A circuit having poor power factor will draw

Question

Examoogle · Accepted Answer

High current

Explanation: A circuit with a poor power factor (low power factor) draws a higher current for the same amount of real power required by the load. This is because the load draws additional reactive power, which increases the total apparent power supplied by the source.

Working Principle: In an AC circuit, the relationship between real power ($P$), apparent power ($S$), and power factor ($\cos\phi$) is defined as $P = VI \cos\phi$. For a constant real power demand ($P$) and fixed voltage ($V$), the current $I = \frac{P}{V \cos\phi}$. As the power factor ($\cos\phi$) decreases, the denominator becomes smaller, necessitating a higher current to maintain the same power delivery.

Key Formulas: $P = VI \cos\phi$ — Relationship between Real Power, Voltage, Current, and Power Factor; $I = \frac{P}{V \cos\phi}$ — Dependence of current on power factor

Key Points: A low power factor signifies a high reactive power component.; Higher current leads to increased $I^2R$ copper losses in the transmission and distribution lines.; Poor power factor reduces the effective capacity of electrical equipment like transformers and alternators.; Power factor improvement is typically achieved by installing capacitor banks in parallel with the load.

Advantages: Higher utilization of source capacity by improving power factor; Reduced transmission line losses

Disadvantages: Increased current causes voltage drops in the supply system; Requires larger cable sizes for the same real power output

Applications: Industrial power distribution systems; Motor load compensation

Additional Info: A power factor of 1 is ideal (unity).. Low current corresponds to a high power factor (close to 1).. Inductive loads (motors, transformers) are the primary cause of lagging power factors.

Answer: D is correct — A circuit with a poor power factor requires more current to deliver the same real power because the source must supply both the useful real power and the wasted reactive power.

Exam Tip: Always remember that in power systems, $S = \sqrt{P^2 + Q^2}$. If $Q$ (reactive power) increases due to a poor power factor, $S$ must increase, forcing $I$ to rise since $S = VI$.

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