Which one of the following statements is NOT true for capacitor used for improvement of power facto

Question

Which one of the following statements is NOT true for capacitor used for improvement  of power factor?

Examoogle · Accepted Answer

Load current decreases

Explanation: A power factor correction capacitor is connected in parallel with the load to supply reactive power locally. It reduces the reactive current component drawn from the supply, thereby reducing the total supply current, but it does not change the internal current flowing through the original load device.

Working Principle: The load typically draws a lagging current consisting of an active component ($I_a = I \cos \phi$) and a reactive component ($I_r = I \sin \phi$). The capacitor draws a leading current $I_c$ which partially or fully cancels $I_r$. Since $I_c$ and $I_r$ are in phase opposition, the net current supplied by the source becomes $I_s = \sqrt{I_a^2 + (I_r - I_c)^2}$, which is less than the original load current $I$. The load itself remains connected to the same terminal voltage, so its current remains unchanged.

Key Formulas: $Q_c = P(	an \phi_1 - 	an \phi_2)$ — Reactive power required for correction; $I_s = \sqrt{I_a^2 + (I_r - I_c)^2}$ — Resultant supply current

Key Points: Capacitors act as a local source of reactive power ($VARs$).; Power factor improvement reduces line losses ($I^2R$ losses).; The voltage across the load remains constant, hence load current remains unchanged.; Total supply current decreases, allowing for higher load capacity on existing feeders.

Advantages: Reduced $I^2R$ losses in the supply lines; Improved voltage regulation of the system; Reduced electricity bills due to penalty avoidance

Disadvantages: Risk of over-correction leading to leading power factor; Potential for resonance if harmonics are present; Initial investment cost for capacitor banks

Applications: Industrial induction motor feeders; Distribution transformers; Large commercial HVAC systems

Additional Info: Option B is false because the capacitor is connected in parallel; therefore, the load current $I_{load} = V/Z_{load}$ remains constant regardless of the parallel capacitor branch current.. Option A is true as pure capacitors draw current that leads the voltage by $90^\circ$.

Answer: B is correct — Adding a capacitor in parallel does not change the impedance or the voltage across the load, hence the load current remains unchanged.

Exam Tip: Remember that power factor correction is a parallel compensation technique; series components are affected, but branch currents of parallel loads remain independent of other parallel branches.

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