# 4.7: Electromotive Force and Internal Resistance

- Page ID
- 5971

The reader is reminded of the following definition from section 4.1:

*Definition*. The potential difference across the poles of a cell when no current is being taken from it is called the *electromotive force *(EMF) of the cell.

I shall use the symbol *E** *for EMF.

*Question*. A 4 \(\Omega\) resistance is connected across a cell of EMF 2 V. What current flows?

The immediate answer is 0.5 A – but this is likely to be wrong. The reason is that a cell has a resistance of its own – its *internal resistance.* The internal resistance of a lead-acid cell is typically quite small, but most dry cells have an appreciable internal resistance. If the external resistance is \(R\) and the internal resistance is \(r\), the total resistance of the circuit is \(R + r\), so that the current that flows is *E*\(/(R + r)\).

Whenever a current is taken from a cell (or battery) the potential difference across its poles *drops* to a value less than its EMF. We can think of a cell as an EMF in series with an internal resistance:

\(\text{FIGURE IV.4}\)

If we take the point A as having zero potential, we see that the potential of the point B will be *E* - \(Ir\), and this, then, is the potential difference across the poles of the cell when a current \(I\) is being taken from it.

Show that this can also be written as \(\frac{ER}{R+r}\).