Current divider calculator

A current splits between two parallel resistors as I₁ = I × R₂ / (R₁ + R₂): the current through one resistor is set by the other one over the sum, so the smaller resistor carries the larger share. One ampere into 100 Ω and 300 Ω puts 750 mA through the 100 Ω branch and 250 mA through the 300 Ω. For more branches each takes its conductance over the total conductance. Enter the total current and the resistors to see the split, the voltage across the set and the power in each branch — or enter the current one branch must carry and get the resistor that arranges it.

Fig 1 — 1.00 A into 2 parallel branches; 75.0 V appears across all of them.

Mode

Total current (mA)

R1 (Ω)

R2 (Ω)

R3 (0 = none) (Ω)

R4 (0 = none) (Ω)

Resistor tolerance (%)

Parallel resistance · voltage across: 75.0 Ω · 75.0 V
R1 = 100 Ω: 750 mA · 56.3 W · 75.0 %
R2 = 300 Ω: 250 mA · 18.8 W · 25.0 %
Total dissipation: 75.0 W
I1 worst case, ±1 % parts: 746 mA to 754 mA

100 Ω dissipates 56.3 W; check the package rating.

How this is calculated

Standard: Kirchhoff's current law; Ohm's law

$V = I \cdot R_{par}, \quad R_{par} = \left(\sum_k \frac{1}{R_k}\right)^{-1}$: Every branch sees the same voltage; this is the voltage.
$I_k = \frac{V}{R_k} = I \cdot \frac{G_k}{\sum G}, \quad G = \frac{1}{R}$: The general rule, for any number of branches.
$I_1 = I \cdot \frac{R_2}{R_1 + R_2}$: Two branches. The other resistor is in the numerator.
$R_2 = \frac{R_1 \cdot I_1}{I - I_1}$: The previous line solved for R2, for a wanted branch current.
$P_k = I_k^2 R_k, \quad \sum P_k = V \cdot I$
$I_{1,max} = I \cdot \frac{R_2(1+t)}{R_1(1-t) + R_2(1+t)}$: Worst case high with tolerance t; the low case swaps the signs.

Assumptions

The total current is fixed by the source. Across a voltage source the branches set the total and the rule does not apply as written.
DC, or branches of equal impedance angle; reactive branches divide as phasors.
Ideal resistors at one temperature. Self-heating in a heavily loaded branch moves current to the cooler one.
Connection resistance is zero. At milliohm shunt values it is not, which is what Kelvin connections are for.
The E-series snap picks the nearest stock value, not the nearest that errs in a chosen direction; the note reports the resulting current.

What sets how a current divides

Parallel branches share one voltage. That is the whole rule: the current entering the set produces V = I × R_parallel across every branch at once, and each branch then carries V / R_k. Written out for two resistors it becomes the form in the textbooks — I₁ = I × R₂ / (R₁ + R₂) — and the thing to notice is that the current through R₁ is set by R₂. The other resistor is in the numerator. That is the mirror image of thevoltage divider, where the output across R₂ has R₂ on top, and it is where the sign errors come from: in a current divider the smaller resistor takes the larger share.

For more than two branches the two-resistor form does not extend directly, and the honest statement is in conductances: each branch takes the fraction G_k / ΣG of the total, where G = 1/R. The tool works that way for any number of branches, so the two-branch formula is a special case of it rather than the other way round.

The precondition matters more than the arithmetic. A current divider divides a current that is already fixed — from a constant-current LED driver, a shunt-regulated line, an inductor in the middle of a cycle. Put two resistors across a voltage source and the "division" is not a division: the resistors set the total between them, and each simply carries V / R. Both calculators on this site are correct; which one describes a circuit depends on what is holding constant.

Worked example: 1 A into 100 Ω and 300 Ω

A 1 A constant-current source feeding a 100 Ω branch in parallel with a 300 Ω branch.

R_par   = (100 × 300) / (100 + 300)   = 75 Ω
V       = 1 A × 75 Ω                   = 75 V   across both branches
I1      = 75 V / 100 Ω                 = 750 mA   (= 1 A × 300 / 400)
I2      = 75 V / 300 Ω                 = 250 mA   (= 1 A × 100 / 400)
P1      = 0.75² × 100                  = 56.3 W
P2      = 0.25² × 300                  = 18.8 W
total   = 75 V × 1 A                   = 75 W     (= P1 + P2)

solve:  want 750 mA through 100 Ω from 1 A
R2      = 100 × 0.75 / (1 − 0.75)      = 300 Ω

The calculator gives 750 mA, 250 mA, 75 V and 75 W, and solves 300 Ω. The classic application is a meter shunt: a 50 µA, 1 kΩ movement made to read 1 A full scale needs R₂ = 1000 × 50 µA / (1 A − 50 µA) = 50.0025 mΩ in parallel — and the "− 50 µA" is why the shunt is not quite V/I of the movement, though it is close enough that most meters ignore it.

Where the current divider stops being valid

The source has to be a current source. If it is a voltage with a series resistance, the total current changes as the branches change, and the division is only part of the answer. Model the whole thing — source resistance in series with the parallel set — with Ohm's law from the top.
DC, or equal-phase branches. With a capacitor or inductor in one branch, the currents no longer share a phase and the shares combine as phasors. The formula still holds for the magnitudes only when every branch has the same impedance angle.
Tolerance acts twice. Because I₁ rises when R₁ falls and when R₂ rises, 1 % parts give a band close to ±2 % on the branch current. For a shunt this is the reason to use a 0.1 % part, or to calibrate.
Self-heating. A branch dissipating watts rises in temperature, its resistance rises with copper's or the film's coefficient, and it hands current to the cooler branch. Low-value shunts are specified with a temperature coefficient for exactly this reason.
Resistance in the connections. At milliohm shunt levels the solder joints and the trace to the paralleled part are resistors too. A four-terminal (Kelvin) shunt exists so the division happens where the sense leads say it does.

Common current divider mistakes

Putting R₁ in the numerator. I₁ = I × R₂/ (R₁ + R₂). The resistor whose current is being found is the one that is not on top.
Expecting the larger resistor to carry more current. It carries less; the shares go as the conductances.
Applying the rule across a voltage source. Two resistors across a battery each carry V / R, and adding a third does not reduce the first two — it adds to the total.
Paralleling LEDs on one resistor and calling it a divider. LEDs are not resistors; the string with the lowest forward voltage takes almost everything. One resistor per string, as theLED resistor calculator insists.
Sizing a shunt from V / I_meter and forgetting the meter's own current comes out of the total. Small at 50 µA; not small when the "meter" is a 20 mA ADC front end.