Active and Reactive Power: why the power company cares about your power factor
Some current does work. Some just sloshes back and forth. You pay for both.
Start here: carrying shopping bags badly
You are carrying shopping home. Held close to your body, the bags are manageable. Held out at arm's length, the same bags are exhausting.
The shopping has not got heavier. You are simply doing extra work that achieves nothing — effort spent fighting the awkward position rather than moving the bags.
Electricity has the same problem. Some current does real work — heating, lighting, turning motors. Some current just sloshes back and forth doing nothing useful, and the wires have to carry it anyway. Power factor is the measure of how much of your current is actually working.
Where the useless current comes from
In a simple resistor, voltage and current rise and fall together — perfectly in step. Every bit of current does work.
But motors and transformers contain coils, and a coil takes energy to build up its magnetic field, then gives it back later in the cycle. That energy travels out to the load and back again, over and over, achieving nothing — but it is real current, and it flows through real wires.
⚡ The three powers
Real power (P)Watts. The part that actually does something. This is what your meter charges you for.
Reactive power (Q)VAR. The part sloshing back and forth. Does no work — but the wires must be thick enough to carry it.
Apparent power (S)VA. What the supply actually has to deliver: volts × amps, ignoring whether it was useful.
power factor = P / S = real work ÷ total effort
A perfect load: pf = 1.0 (all effort useful)
A typical motor: pf = 0.8 (80% useful)
A bad load: pf = 0.5 (half wasted)
Why the electricity company cares
Here is the number that makes it matter. Suppose a factory needs 1840 W of real work from a 230 V supply.
At power factor 1.0: current = 1840 / 230 = 8.0 A
At power factor 0.5: current = 1840 / 115 = 16.0 A
SAME work done. TWICE the current.
The supply company must install wires, transformers and switchgear big enough for 16 A while only being able to bill for the work of 8 A. That is why industrial customers are penalised for poor power factor — they are demanding infrastructure they are not paying for.
The fix, and why it works
Motors are inductive — their current lags behind the voltage. A capacitor does the exact opposite: its current leads.
So you put capacitors across an inductive load and the two cancel, exactly as they did in resonance. The sloshing energy now goes back and forth between the coil and the capacitor locally, instead of travelling all the way back to the power station and out again.
💡 This is why you see banks of capacitors in factories and substations. They are not storing energy for later — they are cancelling the reactive part so the long wires only carry current that does work. The energy still sloshes; it just does its sloshing over a few metres instead of a few kilometres.
In a DC circuit, power is simply VI. In AC it isn't, because voltage and current can be out of step. If the current peaks a quarter-cycle after the voltage, then during part of every cycle power flows backwards from the load to the source. Averaged out, some of the current does real work and some does nothing but circulate — and separating those two is what this topic is about.
The three powers
Active (real) power: P = VI cos φ watts (W)
Reactive power: Q = VI sin φ volt-amperes
reactive (VAR)
Apparent power: S = VI volt-amperes (VA)
Relationship: S² = P² + Q² S = √(P² + Q²)
Power factor: pf = cos φ = P/S
💡 What each power actually means
Active PThe power that does useful work — turns motors, produces heat and light. Consumed only by resistance. This is what your energy meter records in kWh.
Reactive QPower that flows to the load and back again each cycle, storing energy in magnetic or electric fields. Does no net work, but it still occupies capacity in the cables and transformers.
Apparent SThe product of RMS voltage and current — what the equipment must be rated for. Transformers and generators are rated in kVA, not kW, precisely because they must carry the total current whether it does work or not.
A low power factor means you're drawing a lot of current for little useful power. The utility must size its cables and transformers for that current. This is why industrial consumers are penalised for a poor power factor. The fix is to add capacitors: their leading reactive power cancels the motors' lagging reactive power. That's the practical reason this topic exists.
Worked numerical 1 — an inductive load
A load draws 10 A from a 230 V, 50 Hz supply at a power factor of 0.8 lagging. Find P, Q, S and the phase angle.
S = VI = 230 × 10 = 2300 VA = 2.3 kVA
P = VI cos φ = 2300 × 0.8 = 1840 W = 1.84 kW
φ = cos⁻¹(0.8) = 36.87°
sin φ = sin(36.87°) = 0.6
Q = VI sin φ = 2300 × 0.6 = 1380 VAR = 1.38 kVAR (lagging)
Check with the triangle:
S = √(P² + Q²) = √(1840² + 1380²)
= √(3 385 600 + 1 904 400) = √5 290 000
= 2300 VA ✔
Worked numerical 2 — power factor correction
The load above is to be corrected to unity power factor at 50 Hz. What capacitance is needed in parallel?
To reach pf = 1, the capacitor must supply all the
reactive power the load absorbs:
Q_C = 1380 VAR
For a capacitor: Q_C = V²/X_C = V² × ωC
1380 = 230² × (2π × 50) × C
1380 = 52 900 × 314.16 × C
1380 = 16 619 064 C
C = 8.30 × 10⁻⁵ F = 83.0 µF
After correction:
P is UNCHANGED at 1840 W (the useful work is the same)
S = P = 1840 VA
New current I = S/V = 1840/230 = 8 A (was 10 A)
Current dropped 20% for the same useful work — that's the
saving in cable and transformer capacity.
Notice that power factor correction does not reduce P — the useful work is identical. It reduces the current, and therefore the I²R losses in the distribution network. If a question asks whether correction saves energy at the load, the answer is no; it saves losses in the wires.
Worked numerical 3 — R-L series circuit from scratch
A coil of R = 30 Ω and L = 0.127 H is connected to 230 V, 50 Hz. Find the current, power factor, P and Q.
X_L = 2πfL = 2π × 50 × 0.127 = 39.9 ≈ 40 Ω
Z = √(R² + X_L²) = √(30² + 40²) = √(900 + 1600)
= √2500 = 50 Ω
I = V/Z = 230/50 = 4.6 A
Power factor = cos φ = R/Z = 30/50 = 0.6 lagging
φ = cos⁻¹(0.6) = 53.13°
P = I²R = 4.6² × 30 = 634.8 W
(or VI cos φ = 230 × 4.6 × 0.6 = 634.8 W ✔)
Q = I²X_L = 4.6² × 40 = 846.4 VAR
S = VI = 230 × 4.6 = 1058 VA
Check: √(634.8² + 846.4²) = √(403 071 + 716 393)
= √1 119 464 = 1058 VA ✔
Note the shortcut used here: for a series circuit, cos φ = R/Z directly — no need to find the angle first.
💡 Exam angle: a very common 6–8 mark question. Draw the power triangle — it earns marks and prevents sign errors. Remember cos φ = R/Z for series circuits — it is the fastest route to power factor. Then always say whether it is lagging (inductive) or leading (capacitive). Dropping that one word loses a mark even when your number is correct.
Syllabus points
Power triangle; power factor (numerical)
Create a free account to tick topics off, take notes as you read, watch the video lessons and get a day-by-day study plan built around your exam date.