The emf of a cell and its internal resistance
Think of the emf of a source as what it gives to each coulomb for one complete journey round the circuit, counting the part inside the source as well as the part outside: it is work per coulomb. We write it VB and measure it in volts. The charge does not only cross the outside resistor R: it also has to pass through the material of the cell, which has a small resistance of its own, the internal resistance r. So the emf is shared between two parts: the part IR used in the outside circuit and the part Ir used up inside the cell, and VB = IR + Ir. The part Ir is called the lost volts, because it cannot be used outside the cell. A fresh cell has a small r; an old, nearly flat cell has a large r, and that is the real reason why it cannot drive a bulb although its emf may still be close to 1.5 V.
| Part of the circuit | Voltage | What happens |
|---|---|---|
| outside the cell | IR | drives the current through the outside resistor |
| inside the cell | Ir | lost as heat in the cell itself |
| whole circuit | VB | the emf, equal to IR + Ir |
| Part | Resistance (Ω) | Current (A) | Voltage (V) |
|---|---|---|---|
| outside | 8 | 1 | 8 |
| inside | 1 | 1 | 1 |
| whole circuit | 9 | 1 | 9 |
With a grown-up, if you have a fresh and an old 1.5 V torch cell and a small torch bulb, join the bulb to one cell at a time. Before you look, predict which gives the dimmer light and say why (the old cell is expected to have the larger r, so more of its emf is lost inside), then see what happens. Stop if anything feels warm, and never join a wire straight across a cell. Without cells, make a table of IR, Ir and VB for a cell with r = 0.5 Ω and currents of 1 A and 2 A through a 3 Ω resistor.