Question

A circuit draws a power of 550 W from a source of 220 V, 50 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 will have to be connected with it?

• A. $\frac{1}{42\pi }\times {10}^{-2}F$
• B. $\frac{1}{41\pi }\times {10}^{-2}F$
• C. $\frac{1}{5\pi }\times {10}^{-2}F$
• D. $\frac{1}{84\pi }\times {10}^{-2}F$

Answer 1

Answer: (A)

$\frac{1}{42\pi }\times {10}^{-2}F$

As the current lags behind the potential difference, the circuit contains resistance and inductance.

Power $P=V_{rms}\times I_{rms}\times cos\varphi$

here, $I_{rms}=\frac{V_{rms}}{Z}$, where $Z=\sqrt{(R{)}^2+(\omega L{)}^2}$

$\therefore P=\frac{V_{rms}^2}{Z}\times cos\varphi$

or

$Z=\frac{V_{rms}^2}{P}\times cos\varphi$

So, $Z=\frac{(220{)}^2\times 0.8}{550}=70.4\Omega$

Now, power factor $cos\varphi =\frac{R}{Z}orR=Zcos\varphi$

$\therefore R=70.4\times 0.8=56.32\Omega$

Further, $Z^2=R^2+(\Omega L{)}^2$

or $\omega L=\sqrt{(Z^2-R^2)}$

When the capacitor is connected in the circuit,

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

According to the question,

Power factor $cos\varphi =1$

Hence, $R=Zcos\varphi =Z\times 1,R=Z$

$\therefore Z=\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2};{(\omega L-\frac{1}{\omega C})}^2=0$

$\therefore C=\frac{1}{\omega \left(\omega L\right)}=\frac{1}{2\pi \upsilon v\left(42.2\right)}=\frac{1}{2\pi \times 50\times 42.2}=\frac{1}{42\pi }\times {10}^{-2}F$