How this instrument works
Two branches wired across one pair of nodes sit under a single shared voltage, so the only thing deciding how charge splits is how readily each path passes it. Stated in conductance the rule reads plainly — a branch claims the fraction G₁ ⁄ (G₁ + G₂) of everything arriving at the junction. Convert back to resistances, clear the fractions, and the numerator swaps sides: branch one's share comes out written with R₂ on top. That swap is neither a typo nor a sign convention. It is simply what (1 ⁄ R₁) ÷ (1 ⁄ R₁ + 1 ⁄ R₂) collapses to, and putting R₁ there instead is the commonest slip made with this expression.
Nobody is individually credited with the divider — it is a two-line consequence of Kirchhoff's junction rule and Ohm's relation — yet by the 1880s it was quietly load-bearing inside a manufactured product. Moving-coil movements are exquisitely sensitive and hopeless at scale: a fine instrument reaches full deflection on a few tens of microamperes and would be destroyed by a single ampere. Edward Weston's answer, sold from 1888 in his portable direct-reading meters, was to hang a very low resistance across the movement so that the split routed almost everything harmlessly past it. Multi-range instruments refined the idea into the Ayrton shunt of the 1890s, a tapped chain that never leaves the movement unprotected while the range switch travels. Weston also gave those shunts their alloy — manganin, patented in 1892, chosen because its resistance barely stirs with temperature.
The expression assumes a good deal it does not say aloud. It assumes the arriving figure is genuinely fixed, which is true of a current source and false of a battery behind its own internal resistance, where lowering either branch raises the total draw and quietly invalidates your starting number. It assumes exactly two paths meeting at both ends with no third load tapping in between, and that both parts are ohmic and sitting at one temperature. Wiring counts as zero ohms, a fiction that gets expensive in shunt work, where a milliohm-class sense element competes with a few centimetres of copper — which is why such parts carry four terminals, keeping sense leads out of the measured path. Push into the megahertz and each path acquires reactance, so the split turns frequency-dependent and picks up a phase angle.
- Put your source figure into Total current in. The unit menu takes milliamps and kiloamps as well as amps.
- Enter Branch 1 resistance and Branch 2 resistance in ohms, kilohms, or megohms — whichever matches what is printed on the parts.
- Read Current through branch 1 and Current through branch 2. The smaller resistance must show the larger figure; if it does not, the two resistances went in the wrong way round.
- Confirm the pair adds back to your total, then multiply either branch figure by its own resistance to get the voltage sitting across both.
Worked example — 1 A into 100 Ω beside 300 Ω
A bench source is set to push a steady 1.000 A into two resistors wired side by side, one of 100 Ω and one of 300 Ω. Put 1 into Total current in, 100 into Branch 1 resistance, 300 into Branch 2 resistance. Current through branch 1 returns 0.75 A, since 1 × 300 ⁄ 400 = 0.75, and Current through branch 2 returns 0.25 A. Three quarters of the arriving charge goes down the easier path — and notice that its answer was built from its neighbour's 300 Ω, the swap everybody gets backwards.
Two independent checks confirm the pair without redoing the sum. The branch figures add to 1.000 A, as the junction rule demands. Each branch must also show an identical voltage: 0.75 × 100 = 75 V, and 0.25 × 300 = 75 V. That shared 75 V settles the thermal question too, which matters rather more than the arithmetic — branch one burns 0.75² × 100 = 56.25 W while branch two burns only 18.75 W. The smaller part is shedding three times the heat of its larger neighbour, so rate it for what it genuinely sees and never for an even share.
Questions
Why does the formula use the other resistance in the numerator?
Because the split follows conductance, and conductance is the reciprocal of resistance. A branch takes G₁ ⁄ (G₁ + G₂) of what arrives; multiply top and bottom by R₁R₂ and that turns into R₂ ⁄ (R₁ + R₂). Physically it says the easier path wins, something the resistance form can only express by lifting the neighbour into the numerator. If you cannot recall which way round it goes, decide first which side ought to win — the smaller resistance — then check that your expression hands it the bigger figure.
What units do the fields expect?
Amperes for the currents and ohms for the resistances, with prefix menus on every field so a 250 mA source or a 4.7 kΩ branch needs no hand conversion. The ampere is an SI base unit, fixed since the 2019 revision by the elementary charge e = 1.602176634 × 10⁻¹⁹ coulombs rather than by the old force-between-two-wires definition. Because the split is a pure ratio, the two resistances only need to share a scale: 100 Ω beside 300 Ω divides identically to 0.1 kΩ beside 0.3 kΩ. The total is what sets the size of the answer.
What happens if one branch is a dead short?
Everything goes through it. Set Branch 1 resistance to zero and the expression gives I₁ = I·R₂ ⁄ R₂ = I, leaving nothing for its neighbour — an accurate account of a solder bridge across a component. The reverse edge case is equally real: an open path behaves as infinite resistance and draws nothing, so the whole total keeps to the remaining route. Only both resistances at zero has no answer, since the junction then has no basis on which to apportion anything, and this sheet flags that rather than dividing by zero.
Can I use this when the supply is a battery instead of a current source?
Only once you know the total that actually flows with both branches in place. A voltage source holds volts steady, not amps — alter either resistance and the parallel combination shifts, so the total shifts with it. The usual route is to combine the branches into their equivalent parallel resistance, divide the supply voltage by that to get the total, then bring the figure here. If instead you have measured the voltage sitting across the pair, skip the divider altogether: each branch carries that voltage over its own resistance.
How is this different from a voltage divider?
They are duals of one another, and the difference lies in how the parts are wired. Series components share one current and split the voltage, each keeping its own resistance in the numerator. Parallel branches share one voltage and split the current, each taking its neighbour's resistance in the numerator. Both expressions carry an identical R₁ + R₂ underneath and look nearly the same written down, which is precisely why the numerators get confused so often.
How do I handle three or more parallel branches?
Work in conductances: each path takes G ⁄ ΣG of the total, with G = 1 ⁄ R. The neat two-branch form exists only because two reciprocal terms clear so tidily, and it does not generalise to three. You can still reach the answer on this sheet by folding — collapse every path except the one you want into a single equivalent parallel resistance, enter that as the second branch, and read the first. Paths of 100 Ω, 300 Ω and 600 Ω fed 1 A, for example, give a 200 Ω equivalent for the last two and 0.667 A down the 100 Ω route.