Question

$\frac{25}{\pi }\mu F$ capacitor and 3000-ohm resistance are joined in series to an ac source of 200 volt and $50se{c}^{-1}$ frequency.The power factor of the circuit and the power dissipated in it will respectively

• A. 0.6, 0.06 W
• B. 0.06, 0.6 W
• C. 0.6, 4.8 W
• D. 4.8, 0.6 W

Answer 1

The correct option is C 0.6, 4.8 W

z=$\sqrt{R^2+{\left(\frac{1}{2\pi vC}\right)}^2}=\sqrt{(3000{)}^2+\frac{1}{{(2\pi \times 50\times \frac{2.5}{\pi }\times {10}^{-6})}^2}}$

$\Rightarrow Z=\sqrt{(3000{)}^2+(4000{)}^2}=5\times {10}^3\Omega$

So power factor $cos\varphi =\frac{R}{Z}=\frac{3000}{5\times {10}^3}=0.6$ and

Power P=$V_{rms}{i}_{rms}cos\varphi =\frac{V_{rms}^{2}cos\varphi }{Z}\Rightarrow P=\frac{(200{)}^2\times 0.6}{5\times {10}^3}=4.8W$