The meter bridge

A Wheatstone bridge you can slide

In Electric Circuits you met the Wheatstone bridge: four resistors in a diamond, and a galvanometer that reads zero when the ratios of the arms are equal. The meter bridge is the same bridge, built with a wire so that we can change one ratio smoothly.

A uniform wire, $100\ \text{cm}$ long, lies along a scale. A strip of copper above it has two gaps. The unknown resistance $X$ goes in one gap, and a resistance box $R$ in the other. A galvanometer joins the middle of the strip to a sliding contact, the jockey J, which touches the wire. A cell and a key send current through the whole arrangement.

The wire is uniform. Its resistance, $\rho L/A$, is proportional to its length, because $\rho$ and $A$ are the same everywhere. The jockey cuts the wire into a piece of length $l$ on the left and a piece of length $100 - l$ on the right. Their resistances are in the ratio $l : (100 - l)$.

The balance rule

Slide the jockey until the galvanometer shows no deflection. The bridge is balanced. The four arms are $X$, $R$ and the two pieces of wire. The balance condition for any Wheatstone bridge then gives

The length $l$ is measured from the end of the wire that is joined to $X$. Mixing up the two ends is the commonest slip.

A worked example. A $20\ \Omega$ box is in the right gap. The bridge balances with the jockey at $36.0\ \text{cm}$.

1. The lengths. $l = 36.0\ \text{cm}$ and $100 - l = 64.0\ \text{cm}$.
2. The resistance. $X = 20\ \Omega \times \dfrac{36.0}{64.0} = 11.25\ \Omega$.

Notice that the answer does not depend on the emf of the cell, or on the resistance of the galvanometer. At balance no current flows through the galvanometer, so neither of them matters.

Set $X = 12\ \Omega$ and $R = 20\ \Omega$. Find the balance point. Then make $X$ larger and watch the balance point move. Which way does it go, and why?

Why the balance point should be near the middle

The only measurement is $l$, read from a scale to about $0.1\ \text{cm}$. Suppose $l$ is uncertain by $\delta l$. Then $X = Rl/(100 - l)$ is uncertain too. The fractional error is

The bottom line, $l(100 - l)$, is greatest at $l = 50\ \text{cm}$. So the error is smallest there. With $\delta l = 0.1\ \text{cm}$:

• At $50\ \text{cm}$, the error in $X$ is $0.40\ \%$.
• At $20\ \text{cm}$, it is $0.63\ \%$.
• At $10\ \text{cm}$, it is $1.1\ \%$.

So choose $R$ close to $X$ in size. Then the balance point falls between $40\ \text{cm}$ and $60\ \text{cm}$, where the reading is best.

Reducing the errors

• End corrections. The copper strips and the joints add small resistances at the ends of the wire. Then the lengths that count are a little longer than the lengths we read. The balance rule is slightly wrong.
• Interchange. Swap $X$ and $R$ and balance again at $l_2$. Now $X = R\,\dfrac{100 - l_2}{l_2}$. The two values of $X$ err in opposite directions, so their average is closer to the truth.
• Heating. Press the key only while you take a reading. A warm wire changes its resistance.
• The jockey. Tap it on the wire. Do not drag it, or the wire wears unevenly and stops being uniform.

Finding the resistivity of a wire

Suppose $X$ is a wire of length $L$ and diameter $d$. Then $X = \rho L/A$ and $A = \pi d^2/4$, so

Measure $X$ with the meter bridge. Measure $L$ with a scale and $d$ with a screw gauge. The diameter has the largest error, since it enters as a square.

Practise with new numbers

Try it in Python

The program finds $X$ and the error in $X$ for several balance points, with a $20\ \Omega$ box.

Common mistakes

• Turning the ratio upside down. The length $l$ on the side of $X$ goes on top: $\dfrac{X}{R} = \dfrac{l}{100 - l}$.
• Measuring $l$ from the wrong end. It must be from the end where $X$ is joined.
• Saying the answer depends on the cell. At balance no current flows in the galvanometer, so the emf does not matter.
• Forgetting that the error depends on both lengths. Only $l$ is read, but both $l$ and $100 - l$ change together.

Remember these

• Balance: $X = R\,\dfrac{l}{100 - l}$, with $l$ measured from the side of $X$.
• Independent of: the emf of the cell and the resistance of the galvanometer.
• Best accuracy: a balance point near $50\ \text{cm}$.
• Fractional error: $\dfrac{100\,\delta l}{l(100 - l)}$.
• Interchange $X$ and $R$: $X = R\,\dfrac{100 - l_2}{l_2}$. Average the two values.
• Resistivity: $\rho = \dfrac{\pi d^2 X}{4L}$.

Check yourself

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

1. A resistance box of $30\ \Omega$ is in the right gap of a meter bridge. The balance point is $45\ \text{cm}$ from the left end. What is $X$?

2. $X$ and $R$ are interchanged, and the new balance point is at $l_2$. Where was the first balance point, in terms of $l_2$?

3. Why is it better to balance near the middle of the wire?

4. Does the balance point change if the cell is replaced by one with twice the emf?