Question

$\frac{2.5}{\pi }\mu F$ capacitor and $3000$-ohm resistance are joined in series to an cource of $200$ volt and $50{\sec }^{-1}$ frequency The power factor of the circuit and the power dissipated in it will respectively

• A. $0.8,0.7$W
• B. $0.06,0.6W$
• C. $0.6,4.8W$
• D. $4.8,0.6W$

Answer 1

The correct option is D $4.8,0.6W$

We know that capacitive reactance, $X_c=\frac{1}{\omega c}$

${\Rightarrow X}_c=\frac{1}{2\pi \left(50\right)\times \frac{2.5\times {10}^{-6}}{\pi }}$=4000$\Omega$

fro RC circuit, $\tan \varphi =\frac{X_c}{R}$

$\Rightarrow \tan \varphi =\frac{4000}{3000}$ (since R=3000$\Omega$)

$\Rightarrow \varphi =53.13°$

So, Power factor = cos$\varphi$=cos 53.13=0.6

Now,

As we know that impedance(Z)=$\sqrt{R^2+{X_c}^2}$

$\Rightarrow Z=\sqrt{{3000}^2+{4000}^2}=5000$

now phase angle between voltage and current ($\theta$)=${\cos }^{-1}\frac{R}{Z}$

$\theta ={\cos }^{-1}\frac{3000}{5000}=53.13°$

so, power dissipated(P)=$\frac{V^2}{Z}\cos \theta$

$\Rightarrow P=\frac{{200}^2}{5000}\cos 53.13=4.8watt$