Current splits between parallel resistors according to the current divider formula. Enter up to six to get the current, share of the total and power in each branch, with an optional tolerance check. Or size a shunt so one branch carries exactly the current you want.
Optional: adds the worst-case current range of each branch.
Current in R1
6 mA
54.5 % of the total
Assumes the total current is fixed (a current source, or a circuit whose current does not depend on these resistors). If a voltage source drives the branches directly, each current is simply V / R.
Branch by branch
| Branch | Resistance | Current | Share | Power |
|---|---|---|---|---|
| R1 | 1 kΩ | 6 mA | 54.5 % | 36 mW |
| R2 | 3 kΩ | 2 mA | 18.2 % | 12 mW |
| R3 | 2 kΩ | 3 mA | 27.3 % | 18 mW |
Use I_n = I_total × R_eq ÷ R_n, where R_eq is the parallel combination. For two resistors this is I1 = I_total × R2 ÷ (R1 + R2). For example, 11 mA into 1 kΩ, 3 kΩ and 2 kΩ in parallel (R_eq = 545.5 Ω) splits as 6 mA, 2 mA and 3 mA: the smallest resistor takes the largest share.
Resistors in parallel all see the same voltage, V = I_total × R_eq. Ohm’s law then gives the current division equation: each branch current is the total scaled by the ratio of the equivalent resistance to that branch.
For exactly two branches the equivalent resistance cancels out of the current division formula:
Each branch dissipates P_n = I_n² × R_n = V² ÷ R_n. Turned around, the shunt that leaves a chosen current I_b in a branch of resistance R_b is:
Three cases you can reproduce in the calculator above: a textbook split, two mismatched sense resistors and an ammeter shunt.
1 ÷ R_eq = 1/1000 + 1/3000 + 1/2000, so R_eq = 545.5 Ω and the voltage across the group is 11 mA × 545.5 Ω = 6 V. The branches carry 6 mA, 2 mA and 3 mA (54.5 %, 18.2 % and 27.3 % of the total), and the 1 kΩ dissipates 36 mW. This is the textbook example in Lessons in Electric Circuits: the 1 kΩ carries twice the current of the 2 kΩ.
Two nominally equal 100 mΩ resistors share 5 A, but one is really 105 mΩ. The 100 mΩ part carries 2.561 A and dissipates 656 mW; the 105 mΩ part carries 2.439 A and 625 mW. With two truly equal ±1 % parts the worst case is 2.475 A to 2.525 A per resistor: the split error is about the tolerance, so derate each part for slightly more than half the total.
A 1 mA, 500 Ω movement drops 500 mV at full scale. The shunt carries the other 4.999 A, so R_shunt = 500 × 0.001 ÷ (5 − 0.001) = 100.02 mΩ, and it dissipates 2.5 W. A shunt this small should be a four-terminal part: one square of 1 oz copper is already 493 µΩ, about 0.493 % of it.
The share depends only on the ratio R2 / R1. Find your ratio and read the fraction of the total in each resistor.
| R2 / R1 | Share in R1 | Share in R2 |
|---|---|---|
| 0.1 | 9.09 % | 90.9 % |
| 0.25 | 20 % | 80 % |
| 0.5 | 33.3 % | 66.7 % |
| 1 | 50 % | 50 % |
| 2 | 66.7 % | 33.3 % |
| 3 | 75 % | 25 % |
| 4 | 80 % | 20 % |
| 9 | 90 % | 10 % |
| 10 | 90.9 % | 9.09 % |
| 99 | 99 % | 1 % |
Shunt resistance and full-scale dissipation for common ranges, from R_shunt = R_m × I_m ÷ (I_range − I_m). The meter drops 500 mV at full scale on every range.
| Full-scale range | Shunt resistance | Shunt current | Shunt power |
|---|---|---|---|
| 100 mA | 5.05051 Ω | 99 mA | 49.5 mW |
| 1 A | 500.501 mΩ | 999 mA | 500 mW |
| 5 A | 100.02 mΩ | 4.999 A | 2.5 W |
| 10 A | 50.005 mΩ | 9.999 A | 5 W |
| 100 A | 5.00005 mΩ | 99.999 A | 50 W |
The rule assumes the total current is fixed. That holds for a current source (an LED driver, a sensor excitation source, an op-amp current pump) and for any branch point where the rest of the circuit sets the current. If an ideal voltage source drives the branches directly, each current is just V ÷ R and the other branches do not affect it.
Below about 1 Ω the copper matters. One square of 1 oz copper is about 493 µΩ, so a few extra squares of trace in one branch of a pair of 10 mΩ resistors shift the split by several percent. Route paralleled low-value resistors symmetrically, and use four-terminal parts for shunts. Temperature also moves the ratio when the branches are different materials: a copper meter coil rises about 3.93 % in resistance for a 10 °C rise (α ≈ 0.393 %/K), while a manganin shunt barely moves, so the meter reading drifts by about the same amount.
The formula needs linear branches. Diodes and LEDs in parallel do not share by resistance: the part with the lowest forward voltage takes most of the current, and its share grows as it heats. Give each one its own series resistor so the resistors, not the junctions, set the split.
Each branch of a parallel circuit carries I_n = I_total × R_eq / R_n, where R_eq is the parallel combination of all branches. Every branch sees the same voltage, so current divides in inverse proportion to resistance (in direct proportion to conductance).
I1 = I_total × R2 / (R1 + R2) and I2 = I_total × R1 / (R1 + R2): each resistor gets the total times the other resistor over the sum. For 10 A into 15 Ω and 5 Ω, the 15 Ω carries 2.5 A and the 5 Ω carries 7.5 A.
Use I_n = I_total × R_eq / R_n with 1 / R_eq = 1/R1 + 1/R2 + 1/R3. For 11 mA into 1 kΩ, 3 kΩ and 2 kΩ, R_eq = 545.5 Ω and the branches carry 6 mA, 2 mA and 3 mA.
Two or more parallel branches that share one total current. Every branch sees the same voltage, so current division follows conductance: the lowest resistance carries the largest share.
All parallel branches have the same voltage across them, and I = V / R, so the branch with the lowest resistance passes the most current. Half the resistance means twice the current.
A voltage divider uses resistors in series and scales by R_n / R_total; a current divider uses resistors in parallel and scales by R_total / R_n. Both ratios are below one, which is a quick check that you have not flipped the fraction.
R_shunt = R_meter × I_meter / (I_total − I_meter). A 1 mA, 500 Ω movement reading 5 A full scale needs 100.02 mΩ, which drops 500 mV and dissipates 2.5 W at full scale.
Only when their values are equal. At the same voltage, P = V² / R, so the lower-value resistor dissipates more. Two 100 mΩ parts that are really 100 and 105 mΩ split 5 A as 2.56 A and 2.44 A.
Yes, with complex impedances in place of resistances: I_n = I_total × Z_eq / Z_n. The branch currents are then phasors, so for a resistor and a capacitor in parallel the two branch current magnitudes add up to more than the total, because the currents are 90° apart.
A fixed total current into linear branches. If an ideal voltage source drives the branches directly, each current is just V / R and the other branches do not affect it. Parallel diodes and LEDs do not share by resistance, so give each one its own series resistor.
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