E.m.f. measures the energy a source gives to each coulomb

1. A cell does not make charge; the charge is already in the wires, and what the cell supplies is energy.
2. Chemical reactions inside the cell transfer energy to every coulomb that passes through it, and that energy transfer is the work done being counted.
3. The word "force" in the name is misleading, because e.m.f. is an energy per unit charge and is not measured in newtons.
4. E.m.f. describes the whole circuit: the energy handed to one coulomb for a complete trip out of the cell, round the loop and back in again.
5. A larger e.m.f. means each coulomb carries more energy, so the source can drive a larger current through the same resistance.

E.m.f. is measured in volts, and a volt is a joule per coulomb

1. The unit is the volt, symbol $V$, and $1V$ means $1J$ of energy given to each coulomb.
2. So a $1.5V$ cell hands $1.5J$ to every coulomb that passes through it.
3. Cells in series add, so the four $1.5V$ cells in a torch give a total e.m.f. of $6.0V$.
4. Familiar values give you a sense of scale: an AA cell is $1.5V$, a car battery $12V$, and the UK mains $230V$.
5. A voltmeter placed across a cell that is not connected to anything else reads its e.m.f., because no current is flowing and none of the energy is being lost inside the cell.

Calculating e.m.f. from work and charge

1. The equation to recall is $E=\frac{W}{Q}$, where $E$ is the e.m.f. in $V$, $W$ is the work done in $J$ and $Q$ is the charge in $C$.
2. Rearranged it gives $W=EQ$, the energy a source supplies to a known amount of charge.
3. Since $Q=It$, the energy supplied over a period of time can be written as $W=EIt$.
4. Keep work in joules, charge in coulombs and time in seconds, and the answer comes out in volts with no conversion needed.