Current Divider Calculator
Current Divider Calculator
How a current splits between resistors in parallel: the current in every branch, its share as a percentage, the voltage across the group and the power each resistor has to get rid of — for up to six branches, with the two-resistor shortcut and the sign of it that everybody gets wrong.
Current divider
100 mA into two branches, R1 = 100 Ω and R2 = 300 Ω
Current divider formulas
two branches only: I1 = I × R2 ÷ (R1 + R2), I2 = I × R1 ÷ (R1 + R2)
the group’s voltage: V = I × Rp; the power in a branch: Pn = In² Rn = V² ÷ Rn
- I
- the total current arriving at the node where the branches meet
- R p
- the parallel combination of every branch — always smaller than the smallest branch
- R n
- the branch you want the current in. Its own resistance is on the BOTTOM, which is what makes this the dual of the voltage divider
Worked example
100 mA into two branches, R1 = 100 Ω and R2 = 300 Ω
Rp = 1 ÷ (1/100 + 1/300) = 75 Ω
V across the pair = 100 mA × 75 Ω = 7.5 V
I1 = 7.5 V ÷ 100 Ω = 75 mA, which is 75.0% of the total
I2 = 7.5 V ÷ 300 Ω = 25 mA, 25.0% — and 75 + 25 = 100 mA, as Kirchhoff requires
By the shortcut: I1 = 100 mA × 300 ÷ (100 + 300) = 75 mA — note R2 on top
Powers: R1 gets 562.5 mW, R2 gets 187.5 mW
How two branches split one amp
| R1 | R2 | R1 ∥ R2 | R1 takes | R2 takes | I1 ÷ I2 |
|---|---|---|---|---|---|
| 100 Ω | 100 Ω | 50.00 Ω | 50.00% | 50.00% | 1.000 |
| 100 Ω | 300 Ω | 75.00 Ω | 75.00% | 25.00% | 3.000 |
| 100 Ω | 900 Ω | 90.00 Ω | 90.00% | 10.00% | 9.000 |
| 1,000 Ω | 100 Ω | 90.91 Ω | 9.09% | 90.91% | 0.100 |
| 10 Ω | 1,000 Ω | 9.90 Ω | 99.01% | 0.99% | 100.000 |
| 4,700 Ω | 10,000 Ω | 3,197.28 Ω | 68.03% | 31.97% | 2.128 |
Why this is the voltage divider turned inside out
Resistors in parallel all sit between the same two nodes, so they all have the same voltage across them. Feed the group a total current I and that voltage is V = I × Rp, where Rp is the parallel combination. Each branch then takes V ÷ Rn of its own accord, which rearranges to In = I × Rp ÷ Rn. That is the entire derivation, and it is worth doing it that way round rather than memorising a second rule: 100 mA into 100 Ω and 300 Ω gives Rp = 75 Ω, V = 7.5 V, and the branches take 75 mA and 25 mA.
The duality. A voltage divider is the same statement with voltage and current swapped. There, resistors in series share a current and the branch’s OWN resistance goes on top over the series total: Vn = V × Rn ÷ ΣR. Here, resistors in parallel share a voltage and the branch’s own resistance goes on the BOTTOM, under the parallel total. Conductance is the quantity that makes the two identical: in series the voltages divide in proportion to resistance, in parallel the currents divide in proportion to conductance. So the biggest resistor takes the most voltage and the smallest resistor takes the most current, and if you can only remember one thing, remember that.
The shortcut, and the mistake everyone makes. For exactly two branches, Rp ÷ R1 = R2 ÷ (R1 + R2), so I1 = I × R2 ÷ (R1 + R2). The resistor on top is the OTHER one. It looks exactly like the voltage divider formula and it is not, and swapping the two is the single commonest arithmetic error in this part of the subject. The sanity check takes a second: the branch with the smaller resistance must come out with the larger current. With 100 Ω and 300 Ω, the 100 Ω branch gets three quarters of the current, not a quarter. The shortcut is also only a shortcut for two branches — with three or more there is no such cancellation and you must work out Rp first.
Where it actually comes up. Sizing a shunt across an ammeter or a current-sense resistor, where the meter’s own resistance is one branch and the shunt the other. Sharing current between paralleled LEDs or paralleled power resistors, where the point of the calculation is usually to show that the sharing is worse than you hoped: two nominally equal resistors 1% apart split 50.25/49.75, but two paralleled diodes or transistors do not share by resistance at all and need ballast resistors. Working out how much of a transducer’s output current reaches the load rather than leaking through a stray path. And reading any circuit where a branch current, not a node voltage, is the thing you need.
What the model leaves out. The current source is ideal — it delivers I whatever voltage that takes, which a real source will not do once the group’s resistance rises. The resistors are ideal and at one temperature. And the branches are assumed to meet at a point: any resistance in the wiring between them is a resistor in its own right and belongs in the model as one. For the totals and the voltage-driven case — a supply across the group rather than a current into it — the series and parallel resistor calculator is the page you want; it starts from a supply voltage where this one starts from a current. For the voltage-and-current arithmetic on a single resistor, the Ohm’s law calculator.
Frequently asked questions
What is the current divider formula?
I_n = I × R_p ÷ R_n, where R_p is the parallel combination of every branch. For exactly two branches it simplifies to I_1 = I × R_2 ÷ (R_1 + R_2) — the opposite resistor on top.
Why is the other resistor on top in the two-resistor formula?
Because the current in a branch is the shared voltage divided by that branch’s own resistance, so its own resistance is underneath. For two resistors, R_p ÷ R_1 works out to R_2 ÷ (R_1 + R_2), which is where the other resistor comes from. If you use the voltage divider formula by mistake you get the split exactly backwards.
Which branch gets the most current?
The one with the smallest resistance, always. Currents divide in proportion to conductance, which is the reciprocal of resistance. This is the opposite of a series voltage divider, where the largest resistance takes the largest share of the voltage.
Does a current divider work with more than two resistors?
Yes, but only the general form does: work out the parallel combination of all the branches, then I_n = I × R_p ÷ R_n for each one. The two-resistor shortcut has no three-resistor version. This page takes up to six branches.
How do I check my answer?
The branch currents must add up to the total — that is Kirchhoff’s current law, and it is the only check you need. It is also worth confirming that the smallest resistor came out with the largest current.
Can I use this for capacitors or inductors?
Not directly. At a single frequency the same algebra works with impedances instead of resistances, but the currents are then complex numbers and add as phasors rather than arithmetically, so two branch currents can each be larger than the total. This page is resistive and DC.
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References
- Nilsson JW, Riedel SA. Electric Circuits, 11th ed. Pearson, 2019. §3.3 “The Voltage-Divider and Current-Divider Circuits” and §3.4 “Voltage Division and Current Division”. Section titles confirmed against Pearson’s own published front matter for the eleventh edition.
- Horowitz P, Hill W. The Art of Electronics, 3rd ed. Cambridge University Press, 2015. Chapter 1: voltage, current and resistance; resistors in parallel; power in resistors.
- IEC 60063:2015. Preferred number series for resistors and capacitors (the E6, E12, E24, E48, E96 and E192 series). International Electrotechnical Commission.