9478 / 2027
Currents overview

Lesson 2 of 7

Energy per charge

  • I can use potential difference as electrical work done per unit charge, V = W/Q.Syllabus

    Syllabus 9478, 15(c). Use potential difference as electrical work done per unit charge, V = W/Q.

  • I can distinguish source e.m.f. and potential difference using energy transfers.Syllabus

    Syllabus 9478, 15(e). Distinguish source e.m.f. and potential difference using energy transfers.

Make a guess. You are not marked.

A cell has an e.m.f. of 1.5 V. What does this mean?

  1. The cell pushes a current of 1.5 A round any circuit it is connected to.
  2. Each coulomb gives 1.5 J to the external circuit, whatever the current.
  3. The cell transfers 1.5 J to each coulomb of charge passing through it.
Show the answer

The cell transfers 1.5 J to each coulomb of charge passing through it.

E.m.f. is the energy converted from other forms into electrical energy per unit charge passing through the source. 1.5 V means 1.5 J per coulomb.

Potential difference describes energy transferred per charge between two points. A source's e.m.f. describes the energy it supplies per charge, including energy subsequently transferred inside the source.

If electrical work W is done as charge Q passes through a load, the potential difference across its named terminals is:

V = W/Q   or   W = VQ
1 V = 1 J/C = kg m2 s-3 A-1

A load transferring 24.0 J as 4.00 C passes has p.d. 24.0/4.00 = 6.00 V. That is 6.00 J transferred for each coulomb, not 6.00 J regardless of how much charge passes. Current is through a section; voltage is between two points.

A potential level needs a reference

Electric potential is electric potential energy per unit positive test charge, relative to a chosen reference. Suppose point Q is assigned VQ = 0 and point P has VP = +6 V. Their p.d. is VP - VQ = 6 V. For a positive test charge in the given field, this represents a 6 J/C difference in electric potential energy.

Assigning Q a value of +10 V and P a value of +16 V leaves that difference unchanged. Adding the same constant to both potential levels changes the reference, not the p.d. A voltmeter compares its two connection points; it does not report a potential level without a reference.

What a source supplies

Electromotive force, or e.m.f., is the energy a source supplies to the circuit per unit charge through it, by converting another form of energy into electrical energy. It is measured in volts, despite the word force. It is not a mechanical force measured in newtons.

E.m.f. ε = source energy supplied / charge

For example, a discharging cell converts chemical energy. Some energy can be transferred internally before the remainder reaches the external circuit. Its terminal p.d. can therefore be lower than its e.m.f. The symbol ε is used here for e.m.f.; E is also used for it in circuit equations. Define the quantity locally so it is not confused with electric field strength or total energy.

Worked source account

Keep the source and external boundaries distinct

A steady discharging source has e.m.f. 1.60 V, terminal p.d. 1.50 V and current 0.400 A. The source converts 1.60 J per coulomb; 1.50 J per coulomb reaches the external circuit and 0.10 J per coulomb is transferred internally.

Separate conversion inside the source from delivery to the load

For the same steady current, I = 0.400 A, the source e.m.f. is 1.60 V and its terminal p.d. is 1.50 V. All bars use the same power scale. They show energy rates, not current or voltage arrows.

A 0.640-watt source conversion divides into 0.040 watt internally and 0.600 watt externallyThe upper dashed boundary is the discharging source. Its 0.640-watt conversion bar is 256 drawing units wide, and its internal-transfer bar for 0.040 watt is sixteen units wide. A labelled downward energy-delivery arrow leads to the separate external load boundary, where the 0.600-watt bar is 240 units wide. All bars start at the same horizontal reference and share one power scale. The internal and external values add to the source conversion. These are parts of one source energy account at the same current, not three sources or a claim that charge is consumed.Discharging sourceConversion: 0.640 WInternal: 0.040 WDeliveryto loadExternal load: 0.600 W

For each coulomb passing through the source, 1.60 J is supplied, 0.10 J is transferred internally and 1.50 J reaches the external circuit. Current continues through the load.

The total source conversion divides into external terminal output and internal transfer. The three labelled powers form one energy account at the same current; the e.m.f. and terminal p.d. describe different energy transfers per charge.

Current 0.400 A means 0.400 C passes each second. Multiplying each energy-per-charge value by that rate gives:

One steady source: energy per charge and the corresponding power
TransferJ per CPower / W
Total source conversion1.600.640
External circuit1.500.600
Inside the source0.100.040
0.640 W = 0.600 W + 0.040 W

The e.m.f. and terminal p.d. are supplied here; the account does not require an internal-resistance calculation. A voltmeter across the external load measures its p.d., not automatically the source's total energy conversion per charge.

The power equations express these energy-per-charge and charge-per-time ideas directly.

Check your understanding

A discharging source has e.m.f. 1.60 V, terminal p.d. 1.50 V and steady current 0.400 A. Which power account is consistent with these readings?

  1. It converts 0.640 W and also supplies 0.600 W internally.
  2. It converts 0.600 W, and every bit of that reaches the external circuit.
  3. It converts 0.640 W: 0.600 W outside and 0.040 W inside the source.
Show the answer

It converts 0.640 W: 0.600 W outside and 0.040 W inside the source.

E.m.f. counts the source energy supplied per charge; terminal p.d. counts the external transfer per charge. Multiplying each by the same steady current gives this balanced account.

Practise an exam question

Write your answer first, then tick the mark points it makes.

Exam-style question

Define electromotive force (e.m.f.) and potential difference (p.d.) in terms of energy. [2]

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