Electrical power
P = U · I
| Quantity | Symbol | Unit |
|---|---|---|
| Power | P | W |
| Voltage | U | V |
| Current | I | A |
Elec Class
learn electricity hands-on
Lesson · chapter 5 of 15
Electric power is the voltage multiplied by the current: P = U × I, in watts. Multiplied by the running time, it gives the energy used, W = P × t, which the supplier bills in kilowatt-hours.
The result is worked out before your eyes, line by line, just like on the board.
1I'm looking for
2I enter the known values
Formula used P = U · I
3The result
Power P2.3 kW
At 230 V, an appliance carrying 10 A draws a power of 2.3 kW.
The definition, the key points to remember and everyday examples.
Power P = U·I is the energy exchanged per second, in watts. In a resistor it is entirely turned into heat (Joule heating): P = R·I² = U²/R.
Energy W = P·t is measured in joules; the utility bills in kilowatt-hours (1 kWh = 3.6 × 10⁶ J). A machine absorbs a power Pa and returns a useful part Pu: its efficiency is η = Pu / Pa, always below 1.
Power is the effort you put into each push on the pedals; energy is the tiredness built up by the end of the ride. An amateur cyclist can keep up about 150 W: in one hour, that's 0.15 kWh, barely 5 cents of electricity at €0.30/kWh. Where it falls short: your body is about 25 % efficient, so it burns far more energy than it delivers to the pedals.
| Quantity | Symbol | Unit |
|---|---|---|
| Power | P | W |
| Voltage | U | V |
| Current | I | A |
| Quantity | Symbol | Unit |
|---|---|---|
| Power | P | W |
| Resistance | R | Ω |
| Current | I | A |
| Quantity | Symbol | Unit |
|---|---|---|
| Energy | W | J |
| Power | P | W |
| Time | t | s |
In kWh: power in kW × time in hours.
| Quantity | Symbol | Unit |
|---|---|---|
| Efficiency | η | - |
| Useful power | Pu | W |
| Absorbed power | Pa | W |
Four steps to solve a problem, then an example solved line by line.
Power (W), energy (J or kWh) or efficiency (no unit)? "For 3 hours" almost always signals an energy.
P = U × I if you have U and I; P = R × I² or U² / R if you have the resistance; W = P × t for energy.
kW and hours give kWh; W and seconds give joules. Never mix the two.
An efficiency lies between 0 and 1; a 16 A socket at 230 V supplies no more than 3,680 W.
A 2 000 W heater runs at 230 V. What current does it draw, what is its resistance? It heats 3 h a day for 30 days: what energy, and what cost at €0.30/kWh? A motor draws 1 500 W and delivers 1 200 W: what efficiency?
I = P / U = 2 000 / 230 = 8.7 A. R = U² / P = 230² / 2 000 = 26.5 Ω.
W = P · t = 2 kW × 90 h = 180 kWh, i.e. 180 × 0.30 = €54 (in joules: 180 × 3.6 × 10⁶ = 6.48 × 10⁸ J).
η = Pu / Pa = 1 200 / 1 500 = 0.8, i.e. 80 %; the remaining 300 W are lost as heat.
Multiply its power in kW by its time of use in hours. A 1,800 W hair dryer used 10 minutes a day: 1.8 × 10 / 60 = 0.3 kWh a day, or 9 kWh over 30 days.
Annual energy in kWh × price per kWh. A 100 W TV on 4 h a day: 0.1 × 4 × 365 = 146 kWh, or about €44 at €0.30/kWh.
I = P / U. A 3,000 W oven at 230 V draws 3,000 / 230 ≈ 13 A: it needs a socket on a circuit protected at 16 A or more.
Because of the Joule effect: its resistance turns a power P = R × I² into heat. Since the current is squared, doubling the current multiplies the heating by 4; a coiled extension cord, which gets rid of this heat poorly, can melt.
It's its rated power, the power it draws at its rated voltage (230 V for a household appliance). A "6 V, 3 W" lamp connected to 6 V draws 3 W; at a lower voltage, it draws less.
Some energy is always lost: Joule heating in the windings, friction, magnetic losses in the iron. A small motor is often 50 to 80 % efficient, a large industrial motor over 95 %.
The kilowatt is a power (what the appliance draws at each instant), the kilowatt-hour an energy (what it has consumed after some time). A 1 kW appliance running for 1 h uses 1 kWh.
They are two ways of writing P = U·I, replacing U by R·I or I by U/R (Ohm's law). Choose the one whose data you know.
From the simplest to the most complete, each problem comes with a step-by-step solution.
Have students note at home the power of five appliances (label or manual) and how long they are used each day, then calculate their annual energy use and its cost in class; in the lab, check P = U × I on a 6 V, 3 W lamp.
Using Elec Class in class