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Potential Difference, E.M.F. & Electrical Power

Key idea: Distinguish potential difference from e.m.f. using energy transferred per unit charge, then apply P = VI, P = I²R and P = V²/R.

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Apply potential difference, e.m.f. and electrical power relationships.

1. Energy transferred per unit charge

Potential difference and electromotive force are both measured in volts, where

1 V = 1 J C⁻¹.

They describe different energy transfers:

  • E.m.f., E: energy transferred from non-electrical forms to electrical energy by a source per unit charge passing through it.
  • Potential difference, V: electrical energy transferred to other forms per unit charge passing through a component.

In each case,

E = W_source/Q, V = W_component/Q.

E.m.f. is not a force

Despite its name, e.m.f. is an energy-per-charge quantity. Its unit is the volt, not the newton.

2. Reading the energy story

Follow one coulomb of charge around a circuit:

  1. The source supplies energy to it. A source of e.m.f. 6.0 V supplies 6.0 J per coulomb.
  2. Components transfer electrical energy to heating, light, motion or other forms. A p.d. of 4.0 V across a lamp means 4.0 J is transferred per coulomb in the lamp.
  3. Current is not “used up”. Charge continues around the circuit while energy is transferred.

For an ideal source connected directly to a component, the source e.m.f. and component p.d. can have the same numerical value. Their meanings remain different.

3. Electrical power

Power is energy transferred per unit time:

P = W/t.

Using V = W/Q and I = Q/t,

P = (W/Q)Q/t = VI.

For a resistor that obeys V = IR:

P = VI = I²R = V²/R.

Choose the form that uses the quantities given. The unit of power is the watt (W), equal to J s⁻¹.

4. Common Mistakes

  • Defining e.m.f. as “force per unit charge”. That is electric field strength, not e.m.f.
  • Saying current or charge is consumed by a resistor. Energy is transferred; charge is conserved.
  • Using a power form that introduces an unnecessary unknown.
  • Mixing energy in joules with power in watts.

5. Exam Tips

  • State whether energy is supplied to or transferred from the electrical store.
  • Translate volts as joules per coulomb when distinguishing e.m.f. and p.d.
  • Choose the power relation containing the quantities already given.

6. Worked Examples

Modelled example 1

Distinguish e.m.f. from p.d.

Core

Problem

A cell supplies 18 J of electrical energy for 3.0 C. A lamp transfers 12 J from electrical energy for the same charge. Find cell e.m.f. and lamp p.d., then distinguish their meanings.
Study the worked solution

  1. Source energy per charge

    Method

    Divide the energy supplied by charge.

    Reason

    E.m.f. is energy supplied to electrical form per coulomb.

    Working

    E = 18/3.0 = 6.0 V

  2. Component energy per charge

    Method

    Divide the lamp’s transferred energy by charge.

    Reason

    P.d. is electrical energy transferred to other forms per coulomb.

    Working

    Vₗₐₘₚ = 12/3.0 = 4.0 V

Guided practice 2

Electrical energy in kilowatt-hours and joules

About 5 min

Problem

An 80.0 W television operates for 4.00 h. Calculate the energy transferred in kW h and in MJ.

Try this before viewing the solution

Unit: kW h
Unit: MJ

Hints

Hint 1: match the power and time units
Use kilowatts with hours for an answer in kW h.
Hint 2: convert the energy unit
Use 1.00 kW h = 3.60 MJ.
View solution step by step

  1. Express power in kilowatts

    Method

    Convert 80.0 W to 0.0800 kW.

    Reason

    A kilowatt-hour calculation requires power in kilowatts and time in hours.

    Working

    80.0 W = 0.0800 kW

  2. Calculate kilowatt-hours

    Method

    Multiply power by operating time.

    Reason

    Energy equals power multiplied by time.

    Working

    E = (0.0800)(4.00) = 0.320 kW h

  3. Convert to megajoules

    Method

    Use the joule equivalent of one kilowatt-hour.

    Reason

    1.00 kW h = (1000 W)(3600 s) = 3.60 MJ.

    Working

    E = (0.320)(3.60) = 1.15 MJ

Common misconception 3

E.m.f. is not a force

Find and correct the mistake

Learner definition

A learner defines e.m.f. as “the force exerted by a cell per coulomb”. Locate the first error and write the correct energy definition with its unit.

Try this before viewing the solution

First error

View solution step by step

  1. Replace force with energy

    Method

    Define e.m.f. as energy supplied by a source to electrical form per unit charge.

    Reason

    The source raises the electrical energy of the charge passing through it.

    Working

    E = W_supplied/Q.

  2. State the unit

    Method

    Use volts, equivalent to joules per coulomb.

    Reason

    The numerator is energy, not mechanical force.

    Working

    1 V = 1 J C⁻¹.

Examiner practice 4

Select the quickest resistor power formula

3 marks

Examination question

A 22 Ω resistor has 12 V across it. Calculate power dissipation and justify the selected relation. [3 marks]

Try this before viewing the solution

View solution step by step

  1. Select relation

    1 mark

    Method

    Use P = V²/R.

    Reason

    Voltage and resistance are supplied directly.

    Working

    P = V²/R.

  2. Substitute

    1 mark

    Method

    Insert 12 V and 22 Ω.

    Reason

    The values are already in SI units.

    Working

    P = 12²/22.

  3. Evaluate

    1 mark

    Method

    Obtain 6.5 W.

    Reason

    Power is an energy-transfer rate.

    Working

    P = 6.5 W.

Challenge 5

From battery capacity to energy and power

Minimal support

Independent transfer

An emergency-lighting battery is rated 24.0 V and 140 A h. Treat its voltage as constant and its internal resistance as negligible. It discharges steadily in 14.0 h. Calculate the charge delivered, energy delivered, external resistance and output power.

Try this before viewing the solution

Hints

Hint 1: interpret ampere-hour
An ampere-hour is charge: multiply by 3600 s to express it in coulombs.
Hint 2: find the operating current
A steady 140 A h discharge over 14.0 h gives the current directly.
View solution step by step

  1. Convert capacity to charge

    Method

    Express the ampere-hour rating in coulombs.

    Reason

    1 A = 1 C s⁻¹, so one ampere-hour contains 3600 C.

    Working

    Q = (140)(3600) = 5.04 × 10⁵ C

  2. Find delivered energy

    Method

    Use energy per charge.

    Reason

    The constant-voltage idealisation permits E = QV over the discharge.

    Working

    E = (5.04 × 10⁵)(24.0) = 1.21 × 10⁷ J

  3. Find current and resistance

    Method

    Use the capacity per discharge time, then Ohm’s law.

    Reason

    The stated discharge is steady and the ideal source places 24.0 V across the load.

    Working

    I = 140/14.0 = 10.0 A, R = 24.0/10.0 = 2.40 Ω

  4. Find and check power

    Method

    Calculate VI and compare Pt with QV.

    Reason

    Both routes must describe the same energy transfer under the constant-power model.

    Working

    P = (24.0)(10.0) = 240 W; Pt = (240)(14.0 × 3600) = 1.21 × 10⁷ J

7. Mind Stretchers

Mind stretcher 1: Why can e.m.f. and p.d. share a unit?Extension

Explain why e.m.f. and p.d. are both measured in volts even though one describes a source and the other a component.

Show Answer

Both are energy transferred per unit charge, so both have unit J C⁻¹ = V. The distinction is the direction of the energy transfer: a source supplies electrical energy, while a component transfers electrical energy to other forms.

Next: Sinusoidal Alternating Current & Voltage

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Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027