DC bridges: Wheatstone and Kelvin double bridge

Why it matters

Bridges compare an unknown with known standards at a null, so the result depends on resistor ratios rather than on a meter's calibration. That is why the Wheatstone bridge is still the reference method for medium resistance and the heart of strain-gauge and RTD circuits, and why the Kelvin double bridge is the way to measure milliohm shunts and joints accurately.

Key ideas

Wheatstone bridge. Four arms P, Q, R, S form a diamond. A DC source E is connected across one diagonal and a galvanometer across the other. P and Q in series form one leg across the supply and R (unknown) and S (standard, variable) form the other, with P and R joined to the same supply terminal. P and Q are called the ratio arms.

• At balance the galvanometer current is zero, so the potentials of its two terminals are equal and P/Q = R/S.
• The result does not depend on E or on the galvanometer's calibration — only on the ratio arms and the standard, which is the source of the bridge's accuracy.
• Errors and limits: limiting errors of the known arms; sensitivity of the galvanometer (it limits how precisely balance can be detected); heating of the arms by the current; thermo-electric EMFs (reverse the supply and average); lead and contact resistance (limits use below about 1 Ω); leakage (limits use above about 100 kΩ).

Unbalanced bridge. Slightly off balance, the bridge is best analysed by Thevenin's theorem seen from the galvanometer terminals. The galvanometer current gives the deflection, and the bridge sensitivity is the deflection per unit fractional change in R. Sensitivity is greatest when the arms are equal; this is the basis of strain-gauge and RTD bridges used in deflection mode.

Kelvin double bridge. For resistances below about 1 Ω, the unknown R and the standard S are four-terminal resistors joined by a heavy link of resistance r. In a plain Wheatstone bridge r would add to one arm or the other depending on where the galvanometer is tapped. Kelvin's solution adds a second pair of ratio arms p and q across the link, with the galvanometer connected to their junction.

• The exact balance is R = (P/Q)·S + [q·r/(p + q + r)]·(P/Q − p/q).
• With the two ratio pairs ganged so that P/Q = p/q, the second term vanishes and R = (P/Q)·S — independent of r and of lead and contact resistance.
• Thermo-electric EMFs matter at these low voltages, so readings are taken with the current reversed and averaged. Range about 1 μΩ to 1 Ω, accuracy around 0.05–0.2 %.

Formulas

R = S·P / Q (Wheatstone balance) P, Q = ratio arms; S = standard arm; R = unknown (all Ω).

Vth = E·[R/(R + S) − P/(P + Q)] (open-circuit voltage at the galvanometer terminals) Rth = P·Q/(P + Q) + R·S/(R + S) (resistance seen by the galvanometer, source short-circuited) Ig = Vth / (Rth + Rg) E = supply voltage (V); Rg = galvanometer resistance (Ω); Ig in A. Galvanometer deflection = Ig × current sensitivity (e.g. mm/μA).

R = (P/Q)·S + [q·r/(p + q + r)]·(P/Q − p/q) ; with P/Q = p/q: R = (P/Q)·S (Kelvin double bridge) p, q = inner ratio arms (Ω); r = link resistance (Ω).

Worked examples

Example 1 — balanced Wheatstone bridge. The ratio arms are P = 1000 Ω and Q = 100 Ω, and balance is obtained with S = 4.287 Ω. Find R.

1. R = S·P/Q = 4.287 × 1000/100 = 42.87 Ω.

Answer: R = 42.87 Ω. Using P/Q = 10 lets a 4-dial standard measure to four significant figures over a range ten times larger.

Example 2 — unbalanced bridge and smallest detectable change (GATE level). P = Q = 1 kΩ, S = 2 kΩ and R = 2.005 kΩ. E = 10 V. The galvanometer has Rg = 100 Ω and a current sensitivity of 10 mm/μA. Find (a) the galvanometer current and deflection, (b) the smallest change in R the bridge can detect if 1 μA is the least detectable current.

1. Vth = 10 × [2005/4005 − 1000/2000] = 10 × (0.500624 − 0.5) = 6.242 mV.
2. Rth = 1000 × 1000/2000 + 2005 × 2000/4005 = 500 + 1001.2 = 1501.2 Ω.
3. Ig = 6.242 × 10⁻³ / (1501.2 + 100) = 3.898 μA.
4. Deflection = 3.898 × 10 = 39.0 mm.
5. Near balance Ig is proportional to ΔR, so ΔR_min = 5 Ω × (1/3.898) = 1.28 Ω.

Answer: (a) Ig = 3.90 μA, deflection 39.0 mm; (b) about 1.28 Ω (0.064 % of 2 kΩ).

Example 3 — Kelvin double bridge. P = 1000 Ω, Q = 100 Ω, p = 1000 Ω, q = 100 Ω; balance with S = 1.000 mΩ; link r = 0.5 mΩ. Find R.

1. P/Q = p/q = 10, so the correction term is zero.
2. R = (P/Q)·S = 10 × 1.000 mΩ = 10.00 mΩ.

Answer: R = 10.0 mΩ, unaffected by the 0.5 mΩ link (a plain Wheatstone bridge could be in error by up to 5 %).

Common mistakes

• Writing the balance condition with the arms in the wrong positions. Identify the two legs (series pairs across the supply); balance means each leg divides the supply in the same ratio, so P/Q = R/S with P and R on the same supply terminal.
• Forgetting to short the source when finding Rth for the galvanometer.
• Assuming the Kelvin bridge needs no matching: the link term vanishes only when P/Q = p/q.
• Using a Wheatstone bridge for sub-ohm values; lead and contact resistance (often tens of milliohms) appear in the result.
• Ignoring thermo-EMFs in low-resistance work; reverse the current and average.

For GATE IN

Expect NAT questions on balance values, the Thevenin voltage or galvanometer current of a slightly unbalanced bridge, and bridge output for a small fractional change (also seen in sensor bridges). MCQs test why the Kelvin bridge removes link and lead resistance and what limits the Wheatstone bridge at low and high values. Practise the Thevenin method on unequal-arm bridges.

Quick check

1. P = 100 Ω, Q = 10 Ω, S = 25.3 Ω at balance. R?
2. What condition makes the Kelvin bridge balance independent of the link resistance?
3. Does the balance of a Wheatstone bridge depend on the supply voltage?
4. An equal-arm bridge (all arms R) with E = 5 V has one arm changed by +1 %. Approximate open-circuit output? Answers: 1. 253 Ω; 2. P/Q = p/q; 3. no (only its sensitivity does); 4. about E·(ΔR/R)/4 = 12.5 mV (exactly 12.4 mV).