Electrical power and energy

Power in a circuit

$$P = IV$$

• $P$ — power (watts, W)

• $I$ — current (A)

• $V$ — potential difference (V)

• Power is the rate at which energy is transferred, so one watt is one joule per second.

• The relationship makes sense from the definitions: each coulomb carries $V$ joules, and $I$ coulombs pass every second, so $IV$ joules are transferred every second.

Combining with Ohm's law

Substituting $V = IR$ gives two more useful forms:

$$P = I^2R \qquad \text{and} \qquad P = \frac{V^2}{R}$$

• These are not on the resource sheet as separate formulae, but they follow in one step and are perfectly acceptable — show the substitution.

• Choose the form that matches what you were given:

  • know $I$ and $V$ → $P = IV$,
  • know $I$ and $R$ → $P = I^2R$,
  • know $V$ and $R$ → $P = V^2/R$.

• Note the squares. Doubling the current through a fixed resistor quadruples the power dissipated, which is why overloaded wiring heats up so dramatically.

Energy transferred

$$P = \frac{\Delta E}{t} \qquad \Rightarrow \qquad \Delta E = Pt = IVt$$

• $\Delta E$ — energy transferred (J)

• $t$ — time (s)

• Always convert the time to seconds before using this. Minutes and hours are common in questions and are the usual source of error.

Where the energy goes

• In a resistor or a filament lamp, electrical energy becomes heat (and light in the lamp).
  • The energy is transferred in the collisions between drifting electrons and the vibrating lattice ions.
• In a motor, most becomes kinetic energy, with some lost as heat.
• Total energy is always conserved — the electrical energy supplied by the battery equals the sum of the energy transferred in every component.

Power ratings

• A component's rating, such as "$60$ W, $230$ V", states the power it transfers at its stated operating voltage.
• Run it at a different voltage and the power changes, following $P = V^2/R$.
• The fuse rating needed is found from $I = P/V$, then choosing the next standard fuse above that current.