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Current Divider

I1=ItR2R1+R2I_{1} = I_{t} \frac{R_{2}}{R_{1} + R_{2}}

Enter your known values, leave one input blank, and solves for the missing one. Try different units for next level excitement!

Learning zone

Parallel branches share a voltage, so current splits inversely with resistance — and that inversion is why the opposite resistor appears on top. Six amps entering a 4 Ω branch paralleled with a 2 Ω branch sends 6 × 2/6 = 2 A through the 4 Ω path and 4 A through the 2 Ω path: the easier road takes the most traffic.

Electricians meet this as the reason parallel feeders must be identical in size, length and material — a run even slightly shorter hogs current and overheats while its twin loafs, which is exactly what the NEC's paralleling rules exist to prevent. Benchtop electronics meets it as the ammeter shunt: send 99% of the current through a milliohm resistor and measure the small remainder. Note that for more than two branches the tidy two-resistor form fails; use conductance ratios instead.

Current Divider
I1=ItR2R1+R2I_{1} = I_{t} \frac{R_{2}}{R_{1} + R_{2}}
Where
  • I1I_{1}= Current in branch 1
  • ItI_{t}= Total current
  • R1R_{1}= Branch 1 resistance
  • R2R_{2}= Branch 2 resistance
Missing one of these? Work it out first, then come back