An industrial motor draws 100 kVA of apparent power at a power factor of 0.6. A capacitor bank is added that improves the power factor to 1.0. What happens to the real power delivered to the motor?
AReal power increases to 100 kW because the power factor is now 1.0
BReal power decreases to 60 kW because the reactive power is eliminated
CReal power remains at 60 kW — the capacitor corrects the power factor without changing the useful power delivered to the motor
DReal power becomes zero because the capacitor and inductor cancel each other
Real power P = |S|·cos(θ) = 100 kVA × 0.6 = 60 kW before correction. The capacitor adds leading reactive power (Q < 0) that cancels the motor's lagging reactive power (Q > 0). When Q_total → 0, the power factor → 1.0 and |S| → 60 kW. The motor still receives exactly 60 kW of real power — its operation is unchanged. What changed is that the source no longer needs to supply reactive current: the capacitor supplies it locally. This is why power factor correction reduces utility bills without affecting how equipment performs.
Question 2 Multiple Choice
A load draws voltage and current that are perfectly in phase (θ = 0). What does this imply about its reactive power?
AReactive power is maximized because the load is purely resistive
BReactive power Q = 0, because Q = ½Vm·Im·sin(0) = 0
CApparent power equals zero because no phase difference exists
DReactive power is undefined when voltage and current are in phase
Reactive power Q = ½Vm·Im·sin(θ). When θ = 0 (unity power factor), sin(0) = 0, so Q = 0. This is a purely resistive load — all power delivered is real power. Option A is backwards: a purely resistive load minimizes reactive power (Q = 0), not maximizes it. Apparent power |S| = ½Vm·Im is nonzero (option C is wrong) — it equals real power P when Q = 0.
Question 3 True / False
Reactive power is wasted power that has no physical significance for power system design, since it averages to zero over a full cycle.
TTrue
FFalse
Answer: False
While reactive power averages to zero in terms of energy delivered to the load, it still flows back and forth between the source and reactive elements — and that flowing current must be carried by generators, transformers, and transmission lines. These are sized by the current they carry (apparent power |S|), not just real power P. A 0.7 power factor load requires 43% more current capacity than a unity PF load delivering the same real power. Power factor correction exists precisely because reactive power, though not 'consumed,' has real economic costs in equipment sizing.
Question 4 True / False
If voltage leads current by 30° in an AC circuit, the load is capacitive.
TTrue
FFalse
Answer: False
If voltage *leads* current (positive θ, where θ is the angle of voltage minus angle of current), the load is *inductive*, not capacitive. In an inductive load, current lags behind the voltage — equivalently, voltage leads current. In a capacitive load, current leads voltage (negative θ). This distinction is critical for power factor correction: inductive loads (lagging PF) are corrected by adding shunt capacitors; capacitive loads (leading PF) are corrected by adding inductors.
Question 5 Short Answer
Explain why apparent power |S| determines the required rating of electrical equipment like transformers and generators, rather than real power P.
Think about your answer, then reveal below.
Model answer: Generators, transformers, and transmission lines are physically limited by the current they can carry without overheating (I²R losses in conductors and windings). The current magnitude is determined by apparent power: I_rms = |S|/V_rms, regardless of the phase angle. A machine must carry the full current corresponding to |S| even if only the P component does useful work — reactive current heats conductors just as much as resistive current. This is why equipment is rated in volt-amperes (VA or kVA), not watts. Power factor correction reduces |S| toward P by eliminating Q, so the source sees less current for the same delivered real power.
This is why utilities charge large industrial customers a power factor penalty: low power factor means the utility must size equipment for large apparent power even when the real load is moderate. The penalty incentivizes customers to install local reactive compensation (capacitor banks) and reduce reactive demand on the grid.