How do you measure a resistance with a balance? The Wheatstone bridge trick

How to use this simulation

1. Start in Wheatstone bridge mode with the defaults: E = 6 V, P = 10 Ω, Q = 20 Ω, R = 15 Ω, S = 25 Ω. The ratios P/Q = 0.50 and R/S = 0.60 are close but not equal, so the needle leans to one side and the galvanometer reads 2.15 mA.
2. Raise S slowly. The needle creeps towards zero. At S = 30 Ω the two ratios are equal at 0.50, the needle sits at zero and the panel says the bridge is balanced.
3. Now change the battery voltage while the bridge is balanced. The needle does not move. Balance depends on the four resistors only.
4. Push S past 30 Ω and the needle swings the other way, because the current through the galvanometer has reversed.
5. Switch to Metre bridge. The jockey starts at l = 40 cm. Slide it along the wire until the needle returns to zero. Compare the calculated X with the actual X in the readings.
6. Change the known resistance R and find the new balance point again. The balance length moves, but X = R·l ÷ (100 − l) still gives the same unknown.

Weighing resistance

Think of an old two-pan balance. You do not need to know how much one pan weighs in absolute terms. You only need the pointer to rest at zero, because then both sides are equal. A Wheatstone bridge works the same way, except that the thing being balanced is the voltage on two sides of a circuit instead of weight on two pans.

This matters because a plain ammeter-voltmeter method has errors from both meters. A null method needs only one sensitive detector that reads zero, and zero is very exact. That is why the bridge has been the standard way to measure resistance accurately since Samuel Hunter Christie invented it and Charles Wheatstone made it famous in 1843.

In exams this topic produces three kinds of question: state and derive the balance condition, find an unknown resistance from a balanced bridge, and find the unknown from a metre bridge reading. The simulation covers all three.

The four arms and the galvanometer

Four resistors P, Q, R and S form a diamond with corners A, B, C and D. The battery connects across the A–C corners. A sensitive galvanometer connects across the B–D corners, which is the “bridge” that gives the circuit its name.

Current leaves A and has two routes to C: through P then Q (via B), or through R then S (via D). If B and D are at different potentials, some current crosses the galvanometer as well. The needle shows how much and in which direction.

In the default set-up the battery is 6 V, and the galvanometer’s own resistance is 100 Ω. Solving the circuit exactly, B sits at 3.99 V and D at 3.77 V, measured from C. The 2.15 mA that crosses from B to D comes from that small difference.

The balance condition

The bridge is balanced when the galvanometer shows no deflection. That means B and D are at the same potential and no current crosses between them. Now every number follows from simple Ohm’s law.

With no current through the galvanometer, the same current I₁ passes through P and Q in series, and the same current I₂ through R and S in series. B and D being at equal potential means the drops from A are equal: I₁P = I₂R. The drops to C are also equal: I₁Q = I₂S. Divide the first by the second, and I₁ and I₂ vanish.

What happens just off balance

The table keeps P = 10 Ω, Q = 20 Ω and R = 15 Ω, and changes only S. The galvanometer current is a real value from an exact solution of the circuit, and the last column is how far the needle would swing on the simulated dial (it saturates at about 42 degrees).

Below S = 30 Ω, R/S is larger than P/Q and the current flows from B to D (positive). Above it, the current reverses. Exactly at S = 30 Ω it is zero. A small change in S gives a visible needle movement, which is why the bridge can detect tiny differences.

Worked example: find the missing arm

Example 1. In a balanced bridge P = 10 Ω, Q = 20 Ω and R = 15 Ω. Find S.

Use P/Q = R/S, so S = Q × R ÷ P = 20 × 15 ÷ 10 = 30 Ω. That matches the value at which the simulation’s needle sits at zero.

Example 2. A bridge is balanced with P = 5 Ω, Q = 15 Ω and S = 60 Ω. Find R. The ratio P/Q is 0.33, so R = S × P ÷ Q = 60 × 5 ÷ 15 = 20 Ω.

Example 3 (a useful by-product). When the default bridge is balanced at S = 30 Ω, no current crosses B–D, so it behaves as two series pairs in parallel. The equivalent resistance is (10 + 20) × (15 + 30) ÷ (10 + 20 + 15 + 30) = 18.00 Ω, and a 6 V battery drives 0.33 A. The solver agrees: the battery supplies 0.33 A and the galvanometer shows 0.00 mA.

