Poor power factor

Question

Poor power factor

Examoogle · Accepted Answer

all of the above

Explanation: A poor power factor (low $\cos \phi$) implies that a significant portion of the current supplied is reactive current rather than active power. This increases the total current drawn from the source for a given real power requirement, leading to various inefficiencies in the power system.

Working Principle: The active power is given by $P = VI \cos \phi$. To maintain constant $P$ at a low $\cos \phi$, the line current $I = \frac{P}{V \cos \phi}$ must increase. This increased current leads to higher $I^2R$ losses, larger voltage drops across the transmission line impedance $Z = R + jX$ (specifically $V_{drop} \approx I(R \cos \phi + X \sin \phi)$), and forces generators and transformers to operate at higher apparent power ($S = VI$) ratings, effectively reducing their useful load-carrying capacity.

Key Formulas: $P = VI \cos \phi$ — Real Power; $S = VI$ — Apparent Power; $I = \frac{P}{V \cos \phi}$ — Current draw at constant load

Key Points: Low power factor increases the required cross-sectional area of conductors for a specific load.; It leads to inefficient utilization of capital equipment like transformers and generators.; Poor power factor causes higher voltage regulation issues, leading to unstable voltage at the consumer end.

Advantages: Improving power factor reduces electricity bills in industrial tariff structures.; Enhanced power factor improves equipment longevity by reducing thermal stress.

Disadvantages: Requires installation of capacitor banks, which adds cost.; Over-correction can lead to leading power factor, which may cause unwanted voltage rises.

Applications: Industrial motor drives; Transmission and distribution networks

Additional Info: Most industrial loads are inductive (motors, transformers), leading to a lagging power factor.. Option A is correct because the generator/transformer capacity (VA) is limited; higher current for the same watts leaves less room for real work.. Option B is correct because losses are $I^2R$, so higher $I$ exponentially increases copper losses.. Option C is correct because the line impedance causes a larger drop as $I$ increases.

Answer: D is correct — Poor power factor increases current flow, leading to increased copper losses, higher voltage drops, and reduced effective power capacity of electrical infrastructure.

Exam Tip: Remember that power factor correction using shunt capacitors provides reactive power locally, which prevents the need to transmit this current through the entire grid, minimizing losses.

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