K323 / 2027
Current of electricity overview

Lesson 2 of 4

E.m.f. and potential difference

  • I can define a source e.m.f. as work per charge around the complete circuit, measured in volts.Syllabus

    Syllabus K323, 14(d). Define a source e.m.f. as work per charge around the complete circuit, measured in volts.

  • I can calculate total e.m.f. for sources arranged in series.Syllabus

    Syllabus K323, 14(e). Calculate total e.m.f. for sources arranged in series.

  • I can define potential difference across a component as work per charge through it, measured in volts.Syllabus

    Syllabus K323, 14(f). Define potential difference across a component as work per charge through it, measured in volts.

Voltage describes energy transferred per coulomb of charge. The source and the component have different roles in that energy transfer.

Work done is an energy transfer, measured in joules, J. Charge is measured in coulombs, C. Current measures coulombs per second; voltage measures joules per coulomb.

Electromotive force: e.m.f.
The e.m.f. of a source is the work done per unit charge by the source in driving charge around a complete circuit. It is measured in volts. A 1.5 V source supplies 1.5 J per coulomb in this energy account.
Potential difference: p.d.
The p.d. across a component is the work done per unit charge in driving charge through the component. It is measured in volts. A p.d. of 3.0 V means 3.0 J is transferred in that component for each coulomb passing through it.
1 volt = 1 joule per coulomb1 V = 1 J/C. Potential difference = work done / charge; work done = potential difference x charge.

Small voltages may be given in millivolts: 1 mV = 0.001 V. Divide a reading in millivolts by 1000 to convert it to volts.

A cell transfers energy from its chemical store into the electrical pathway. In a resistor, energy is transferred to internal stores; in a motor, part can be transferred mechanically to a load. Charge carries on through the circuit while energy is transferred. The energy stores and pathways account helps keep these two ideas separate.

The name electromotive force does not mean a force measured in newtons. E.m.f. is an energy-per-charge quantity, just as p.d. is. Neither quantity measures how many coulombs pass each second.

Energy per coulomb

Find a component's potential difference

A component transfers 60 J when 12 C passes through it.

P.d. = 60/12 = 5.0 J/C = 5.0 V.

For a separate component with p.d. 3.0 V, passing 8.0 C transfers work of 3.0 x 8.0 = 24 J.

These are supplied energy and charge amounts. A temperature rise alone would not measure every energy transfer unless the heated mass, its properties and other transfers were also accounted for.

Try it step by step

Work it out in steps: Charge Q = It in C, then Energy transferred E = VQ in J.

  1. Problem 1. A current of 0.20 A flows through a resistor for 50 s. The p.d. across the resistor is 6.0 V. How much energy is transferred in the resistor?
  2. Problem 2. A current of 1.5 A flows through a heater for 40 s. The p.d. across the heater is 12 V. How much energy is transferred?
  3. Problem 3. A current of 0.40 A flows through a lamp for 25 s. The p.d. across the lamp is 3.0 V. How much energy is transferred?
Show the answer

Problem 1.

Charge Q = It in C: Q = It = 0.20 x 50 = 10 C

Energy transferred E = VQ in J: E = VQ = 6.0 x 10 = 60 J (2 sf)

Problem 2.

Charge Q = It in C: Q = 1.5 x 40 = 60 C

Energy transferred E = VQ in J: E = 12 x 60 = 720 J (2 sf)

Problem 3.

Charge Q = It in C: Q = 0.40 x 25 = 10 C

Energy transferred E = VQ in J: E = 3.0 x 10 = 30 J (2 sf)

Measure voltage across two points

A voltmeter, shown as a circle containing V, connects in parallel across the component: one lead at each of its terminals. This measures the p.d. between those two points. It is not inserted into the main current path as an ammeter is.

For a positive d.c. reading, connect the meter's positive lead to the higher-potential side and its negative lead to the lower-potential side. Reversing the leads reverses the sign shown by a suitable digital meter; it does not mean the component has changed its resistance.

