Engineer's Notebook §1.2

Electrical power

Power is the rate at which a circuit converts electrical energy into heat, light or work. It is voltage times current:

P = VI
P
power
W
V
voltage
V
I
current
A
R
resistance
Ω
E
energy
J, kWh
t
time
s, h
cos φ
power factor
—
VL
line voltage
V
IL
line current
A
A graph with current along the bottom and voltage up the side. The shaded rectangle from the origin to the operating point has an area equal to the power. CURRENT I, AVOLTAGE V, V 400 32 P = 1.15 kW
Fig. 1.2-1Axes run to 400 V and 32 A. The shaded area is the power.
R = V/I46 Ω
E = Pt1.15 kWh
P = VI1.15 kW

Worked example

Uses the values in Fig. 1.2-1. Change them and the working updates.

Given
  • Vsupply voltage230 V
  • Icurrent drawn5 A
  • ttime switched on1 h
Find

The power P, and the energy used in time t.

Solution
P = VI = 230 V × 5 A = 1,150 W E = Pt = 1.15 kW × 1 h = 1.15 kWh
Answer
P = 1.15 kW

The equivalent resistance is V/I = 46 Ω, so P = I²R gives the same 1.15 kW.

Related forms

  • Current and resistance

    P = I²R

  • Voltage and resistance

    P = V²R

  • Three-phase supply

    P = √3 VLIL cos φ

Assumptions
  • Direct current, or AC with RMS values and a resistive load (power factor 1).
  • For loads with a power factor below 1, such as motors, multiply by cos φ to get the real power.
  • Voltage and current stay steady for the whole time t.

Derivation
  1. Voltage is energy per unit charge: V = E/q.
  2. Current is charge per unit time: I = q/t.
  3. Multiply them: VI = E/t, which is power.
  4. With Ohm's law, V = IR, this becomes P = I²R = V²/R.

One kilowatt-hour is 1 kW for one hour, or 3.6 MJ.