Question

A circuit draws a power of $550\text{W}$ from a source of $220\text{V}$, $50\text{Hz}$. The power factor of the circuit is $0.8$ and the current lags in phase behind the potential difference. To make the power factor of the circuit as $1.0$, what capacitance is needed to be connected in series with it.

• A. $10\mu \text{F}$
• B. $35\mu \text{F}$
• C. $75\mu \text{F}$
• D. $100\mu \text{F}$

Answer 1

The correct option is C $75\mu \text{F}$

Given that current lags behind voltage, so the circuit must contain inductance and resistance initially.

$P_{avg}=\frac{E_{rms}^2}{Z}\times \cos \varphi$

$Z=\frac{E_{rms}^2\times \cos \varphi }{P_{avg}}$

Power factor $=0.8=\cos \varphi$

$Z=\frac{(220{)}^2\times 0.8}{550}$

$\therefore Z=70.4\Omega$

As, $\cos \varphi =\frac{R}{Z}$

$R=Z\cos \varphi =70.4\times 0.8=56.32\Omega$

Initially for $LR$ circuit

$\Rightarrow Z^2=R^2+{\omega }^2{L}^2$

$\Rightarrow {\omega }^2{L}^2=Z^2-R^2$

$\Rightarrow \omega L=\sqrt{Z^2-R^2}=42.24\Omega$

Given that by applying capacitor $C$ to make $\cos \varphi =1$

So $\cos \varphi =1\Rightarrow \frac{R}{Z_{final}}=1$

$\Rightarrow R=Z_{final}$

$\Rightarrow R=\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2}$

$\Rightarrow \omega L=\frac{1}{\omega C}$

$\Rightarrow C=\frac{1}{{\omega }^{2}L}=\frac{1}{\omega \left(\omega L\right)}=\frac{1}{2\pi f\left(\omega L\right)}$

$\Rightarrow C=\frac{1}{2\times 3.14\times 50\times 42.2}$

$\therefore C\approx 75\mu \text{F}$

Hence, option $\left(\text{C}\right)$ is correct.