The Potentiometer

A ruler for volts

A potentiometer is a long uniform wire AB, often several metre lengths joined in series along a scale. A driver cell sends a steady current through it. The wire is uniform, so every centimetre has the same resistance and the same drop. The potential falls steadily from A to B, and the drop from A to any point is proportional to the length.

The drop per metre is the potential gradient, \(k = V_{AB}/L\), in volts per metre. The drop across a length \(l\) from A is then \(V = kl\). The wire becomes a ruler for volts: each centimetre stands for a fixed, known voltage.

Take a wire \(4\,\text{m}\) long of resistance \(8\,\Omega\), driven by a \(2\,\text{V}\) cell of negligible internal resistance. With nothing else in the driver circuit, the whole \(2\,\text{V}\) falls across the wire, and \(k = 2/4 = 0.5\,\text{V/m}\). Put \(8\,\Omega\) in series with the wire. The current falls to \(2/16 = 0.125\,\text{A}\), the drop across the wire to \(1\,\text{V}\), and the gradient to \(0.25\,\text{V/m}\).

Balancing a cell against the wire

Join the positive terminal of a test cell to A, the end where the driver's positive terminal is. Join its negative terminal through a galvanometer to the jockey. Slide the jockey. At one point J the drop along AJ equals the cell's voltage. There the two push equally and oppositely round the small loop, and the galvanometer rests. At that null, \(\mathcal E = kl\).

In the figure above the jockey stops exactly under the point where the two lines cross. With no series resistance a \(1.2\,\text{V}\) cell balances at \(2.4\,\text{m}\). Short of the null the needle leans left: the test cell drives current round its own loop. Beyond the null it leans right: the wire's larger drop drives current backwards through the cell.

Now take the chip with \(8\,\Omega\) in series. The whole wire now drops only \(1\,\text{V}\). A \(1.2\,\text{V}\) cell would need \(1.2/0.25 = 4.8\,\text{m}\), longer than the wire, so there is no null at all. The galvanometer deflects the same way wherever the jockey touches.

Values used in this lesson: the Class 12 practical's \(4\,\text{m}\) potentiometer wire and the cells of the workbook's second sheet. The driver is \(2\,\text{V}\) with negligible internal resistance.

Why it reads the emf

At the null no current flows through the test cell. So there is no drop \(Ir\) inside it, and the length balances the emf itself. A voltmeter cannot do this. To deflect, a voltmeter must draw some current \(I\) from the cell, and then it reads \(\mathcal E - Ir\). A potentiometer measures by refusing to load the source.

Comparing two emfs

Balance one cell, then another, without touching the driver circuit. The gradient \(k\) is the same for both, so \(\mathcal E_1 = kl_1\) and \(\mathcal E_2 = kl_2\). Divide: \(\mathcal E_1/\mathcal E_2 = l_1/l_2\). The current in the wire, its resistance and the galvanometer's resistance all cancel. What remains is a ratio of two lengths on a metre scale.

In Fig 5.1 cell 1 balances at \(3\,\text{m}\) and cell 2 at \(2\,\text{m}\), so \(\mathcal E_1 : \mathcal E_2 = 3 : 2\). If cell 1 is a standard of known emf, the other follows. A \(1.5\,\text{V}\) cell that balances at \(300\,\text{cm}\) fixes the scale, and a cell balancing at \(250\,\text{cm}\) on the same wire has \(1.5 \times 250/300 = 1.25\,\text{V}\).

Finding the internal resistance

A potentiometer can reach inside a cell. Balance the cell twice. First, with the key \(K_2\) open and nothing across the cell, the null at \(l_1\) gives \(\mathcal E = kl_1\). Then close \(K_2\), which joins a known resistance \(R\) across the cell. The cell now drives a current through \(R\), and its terminal voltage \(V = \mathcal E R/(R + r)\) balances at \(l_2\).

Take the third chip of the first figure of this lesson. A cell of \(1.5\,\text{V}\) and \(r = 2\,\Omega\) balances at \(3\,\text{m}\) when idle. With \(10\,\Omega\) across it the terminal voltage is \(1.5 \times 10/12 = 1.25\,\text{V}\), so the null moves towards A, to \(2.5\,\text{m}\). The dashed line there marks the emf; the red line has dropped to the terminal voltage.

For the cell above, \(r = 10 \times (3 - 2.5)/2.5 = 2\,\Omega\), as it should be. The shift of the null shows the size of \(r\) at a glance. An ideal cell would not move its null at all.

What moves the null

Every question about the null reduces to \(l = V/k\). Raise the driver's emf, and the current and the gradient rise, so the same cell balances at a shorter length. Add resistance in series with the wire, and the gradient falls, so the null moves further from A. The cell's own emf moves the null in proportion.

Two faults give no null at all. If the cell's voltage is larger than the drop across the whole wire, no point on the wire can match it. If the test cell's negative terminal is at A, the wire's drop and the cell's emf add round the loop instead of opposing. In both cases the galvanometer deflects the same way wherever the jockey touches.

Making it sensitive

A potentiometer is sensitive when a small change of voltage moves the null a long way. The null moves by \(\delta l = \delta V/k\), so a smaller gradient is better. Use a longer wire, or add resistance in series with it, keeping the full drop just above the voltage to be measured. Halving \(k\) from \(0.5\) to \(0.25\,\text{V/m}\) makes a change of \(0.01\,\text{V}\) move the null \(4\,\text{cm}\) instead of \(2\,\text{cm}\).