Question

In an ac circuit, power factor improvement is achieved by

(a)Connecting a resistor in parallel with the inductive load.

(b)Connecting an inductor in parallel with the inductive load.

(c)Connecting a capacitor in parallel with the inductive load.

Short answer

Answer

Therefore the correct option is (c) Connecting a capacitor in parallel with the inductive load.

Step-by-step solution

Draw a power triangle

Consider that the inductive load has reactive resistance voltage $X_L$ and resistance $R$. When capacitor is not connected, the source supplies $Q_L$ reactive power and $P$ active power to the load. The apparaent power for this condition is $S_1$ and the power factor is $p.f1$.

Connecting a capacitor in parallel.

Consider as a a capacitor is connected in parallel across the inductive load, it will supply a reactive power$Q_c$to the load and source will supply reactive power$Q_S$.

From the diagram,

${\mathbf{Q}}_{\mathbf{S}}\mathbf{=}{\mathbf{Q}}_{\mathbf{L}}\mathbf{-}{\mathbf{Q}}_{\mathbf{C}}$ …… (1)

That is, reactive power supplied from source decreases by connecting a capacitor, the apparent power decrease to $S_2$, power angle decreases and hence power factor increases.

Therefore, from the diagram the power factor increases with a capacitor.

Thus the option (c) is correct.

Explanation for option (a)

By connecting a resistor in parallel with inductive load, it decreases the overall impedance, but doesn’t affect the power angle.

Because, current through the resistance will be in phase with the supplied voltage.

Thus the option (a) is incorrect.

Explanation for option (b)

Consider the inductor is connected in parallel across the load, it will aborbs the positive reactive power instead of supplying.

So, the power angle will increase, hence power factor decreases.

Thus the option (b) is incorrect.