Physics · 4 Electricity and magnetism

4.2.5 Electrical energy and electrical power

Before you start

$I = Q / t$ and $V = W / Q$ from 4.2.2 and 4.2.3.
$R = V / I$ from 4.2.4.
That power is energy transferred per unit time, measured in watts.

Syllabus objectives covered (4)

Understand that electric circuits transfer energy from a source of electrical energy, such as an electrical cell or mains supply, to the circuit components and then into the surroundings
Recall and use the equation for electrical power P = IV
Recall and use the equation for electrical energy E = IVt
Define the kilowatt-hour (kW h) and calculate the cost of using electrical appliances where the energy unit is the kW h

Part 1 of 3

Where the energy goes

A circuit does not create energy or use it up. It transfers energy from a source to components, and from components into the surroundings.
The source is a cell, a battery or the mains supply. It transfers energy to the charge, which carries it round the circuit.
Each component takes some of that energy and transfers it to another form: a lamp to light and thermal energy, a motor to kinetic energy, a heater to thermal energy.
Eventually almost all of it ends up as thermal energy in the surroundings, spread out and no longer useful.
So the correct language is 'energy is transferred', not 'energy is used up' or 'lost'. Examiners mark on that wording, and it also happens to be what is physically true.

Part 2 of 3

Power and energy in a circuit

Power is the rate of energy transfer — how many joules per second. In a circuit it is given by:
$P = I V$
$P$ is in watts, $I$ in amps, $V$ in volts. One watt is one joule per second.
You can see where the equation comes from. Each coulomb carries $V$ joules, and $I$ coulombs pass every second, so $I V$ joules are transferred per second.
Energy is power multiplied by time, which gives the second equation:
$E = I V t$
$E$ is in joules and $t$ must be in seconds. As with $I = Q / t$ , a question quoting minutes or hours is testing the conversion.
Combining $P = I V$ with $V = I R$ gives $P = I^{2} R$ , which you met for power losses in cables. It is not a separate law, just a substitution.
Neither $P = I V$ nor $E = I V t$ is given to you in Paper 4. They must be recalled.

Part 3 of 3

The kilowatt-hour

The joule is inconveniently small for household electricity. Running a 2 kW heater for an hour transfers 7 200 000 J, which is an awkward number for a bill.
So electricity companies use a bigger unit. The kilowatt-hour is the energy transferred by a device of power 1 kW operating for 1 hour.
It is a unit of energy, not power, despite having 'watt' in the name. This trips people up constantly — 'kilowatt-hour' means kilowatts multiplied by hours.
The calculation is deliberately simple, but only if the units are right:
$energy (kW h) = power (kW) \times time (h)$
Power must be in kilowatts and time in hours. A 500 W appliance is 0.5 kW; 30 minutes is 0.5 h.
Then the cost follows directly:
$cost = energy (kW h) \times cost per kW h$
Notice the two equations use different unit systems. $E = I V t$ needs joules, watts and seconds; the kW h calculation needs kilowatt-hours, kilowatts and hours. Never mix them in one line of working.

Worked examples

Example 1. An electric heater is connected to a $230$ V mains supply and draws a current of $8.0$ A. Calculate its power.

Show the working

Current and p.d. are given and power is wanted, so use $P = I V$ directly.
Substitute.
$P = 8.0 \times 230 = 1840 W$
Give the unit — the watt.
It is often worth also quoting the value in kilowatts, $1.84$ kW, since a later part will usually want the cost.

Answer: $P = 1840$ W, or $1.84$ kW.

Example 2. A lamp carries a current of $0.50$ A at a p.d. of $12$ V for $4.0$ minutes. Calculate the energy transferred.

Show the working

Energy is wanted from current, p.d. and time, so use $E = I V t$ .
Convert the time to seconds first: $4.0 \times 60 = 240$ s.
Substitute all three values.
$E = 0.50 \times 12 \times 240 = 1440 J$
The unit is the joule. Since the time was in seconds, the answer is in joules — the units are consistent.

Answer: $E = 1440$ J.

Example 3. A $2.5$ kW immersion heater is used for $45$ minutes each day. Electricity costs $18$ cents per kW h. Calculate the cost of using the heater for one week.

Show the working

This is a kW h question, so work in kilowatts and hours throughout and do not convert to joules.
Convert the time: $45$ minutes $= 45 / 60 = 0.75$ h.
Energy per day = power × time.
$E = 2.5 \times 0.75 = 1.875 kW h$
For a week, multiply by 7.
$E = 1.875 \times 7 = 13.125 kW h$
Now the cost: multiply the energy by the price per kW h.
$cost = 13.125 \times 18 = 236.25 cents$
Give the answer sensibly for money — $$ 2.36$ , and note that the accuracy rules do not require you to round money to 3 significant figures.

