Electrical Power (AQA GCSE Combined Science: Synergy: Physical Sciences): Revision Note

Calculate the potential difference across a 48 W electric motor with a current of 4 A.

Answer:

Step 1: List the known quantities

• Power, P = 48 W

• Current, I = 4 A

Step 2: Write down the relevant equation

P = VI

Step 3: Rearrange for potential difference, V

$V=\frac{P}{I}$

Step 4: Substitute in the values

$V=\frac{48}{4}$ = 12 V

Remember: Power is just energy per second.

Always check the unit of power in the question. If it is given in kW or mW, convert it to W before you substitute, using 1 kW = 1000 W and 1 mW = 0.001 W. If you forget, your answer will be 1000 times too big or too small and you'll lose the final mark.

Match each number to its unit before you substitute. A 24 V power supply has a potential difference of 24 V, not a power of 24 W, so don't use it as the power in your calculation.

Calculate the power through a 20 Ω resistor with a current of 6 A through it.

Answer:

Step 1: List the known quantities

• Resistance, R = 20 Ω

• Current, I = 6 A

Step 2: Write down the relevant equation

P = I²R

Step 3: Substitute in the values

P = (6)² × 20 = 720 W

A kettle element has a power of 2420 W and a current of 11.0 A. Calculate its resistance.

Answer:

$R=\frac{P}{I^2}=\frac{2420}{{11.0}^2}=20\mathrm{\Omega }$

Both equations are given on the Physics Equations Sheet, so there is no need to memorise them. Pick the equation that uses the two quantities you are given: $P=VI$ for potential difference and current, $P=I^{2}R$ for current and resistance.

A common mistake is to square the resistance instead of the current, so make sure you square the current.

The equation $P=\frac{V^2}{R}$ is not on the Physics Equations Sheet, and it only works when the resistance stays the same. For a filament lamp the resistance rises as the current rises, so doubling the potential difference won't quadruple the power.