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Circuits

Electric power

Electric power is the rate of using electrical energy, P = VI. A bulb on 220 V drawing 0.5 A uses P = 220 × 0.5 = 110 W, and in one hour 0.11 kWh. One kWh is 3.6 × 10⁶ J.

Ohm's law

V = IR at a fixed temperature. Pick what to find and type the other two. Add a time to get the heat, H = I²Rt, and the energy in kWh.

Resistance R440 Ω
Power110 W
Heat396000 J
Energy0.11 kWh
  1. R = V / I = 220 / 0.5 = 440 Ω
  2. Power: P = VI = 220 × 0.5 = 110 W. The same as I²R and V²/R.
  3. Heat in 3600 s: H = I²Rt = 0.5² × 440 × 3600 = 396000 J
  4. In units of electrical energy: 396000 J / (3.6 × 10⁶ J per kWh) = 0.11 kWh
Derivation

Why it works

  1. The energy used in a time t is W = VIt, from the heating effect.
  2. Power is energy per unit time: P = W/t = VI.
  3. With V = IR, P = I²R. With I = V/R, P = V²/R.
  4. The SI unit is the watt: 1 W = 1 V × 1 A. 1 kW = 1000 W.
  5. The commercial unit of energy is the kilowatt hour: 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J.
Questions

Electric power: quick answers

What is 1 kWh?

The energy used by a 1000 W appliance in one hour: 1 kWh = 3.6 × 10⁶ J. It is the "unit" on an electricity bill.

What is the power of a bulb on 220 V drawing 0.5 A?

P = VI = 220 × 0.5 = 110 W. Its resistance while lit is R = V/I = 440 Ω.

Which power formula should I use?

All three are the same law. Use P = VI when you know the voltage and current, I²R in series (same current) and V²/R in parallel (same voltage).

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