The big idea: A kettle boils in a minute while a phone charger trickles energy in over hours — both draw electricity, but the kettle turns it into heat far faster. That rate of transferring electrical energy is its power.
Whenever a current flows through a component, the component transfers energy every second.
Unit: the watt (W) — 1 watt = 1 joule of energy every second.
A cell drives a current I through the resistor R. The voltage V across R times the current I gives the rate it turns electrical energy into heat: P = IV.
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P = I × V. Cover the one you want: two side by side → multiply (P = IV); one above the other → divide (I = P ÷ V, V = P ÷ I).
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Spot it: Bigger current or bigger voltage → more power.
The simplest form is P = I × V (current times voltage). A 12 V supply driving 2 A delivers 12 × 2 = 24 W.
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Power is current × voltage. Using Ohm's law (V = IR) you can swap V or I out, giving three equal forms of the same equation:
- electrical power (watts, W)
- current (amperes, A)
- potential difference / voltage (volts, V)
- resistance (ohms, Ω)
Which form do I use?: Choose the form using the two quantities you already know, so you don't have to find a third first:
| You know… | Use this form | Why |
|---|---|---|
| I and V | P = IV | both are given directly |
| I and R | P = I²R | no need to find V first |
| V and R | P = V²/R | no need to find I first |
Ohm's law links them: All three forms come from P = IV combined with V = IR (resistance). V = IR is also given in the data booklet.
- potential difference / voltage (volts, V)
- current (amperes, A)
- resistance (ohms, Ω)
A 24 Ω heater element is connected across a 12 V supply. Find the power it dissipates.
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How this is tested — power is usually a 'determine' question, and almost always a ratio, not a one-off number:
Paper 1A
- Compare the power of two components when something changes — a wire twice as long, or a switch opened/closed that changes V or I.
Paper 2
- Plug numbers into P = IV / I²R / V²/R.
- Find the energy E = Pt and the cost of running an appliance.
The classic trap: Picking the wrong form. If the voltage stays fixed, use P = V²/R (P ∝ 1/R); if the current stays fixed, use P = I²R (P ∝ R). They pull opposite ways.
Resistance of a wire: For a wire of the same metal and thickness, resistance is proportional to its length: a wire twice as long has twice the resistance (R ∝ L). Combine that with the right power form to get the ratio.
Two heating wires are the same metal and thickness, each connected across the same 6.0 V supply. Wire 2 is twice as long as wire 1. Determine the ratio of the power in wire 2 to that in wire 1.
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See the mark-by-mark plan — for / against / judgement, with marking guidance — in study mode.