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Circuits, electronics and noise
Current division between two parallel resistors
Part of a total current I that flows through R₁ when it splits between two parallel resistors R₁ and R₂.
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- ELN-01
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- v1
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- Circuits, electronics and noise
- Result
- I1 [A]
When a current I reaches two resistors in parallel, it splits in inverse proportion to their resistances: the branch through R₁ carries I R₂/(R₁ + R₂). The larger current goes through the smaller resistance. The two branch currents add up to I, and both branches have the same voltage across them.
Part of a total current I that flows through R₁ when it splits between two parallel resistors R₁ and R₂. The larger share goes through the smaller resistance.
Calculator
- Current through R₂0.004 A
- Voltage across the pair16 V
- Ratio of currents I₁ : I₂ (equals R₂ : R₁)2
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- Of 0.012 A, 0.008 A flows through 2,000 Ω and 0.004 A through 4,000 Ω, with 16 V across both.
- The branch with half the resistance takes twice the current, so a small shunt resistor carries almost all of the current.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (A) | Local step | Change in result |
|---|---|---|---|---|---|---|
| I Total current entering the pair | 1e-4 to 1 A | 6.667e-5 | 0.6667 | 0.6666 | +10% | +10 % |
| R1 Resistance of branch 1 | 10 to 1,000,000 Ω | 4.781e-5 | 0.01197 | 0.01192 | +10% | −3.23 % |
| R2 Resistance of branch 2 | 10 to 1,000,000 Ω | 5.97e-5 | 0.01198 | 0.01192 | +10% | +3.13 % |
Sweep of Resistance of branch 2
| Resistance of branch 2 (Ω) | Current through R₁ (A) |
|---|---|
| 10 | 5.97e-5 |
| 42.17 | 2.478e-4 |
| 177.8 | 9.798e-4 |
| 749.9 | 0.003272 |
| 3,162 | 0.007351 |
| 13,340 | 0.01043 |
| 56,230 | 0.01159 |
| 237,100 | 0.0119 |
| 1,000,000 | 0.01198 |
Sweep one variable
Variables and how each one contributes
Current through R₁
What the equation gives you. Everything else feeds it.
Total current entering the pair
The total current entering the parallel pair, in A. Scales both branch currents in direct proportion. Raising it by 10 % (to 0.0132) moves the result by +10 %, from 0.008 to 0.0088 A.
Resistance of branch 1
The resistance of branch 1, in Ω. A larger R₁ pushes more of the current into branch 2, so I₁ falls. Raising it by 10 % (to 2,200) moves the result by −3.23 %, from 0.008 to 0.007742 A.
Resistance of branch 2
The resistance of branch 2, in Ω. A larger R₂ pushes more of the current into branch 1, so I₁ rises. Raising it by 10 % (to 4,400) moves the result by +3.13 %, from 0.008 to 0.00825 A.
Worked example with the values above
- Add the resistances: R₁ + R₂ = 2,000 + 4,000 = 6,000 Ω.
- Share going to branch 1: R₂/(R₁ + R₂) = 0.66667.
- I₁ = 0.012 × 0.66667 = 0.008 A.
Assumptions and limits
- Ideal sources and linear resistors with positive values, all in base SI units.
- The circuit is exactly as described: no extra loads, and instruments that draw no current.
Common mistakes
- Forgetting a load across a divider, which lowers the output.
- Using the resistor on the wrong side of the divider or the wrong branch for the current share.
- Mixing kΩ and Ω in one calculation.
Statistics and uncertainty
The split depends only on the ratio of the two resistances, so equal percentage errors in both resistors cancel, while the absolute tolerance of the current meter limits the check.
Where it is used
- Sensor interfaces, bias networks, bridge circuits and mixers.
- Reducing a network to simpler series and parallel parts.
Maths you need
- Ohm’s law: V = I R
- Kirchhoff’s current law: currents into a node add to zero
- Series and parallel resistors: R₁ + R₂ and R₁R₂/(R₁ + R₂)