What a current divider is and when you need one
A current divider is any group of components wired between the same two nodes. Current arriving at the first node has to leave through one branch or another, and how much goes down each path is fixed entirely by the branch resistances. You reach for the divider rule whenever you know the total current but need a single branch current: sizing an ammeter shunt, working out how much fault current returns through a parallel ground path, checking that four paralleled power resistors in a dummy load share evenly, or answering a circuit-analysis question that hands you the source current rather than the source voltage.
The physical statement behind it is Kirchhoff's current law: charge does not pile up at a node, so the currents leaving equal the current arriving. The second statement is that parallel elements share one voltage. Put those together and each branch current is that shared voltage divided by that branch's resistance. Nothing else enters the calculation — not the source voltage, not what sits upstream, not the wire between the node and the resistor, provided that wire's resistance is small compared with the branch it feeds.
The dual of this problem is the voltage divider, which splits a voltage between series elements. Series and parallel are mirror images: in series the current is common and the voltage splits in proportion to resistance; in parallel the voltage is common and the current splits in inverse proportion to resistance. If you are combining resistances rather than splitting current, use the parallel resistance calculator or the series resistance calculator.
The formula explained, term by term
Start with what the branches have in common. Because every branch spans the same pair of nodes, they all see one voltage V. The group as a whole behaves like a single resistor of value R_eq, defined by adding conductances rather than resistances:
1/R_eq = 1/R₁ + 1/R₂ + 1/R₃
Conductance, measured in siemens, is the reciprocal of resistance, and it is the natural quantity here because conductances in parallel simply add — each extra branch opens another path. The node voltage then follows from Ohm's law applied to the group: V = I × R_eq. Ohm's law applied to one branch gives that branch's current:
I₁ = V / R₁ = I × R_eq / R₁
Read that last expression as a fraction of the total. The factor R_eq / R₁ lies between 0 and 1, because R_eq is always smaller than the smallest branch resistance. Add the factors for all branches and you get exactly 1, which is Kirchhoff's current law reappearing as a bookkeeping check.
With only two branches the algebra collapses to the form most people memorise. Substituting R_eq = R₁R₂/(R₁+R₂) gives I₁ = I × R₂/(R₁+R₂). Notice the opposite resistance sits on top. That inversion is the most common slip in the whole topic: in a voltage divider the resistor you are asking about appears in the numerator, and in a current divider it does not.
Everything above assumes pure resistance. At DC, or at frequencies low enough that reactance is negligible, that is exactly right. When branch impedance is complex you replace each R with an impedance and do the arithmetic with phasors — see the RLC impedance calculator for how magnitude and angle combine.
Worked example: 6 A into 2 Ω, 3 Ω and 6 Ω
A current source pushes 6 A into a node feeding three parallel resistors of 2 Ω, 3 Ω and 6 Ω. Find each branch current.
- Add the conductances. 1/2 + 1/3 + 1/6 = 0.500000 + 0.333333 + 0.166667 = 1.000000 S.
- Invert to get the equivalent resistance. R_eq = 1 / 1.000000 = 1.0000 Ω. It is smaller than the smallest branch, 2 Ω, as it must be.
- Find the node voltage. V = I × R_eq = 6 × 1.0000 = 6.0000 V. Every resistor has 6 V across it.
- Divide that voltage by each resistance. I₁ = 6 / 2 = 3.0000 A. I₂ = 6 / 3 = 2.0000 A. I₃ = 6 / 6 = 1.0000 A.
- Check the sum. 3 + 2 + 1 = 6 A, the current you started with.
The shares are 3/6 = 50.00%, 2/6 = 33.33% and 1/6 = 16.67%. They are exactly the conductance shares 0.5/1, 0.3333/1 and 0.1667/1. Power follows from V²/R: 36/2 = 18 W, 36/3 = 12 W and 36/6 = 6 W, totalling 36 W — the same as I × V = 6 × 6. Those three resistors need serious wattage ratings, which the calculator flags for you.
Now the two-branch shortcut on the same numbers. Delete the 6 Ω branch and keep 2 Ω and 3 Ω with 6 A arriving. I₁ = 6 × 3/(2+3) = 6 × 0.6 = 3.6 A, and I₂ = 6 × 2/5 = 2.4 A. The 2 Ω branch takes more, because for one shared voltage the smaller resistance passes the larger current. Removing the third path also raises R_eq from 1.0000 Ω to 2×3/5 = 1.2000 Ω, and the node voltage rises with it to 6 × 1.2 = 7.2 V.
How to read the result
Look first at the share each branch takes. A branch's share of the total equals its share of the total conductance, so a branch of half the resistance takes twice the current of the branch it is compared with. If your design intends four paralleled resistors to share a load equally, a share that departs from 25% by more than the resistor tolerance points at a wiring or measurement problem rather than a rounding one — in a 100 W load bank, four watts of imbalance is the difference between a part that survives and one that does not.
Next look at the equivalent resistance. It is always lower than the smallest branch, and adding a branch always lowers it further. That matters upstream: if the parallel group is fed from a real source rather than an ideal current source, a lower R_eq changes the loop current, and the node voltage reported here assumes the stated total current genuinely arrives.
