Current Divider
Finds how a total current splits between two resistors in parallel, and their equivalent resistance. Use it to size a meter shunt, to see how load sharing works between two parallel paths, or to check how much current a bypass resistor diverts from a component.
Equations
I1 = Itotal·R2/(R1 + R2)
I2 = Itotal·R1/(R1 + R2)
Req = R1·R2/(R1 + R2)
Itotal is the current flowing into the pair, R1 and R2 the two branch resistances, I1 and I2 the branch currents and Req the resistance of the pair seen from outside [1]. Both branches share one voltage, Itotal·Req, so each carries that voltage divided by its own resistance. Notice that each branch current uses the other branch’s resistance on top: the lower resistance takes the larger share. The two branch currents always add up to Itotal.
Limits and common mistakes
- The page handles exactly two branches. For more, find the shared voltage, V = Itotal·Req with Req from the Series/Parallel Resistance calculator, and divide it by each branch’s resistance.
- Both resistances must be positive. A short (0 Ω) would take all the current.
- Swapping the resistances in the formula is the usual slip. If the larger resistor seems to carry more current, check the formula again.
- The split holds only when the total current is fixed, as from a current source or a series path. If a fixed voltage feeds the pair, work out each branch current directly as V/R.
- Branch resistance includes everything in the branch: wiring, contacts and any meter’s internal resistance. For a low-ohm shunt these can matter.
Worked example
An analog meter reads 50 mA at full scale and has an internal resistance of 50 Ω. A 0.5 Ω shunt goes across it to extend the range. When 5 A flows into the pair, will the meter stay on scale?
- Enter Total current (Itotal) 5 A.
- Enter R1 50 Ω for the meter and R2 0.5 Ω for the shunt.
- Press Solve. The Solution shows:
Current through R1 (I1): 0.049505 A
Current through R2 (I2): 4.9505 A
Equivalent resistance (Req): 0.49505 Ω
Check by hand: I1 = 5 × 0.5/50.5 = 0.0495 A, just under the 50 mA full scale. Full scale now stands for 0.05 × 50.5/0.5 = 5.05 A. The two currents add to 5 A. The shunt carries 4.95 A and dissipates I2R = 4.952 × 0.5 ≈ 12 W, so it must be a power resistor.
