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.

S = √(P² + Q²)

Power factor = P/S

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

I_L = P/(√3 V_Lcos φ)

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.

Quick Trick

PF ↑ → current ↓ → losses and kVA burden ↓.

Why Other Options Are Wrong

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.

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.

S = √(P² + Q²)

Power factor = P/S

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

I_L = P/(√3 V_Lcos φ)

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.

Quick Trick

PF ↑ → current ↓ → losses and kVA burden ↓.

Why Other Options Are Wrong

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.

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.

S = √(P² + Q²)

Power factor = P/S

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

I_L = P/(√3 V_Lcos φ)

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.

Quick Trick

PF ↑ → current ↓ → losses and kVA burden ↓.

Why Other Options Are Wrong

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.