Previous Year Question
Correct answer: D
Explanation
D — It minimizes the reactive power burden on generators and transmission equipment. Local compensation reduces the reactive current carried upstream.
Formula
Step-by-step solution
Correct answer: D — It minimizes the reactive power burden on generators and transmission equipment.
Why low power factor burdens the network
An inductive load draws active power P to perform useful work and reactive power Q to maintain its electromagnetic fields. The supply equipment must carry the resulting apparent power S, even though reactive power does not represent net energy converted into useful work over a full AC cycle.
Power-factor improvement supplies part of the reactive requirement locally, for example through a capacitor bank. This reduces the net reactive power demanded from the upstream network while the useful active-power requirement can remain unchanged.
Effect on current and equipment capacity
At fixed active power and voltage, a higher power factor means a smaller line current. The lower current reduces I²R losses, reduces the current-related voltage drop and releases some of the kVA capacity of generators, transformers and lines.
For illustration, a 100 kW load requires 125 kVA at 0.8 power factor but only about 105.3 kVA at 0.95 power factor. Its active-power requirement remains 100 kW; the supply carries less apparent power because the reactive component is smaller. This is the meaning of reducing the reactive-power burden in option D. Compensation should match the load rather than create unnecessary leading VARs.
Other options
A transients are not the purpose. B power-factor improvement increases usable system capacity rather than decreasing it. C it tends to reduce, not increase, transmission cost for a required power transfer. D correctly states the reactive-burden benefit.
Common mistake
Saying power-factor correction reduces the load’s real kW demand. It mainly reduces reactive demand and line current for the same useful real power.
Exam tip
Fixed kW: current is inversely proportional to power factor.
Quick method
PF ↑ → current ↓ → losses and kVA burden ↓.