Ohm's law, the metre bridge and the potentiometer

Aim

To find the resistance and the resistivity of a wire with Ohm's law and with a metre bridge. To compare two cells and find the internal resistance of a cell with a potentiometer.

Ohm's law and the resistivity of a wire

The current in a metal wire is in proportion to the voltage across it, as long as its temperature stays the same:

The resistance $R$ depends on the wire. For a wire of length $l$ and diameter $d$, made of a material of resistivity $\rho$,

The apparatus. The wire (constantan), a battery and a key, and a rheostat. An ammeter is connected in series with the wire. A voltmeter is connected in parallel, across the wire only.

The steps.

1. Measure the length $l$ of the wire with a metre scale, between the two points where the leads are fixed.
2. Measure the diameter $d$ with a screw gauge at several places. Correct each reading for the zero error.
3. Close the key. Use the rheostat to change the current in steps. Read $V$ and $I$ each time.
4. Open the key between readings, so that the wire does not heat.

The readings. The wire has $l = 100.0$ cm and a mean diameter of $0.340$ mm.

The graph of $V$ against $I$ is a straight line through the origin, and its slope is $R = 5.60\ \Omega$. Then

Put in the numbers: $\rho = \dfrac{5.60\ \Omega \times \pi}{4 \times 1.000\ \text{m}}$ $\times (0.340 \times 10^{-3}\ \text{m})^2$ $= 5.1 \times 10^{-7}\ \Omega\,\text{m}$.

The error is $\dfrac{\Delta\rho}{\rho} = \dfrac{\Delta R}{R} + 2\dfrac{\Delta d}{d} + \dfrac{\Delta l}{l}$ $= 0.1\% + 5.9\% + 0.1\%$ $= 6.1\%$. So $\rho = (5.1 \pm 0.3) \times 10^{-7}\ \Omega\,\text{m}$. This agrees with the value of about $4.9 \times 10^{-7}\ \Omega\,\text{m}$ for constantan. The diameter again gives nearly all of the error.

A hidden error. The voltmeter takes a little current itself. The ammeter then reads the current of the wire and of the voltmeter together. So $V/I$ is the resistance of the wire and the voltmeter in parallel, which is a little too small: $\dfrac{R R_V}{R + R_V}$. For $R = 5.60\ \Omega$ and a voltmeter of $1000\ \Omega$, it is $0.6\%$ too low. A voltmeter of high resistance keeps this small.

Precautions. Use small currents. A wire heats up, and its resistance changes. Do not touch the wire. Check the zero of the meters. Join the meters with the right terminals.

The metre bridge

A metre bridge is a Wheatstone bridge made with a wire of length $100$ cm and uniform cross-section. The wire $AC$ lies along a metre scale. Two gaps in a thick copper strip hold the unknown resistance $R$ (left, near $A$) and a known resistance $S$ from a resistance box (right, near $C$). A galvanometer joins the middle of the strip to a jockey that touches the wire.

Move the jockey until the galvanometer shows no deflection. The bridge is balanced, and no current flows through the galvanometer. The resistances of the two lengths of wire, $AJ = l$ and $JC = 100 - l$, are in proportion to the lengths, so

The steps. Take $S = 4, 5, 6, 7\ \Omega$ in turn. For each, press the jockey at a point, note the deflection, move it, and find the balance length $l$. Tap the jockey. Do not slide it along the wire.

The readings.

The mean is $R = 5.60\ \Omega$, the same as the value from Ohm's law.

Why choose $S$ for a balance near $50$ cm? A mistake $\Delta l$ in reading $l$ changes the length $100 - l$ by the same amount in the other direction. So

The product $l(100 - l)$ is greatest at $l = 50$ cm, so the error is least there. For $\Delta l = 0.1$ cm, the error is $0.40\%$ at $50$ cm, $0.44\%$ at $35$ cm and $0.63\%$ at $20$ cm.

Set $R = 6$ and $S = 6$. Where does the bridge balance? Then set $S = 1$. What happens to the error?

Other errors. The wire may not be uniform. The strip and the joints have a small resistance, so the wire seems a little longer or shorter at the ends. These end corrections are reduced by interchanging $R$ and $S$ and taking the mean of the two results. Do not let a current flow for long, since it heats the wire.

