Case 1
- Total current
- 2A
- Resistance R1
- 100Ω
- Resistance R2
- 300Ω
1.500000A
Open with these valuesChange any value to update the result.
1.500000A
Current through R1: 1.500000 AYour current valid inputs, with one value varied across the range.
Held fixed: Total current 2.000 A, Resistance R1 100.000 Ω.
| Resistance R2 (Ω) | Current through R1 (A) |
|---|---|
| 200.000 | 1.333333 |
| 300.000Your value | 1.500000 |
| 400.000 | 1.600000 |
Two 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.
1.500000A
Open with these values3.405797A
Open with these values9.000000A
Open with these valuesI₁ = I × R2 ÷ (R1 + R2)
| 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 |
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.
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.
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.
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.
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