Topic 2 of 4
E.m.f. and potential difference
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.
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.
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.
Optional check A component transfers 60 J of energy when 12 C passes through it. What is the potential difference across it?
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
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
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.
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.