The potentiometer

A potentiometer is a long wire of uniform cross-section, say $10$ m, through which a steady current from a driver cell flows. The potential drop in the wire is in proportion to the length. Between $A$ and a point at the distance $l$, it is $V = kl$, where $k$ is the potential gradient in V/cm.

Join a cell of emf $E$ between $A$ and the jockey, with its positive terminal at the end $A$ that is also the positive end of the driver cell. Move the jockey until the galvanometer shows no deflection. The cell then gives no current, and its emf equals the potential drop of the wire between $A$ and the jockey: $E = kl_1$.

Comparing two cells. Find the balance lengths $l_1$ and $l_2$ of two cells in the same circuit, with the same $k$. Then

For a Daniell cell the balance length is $540$ cm. For a Leclanché cell it is $750$ cm. So $\dfrac{E_2}{E_1} = \dfrac{750}{540} = 1.39$. This agrees with the ratio of the standard emfs, $1.5\ \text{V}/1.08\ \text{V} = 1.39$.

The internal resistance. Join a resistance box $R$ across the cell through a key. A current $I = \dfrac{E}{R + r}$ now flows, and the terminal voltage is $V = IR = E\,\dfrac{R}{R + r}$. Find the new balance length $l_2$, with $V = kl_2$. With $E = kl_1$,

The readings. The Leclanché cell has $l_1 = 750.0$ cm.

A graph of $\dfrac{l_1}{l_2}$ against $\dfrac{1}{R}$ is a straight line with the slope $r$, and an intercept of $1$.

Make $R$ very large. What does $l_2$ approach? What does that say about the terminal voltage of a cell on open circuit?

Why a potentiometer and not a voltmeter? A voltmeter takes a current, so it reads the terminal voltage, which is lower than the emf. At the balance point a potentiometer takes no current from the cell, so it measures the emf itself.

Precautions. The driver cell must have a larger emf than the test cell, or the balance point is not on the wire. All positive terminals go to the same end. Put a high resistance in series with the galvanometer at first, and remove it for the final balance. Do not pass current for long. Do not rub the jockey on the wire.

Practise with new numbers

Try it in Python

The first program finds the resistance of the wire from the table of $V$ and $I$ and the resistivity with its error.

The second program finds the resistance from the metre bridge readings, and shows how the error depends on the balance point.

The third program finds the internal resistance of the cell from the straight line of $l_1/l_2$ against $1/R$.

Common mistakes

• Swapping $R$ and $S$ in the formula. $R$ is next to the length $l$. If $R$ is in the left gap, $R/S = l/(100 - l)$.
• Measuring $l$ from the wrong end. $l$ is measured from the end of the wire next to $R$.
• Taking $r = R(l_1 - l_2)/l_1$. The loaded length $l_2$ is the divisor.
• A driver cell weaker than the test cell. The balance point lies beyond the end of the wire.
• Joining the test cell the wrong way round. The positive terminals of both cells must go to the same end.
• Forgetting the resistance of the voltmeter. It lowers the resistance that is measured.

Remember these

• $\rho = \dfrac{R\pi d^2}{4l}$, and $\dfrac{\Delta\rho}{\rho}$ has the term $2\dfrac{\Delta d}{d}$.
• Metre bridge: $\dfrac{R}{S} = \dfrac{l}{100 - l}$. The error is least at $l = 50$ cm. Interchange $R$ and $S$ to reduce end errors.
• Potentiometer: $V = kl$. $\dfrac{E_1}{E_2} = \dfrac{l_1}{l_2}$ and $r = R\,\dfrac{l_1 - l_2}{l_2}$.
• A potentiometer takes no current from the cell at the balance point.

Check yourself

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

1. A metre bridge balances at $40.0$ cm with a known resistance of $9.0\ \Omega$ in the right gap. What is the unknown resistance in the left gap?

2. Two cells balance at $480$ cm and $360$ cm on a potentiometer wire. What is the ratio of their emfs?

3. A cell balances at $600$ cm on open circuit and at $540$ cm with $9.0\ \Omega$ across it. What is its internal resistance?

4. Why is the potentiometer wire made long?

References

• NCERT, Physics Class XII, Chapter 3 (Current Electricity), the sections on Ohm's law, the Wheatstone bridge, the metre bridge and the potentiometer.
• NCERT, Laboratory Manual: Physics, Class XII, the experiments on Ohm's law, the metre bridge and the potentiometer.
• OpenStax University Physics Volume 2, Chapter 9 (Current and Resistance), the section on resistivity.