}{\omega C}=\frac{1}{100\times 1\times {10}^{-6}}={10}^4\Omega$

ac ammeter reads the arms value of current . Therefore, the reading of the ammeter is $I_{rms}=\frac{V_{rms}}{X_C}=\frac{200V}{{10}^4\Omega }=20\times {10}^{-3}A=20mA$

The average power consumed in the circuit,
$P=I_{rms}{V}_{rms}\cos \varphi$

In an pure capacitive circuit , the phase difference between
current and voltage is $\frac{\pi }{2}$
$\therefore cos\varphi =0\therefore P=0.$