Power

Power is the rate of energy transfer. Master P = VI and its Ohm's-law forms, efficiency, power ratings, circuit loading and an introduction to AC real, reactive and apparent power.

Power tells you how fast energy is being converted: a 2 kW kettle turns electrical energy into heat twice as fast as a 1 kW one. It is the number on every appliance label and the quantity that decides wire sizes, breaker ratings, heat-sink sizes and battery life.

In DC circuits power is simply voltage times current. In AC circuits the voltage and current can be out of step, which splits power into real, reactive and apparent parts — an idea that runs through the whole of power engineering.

An analogy

A waterfall's power depends on both its height (voltage) and how much water goes over per second (current). A tall trickle and a short torrent can deliver the same power.

Power = voltage × current

Each coulomb delivers V joules, and I coulombs arrive per second, so energy arrives at V·I joules per second — watts. Combine with Ohm's law (V = IR) to get the two other forms: I²R (useful when you know the current, e.g. cable losses) and V²/R (useful when you know the voltage, e.g. a heater on mains).

Power ratings

Every component has a maximum power it can dissipate. A 0.25 W resistor asked to dissipate 0.5 W will overheat. The standard practice is to derate: choose a part rated at least 2× the expected dissipation.

Efficiency and losses

No conversion is perfect. Losses (heat) = P_in − P_out. A 90 %-efficient 500 W power supply draws 556 W and must get rid of 56 W as heat. Efficiency compounds: two 95 % stages in series give 0.95² = 90.25 %.

AC power: real, reactive and apparent

With inductive or capacitive loads the current is shifted in time from the voltage by an angle φ. Only the in-phase part of the current does useful work (real power P, watts). The out-of-phase part sloshes energy back and forth (reactive power Q, vars). Their vector sum is apparent power S = V·I (VA), which is what cables and transformers must be sized for.

Power factor pf = P/S = cos φ. A motor running at pf = 0.7 draws 43 % more current than it would at pf = 1 for the same useful power.

Instantaneous power and non-sinusoidal loads

p(t) = v(t)·i(t). For sinusoids it pulses at twice the line frequency around its average P. Rectifier-fed loads draw distorted, spiky current, so their power factor has a displacement part (cos φ) and a distortion part; true pf = P / (V_rms·I_rms) accounts for both. Standards such as IEC 61000-3-2 limit harmonic current, which is why power supplies above 75 W include active power-factor correction (PFC).

Formulas

Worked example: Kettle on the mains

Given: P = 2.2 kW; V = 230 V (resistive load)

Find: Current drawn and element resistance.

Answer: I ≈ 9.6 A, R ≈ 24 Ω

Worked example: Motor at low power factor

Given: P = 7.5 kW; V = 400 V single-phase equivalent; pf = 0.75

Find: Apparent power, current and reactive power.

Answer: S = 10 kVA, I = 25 A, Q ≈ 6.6 kvar

Common mistakes

Applications

Key takeaways

Going further