A current divider splits a total current between parallel branches in inverse proportion to their resistance: Iₙ = Itotal × Rp ÷ Rₙ. For two resistors that becomes I₁ = I × R₂ ÷ (R₁ + R₂). The smaller resistor always carries more current.
Enter the total current (or the voltage across the group) and the branch resistances in Ω, kΩ or MΩ to get each branch current, its share and its power, plus the equivalent resistance from the parallel resistor formula. This is a learning and design aid for resistor circuits; for building or mains wiring, follow your electrical code and consult a qualified electrician.
How to use this calculator
- Choose what you know: the total current flowing into the parallel group, or the voltage across it. Pick A or mA (or V or mV).
- List the branch resistances, separated by commas or spaces — two or as many as you like — and pick Ω, kΩ or MΩ for the whole list.
- Read the current through each resistor, starting with R1, with its percentage of the total and the power it dissipates.
- Check the equivalent resistance and the voltage across the branches, which is the same for every branch.
- Use Share to save or send the result.
How it's calculated
The calculator first finds the equivalent parallel resistance, then the shared voltage, then each branch current:
1/R_p = 1/R₁ + 1/R₂ + … + 1/Rₙ(OpenStax College Physics 2e §21.1)V = I_total × R_p— each resistor in parallel has the same full voltage applied to it.Iₙ = V / Rₙ = I_total × R_p / Rₙ(Ohm's law)
Here Itotal is the current entering the parallel group, Rₙ is the resistance of branch n, Rp is the equivalent resistance and V is the voltage across the group. For exactly two resistors, Rp = R₁R₂ ÷ (R₁ + R₂), which simplifies to the familiar form in MIT OpenCourseWare 6.061: I₂ = I × R₁ ÷ (R₁ + R₂). Notice the other resistor goes on top.
Worked examples
10 A through 2 Ω and 3 Ω
Rp = 2 × 3 ÷ 5 = 1.2 Ω, so V = 10 × 1.2 = 12 V. I₁ = 12 ÷ 2 = 6 A and I₂ = 12 ÷ 3 = 4 A. Check: 6 + 4 = 10 A.
22 A through 200, 300 and 100 Ω
Rp = 54.55 Ω and V = 1,200 V. The branches carry 6, 4, 12 A — the 100 Ω branch takes half the total. This matches the Oregon State ENGR 102 example.
OpenStax Example 21.2: 12 V across 1, 6 and 13 Ω
Rp = 0.804 Ω, so the source supplies 14.92 A. Dividing that current gives 12, 2, 0.92 A, the same values OpenStax reports, and power P = I²R in each branch of 144, 24, 11.1 W.
50 mA shared by 100 Ω and 1 kΩ
The 100 Ω branch takes 45.45 mA and the 1 kΩ branch 4.55 mA. When one resistor is ten times the other, it carries one-eleventh of the total.
Parallel resistor formula
Every current divider problem starts with the parallel resistor formula, because the equivalent resistance sets the shared voltage. OpenStax University Physics Vol. 2 §10.2 gives it for any number of resistors:
R_p = (1/R₁ + 1/R₂ + … + 1/Rₙ)⁻¹
and notes that the result is smaller than any individual resistance in the combination. Three shortcuts cover most homework and bench work:
- Two resistors, product over sum: Rp = R₁R₂ ÷ (R₁ + R₂). MIT OpenCourseWare 6.061 (eq. 7–8) writes this as R₁ ∥ R₂, the “parallel” symbol you will see on schematics and in textbooks.
- Equal resistors: N identical resistors R in parallel give R ÷ N. OpenStax's worked circuit combines two 10 Ω resistors into 5 Ω, then adds a 7 Ω resistor in series (series resistances simply add, MIT eq. 6) for 12 Ω total.
- Very unequal resistors: when one resistor is ten or more times the other, Rp is only slightly below the smaller one. 100 Ω in parallel with 10 kΩ is 99.01 Ω.