Answer: $13.125$ kW h in a week, costing $236$ cents, or about $$ 2.36$ .

Where people go wrong

Mistake: Treating the kilowatt-hour as a unit of power.

Instead: It is a unit of energy — kilowatts multiplied by hours.

The name contains 'watt', which suggests power, but the 'hour' is what turns it into energy.

Mistake: Using watts and seconds in a kW h calculation.

Instead: Use kilowatts and hours for kW h; use watts and seconds for joules.

Mixing the systems gives an answer out by a factor of 3 600 000, which is the most damaging error in this subtopic.

Mistake: Leaving time in minutes in $E = I V t$ .

Instead: Convert to seconds.

The joule is defined as a watt-second, so the equation only works with seconds.

Mistake: Saying a lamp 'uses up' or 'loses' energy.

Instead: The lamp transfers energy to light and thermal energy in the surroundings.

Energy is conserved. The syllabus objective is specifically about transfer, and the wording is marked.

Mistake: Converting a 500 W appliance to 500 kW, or forgetting to convert at all.

Instead: 500 W is 0.5 kW — divide by 1000.

The prefix kilo means a thousand, so the number in kilowatts is smaller than the number in watts.

Check yourself

State the equation for electrical power and the unit of each quantity.
Answer
$P = I V$ , with $P$ in watts, $I$ in amps and $V$ in volts.
A device draws $3.0$ A from a $12$ V supply. What is its power?
Answer
$P = I V = 3.0 \times 12 = 36$ W.
How much energy does that device transfer in $2.0$ minutes?
Answer
$t = 120$ s, so $E = I V t = 3.0 \times 12 \times 120 = 4320$ J.
Define the kilowatt-hour.
Answer
The energy transferred by a device of power 1 kW operating for 1 hour.
A $100$ W lamp is left on for $10$ hours. How many kilowatt-hours does it use?
Answer
$100$ W is $0.1$ kW, so energy $= 0.1 \times 10 = 1.0$ kW h.
Describe the energy transfers when a battery lights a lamp.
Answer
The battery transfers energy to the charge in the circuit; the lamp transfers that energy to light and thermal energy, which pass into the surroundings.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. An electric kettle is rated at $2.2$ kW and is connected to a $230$ V mains supply.

(a) Calculate the current drawn by the kettle. 3 marks

Mark scheme

C1 2.2 kW converted to 2200 W
C1 P = IV rearranged to I = P / V
A1 9.6 A (accept 9.57); unit required

(b) The kettle is used for $3.0$ minutes. Calculate the energy transferred, in joules. 3 marks

Mark scheme

C1 time converted to 180 s
C1 E = Pt or E = IVt used
A1 396 000 J (accept 3.96 × 10⁵ J); unit required

(c) The kettle is used for a total of $30$ minutes a day. Electricity costs $20$ cents per kW h. Calculate the daily cost. 3 marks

Mark scheme

C1 time converted to 0.5 h
C1 energy = 2.2 × 0.5 = 1.1 kW h
A1 22 cents; unit required

Question 2. A student compares two lamps: lamp A is rated $60$ W and lamp B is rated $11$ W. Both produce a similar amount of light.

(a) Explain, in terms of energy transfer, why lamp A costs more to run. 2 marks No calculator

Mark scheme

B1 lamp A transfers more energy per second / has a higher power
B1 more of the energy is transferred to thermal energy in the surroundings rather than to light, owtte

(b) Calculate the energy, in kW h, saved in $1000$ hours by using lamp B instead of lamp A. 3 marks

Mark scheme

C1 power difference = 49 W = 0.049 kW
C1 energy = power × time used
A1 49 kW h; unit required

Question 3. A $12$ V motor draws a current of $2.5$ A.

(a) Calculate the power of the motor. 2 marks

Mark scheme

C1 P = IV used
A1 30 W; unit required

(b) Determine how long the motor must run to transfer $5400$ J. 2 marks

Mark scheme

C1 E = Pt rearranged to t = E / P; ecf from (a)
A1 180 s (accept 3.0 minutes); unit required

(c) State one energy transfer that takes place in the motor other than the useful one. 1 mark No calculator

Mark scheme

B1 electrical energy to thermal energy (accept: to sound) in the motor and surroundings

Words to know

Power: The rate of energy transfer, measured in watts. In a circuit, $P = I V$ .
Watt (W): One joule per second.
Kilowatt-hour (kW h): The energy transferred by a device of power 1 kW operating for 1 hour. A unit of energy, not power.
Energy transfer: The movement of energy from a source, through components, and into the surroundings. Energy is never created or destroyed.

All Physics lessons Next: 4.3.1 Circuit diagrams and circuit components