Electricity and magnetism · GCSE Physics
Electrical power
Teacher-written GCSE Physics revision on electrical power: P = VI, P = I²R and P = V²/R, energy E = Pt = VIt, and how heating in cables links power to the national grid.
Electrical power is the rate of energy transfer: P = VI. One watt is one joule per second. Heating in a resistor is I²R — that is why high current wastes energy in cables.
The important bits
What you need to know
- 1
Power P = VI, with P in watts, V in volts, I in amperes. This is the headline electrical power equation.
- 2
Combining with V = IR gives P = I²R and P = V²/R. Use I²R when you know current and resistance; use V²/R when you know potential difference and resistance.
- 3
Energy transferred electrically E = Pt = VIt, in joules if t is in seconds. On bills, energy in kWh = power in kW × time in hours.
- 4
A fuse rating is chosen a little above the normal operating current: I = P/V. A 230 V, 500 W lamp needs about 2.2 A, so a 3 A fuse, not 13 A.
- 5
Heating in cables is I²R. For a given power from a power station, raising V lowers I, which cuts heating losses. That is the physics of the national grid.
- 6
Appliances convert electrically transferred energy into other stores: thermal in a heater, kinetic in a motor, light and thermal in a lamp. Efficiency still applies.
- 7
Power is not the same as potential difference. A 12 V car headlamp can have a high current and therefore a high power; a 230 V phone charger may draw little current.
- 8
Check units: milliamperes to amperes, kilowatts to watts, minutes to seconds, unless you are deliberately using kWh.
Quotations worth analysing
Short evidence. Real method.
“P = VI = I²R = V²/R”
Three costumes of the same idea. Pick the one that matches the quantities you have, then substitute with SI units.
“Energy transferred = power × time”
Time in seconds gives joules. Time in hours with power in kilowatts gives kilowatt-hours on a bill.
Go deeper
Choose the form of P that matches the data
If the question gives the current through a heating element and its resistance, P = I²R is faster than finding V first. If it gives mains voltage and the resistance of the element, P = V²/R avoids an extra step. P = VI is the safest when you have a rating plate: 230 V, 2.0 kW means I = P/V = 2000 / 230 ≈ 8.7 A, so a 13 A fuse is appropriate and a 3 A fuse would blow in normal use. Write the form you are using, substitute, then check the size: a phone is a few watts, a kettle about 2 kW, an immersion heater 3 kW. An answer of 2 000 000 W for a kettle means you used 230 V and 13 A without converting, or mixed kW and W.
Go deeper
I²R is the thread that ties fuses to the grid
Heating in a wire is I²R. A fuse is a thin wire designed to melt at a stated current, breaking the live connection. The grid raises V so that for the same power, I falls, so I²R losses in the cables fall. That is why transmission is at hundreds of kilovolts, not at 230 V. Energy transferred to a home is still P × t; the bill uses kilowatt-hours because a joule is too small to price. None of this changes the circuit rules: live to the fuse and switch, then to the appliance. In evaluations, “high voltage is dangerous” is true at the pylons, but the reason for high voltage is lower current and less heating, not danger for its own sake.
See the idea in action
A 230 V heater has a resistance of 21.1 Ω. Power P = V²/R = (230)² / 21.1 ≈ 2500 W, or 2.5 kW. Current I = V/R = 230 / 21.1 ≈ 10.9 A, so a 13 A fuse is suitable. Energy transferred in 15 minutes: t = 900 s, E = Pt = 2500 × 900 = 2.25 × 10⁶ J. In kilowatt-hours: 2.5 kW × 0.25 h = 0.625 kWh. If the same 2.5 kW were transmitted along a cable at only 230 V instead of 400 kV, the current would be much larger and I²R heating in the cable would waste far more of the generated energy.
Exam technique
Turn knowledge into marks
Show P = VI or I²R or V²/R with units, then E = Pt with time in seconds. For fuse choice, calculate I = P/V and pick the next standard fuse above that current. For the grid, say higher V, lower I, less I²R heating.
Common mistakes
Do not give these marks away
- 01
Using P = VI with V in kilovolts and I in milliamperes without converting.
- 02
Choosing a 13 A fuse for every appliance, including a 2 A lamp.
- 03
Leaving time in minutes in E = Pt, or quoting power in joules.
A 12 V heater draws 4.0 A. What is its power, and how much energy does it transfer in 30 s?
A3.0 W and 90 J
B48 W and 1440 J
C48 W and 1.6 J
D16 W and 480 J
Show the answer
48 W and 1440 J. P = VI = 12 × 4.0 = 48 W. E = Pt = 48 × 30 = 1440 J. 3.0 W is V/I; 16 W is a confused mix of numbers.
Quick questions
If this is the bit you searched
Which power equation should I use?
P = VI if you have potential difference and current. P = I²R if you have current and resistance. P = V²/R if you have potential difference and resistance.
How do you choose a fuse?
Find the normal current from I = P/V, then choose the next standard fuse above that value (3 A, 5 A or 13 A in many UK plugs) so it does not blow in normal use but still protects the cable.
Why does a high current heat a cable?
Power dissipated as heating is I²R. Doubling the current quadruples the heating in the same resistance.
What is the difference between a joule and a kilowatt-hour?
Both are energy. 1 kWh = 3.6 million J. Use joules with watts and seconds; use kWh with kilowatts and hours on bills.