For three or more resistors you can also combine two at a time with product over sum and repeat, which is how you reduce a series-parallel network step by step. The table below gives the equivalent resistance of common standard-value pairs.
| R₁ | R₂ | R₁ ∥ R₂ | Current share in R₁ |
|---|---|---|---|
| 10 Ω | 10 Ω | 5 Ω | 50% |
| 100 Ω | 100 Ω | 50 Ω | 50% |
| 100 Ω | 220 Ω | 68.75 Ω | 68.8% |
| 220 Ω | 330 Ω | 132 Ω | 60% |
| 330 Ω | 470 Ω | 193.88 Ω | 58.8% |
| 470 Ω | 1 kΩ | 319.73 Ω | 68% |
| 1 kΩ | 1 kΩ | 500 Ω | 50% |
| 1 kΩ | 2.2 kΩ | 687.5 Ω | 68.8% |
| 4.7 kΩ | 10 kΩ | 3.197 kΩ | 68% |
| 10 kΩ | 10 kΩ | 5 kΩ | 50% |
Current divider from a known voltage
Often you know the supply voltage rather than the total current. Because each parallel branch sits across the full voltage, you can skip the divider ratio and use Ohm's law on each branch. Example: 9 V across 4.7 kΩ and 10 kΩ. Rp = 4,700 × 10,000 ÷ 14,700 = 3,197.3 Ω, so the supply delivers 2.815 mA. The 4.7 kΩ branch takes 9 ÷ 4,700 = 1.915 mA and the 10 kΩ branch 9 ÷ 10,000 = 0.9 mA. The two methods always agree; use whichever quantity you actually measured. Choose “Voltage across the parallel group” in the calculator to work this way, and pick kΩ so you can type 4.7, 10.
Two-resistor split by ratio
How 1 A divides between R₁ and a larger R₂. Multiply by your actual current.
| R₂ ÷ R₁ | Current in R₁ | Current in R₂ | R₁ share |
|---|---|---|---|
| 1 | 0.5 A | 0.5 A | 50% |
| 2 | 0.6667 A | 0.3333 A | 66.7% |
| 3 | 0.75 A | 0.25 A | 75% |
| 4 | 0.8 A | 0.2 A | 80% |
| 5 | 0.8333 A | 0.1667 A | 83.3% |
| 9 | 0.9 A | 0.1 A | 90% |
| 10 | 0.9091 A | 0.0909 A | 90.9% |
| 100 | 0.9901 A | 0.0099 A | 99% |
Where current dividers are used
The classic example is the ammeter shunt. OpenStax College Physics 2e §21.4 describes turning a sensitive galvanometer into an ammeter by placing a small shunt resistance in parallel with it, so most of the current bypasses the meter. In their example, a 25 Ω galvanometer that reads full scale at 50 μA is paired with a 0.00125 Ω shunt for a 1.0 A range. Running 1 A through that pair in this calculator puts 50 μA through the meter and 0.99995 A through the shunt — the divider rule at work. The same idea shows up whenever parallel paths share a supply, from paralleled resistors that spread heat to LED strings on separate branches.
Current divider vs voltage divider
In series, the same current flows through every resistor and the voltage divides; the larger resistor gets the larger share of the voltage. In parallel it's the reverse: every branch sees the same voltage and the current divides, with the smaller resistor taking the larger share. If you remember one thing, remember that current prefers the path of lower resistance but still flows through every branch.
Common mistakes
- Putting the wrong resistor on top. In the two-resistor formula, the current in R₁ uses R₂ in the numerator. Getting this backwards gives the larger current to the larger resistor.
- Adding resistances as if in series. Parallel resistance is the reciprocal of the sum of reciprocals, not the sum.
- Mixing units. Keep all resistances in ohms (1 kΩ = 1,000 Ω) and current in amps.
- Forgetting power ratings. The branch with the most current may also dissipate the most power. Check P = I²R against each resistor's wattage rating.
