Q. $\frac{2.5}{\pi}\mu F$ capacitor and 3000-$ohm$ resistance are joined in series to an ac source of $200\ volt$ and $50sec^{- 1}$ frequency. The power factor of the circuit and the power dissipated in it will respectively be
A) 0.6, 0.06 $W$
B) 0.06, 0.6 $W$
C) 0.6, 4.8 $W$
D) 4.8, 0.6 $W$
Explanation
(c) $Z = \sqrt{R^{2} + \left( \frac{1}{2\pi vC} \right)^{2}} = \sqrt{(3000)^{2} + \frac{1}{\left( 2\pi \times 50 \times \frac{2.5}{\pi} \times 10^{- 6} \right)^{2}}}$
$\Rightarrow Z = \sqrt{(3000)^{2} + (4000)^{2}} = 5 \times 10^{3}\Omega$
So power factor $\cos\phi = \frac{R}{Z} = \frac{3000}{5 \times 10^{3}} = 0.6$ and power
$P = V_{rms}i_{rms}\cos\phi = \frac{V_{rms}^{2}\cos\phi}{Z} \Rightarrow P = \frac{(200)^{2} \times 0.6}{5 \times 10^{3}} = 4.8W$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.