Then look at the power column. Dissipation is why most current-divider designs fail in hardware rather than on paper. A shunt carrying 9.95 A through 0.005 Ω dissipates 9.95² × 0.005 = 0.495 W — modest, but its resistance drifts as it warms, and that drift changes the split. Metal-foil and manganin shunts exist precisely because their temperature coefficient is small. For the resistor-heating side of the problem, the electrical power calculator works the same numbers from any two of volts, amps, ohms and watts.
Finally, sanity-check the whole group with Ohm's law: V divided by R_eq must return the total current you entered. If you would rather solve the same circuit from a known source voltage, the Ohm's law calculator gets you the total current first.
How two parallel resistors split a current
| R₂ (Ω) | R₂ / R₁ | Share in R₁ | Share in R₂ | R_eq (Ω) |
|---|---|---|---|---|
| 10 | 0.1 | 9.09% | 90.91% | 9.09 |
| 25 | 0.25 | 20.00% | 80.00% | 20.00 |
| 50 | 0.5 | 33.33% | 66.67% | 33.33 |
| 100 | 1.0 | 50.00% | 50.00% | 50.00 |
| 200 | 2.0 | 66.67% | 33.33% | 66.67 |
| 400 | 4.0 | 80.00% | 20.00% | 80.00 |
| 1000 | 10.0 | 90.91% | 9.09% | 90.91 |
The share column and the R_eq column carry the same digits because R_eq = R₁ × R₂/(R₁+R₂) and R₁ is 100 Ω in every row. The pattern is scale-free: only the ratio R₂/R₁ sets the split.
Mistakes that put the current in the wrong branch
- Putting the branch's own resistance on top. For two branches the numerator is the other resistance: I₁ = I × R₂/(R₁+R₂), not R₁/(R₁+R₂). Reversing it produces an answer that is still plausible, which is why the error survives review.
- Extending the two-resistor shortcut to three branches. R₂/(R₁+R₂+R₃) is not the share of a three-way divider. Go back to I × R_eq/R₁, which is correct for any number of branches.
- Ignoring the resistance of wiring and connections. When branch resistances are milliohms — shunts, busbars, paralleled battery cables — a few milliohms of lug and cable resistance is a large fraction of the branch and shifts the split badly. Enter the measured branch resistance including its leads.
- Forgetting that resistance moves with temperature. Copper changes about 0.39% per °C. Two nominally identical paralleled paths at different temperatures do not share equally, and for copper the hotter path takes the smaller share, which is self-stabilising; for materials with a negative coefficient it is not.
- Applying the DC rule to an AC circuit with reactance. With inductance or capacitance present the branch currents differ in phase, so their magnitudes do not add arithmetically and the resistive ratio no longer predicts the split.
- Assuming the source really delivers the stated current. A voltage source behind a source resistance is not a current source. Adding a branch lowers R_eq, which changes the node voltage and the total current — solve the loop first, then divide.
Key terms
- Conductance (G)
- The reciprocal of resistance, measured in siemens (S). Conductances in parallel add directly, which is why the divider formula is cleaner written in terms of G.
- Equivalent resistance (R_eq)
- The single resistance that would draw the same current from the same node voltage as the whole parallel group. Always smaller than the smallest branch.
- Shunt
- A low-value precision resistor deliberately placed in parallel with a meter movement or sense circuit so that a known small fraction of the current flows through the measuring element.
- Kirchhoff's current law
- The sum of currents entering a node equals the sum leaving it. It is conservation of charge, and it is why the branch shares total 100%.
Where the current divider sits among the other tools
The divider rule is a shortcut, not a separate law. Any circuit it solves can also be solved by node-voltage analysis, and for anything more tangled than one parallel group — bridges, multiple sources, dependent sources — node analysis is the method that scales. The value of the divider rule is that it removes an unknown: you never have to find the source voltage to get a branch current.
In instrumentation the rule is the design equation for an ammeter shunt. A moving-coil movement that reads full scale at 50 mA across 1 Ω becomes a 10 A ammeter when you parallel it with a shunt sized so that the rest bypasses the movement. Rearranged, the shunt resistance is R_shunt = R_movement × I_movement / (I_total − I_movement), which is the two-branch divider solved for a resistance instead of a current. Modern current-sense amplifiers use the same idea with a much smaller shunt and gain applied afterwards.
In power work the same arithmetic explains paralleled conductors and paralleled transformers. The National Electrical Code permits paralleled conductors only under strict conditions on length, size, material and termination, and the physical reason is this formula: unequal branch impedance means unequal current, so one conductor of a parallel set can be overloaded while the set as a whole looks fine. If you are sizing those conductors, start with the wire size and ampacity calculator and then confirm the split here.
The rule also governs unintended paths. Current returning through a grounding conductor divides between that conductor and every other metallic path in parallel with it — conduit, water pipe, building steel. Objectionable current on a water pipe is a current divider working exactly as physics requires, on a circuit nobody drew.