The metre bridge

A metre bridge is a Wheatstone bridge built on a bench. Instead of two fixed resistors Q and S sharing a ratio, it uses a single uniform wire, 100 cm long, stretched along a metre scale. The unknown resistance X sits in the left gap and a resistance box R in the right gap. The galvanometer connects from the point between the gaps to a sliding contact, the jockey, which you press on the wire.

The two pieces of wire on either side of the jockey play the part of the other two arms. Because the wire is uniform, resistance is proportional to length. If the jockey is at l cm from the left end, the left piece has resistance proportional to l and the right piece to (100 − l). Slide the jockey until the galvanometer shows zero: the balance length.

The balance condition P/Q = R/S becomes X/R = l/(100 − l). The wire’s resistance per centimetre cancels, so you never need to know it.

Metre bridge examples

Example 4. The resistance box reads R = 10 Ω and the needle is at zero when the jockey is at l = 60 cm. The other piece is 100 − 60 = 40 cm. So X = 10 × 60 ÷ 40 = 15.00 Ω.

Example 5. With R = 12 Ω the balance comes at l = 45 cm. Then X = 12 × 45 ÷ 55 = 9.82 Ω. Notice X is smaller than R because the balance point is less than 50 cm.

The reverse problem: the simulation defaults have X = 15 Ω and R = 10 Ω. Balance should come at l = 100 X ÷ (X + R) = 60.0 cm. At the starting jockey position of 40 cm the galvanometer shows -11.07 mA, and the formula would wrongly give X = 6.67 Ω. Moving the jockey to the balance point removes that error.

The table fixes R = 10 Ω and shows how X depends on the balance length. Equal lengths on both sides give X = R. The values rise steeply beyond 70 cm.

Why the balance point should be near the middle

Every reading of l has a small error. Suppose you can read the scale to 0.1 cm. The relative error in X is then 0.1 × (1/l + 1/(100 − l)) × 100 per cent.

At l = 50 cm that is 0.40 %. At l = 25 cm it is 0.53 %. At l = 10 cm or 90 cm it is 1.11 % or 1.11 %. The error is smallest in the middle. So in a real experiment you choose R so that the balance falls between about 40 and 60 cm.

Sources of error and precautions

These are standard viva and practical-exam points.

• Connecting wires and copper strips have a little resistance. This end resistance is more important at the ends of the wire, another reason to balance near the middle.
• Do not leave the key closed for long. Heating changes the resistance of the wire and of X.
• Press the jockey lightly. Dragging it along the wire changes the wire’s thickness and ruins its uniformity.
• Take readings with R and X interchanged and average X. This cancels the end-error.
• The galvanometer is delicate. Use a high series resistance first and remove it only when you are near balance.

Where bridges are used

A strain gauge on a bridge or an aeroplane wing changes its resistance by a tiny amount when stretched. A bridge circuit detects that small change as an imbalance and turns it into a voltage. Many weighing scales, pressure sensors and thermometers work the same way. A light-dependent or temperature-dependent resistor placed in one arm turns a bridge into a sensor: when its resistance changes, the bridge unbalances, and the amount tells you how much it changed.

The same pattern shows up in Kirchhoff’s laws. A bridge is a two-loop circuit with a branch across the middle, and the Kirchhoff’s laws animation lets you solve such a circuit step by step.

Common mistakes

Four mistakes cost the most marks.

• Writing the condition with the wrong arms, such as P/R = Q/S. The ratio is of the two arms on the same side of the battery, P to Q and R to S.
• Using the lengths the wrong way round in the metre bridge. Check which gap holds the unknown. In this animation X is on the left, next to l.
• Thinking the battery voltage or galvanometer resistance affects balance. It does not.
• Forgetting that X = R·l ÷ (100 − l) needs the length measured from the end of the wire next to X.

Exam corner

A full answer to “derive the balance condition” has a labelled diagram, a clear statement that no current flows through the galvanometer, two equations for the equal potential drops, and the division step. That is four to five marks.

Metre bridge numericals normally give R and l and ask for X, and sometimes ask for the new balance length if R changes. Practise both directions with the sliders until the formula feels natural.

Revision list

Check yourself without looking.

• A balanced bridge has no current in the galvanometer: V_B = V_D.
• Balance condition: P/Q = R/S.
• Balance does not depend on the battery voltage or the galvanometer resistance.
• Metre bridge: X = R · l ÷ (100 − l), with l from the end next to X.
• Resistance of the wire is proportional to its length because it is uniform.
• Best accuracy when the balance point is near 50 cm.
• Interchange R and X and average to remove end errors.

Frequently asked questions