Potential difference, power and e.m.f.
Key idea: H2 Physics lessons on drift velocity, electrical energy, sinusoidal supplies and half-wave rectification.
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The core idea
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Big question: What does each coulomb gain and lose around a circuit?
Potential difference is energy transferred from each coulomb in a component, V = W/Q; e.m.f. is energy supplied per coulomb by a source. Electrical power P = VI is the transfer rate. Following one coulomb around a complete loop makes conservation clear: source gains equal component transfers.
Follow energy carried by each coulomb
Potential difference is energy transferred from electrical form per unit charge in a component: V = W/Q. E.m.f. is energy supplied per unit charge by a source. Both are measured in volts, but they describe opposite sides of the energy account.
Charge is not used up as energy is transferred. Following one coulomb around a complete circuit keeps charge flow separate from the joules supplied or transferred at each component.
Check your understanding: A source supplies 18 J to 3.0 C. What is its e.m.f.?
ε = W/Q = 18/3.0 = 6.0 V.
Derive and choose the electrical power equations
Power is energy transferred per unit time. Since V = W/Q and I = Q/t, P = W/t = VI. For a resistor at the stated operating point, V = IR then gives P = I²R and P = V²/R.
Every voltage and current must refer to the same component. The three power forms are equivalent only when the V, I and R describe that one operating point.
Check your understanding: A 6.0 Ω resistor carries 2.0 A. Find its power in two ways.
V = IR = 12 V, so P = VI = 24 W and P = I²R = 24 W.
Key ideas to keep
- E.m.f. is not a force and is measured in volts.
- Potential difference describes transfer per charge, not charge flow rate.
- Choose P = VI, I²R or V²/R only when its variables describe the same component.
See the reasoning
Worked example
Connect voltage, current, power and energy
Question: A 12 V heater takes 2.5 A. Find its power, resistance and energy transferred in 3.0 min.
Step 1: Find the transfer rate
Why: The p.d. and current refer to the same heater.
Working: P = VI = 12(2.5) = 30 W.
Step 2: Find operating resistance
Why: The heater's V and I define its resistance at this point.
Working: R = V/I = 12/2.5 = 4.8 Ω; I²R also gives 30 W.
Step 3: Convert power to energy
Why: Power is energy per second and the time must be in seconds.
Working: t = 180 s, so W = Pt = 30(180) = 5.4×10³ J.
Answer: Power is 30 W, resistance is 4.8 Ω and energy transferred is 5.4 kJ.
Check: V²/R = 12²/4.8 = 30 W, confirming the same operating point.
Use a hint if needed
Practise with support
Try this
A 6.0 Ω resistor carries 2.0 A. Find p.d. and power using two power forms.
Hint: Use Ohm's law only to connect the supplied operating values.
Check your answer
V = IR = 12 V. P = VI = 24 W and I²R = 24 W.
Now work without the hint
Practise independently
Your turn
Distinguish e.m.f. from p.d. using energy, then derive all three resistor-power forms.
Check your answer
E.m.f. is energy supplied by a source per unit charge; p.d. is energy transferred from electrical form per unit charge in a component. P = W/t = (W/Q)(Q/t) = VI. With V = IR, P = I²R = V²/R.
Avoid these traps
Common mistakes
Common mistake
E.m.f. and p.d. are interchangeable names for voltage.
What is wrong with this reasoning?
Show better thinking
Both have units J C⁻¹, but e.m.f. describes energy supplied per charge and p.d. describes energy transferred from electrical form per charge.
Common mistake
Current is used up when a component transfers energy.
What is wrong with this reasoning?
Show better thinking
Charge flow is conserved at a steady junction; the component transfers energy, not current.
Write for the examiner
Exam guidance
Name the component and energy transfer before substituting into a power equation.
Exam-style practice [7 marks]
A cell supplies 9.0 J to 1.5 C. A lamp transfers 6.0 J from that charge in 3.0 s. Find the cell e.m.f., lamp p.d., current and lamp power. Explain the energy distinction between e.m.f. and p.d.
Plan before you answer
- Divide each energy by the same charge.
- Use charge per time for current.
- Calculate power and state both definitions.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
The cell e.m.f. is 9.0/1.5 = 6.0 V and the lamp p.d. is 6.0/1.5 = 4.0 V. Current is 1.5/3.0 = 0.50 A, so lamp power is VI = 4.0(0.50) = 2.0 W, also 6.0/3.0. E.m.f. describes energy supplied by a source per coulomb; p.d. describes energy transferred from electrical form per coulomb in a component.
Come back in three days
Check what stayed with you
Recall question 1
Define potential difference using energy.
Check the answer
Energy transferred from electrical form per unit charge.
Recall question 2
Define e.m.f. using energy.
Check the answer
Energy supplied by a source per unit charge.
Recall question 3
State the three resistor power equations.
Check the answer
P = VI = I²R = V²/R for one stated operating point.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Topic 15 states no explicit exclusions. q in I = nAvq is treated as carrier-charge magnitude when calculating current magnitude; electron drift is opposite conventional current. E.m.f. is energy supplied per unit charge, while p.d. is energy transferred from electrical form per unit charge. The peak/√2 and half-maximum-power results are restricted to sinusoidal waveforms and a resistive load; a single ideal diode gives unsmoothed half-wave rectification.
- GCE A-Level H2 PhysicsTopic 15(c) / Topic 15(d) / Topic 15(e) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027