3.5.1.4
Electrical power
Circuits & Potential Dividers — AQA A-Level Physics
Key Definition
Electrical power — The rate of energy transfer by an electrical component. Measured in watts (W).
- PowerThe rate of energyThe capacity to do work. Measured in joules (J). transfer. Measured in watts (W). can be calculated three ways, depending on known quantities.
- All three forms come from combining $P = IV$ with $V = IR$.
$$P = IV$$
$$P = I^{2}R$$
$$P = \frac{V^{2}}{R}$$
- Use $P = IV$ when you know currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). and p.d.
- Use $P = I^2R$ when you know currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). and resistanceThe opposition to currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). flow. The ratio of potential difference to current. Measured in ohms (Ω)..
- Use $P = V^2/R$ when you know p.d. and resistanceThe opposition to current flow. The ratio of potential difference to current. Measured in ohms (Ω)..
Key Definition
Energy transferred — The total energy delivered to a component over time t. Measured in joules (J).
$$E = IVt$$
- The kilowatt-hourA unit of energyThe capacity to do work. Measured in joules (J). equal to 3.6 MJ. The energyThe capacity to do work. Measured in joules (J). transferred by a 1 kW device in 1 hour. (kWh) is a unit of energy used by electricity suppliers.
- 1 $kWh = energy transferred$ by a 1 kW device running for 1 hour.
- 1 $kWh = 3.6 x 10^6 J$.
$$1\;\text{kWh} = 3.6 \times 10^{6}\;\text{J}$$
Worked Example
A 47 ohm resistor carries a current of 0.20 A. Calculate the powerThe rate of energy transfer. Measured in watts (W). dissipated.
Show Solution
1
Known
$I = 0.20\;\text{A}, R = 47 \;\Omega.$
2
Use $P = I^2 R = (0.20)^2 x 47$.
3
$P = 0.04 \times 47 = 1.88\;\text{W}.$
4
$P = 1.9\;\text{W} (2\;\text{s.f.}).$
Answer
1.9 W
Common Mistake
MEDIUM
Students often: Confusing $P = I^2R$ and $P = V^2/R$ by swapping R and 1/R.
Instead: Write out $V = IR first. Substitute$ into $P = IV$ to derive whichever form you need.
Instead: Write out $V = IR first. Substitute$ into $P = IV$ to derive whichever form you need.
Examiner Tips and Tricks
- AQA frequently asks you to 'show that' the powerThe rate of energy transfer. Measured in watts (W). dissipated is a given value.
- You must show the substitution clearly, not just state the formula.