Question

$\frac{2.5}{\pi }\mu F$ capacitor and $3000$ ohm resistor 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 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

Answer 1

Answer: (C)

$Z=\sqrt{R^2+{\left(\frac{1}{2\pi vC}\right)}^2}$
$=\sqrt{(3000{)}^2+\frac{1}{{\left(2\pi \times 50\times \frac{2.5}{\pi }\times 10^{-6}\right)}^2}}$
$Z=\sqrt{(3000{)}^2+(4000{)}^2}=5\times 10^3\Omega$
So, power factor, $\cos \varphi =\frac{R}{Z}=\frac{3000}{5\times 10^3}=0.6$
Now, $P=V_{mm}{i}_{ms}\cos \varphi =\frac{V_{ms}^2\cos \varphi }{Z}$
$\Rightarrow P=\frac{(200{)}^2\times 0.6}{5\times 10^3}=4.8W$