The Wheatstone bridge

Four resistors and a galvanometer

Narrated video. Transcript:

Here is a neat use of Kirchhoff’s rules. Four resistors are joined in a diamond. Call the corners A on the left, B at the top, C on the right and D at the bottom.

R₁ joins A to B, and R₂ joins B to C. R₃ joins A to D, and R₄ joins D to C. A battery is connected across A and C. This is called the battery arm.

Across B and D we connect a galvanometer, a sensitive meter that shows even a tiny current. This is the galvanometer arm. The whole arrangement is a Wheatstone bridge.

From A, the current splits. Part goes along the top, through R₁ and R₂; part goes along the bottom, through R₃ and R₄. In general some current also crosses from B to D, and the galvanometer needle swings.

But for one special choice of the resistances, the needle stays exactly at zero. No current flows through the galvanometer. We say the bridge is balanced.

Let us find that condition. With I_g zero, the same current I₁ flows through R₁ and R₂, and the same current I₃ flows through R₃ and R₄.

Loop rule round the loop A, B, D, back to A. With no current in the galvanometer, it gives I₁R₁ minus I₃R₃ equals zero. So I₁R₁ equals I₃R₃.

Loop rule round B, C, D, back to B gives I₁R₂ equals I₃R₄. In words: B and D are at the same potential, so the drops from A to B and from A to D are equal, and so are the drops onwards to C.

Divide the second equation by the first. The currents cancel, and we get the balance condition: R₂ over R₁ equals R₄ over R₃.

Measuring an unknown resistance

Narrated video. Transcript:

The balance condition lets us measure an unknown resistance. Put the unknown in the place of R₄. Choose fixed resistances R₁ and R₂, and make R₃ a variable resistance box.

R₁ and R₂ are called the ratio arms. Now change R₃, step by step, and watch the galvanometer.

If R₃ is too small, the needle swings one way. If it is too large, it swings the other way. Somewhere in between, it reads zero.

At balance, R₄ equals R₃ times R₂ over R₁. Worked example: R₁ is 100 ohm and R₂ is 10 ohm, so the ratio R₂ over R₁ is one-tenth.

The bridge balances when R₃ is 362 ohm. So R₄ equals 362 × 10 over 100, which is 36.2 ohm. The small ratio lets us read the unknown to a tenth of an ohm, even though the box only goes in steps of one ohm.

Why is this method so good? At balance the galvanometer carries no current, so we never need to read an exact current from it. We only need to see when the needle is at zero.

So the answer depends only on the known resistances, not on the emf of the battery or on the resistance of the galvanometer.

If the bridge is not balanced, we must use Kirchhoff’s rules for the whole network. Take R₁ 100, R₂ 10, R₃ 60 and R₄ 5 ohm, a 10 volt battery, and a galvanometer of 15 ohm.

Here R₂ over R₁ is 0.1 but R₄ over R₃ is about 0.083, so the bridge is not balanced. Solving the loop equations gives a galvanometer current of about 4.87 milliampere, flowing from B to D.

Try it: balance the bridge

Interactive simulation. Things to try:

• R₄ is hidden. Change R₃ and watch the galvanometer needle. Which way does it swing when R₃ is too small?
• Find the R₃ where the needle reads zero. Then R₄ = R₃ × R₂/R₁.
• Try the ratio R₂/R₁ = 0.1 and balance again. Can you now pin down R₄ more finely?
• At balance, compare the potentials of B and D in the readouts.

Find the unknown

In a Wheatstone bridge, R₁ = 100 Ω and R₂ = 10 Ω. The galvanometer shows no deflection when R₃ = 450 Ω. What is the unknown resistance R₄?

1. 4500 Ω
2. 45 Ω
3. 450 Ω
4. 4.5 Ω

An unbalanced bridge

A bridge has R₁ (AB) = 100 Ω, R₂ (BC) = 10 Ω, R₃ (AD) = 60 Ω and R₄ (DC) = 5 Ω. A 10 V battery is across A and C and a 15 Ω galvanometer across B and D. What current flows through the galvanometer?

1. Zero, because the bridge is balanced
2. About 4.87 mA
3. About 0.67 A
4. About 48.7 mA

Summary

• Balance condition: R₂/R₁ = R₄/R₃
• Unknown resistance: R₄ = R₃ × R₂/R₁

• A Wheatstone bridge is four resistors in a diamond, with a battery across one diagonal (A–C) and a galvanometer across the other (B–D).
• It is balanced when no current flows through the galvanometer: then B and D are at the same potential.
• The balance condition follows from the loop rule: I₁R₁ = I₃R₃ and I₁R₂ = I₃R₄.
• To measure an unknown R₄, keep the ratio arms R₁, R₂ fixed and vary R₃ until the needle reads zero. The result does not depend on the battery or the galvanometer.
• An unbalanced bridge is solved with Kirchhoff’s rules for the whole network.