10.22–10.27Energy transfers in everyday appliances
Everyday electrical appliances are designed to bring about energy transfers. The amount of energy an appliance transfers depends on how long the appliance is switched on for and the power of the appliance.
Work is done when charge flows in a circuit. The amount of energy transferred by electrical work can be calculated using the equations:
energy transferred = power × time E = P t energy transferred = charge flow × potential difference E = Q V
energy transferred, E, in joules, J; power, P, in watts, W; time, t, in seconds, s; charge flow, Q, in coulombs, C; potential difference, V, in volts, V.
You should be able to recall and apply both equations.
You should be able to describe how different domestic appliances transfer energy from batteries or ac mains to the kinetic energy of electric motors, or the energy of heating devices.
You should be able to explain how the power of a circuit device is related to the potential difference across it and the current through it, and to the energy transferred over a given time. You should also be able to describe, with examples, the relationship between the power ratings for domestic electrical appliances and the changes in stored energy when they are in use.
The power rating printed on an appliance tells you how much energy it transfers each second. A 2 kW kettle transfers 2000 J every second, so it fills the thermal store of the water quickly; a 40 W lamp transfers 40 J every second. Over the same time, the higher-rated appliance brings about a much larger change in the stored energy.
Worked example
A 2000 W kettle is switched on for 3.0 minutes. Calculate the energy transferred.
- Convert the time to seconds: 3.0 × 60 = 180 s.
- Write the equation: E = P t.
- Substitute: E = 2000 × 180.
- = 360 000.
- Give the unit: joules.
Answer: 360 000 J, or 360 kJ
Worked example
A charge of 45 C flows through a lamp connected across a 12 V supply. Calculate the energy transferred.
- You are given charge and potential difference, so use E = Q V.
- Substitute: E = 45 × 12.
- = 540.
- Give the unit: joules.
Answer: 540 J
10.28–10.31Power
You should be able to explain how the power transfer in any circuit device is related to the potential difference across it and the current through it, and to the energy changes over time:
power = potential difference × current P = V I
power = current² × resistance P = I² R
power, P, in watts, W; potential difference, V, in volts, V; current, I, in amperes, A; resistance, R, in ohms, Ω.
You should be able to recall and apply both equations.
Which of the two to use is decided by what you are given. If you have the potential difference and the current, use P = V I. If you have the current and the resistance, use P = I² R. The two are consistent with each other, because V = I R — substituting that into P = V I gives P = I R × I, which is P = I² R.
The squared term in P = I² R matters. Doubling the current through a component quadruples the power dissipated in it, which is why a small increase in current can make a wire dangerously hot — and why the National Grid works the way it does.
Worked example
A kettle draws 9.0 A from the 230 V mains. Calculate its power.
- You are given potential difference and current, so use P = V I.
- Substitute: P = 230 × 9.0.
- = 2070.
- Give the unit: watts.
Answer: 2070 W, or about 2.1 kW
Worked example
A current of 3.0 A flows through a 15 Ω heating element. Calculate the power dissipated.
- You are given current and resistance, so use P = I² R.
- Square the current first: 3.0² = 9.0.
- Substitute: P = 9.0 × 15.
- Give the unit: watts.
Answer: 135 W
Common mistake
In P = I² R only the current is squared, and it must be squared before multiplying by the resistance. Writing (I × R)² gives a completely different — and much larger — answer.
Before you practise
- A current in a resistor transfers energy to its thermal store, because electrons collide with the ions in the lattice.
- Useful in a kettle, a toaster and a fuse; wasteful in a phone charger or a transmission cable.
- energy transferred = charge × potential difference, and charge = current × time.
- power = energy transferred ÷ time.
- power = current × potential difference, and power = (current)² × resistance.
- Pick the power equation that uses the two quantities the question gives you.
- Halving the current quarters the power dissipated in a cable — which is why the grid transmits at high voltage.