Question

An inductive load is connected to 230 V, 50 Hz AC source, which draws 2 kW real power at a power factor of 0.80 lagging. The value of capacitor required in parallel to load such that combined power factor improves to 0.90 (lagging) will be ________ $\mu$F. (Answer up to two decimal places)

• A. 31.98

Answer 1

The correct option is A 31.98
Given, $P_L=2000W$

$S_L=\frac{2000}{0.8}=2500VA$

$Q_{L1}=\sqrt{S_L^2-P_L^2}=\sqrt{{\left(2500\right)}^2-{\left(2000\right)}^2}=1500VAR$

${\varphi }_1={\cos }^{-1}\left(0.80\right)=36.869{}^{\circ }$

${\varphi }_2={\cos }^{-1}\left(0.90\right)=25.841{}^{\circ }$

$Q_{L2}=P_L\tan {\varphi }_2=2000\tan 25.841{}^{\circ }=968.604VAR$


Reactive power supplied by the capacitor,

$Q_C=1500-968.604$

$=531.396VAR$

We know, $Q_C=\frac{V^2}{\left(\frac{1}{\omega C}\right)}=V^2\omega C$

$\therefore C=\frac{Q_C}{V^2\omega }$

$=\frac{531.396}{{\left(230\right)}^2\times 2\times \pi \times 50}=31.975\mu F$

$=31.98\mu F$