Electrical power

Key idea: Electrical power is the rate at which a component transfers energy. P = VI is general; the resistance forms follow when V = IR applies at that operating point.

GCE A-Level H1 Physics 2027

H1 Physics 8867 · Lesson 3 of 3

Check your understanding

By the end of this lesson, you should be able to

Use P = VI for electrical energy transfer.
Select P = I²R or P = V²/R for a resistive component.
Check that voltage, current and resistance refer to the same operating condition.

Learn the idea

Big question: How do charge flow rate and energy per charge combine to give electrical power?

Derive P = VI

Each coulomb transfers V joules and I coulombs pass each second. Multiplying gives VI joules per second, so P = VI.

The voltage must be across the same component through which the stated current flows. Using source e.m.f. with a component current can give source power, not necessarily the component's useful transfer rate.

Check your understanding: A component has 5 V across it and 2 A through it. What does 10 W mean?

It transfers 10 J of energy each second.

Choose resistance forms with care

Substituting V = IR gives P = I²R or P = V²/R. These forms use the resistance at the stated operating condition; a hot filament may not have its cold resistance.

A power rating describes a particular operating voltage. Changing voltage changes current and power, often strongly because power contains a square in the resistive forms.

Check your understanding: At fixed resistance, what happens to power if voltage doubles?

It becomes four times as large because P = V²/R.

Key ideas

P = VI uses the p.d. across and current through the same component.
Choose the form with the two known quantities.
Power is measured in watts, where 1 W = 1 J s⁻¹.

Relationships to know

P = VI
P = I²R
P = V²/R

Follow the reasoning

Worked example

Use a rating to predict changed power

Question: A resistive heater is rated 1.15 kW at 230 V. Assume its resistance stays constant. Find its operating current, resistance and power at 200 V.

Step 1: Use the rated operating point
Why: Voltage and power at the rating determine current and hot resistance.
Working: I = P/V = 1150/230 = 5.00 A; R = V²/P = 230²/1150 = 46.0 Ω.
Step 2: Apply the constant-resistance assumption
Why: The lower-voltage power can use the same 46.0 Ω only because the question permits it.
Working: P = 200²/46.0 = 870 W.
Step 3: Compare sensibly
Why: Lower voltage must give lower current and power for fixed positive resistance.
Working: Pnew/Prated = (200/230)² = 0.756.

Answer: Rated current is 5.00 A, resistance 46.0 Ω and power at 200 V about 870 W.

Check: The 13% voltage decrease produces about a 24% power decrease because of the square relationship.

Now try it with support

Practise with support

A resistor has 9.0 V across it and transfers 18 W. Find its current and resistance.

Hints

Use P = VI first.
Then use either V = IR or V²/P.

View the guided answer

I = 18/9.0 = 2.0 A. R = 9.0/2.0 = 4.5 Ω.

Your turn

Practise independently

A heater rated 230 V, 1.15 kW is resistive. Calculate its current and operating resistance.

Check your answer

I = P/V = 1150/230 = 5.00 A. For a resistive heater, R = V²/P = 230²/1150 = 46.0 Ω.

Common mistakes and exam guidance

Watch out for

Using V²/R when the quoted resistance is not for the same operating temperature.
Substituting source e.m.f. instead of the p.d. across the component.

In an exam

Write the chosen power equation before substituting so the known quantities are visible.
Use a quick cross-check with a second form when V, I and R are all available.

Put the ideas together

Exam-style practice [6 marks]

A motor has 24 V across it and takes 3.5 A. It delivers 62 W of useful mechanical power. Calculate electrical input power, efficiency, useful energy in 4.0 min and energy transferred to less useful stores.

Plan before you answer

Find electrical power using the motor's own V and I.
Use the useful-to-input ratio.
Convert time before calculating energies.

View the marking points and model answer

Marking points

Finds input power 84 W.
Finds efficiency 62/84 = 73.8%.
Uses 240 s.
Finds useful energy 1.488 × 10⁴ J.
Finds less useful power 22 W.
Finds less useful energy 5.28 × 10³ J.

Model answer

Input power = VI = 24(3.5) = 84 W. Efficiency = 62/84 = 0.738 = 73.8%. In 240 s, useful energy = 62(240) = 1.488 × 10⁴ J. The less useful transfer rate is 84 − 62 = 22 W, so that energy is 22(240) = 5.28 × 10³ J.

Finish from memory

Three-question recap

Derive P = VI in one sentence.
Check
V joules are transferred per coulomb and I coulombs pass per second, giving VI joules per second.
When is P = V²/R useful?
Check
When p.d. and resistance at the same operating point are known.
What is the SI unit of power?
Check
The watt, W, equal to J s⁻¹.

Try this next

Use component I–V behaviour to decide when resistance is constant and when it changes.

Check the whole topic Return to Currents Review Potential difference and electromotive force

Continue with the next resource in this course.

Course and syllabus information

Course: GCE A-Level H1 Physics
Edition: GCE A-Level H1 Physics 2027