Power in AC circuits, power factor and wattless current
Average AC power is P = VI cos φ, where cos φ is the power factor. Current at 90° to the voltage does no work: it is wattless current.
Deriving the average power
- 1Let v = vₘ sin ωt and i = iₘ sin(ωt + φ).
- 2Instantaneous power: p = vi = vₘiₘ sin ωt sin(ωt + φ).
- 3Using 2 sin A sin B = cos(A − B) − cos(A + B): p = (vₘiₘ/2)[cos φ − cos(2ωt + φ)].
- 4The average of cos(2ωt + φ) over a full cycle is zero.
- 5Average power = (vₘiₘ/2) cos φ = (vₘ/√2)(iₘ/√2) cos φ.
- 6★So, P = VI cos φ, where V and I are rms values and cos φ is the power factor.
★ key step
Power factor in different circuits
| Circuit | Phase angle φ | Power factor cos φ | Average power |
|---|---|---|---|
| Pure resistor | 0 | 1 | P = VI, the largest |
| Pure inductor or pure capacitor | π/2 | 0 | Zero |
| Series LCR | tan φ = (X_C − X_L)/R | R/Z | P = VI cos φ = I²R |
| Series LCR at resonance | 0 | 1 | Largest: P = VI = V²/R |
Wattless current
Draw the current phasor I at an angle φ to the voltage phasor V. Split I into two parts. I cos φ lies along V: it is the active part, and it does work. I sin φ is at right angles (π/2) to V. The power linked with it is V × I sin φ × cos 90° = 0. So I sin φ is called the wattless current. In a pure inductor or capacitor (φ = π/2), the whole current is wattless. This is why a choke (an inductor with very small resistance) is used to reduce the current in AC circuits. It limits the current without wasting energy as heat, unlike a resistor.
Why a low power factor is costly
- For a fixed power P = VI cos φ at a fixed supply voltage V, a small cos φ means a large current I.
- A larger current wastes more energy as heat, I²R, in the transmission lines.
- So power companies want the power factor close to 1.
- Factories with many motors (inductive loads) add capacitors in parallel. Their leading current cancels much of the lagging current and raises the power factor.
Worked example: power and wattless current
A motor draws an rms current of 5 A from a 220 V (rms) supply. Its power factor is 0.6. Find (a) the average power, (b) the active current and (c) the wattless current.
- (a) P = VI cos φ = 220 V × 5 A × 0.6 = 660 W.
- (b) Active current = I cos φ = 5 A × 0.6 = 3 A.
- (c) cos φ = 0.6, so sin φ = √(1 − 0.36) = √0.64 = 0.8.
- Wattless current = I sin φ = 5 A × 0.8 = 4 A.
(a) 660 W; (b) 3 A; (c) 4 A
Large factories and buildings in India install capacitor banks to keep their power factor close to 1. A low power factor draws extra current and wastes energy in the wires, and electricity boards may charge a penalty for it.
2 marks: “Define power factor. Give its value for a pure resistor and for a pure inductor.” cos φ, the ratio of the average power to VI; for series LCR it equals R/Z. It is 1 for a pure resistor and 0 for a pure inductor.
2 marks: “What is wattless current? Why is a choke coil preferred to a resistor for controlling the current in an AC circuit?” I sin φ, the part of the current at π/2 to the voltage, which uses no power. A choke has a power factor near zero, so it reduces the current with almost no heat loss.
3 marks: “Derive an expression for the average power in an AC circuit in which the current leads the voltage by φ.” p = vₘiₘ sin ωt sin(ωt + φ) = (vₘiₘ/2)[cos φ − cos(2ωt + φ)]; the second term averages to zero; P = VI cos φ.
Real power = VI cos φ: only the part of the current in step with the voltage pays the bill; I sin φ is the wattless passenger.