Case 1
- Total current
- 2A
- Resistance R1
- 100Ω
- Resistance R2
- 300Ω
1.500000A
Open with these valuesCurrent Divider Calculator
1.500000A
Result: 1.500000 ATwo resistors in parallel share the current in inverse proportion to their resistance — the smaller one takes more. The branch through R1 carries I × R2 ÷ (R1 + R2), so 2 A across 100 Ω and 300 Ω splits into 1.5 A and 0.5 A.
Worked examples
Case 2
- Total current
- 5A
- Resistance R1
- 220Ω
- Resistance R2
- 470Ω
3.405797A
Open with these valuesCase 3
- Total current
- 10A
- Resistance R1
- 10Ω
- Resistance R2
- 90Ω
9.000000A
Open with these valuesHow it's calculated
I₁ = I × R2 ÷ (R1 + R2)
- StepEnter the total current arriving at the parallel pair.
- StepEnter both resistances in ohms.
- ResultRead the current through R1; the rest goes through R2.
Reference table
| Current, R1, R2 | How it splits | Current in R1 |
|---|---|---|
| 0.5, 1000, 2000 | 1 kΩ beside 2 kΩ — two thirds | 0.333333 |
| 1, 100, 100 | Equal branches, an even split | 0.500000 |
| 2, 100, 300 | R2 three times R1 — three quarters | 1.500000 |
| 3, 47, 68 | A common pair, roughly 59 per cent | 1.773913 |
| 5, 220, 470 | 220 Ω beside 470 Ω | 3.405797 |
| 10, 10, 90 | R1 nine times smaller — nine tenths | 9.000000 |
Questions
How does the current divider rule work?
The current through one branch is the total times the opposite resistance divided by the sum of both. The opposite resistance in the numerator is what makes the smaller resistor carry the larger share.
Why does R2 appear in the formula for R1?
Both branches sit across the same voltage, so each current is that voltage divided by its own resistance. Writing that out in terms of the total current leaves the other resistance on top.
What is the current through R2?
The total minus the answer shown here. With 2 A across 100 Ω and 300 Ω, R1 takes 1.5 A and R2 takes 0.5 A.
Does this work for more than two resistors?
Not in this form. For three or more branches, replace R2 with the parallel combination of every other branch, or work from the shared voltage instead.
Sources and last check
Information, not professional advice.