Kirchhoff's Laws & the Wheatstone Bridge

Set a supply, four arms and a detector. The page solves the bridge with two node equations and shows what the detector reads, where the balance point is, and why neither the supply nor the detector appears in it.

Worked example — Kirchhoff's Laws & the Wheatstone Bridge

Potential at C, from the node equation there	6.6667 V
Potential at D, from the node equation there	6.6667 V
Voltage across the detector	0.0000 mV
Current through the detector	0 A
How far from balance, as R₁R₄ − R₂R₃	0.00
What R₄ would have to be	300.0000 Ω
Current out of the supply	0.055556 A
Current unaccounted for at the two nodes	5.90 × 10⁻¹⁷ A

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Kirchhoff's Laws & the Wheatstone Bridge

Two conservation laws, and everything on this page follows from them. Kirchhoff's current law says the currents arriving at a junction add up to the currents leaving it, because charge does not collect anywhere. Kirchhoff's voltage law says the voltage changes around any closed loop add up to zero, because a charge brought back to where it started has gained nothing. Applied to a circuit with n unknown node potentials they give n equations, and that is enough whatever the shape. A Wheatstone bridge is the smallest circuit where you need them: four arms between a supply and a detector across the middle, balanced when R₁/R₂ = R₃/R₄.

What each symbol means

E: the supply, in volts. It sets every current in the circuit and appears nowhere in the balance condition
V_C: the potential at the left-hand midpoint, in volts. One of the two unknowns, and the node equation there is one of the two equations
V_D: the potential at the right-hand midpoint. The other unknown
I_g: the detector current, in amperes. Positive when it runs from C to D; zero, and only zero, at balance
R_g: the detector's own resistance, in ohms. It changes how much current flows when the bridge is off balance, and not where the balance is

Assumes: Ideal wires, linear resistors, steady DC, and a supply with no internal resistance of its own — which the page before this one spent its whole length saying is never true. That assumption is harmless here for the reason this page is about: an internal resistance sits in series with the supply, changes every current in the circuit, and moves the balance point not at all. The detector is treated as a plain resistance, which a moving-coil galvanometer is and a modern electronic one is not.
Breaks when: There is one more way to simplify a bridge and it is worth meeting before someone else raises it: a delta-to-star transform rewrites any three resistors meeting at a triangle as three meeting at a point, and applying it to either half of a bridge does leave something the series and parallel rules can finish. So the honest claim is the one insight 1 makes — no two of these five are in series or in parallel — rather than that the circuit is beyond simplification. The transform is a change of variables that gives the right answer and tells you nothing about which branch the current went down; Kirchhoff gives you every branch current at once. On alternating current, where an arm has reactance as well as resistance and balance needs two conditions rather than one — magnitude and phase — which is what an impedance bridge is for. On anything non-linear: a diode or a lamp filament in an arm has no single resistance to put in the ratio. And on the accuracy question, in a way worth naming: a bridge measurement is only as good as the three arms you think you know, because R₃ = R₁R₄/R₂ inherits all three of their errors. The detector contributes nothing to the accuracy at all — it only has to be able to tell zero from not-zero, which is a far easier instrument to build than one that measures.

Five resistors, and no two of them are in series 🖖

Everything the last three pages taught is spent here. Look at the circuit: node C has three wires on it — R₁, R₂ and the detector — and a series pair needs a junction that nothing else touches, so R₁ and R₂ are not in series. Node D is the same. No two of the five share both of their ends either, so nothing is in parallel. The combination rules have nothing to take hold of. Now press Detector removed: with the galvanometer gone, node C has two wires and the tool reports R₁+R₂ and R₃+R₄ at once. The bridge is irreducible because of the detector, which is exactly why one is put there. What finishes the job instead is two conservation laws — charge at a node, energy round a loop — and two equations in two unknowns.

Zero stays zero whatever the battery is doing 🖖

Press Balanced and read the detector column of the table below. It is zero at 1 V and at 100 V; it is zero with a 5 Ω detector and with a 500 Ω one. The supply current in the next column runs from 0.005556 A to 0.555556 A across those same six rows, a factor of a hundred, and the null does not move. That is what makes a bridge an instrument rather than a calculation. The two quantities hardest to pin down — how much the source really delivers, and how much the meter itself loads the circuit — drop out of a reading of zero. An ohmmeter has to know both of them and is only as good as it knows them.

