Question

In an AC circuit, a capacitor of capacitance $C=\left(25/\pi \right)\mu F$ and a resistor $R=300\Omega$ are connected in series with an AC source of 200 V and 50 $s^{-1}$ frequency. The power dissipated (in Watt) in circuit will be

Answer 1

Given,
$C=\frac{25}{\pi }\times {10}^{-6}F,R=300$,

$V_{rms}=200V,f=50s^{-1}$

Then capacitive reactance,

$X_C=\frac{1}{wc}=\frac{1}{2\pi fc}$

$\Rightarrow X_C=\frac{{10}^6}{2\pi \left(50\right)}\times \frac{\pi }{25}=\frac{{10}^6}{2500}$

Impedance,
$Z=\sqrt{R^2+X_C^2}=\sqrt{(300{)}^2+{\left(\frac{{10}^6}{2500}\right)}^2}$

$\Rightarrow Z=\sqrt{(300{)}^2+(400{)}^2}$
$\Rightarrow Z=500\Omega$

Then power facator,
$\Rightarrow cos\varphi =\frac{R}{Z}=\frac{300}{500}=\frac{3}{5}$

Power dissipated
$P=\frac{V_{rms}^2}{Z}\times cos\varphi$

$\Rightarrow P=\frac{200V\times 200V}{500}\times \frac{3}{5}$

$\Rightarrow$ P = 48 W

Hence, power dissipated in the circuit is 48 W.