The meter bridge and the potentiometer

The meter bridge

Narrated video. Transcript:

This lesson is an extra topic. The meter bridge and the potentiometer are no longer in the current NCERT syllabus, but they are still used in school practicals and asked in competitive exams.

The meter bridge is a practical Wheatstone bridge. A uniform wire, exactly 1 metre long, is stretched along a metre scale between two thick copper strips, A and C.

The copper strips have two gaps. In the left gap goes a resistance box R. In the right gap goes the unknown resistance S. The middle strip between them is point B.

A cell is connected across A and C. A galvanometer joins B to a jockey, D: a sliding contact we can touch anywhere on the wire.

Slide the jockey along. Near A the needle swings one way, near C the other way. At one point, the null point, the needle reads zero.

The wire is uniform, so its resistance is proportional to length. The piece of length l₁ from A to D has resistance R_cm l₁, and the rest has R_cm times 100 minus l₁, where R_cm is the resistance per centimetre.

These two pieces are the other two arms of a Wheatstone bridge. So at balance, R over S equals R_cm l₁ over R_cm times 100 minus l₁. R_cm cancels.

So S equals R times 100 minus l₁, over l₁. Example: R is 10 ohm and the null point is at 40 centimetres. Then S equals 10 × 60 over 40, which is 15 ohm.

Choose R so that the null point falls near the middle of the wire. Then a small error in reading l₁ gives the smallest error in S.

The potentiometer: comparing emfs

Narrated video. Transcript:

The potentiometer is a long uniform wire, often 4 to 10 metres, laid in straight lengths side by side. A driver cell and a rheostat send a steady current I along it.

Since the wire is uniform, the potential falls steadily along it. The fall per unit length is φ. So the potential difference between A and a point at length l is φ times l.

Now take a cell of emf ε₁. Join its positive plate to A, the same side as the driver cell’s positive plate, and its negative plate through a galvanometer to the jockey.

Slide the jockey. At the null point, the potential difference along the wire exactly balances the cell, and no current flows through the galvanometer. Then ε₁ equals φ l₁.

Because no current flows through the cell at balance, there is no drop across its internal resistance. The potentiometer measures the emf itself, not the terminal voltage. An ordinary voltmeter cannot do that.

Now switch a two-way key to a second cell, ε₂, and find its null point, l₂. Then ε₂ equals φ l₂.

Divide: ε₁ over ε₂ equals l₁ over l₂. We never need to know φ, as long as the current in the wire stays the same.

Example: φ is 0.005 volt per centimetre. A 1.5 volt cell balances at 300 centimetres, and a 1.0 volt cell at 200 centimetres. The ratio, 300 over 200, is 1.5, exactly the ratio of the emfs.

The emf to be measured must be less than the potential difference across the whole wire, or there is no null point. A longer wire with a smaller φ makes the potentiometer more sensitive.

The potentiometer: internal resistance

Narrated video. Transcript:

A potentiometer can also find the internal resistance r of a cell. Connect the cell to the wire as before, and put a resistance box R across the cell through a key K₂.

First keep K₂ open. No current flows from the cell at balance, so the null point at l₁ gives its emf: ε equals φ l₁.

Now close K₂. The cell drives a current I through R, and its terminal voltage drops to V equals I R, which is less than ε. The null point moves closer to A, to l₂. So V equals φ l₂.

Divide: ε over V equals l₁ over l₂.

But the current is I equals ε over R plus r, and V equals I R. So ε over V equals R plus r over R.

Put the two together: R plus r over R equals l₁ over l₂. Rearranging, r equals R times l₁ over l₂ minus 1.

Example: with K₂ open the null point is at 300 centimetres. With a 5 ohm resistance across the cell, it moves to 250 centimetres.

Then r equals 5 times 300 over 250 minus 1. 300 over 250 is 1.2, so r equals 5 × 0.2, which is 1 ohm.

Check: if ε is 1.5 volt, the current is 1.5 over 6, or 0.25 ampere, and the terminal voltage is 0.25 × 5 = 1.25 volt. That is φ times 250 centimetres, with φ equal to 0.005 volt per centimetre.

Try it: meter bridge

Interactive simulation. Things to try:

• Drag the jockey along the wire (or use the slider) until the galvanometer reads zero.
• At the null point, check that S = R(100 − l₁)/l₁ gives 15 Ω.
• Change R. Which R puts the null point closest to the middle of the wire?
• Try the hidden-S challenge: balance, then work out S yourself.

Try it: compare two emfs

Interactive simulation. Things to try:

• With the key on cell 1, find the null point and record l₁. Switch to cell 2 and record l₂.
• Compare l₁/l₂ with ε₁/ε₂.
• Change the rheostat. Both null points move, but does the ratio l₁/l₂ change?
• Make the rheostat large. What happens when φ × 400 cm is less than the emf?

Try it: internal resistance

Interactive simulation. Things to try:

• With K₂ open, find the null point l₁. This balances the emf.
• Close K₂ and find the new null point l₂. It is closer to A: the terminal voltage is smaller.
• Work out r = R(l₁/l₂ − 1) and compare with the cell’s r.
• Use a smaller R. Does l₂ move further, and do you still get the same r?

Meter bridge reading

In a meter bridge, a 20 Ω resistance is in the left gap and an unknown resistance S in the right gap. The null point is 40 cm from the left end. What is S?

1. 13.3 Ω
2. 8 Ω
3. 30 Ω
4. 50 Ω

Second cell

In a potentiometer, a cell of emf 1.25 V gives a null point at 35.0 cm. When it is replaced by another cell, the null point shifts to 63.0 cm. What is the emf of the second cell?

1. 0.69 V
2. 2.25 V
3. 1.80 V
4. 1.25 V

How big is r?

With K₂ open, a cell balances at 76.3 cm of a potentiometer wire. With a 9.5 Ω resistance across the cell, it balances at 64.8 cm. What is the internal resistance of the cell?

1. 11.2 Ω
2. 8.1 Ω
3. 0.18 Ω
4. 1.7 Ω

Summary

• Meter bridge: S = R (100 − l₁)/l₁
• Potentiometer: V = φ l; ε₁/ε₂ = l₁/l₂
• Internal resistance: r = R (l₁/l₂ − 1)

• Extra topic: not in the current NCERT syllabus, but used in practicals and competitive exams.
• The meter bridge is a Wheatstone bridge whose two ratio arms are the two parts of a uniform 1 m wire.
• In a potentiometer, a steady current makes the potential fall uniformly along the wire, φ per unit length.
• At the null point no current is drawn from the test cell, so a potentiometer measures the emf itself.
• The emf being measured must be less than the potential difference across the whole wire; a smaller φ (longer wire) gives better sensitivity.