- Assuming an ideal source. The divider rule assumes the total current is known. If you know the source voltage instead, the current is V ÷ Rp (plus any source resistance in series).
When this calculator doesn't apply
This calculator only handles plain resistance. Circuits with capacitors or inductors on AC, and non-linear parts such as diodes and LEDs, need a different analysis — the simple ratio above won't give the right branch currents.
For household circuits, conductor size depends on current, length and code rules — see the wire size calculator. For battery capacity in watt-hours, use the mAh to Wh converter.
Frequently asked questions
What is the current divider formula?
For two resistors in parallel, I₁ = I × R₂ / (R₁ + R₂) and I₂ = I × R₁ / (R₁ + R₂). For any number of resistors, Iₙ = I × R_p / Rₙ, where 1/R_p = 1/R₁ + 1/R₂ + …
How is 10 A split between 2 Ω and 3 Ω?
6, 4 A — the smaller resistor takes the larger share of the current.
How is 22 A split between 200, 300 and 100 Ω?
6, 4, 12 A. The equivalent resistance is 54.545 Ω.
Does more current flow through the smaller resistor?
Yes. Every parallel branch has the same voltage, so by Ohm's law (I = V/R) the branch with the lowest resistance carries the most current. Current divides in inverse proportion to resistance.
What is the current divider rule with conductance?
Using conductance G = 1/R, each branch gets Iₙ = I × Gₙ / (G₁ + G₂ + …). Current divides in direct proportion to conductance, which is often easier with three or more branches.
Is the equivalent resistance always smaller than the smallest resistor?
Yes. OpenStax notes that total parallel resistance is less than the smallest individual resistance. For 1, 6 and 13 Ω it is 0.804 Ω.
What is the difference between a current divider and a voltage divider?
A current divider is resistors in parallel: they share the same voltage and split the current. A voltage divider is resistors in series: they carry the same current and split the voltage, with the larger resistor taking the larger share.
What is the parallel resistor formula?
1/R_total = 1/R₁ + 1/R₂ + … + 1/Rₙ, so R_total is the reciprocal of the sum of the reciprocals (OpenStax University Physics Vol. 2 §10.2). For exactly two resistors it simplifies to product over sum: R = R₁R₂ ÷ (R₁ + R₂), which MIT OpenCourseWare writes as R₁ ∥ R₂.
How do you calculate two resistors in parallel?
Multiply them and divide by their sum. 4.7 kΩ and 10 kΩ in parallel: 4,700 × 10,000 ÷ 14,700 = 3,197.3 Ω. The answer is always below the smaller of the two.
What is the resistance of two equal resistors in parallel?
Half of one of them. OpenStax combines two 10 Ω resistors in parallel into 5 Ω. In general, N equal resistors R in parallel give R ÷ N — four 100 Ω resistors make 25 Ω.
How do I find branch currents if I only know the voltage?
Every parallel branch has the full voltage across it, so each branch current is simply V ÷ Rₙ, and the total is V ÷ R_p. Switch the calculator to “Voltage across the parallel group”. 9 V across 4.7 kΩ and 10 kΩ gives 1.915, 0.9 mA.
How much power does each branch use?
Use P = I²R for each branch. In OpenStax's 12 V example with 1, 6 and 13 Ω, the branches dissipate 144, 24, 11.1 W.
Sources & method
- OpenStax College Physics 2e, §21.1 — resistors in series and parallel
- OpenStax College Physics 2e, §20.4 — electric power and energy (P = IV = I²R = V²/R)
- OpenStax College Physics 2e, §21.4 — DC voltmeters and ammeters (shunt resistor)
- OpenStax College Physics 2e, §21.3 — Kirchhoff's rules (junction rule)
- OpenStax University Physics Vol. 2, §10.2 — resistors in series and parallel
- MIT OpenCourseWare 6.061, Chapter 1 — current divider (Eq. 5)
- Oregon State University ENGR 102 — voltage and current dividers
- NIST Office of Weights and Measures — Metric (SI) prefixes (milli, kilo, mega)
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