Select a voltage range that includes the expected reading. For example, a 0-5 V range is unsuitable for a 6.0 V p.d. Once a suitable range is established, finer resolution helps distinguish nearby readings. For an analogue scale, read at eye level to reduce parallax.

A voltmeter connected across a source with no other external circuit can give an estimate of its e.m.f. when the meter draws negligible current. For the ideal-source model, the source's terminal p.d. equals its e.m.f. The combined meter diagram instead measures the p.d. across a resistor.

Check your understanding

A component transfers 60 J of energy when 12 C passes through it. What is the potential difference across it?

  1. 0.20 V: divide the charge by the energy.
  2. 5.0 V: each coulomb transfers 5.0 J in the component.
  3. 720 V: multiply the energy by the charge to get the energy per coulomb.
Show the answer

5.0 V: each coulomb transfers 5.0 J in the component.

Potential difference is work done per unit charge: V = 60 / 12 = 5.0 J/C = 5.0 V. It is an energy-per-charge quantity, not a rate.

Add sources with their polarities

For ideal sources connected in series, the total e.m.f. is their signed sum. Choose a direction through the connected sources. Crossing a cell from its negative terminal to its positive terminal adds its e.m.f.; crossing it from positive to negative subtracts it.

Add the voltage rises and falls in a stated direction

These are open arrangements of ideal cells. A long plate marks a positive terminal and a short plate a negative terminal. The separate arrow sets the calculation reference from A to B.

Three cells aiding one another

Three cells aiding one anotherFrom A to B, each cell is crossed from its negative short plate to its positive long plate. Each contributes a 1.5 V rise, so B is 4.5 V higher than A in this ideal open arrangement. The terminals A and B are open, with no return wire. The arrow below is a reference for adding source contributions, not an arrow of current in a closed circuit.1.5 V-+1.5 V-+1.5 V-+ABReference: A to B

1.5 + 1.5 + 1.5 = 4.5 V.

From A to B, each cell is crossed from its negative short plate to its positive long plate. Each contributes a 1.5 V rise, so B is 4.5 V higher than A in this ideal open arrangement.

The final cell is reversed

The final cell is reversedFrom A to B, the first two cells each give a 1.5 V rise. The last cell is crossed from positive to negative and gives a 1.5 V fall. B is therefore 1.5 V higher than A in this ideal open arrangement. The terminals A and B are open, with no return wire. The arrow below is a reference for adding source contributions, not an arrow of current in a closed circuit.1.5 V-+1.5 V-+1.5 V+-ABReference: A to B

1.5 + 1.5 - 1.5 = 1.5 V.

From A to B, the first two cells each give a 1.5 V rise. The last cell is crossed from positive to negative and gives a 1.5 V fall. B is therefore 1.5 V higher than A in this ideal open arrangement.

Use the same A-to-B direction in both source combinations. Three aiding 1.5 V cells give +4.5 V. Reversing the last cell changes its contribution to -1.5 V, giving a net +1.5 V. These are open source combinations, not complete circuits.

Keep the reference direction

One of three cells is reversed

Going from A to B through three 1.5 V cells:

  • All aiding: total e.m.f. = 1.5 + 1.5 + 1.5 = 4.5 V.
  • Last cell opposing: total e.m.f. = 1.5 + 1.5 - 1.5 = 1.5 V.

B is at higher potential than A in both illustrated combinations, but by different amounts. Counting the cells without checking their terminal signs would miss the difference.

Check your understanding

Going from A to B through three ideal 1.5 V cells in series, the first two are crossed from negative to positive and the last from positive to negative. What is the net e.m.f. in the A-to-B direction?

  1. 0 V
  2. +1.5 V
  3. +4.5 V
Show the answer

+1.5 V

The first two sources aid the chosen direction and the third opposes it: +1.5 + 1.5 - 1.5 = +1.5 V. Cell polarity determines the sign of each contribution.

Practise an exam question

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

Exam-style question

A cell is connected to a lamp. Distinguish between the electromotive force (e.m.f.) of the cell and the potential difference (p.d.) across the lamp. [2]

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