Side-by-side comparison
| Parameter | Real Power | Reactive Power |
|---|---|---|
| Symbol and Unit | P — measured in Watts (W) or kW | Q — measured in Volt-Ampere Reactive (VAR) or kVAR |
| Physical Meaning | Rate of actual energy conversion (mechanical work, heat, light) | Energy oscillating between source and reactive element (L or C); no net energy consumed |
| Formula | P = V × I × cos φ (W) | Q = V × I × sin φ (VAR) |
| Power Factor Angle | P = S × cos φ; P is adjacent side of power triangle | Q = S × sin φ; Q is opposite side |
| Effect of Inductive Load | Unchanged — real power consumed by resistance | Positive Q — inductors absorb reactive power (lagging PF) |
| Effect of Capacitive Load | Unchanged | Negative Q — capacitors generate reactive power (leading PF) |
| Apparent Power Relationship | S² = P² + Q²; P = S × cos φ | Q = S × sin φ; Q = P × tan φ |
| Billing Impact | Billed directly — kWh consumed | Penalized if kVAR excessive — low PF tariff (TNEB below 0.85) |
| Correction Method | Increase load efficiency to reduce P for same output | Add shunt capacitor bank (kVAR compensation) to cancel inductive Q |
| Measurement Instrument | Wattmeter (electrodynamometer type) | VAR meter or power factor meter |
Key differences
Real power P (watts) represents energy permanently transferred from source to load per second — resistors convert it to heat, motors to mechanical work, lamps to light. Reactive power Q (VAR) represents energy oscillating between the source and the magnetic field of inductors (or electric field of capacitors) at twice the supply frequency — it does no net work. The apparent power S = √(P² + Q²) is what the generator and transmission lines must carry regardless. At PF = 0.7, the same cable carrying 70 kW of real power also carries 71.4 kVAR of reactive current, heating the conductors and wasting cable capacity. A 71.4 kVAR capacitor bank at the load terminates eliminates this, reducing S to equal P and recovering full cable capacity.
When to use Real Power
Reduce reactive power by installing shunt capacitor banks at the load terminals — a factory running 200 kW at PF=0.75 lagging needs 176 kVAR of capacitance (Q_c = P × tan(arccos 0.75) − P × tan(arccos 0.95)) to improve PF to 0.95 and avoid TNEB penalty tariff.
When to use Reactive Power
Maximize real power efficiency by selecting high-efficiency motors (IE3 rating), LED lighting, and resistive heating elements that have power factor near unity — these reduce apparent power demand and eliminate reactive current circulation in the distribution network.
Recommendation
Real power pays the bills; reactive power costs you in penalized tariffs and underutilized cable capacity. Always target PF above 0.90 in industrial installations using capacitor banks. For GATE: know the power triangle cold — draw it, label P, Q, S, and φ, and derive any one quantity given two others in under 30 seconds.
Exam tip: Examiners draw the power triangle and ask you to calculate the capacitor kVAR needed to improve PF from 0.7 to 0.95 for a 100 kW load — apply Q_c = 100×(tan(cos⁻¹ 0.7) − tan(cos⁻¹ 0.95)) = 100×(1.020 − 0.329) = 69.1 kVAR.
Interview tip: Interviewers at PGCIL, NTPC, and distribution companies ask why reactive power causes voltage drop in distribution lines — explain that reactive current in line impedance creates a voltage drop component (IX term) that reduces receiving-end voltage, separate from the IR drop from real current.