Only the ratio matters, so pick arms you can afford 🖖

Balance is R₁/R₂ = R₃/R₄, and nothing else about the four resistances is in it. Press Balanced, then Ten times the arms: R₁ and R₂ go from 100 and 200 Ω to 1,000 and 2,000, the detector still reads 0 A, and the supply current drops from 0.055556 A to 0.025556 A. Same measurement, less than half the current. That is what a ratio arm is for — choose the pair that suits the current you can spare or the heating you can tolerate, and the unknown comes out the same either way.

Problem solved in full

Measuring a resistor you cannot read the value of 6 steps
R₃ is unknown. R₁ = 100 Ω and R₂ = 200 Ω are standards good to 0.01%, and R₄ is a decade box you turn until the detector reads zero. It nulls at R₄ = 300.0000 Ω. Find R₃, and say how well you know it.

Open this problem in the tool
1. At the null no current crosses the detector, so C and D are at the same potential. That is the only fact the whole method needs.
  
  $I_{g} = 0 ⟹ V_{C} = V_{D}$
2. With nothing flowing through R_g, R₁ and R₂ are a plain two-resistor divider, and so are R₃ and R₄. Each midpoint is the supply times its lower arm over its pair.
  
  $V_{C} = E \frac{R _{2}}{R _{1} + R _{2}}, V_{D} = E \frac{R _{4}}{R _{3} + R _{4}}$
3. Set the two midpoints equal and the supply cancels off both sides immediately.
  
  $\frac{R _{2}}{R _{1} + R _{2}} = \frac{R _{4}}{R _{3} + R _{4}}$
4. Cross-multiply and it collapses to a product of pairs: R₁R₄ = R₂R₃.
  
  $R_{2} R_{3} + R_{2} R_{4} = R_{1} R_{4} + R_{2} R_{4} ⟹ R_{1} R_{4} = R_{2} R_{3}$
5. So R₃ = R₁R₄/R₂ = 100 × 300.0000 ÷ 200 = 150.0000 Ω. Read back what is in that line — three resistances and nothing else. No supply voltage, no detector resistance, no current.
  
  $R_{3} = \frac{R _{1} R _{4}}{R _{2}} = \frac{100 \cdot 300 . 0000}{200} = 150.0000 Ω$
6. Now the accuracy, which is the part a textbook skips. Fractional errors add through a product and a quotient, so R₃ carries the errors of all three known arms and none of the detector's.
  
  $\frac{Δ R _{3}}{R _{3}} \leq \frac{Δ R _{1}}{R _{1}} + \frac{Δ R _{4}}{R _{4}} + \frac{Δ R _{2}}{R _{2}} = 3 \times 0 . 01% = 0 . 03%$
Answer

R₃ = 150.0000 Ω, to about 0.03%. The three standards are 0.01% each and their fractional errors add in the worst case; the detector contributes nothing, because it was never asked for a number — only for the difference between zero and not zero. That is the trade the whole instrument is built on, and it is why a bridge beat every direct method for a century and a half: it is far easier to build something that can tell when a current is absent than something that can tell you accurately how big one is.

$R_{3} = \frac{R _{1} R _{4}}{R _{2}} = 150.0000 Ω \pm 0 . 03%$

Learning path

What Ohm’s law leaves out

References (3)

The laws themselves, published while Kirchhoff was still a student. The paper is about current spreading through a plane conductor; the two rules this page uses are stated on the way: G. Kirchhoff, "Ueber den Durchgang eines elektrischen Stromes durch eine Ebene, insbesondere durch eine kreisförmige." Annalen der Physik 140(4), 1845, pp. 497-514. DOI 10.1002/andp.18451400402
The bridge, from the Bakerian Lecture that made it famous — and in which Wheatstone credits Christie rather than claiming it: C. Wheatstone, "XIII. The Bakerian lecture. — An account of several new instruments and processes for determining the constants of a voltaic circuit." Philosophical Transactions of the Royal Society of London 133, 1843, pp. 303-327. DOI 10.1098/rstl.1843.0014
Christie's paper of ten years earlier, where the diamond of four resistances and a null detector appears first. The bridge carries the wrong name and has done since 1843: S. H. Christie, "VIII. The Bakerian Lecture. — Experimental determination of the laws of magneto-electric induction in different masses of the same metal, and of its intensity in different metals." Philosophical Transactions of the Royal Society of London 123, 1833, pp. 95-142. DOI 10.1098/rstl.1833.0011

By the 42Math team · Reviewed against published sources. Corrections welcome.