The electromotive force, written as ε, is the energy supplied by a source (like a battery) to each coulomb of charge that moves through it.
Mathematically:
\$\varepsilon = \frac{W}{Q}\$
where W is the work done (energy) in joules and Q is the charge in coulombs.
So, e.m.f. is energy per unit charge.
| Source | e.m.f. (V) |
|---|---|
| AA Battery | 1.5 |
| Car Battery | 12 |
| Solar Cell (open‑circuit) | 0.5–0.6 |
Think of a battery as a water pump and the circuit as a pipe system.
The pressure difference created by the pump pushes water through the pipe.
Similarly, the e.m.f. creates a “pressure” (voltage) that pushes electrons around the circuit.
Just as the pump’s pressure determines how fast water flows, the e.m.f. determines the current for a given resistance.
A 1.5 V AA battery is connected to a 3 Ω resistor.
What is the current and the power delivered by the battery?
Using Ohm’s law:
\$I = \frac{\varepsilon}{R} = \frac{1.5}{3} = 0.5\ \text{A}\$
Power:
\$P = \varepsilon I = 1.5 \times 0.5 = 0.75\ \text{W}\$
- e.m.f. (ε) = energy per unit charge (J C⁻¹).
- Measured in volts (V).
- Drives current in a circuit just like a pump drives water.
- Use ε = V + IR for circuits with resistive loads.
- Keep the exam tips handy to avoid common mistakes!