The potentiometer

A voltmeter that draws no current

How do you measure the emf of a cell? A voltmeter across it reads less than the emf, because the voltmeter takes a current and the cell loses $Ir$ inside. A potentiometer solves this. At the moment of reading, it takes no current at all from the cell.

The idea

Take a long, uniform wire AB. Pass a steady current $I$ through it from a driver cell, with a rheostat in series. The wire is uniform, so the potential falls by the same amount along every centimetre. The fall for each unit of length is the potential gradient:

Here $R_w$ is the resistance of the wire and $L$ its length. So $k$ is the potential difference across the whole wire, divided by its length. The potential difference between A and a point at a distance $l$ is then

Measuring an emf

Join the cell under test so that its positive end goes to A. Send its other end through a galvanometer to a sliding contact, the jockey J. Slide J along the wire until the galvanometer reads zero. This is the balance point, at a distance $l$ from A.

At balance no current flows through the cell, so there is no lost voltage inside it. The potential difference between A and J equals the emf:

Two cells, balanced one after the other without changing the current, give

A worked example. The wire is $500\ \text{cm}$ long and has $12\ \Omega$. A $4.0\ \text{V}$ driver cell, with a rheostat of $28\ \Omega$, is in series with it.

1. The current. $I = \dfrac{4.0\ \text{V}}{12\ \Omega + 28\ \Omega} = 0.10\ \text{A}$.
2. The gradient. The potential difference across the wire is $0.10\ \text{A} \times 12\ \Omega = 1.2\ \text{V}$. So $k = \dfrac{1.2\ \text{V}}{500\ \text{cm}} = 2.4\ \text{mV/cm}$.
3. A cell of $1.08\ \text{V}$. $l = \dfrac{1080\ \text{mV}}{2.4\ \text{mV/cm}} = 450\ \text{cm}$.
4. A cell of $1.35\ \text{V}$. It would need $562.5\ \text{cm}$. That is longer than the wire, so it cannot be balanced.

Two things must be true for a balance. The driver cell must have a larger emf than the cell under test. And the positive ends of both cells must face the same end of the wire.

Why a voltmeter reads less

A cell of $E = 1.40\ \text{V}$ and $r = 2.5\ \Omega$ is joined to a voltmeter of $60\ \Omega$. The current is $\dfrac{1.40\ \text{V}}{62.5\ \Omega} = 0.0224\ \text{A}$. The voltmeter reads $V = 0.0224\ \text{A} \times 60\ \Omega = 1.344\ \text{V}$. That is $0.056\ \text{V}$ too low.

The potentiometer reads the true $1.40\ \text{V}$. At balance the cell supplies no current, so $Ir = 0$.

Finding the internal resistance

First balance the cell open, with no load. The balance length is $l_1$, and $E = kl_1$.

Then join a resistor $R$ across the cell. A current $I = \dfrac{E}{R + r}$ now flows through it, and its terminal voltage drops to $V = IR$. Balance again, at $l_2$. Now $V = kl_2$.

Divide the two results:

Solving for $r$ gives

A worked example. A cell balances at $l_1 = 275.0\ \text{cm}$ when open. With $R = 8.0\ \Omega$ across it, the balance point moves to $l_2 = 220.0\ \text{cm}$. Then $r = 8.0\ \Omega \times \dfrac{55.0}{220.0} = 2.0\ \Omega$.

Start with the defaults. The cell balances at $450\ \text{cm}$. Raise the rheostat to $50\ \Omega$. What happens to the gradient? Where does the cell balance now? Is a bigger rheostat always better?

Making the potentiometer more sensitive

A potentiometer is sensitive when a small change in $E$ moves the jockey a long way. That needs a small gradient $k$. It also makes the balance length large, so a fixed error $\delta l$ is a smaller fraction of $l$:

To lower $k$:

• Add resistance to the rheostat. This lowers the current in the wire.
• Use a longer wire. For the same potential difference across it, the gradient is smaller.

We can only go so far. The balance point must stay on the wire. The best setting puts it near the far end.

Practise with new numbers

Try it in Python

The program finds the gradient for the worked example, tests two cells for a balance, and finds an internal resistance.

Common mistakes

• Saying the cell supplies a current at balance. At balance the galvanometer shows zero, so the cell supplies none.
• Connecting the cell the wrong way. If the positive ends face opposite ends of the wire, there is no balance.
• Using a driver cell weaker than the cell under test. The potential across the whole wire must exceed $E$.
• Putting $l_1$ on the bottom. In $r = R\,\dfrac{l_1 - l_2}{l_2}$ the closed-circuit length $l_2$ is on the bottom.

Remember these

• Gradient: $k = \dfrac{IR_w}{L}$. Potential: $V = kl$.
• Balance: $E = kl$, and $\dfrac{E_1}{E_2} = \dfrac{l_1}{l_2}$.
• Internal resistance: $r = R\,\dfrac{l_1 - l_2}{l_2}$.
• True emf: no current is drawn at balance, so a potentiometer measures $E$, not $E - Ir$.
• More sensitive: a smaller gradient, with the balance still on the wire.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A potentiometer wire has a gradient of $2.0\ \text{mV/cm}$. A cell balances at $330\ \text{cm}$. What is its emf?

2. Two cells balance at $240\ \text{cm}$ and $360\ \text{cm}$. What is the ratio of their emfs?

3. Why does a voltmeter read less than the emf of a cell, but a potentiometer reads the true emf?

4. What happens if the driver cell has an emf smaller than the cell under test?