Parallel Current Divider Calculator
Computes the current in each branch of a parallel current divider from the total current and two resistances.
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Current Divider in a Parallel Circuit
Put two resistors in parallel and the total current splits between them in inverse proportion to their resistances. The current through R₁ comes out to I₁ = I_total · R₂/(R₁+R₂), and through R₂ it is I₂ = I_total · R₁/(R₁+R₂). The detail that trips people up is that the opposite resistor sits in the numerator. Take I_total = 2 A flowing into R₁ = 100 Ω paralleled with R₂ = 200 Ω: you get I₁ = 2 · 200/300 ≈ 1.33 A, leaving I₂ ≈ 0.67 A.
It all falls out of Kirchhoff's Current Law (the currents flowing into a node sum to zero) combined with Ohm's Law, since both branches see the same voltage. Once you have more than two branches, switch to the general form Iₖ = I_total · (G_k / ΣG), where G = 1/R is conductance. Whatever the count, the smaller resistor hogs the bigger slice of current. For the full treatment, Boylestad covers it in Introductory Circuit Analysis chapter 6, and Sedra & Smith have an appendix on network theorems in Microelectronic Circuits.
Applications
The current-divider rule shows up all over the place. Ammeter shunts rely on it, where a low-value shunt routes most of the current around a sensitive movement. So does current sensing in motor drives, whether Hall-effect or shunt-based, along with ground-loop analysis. It matters too when you parallel MOSFETs or IGBTs for high-power switching, because mismatched R_DS(on) leaves one device carrying more than its share. And it underpins protective relay coordination under IEEE C37 / IEC 60909.
FAQ
Why does the opposite resistor appear in the numerator? Think of it this way: the more resistance the other branch has, the more current gets pushed through the one you care about. On paper it drops out of I₁ = V/R₁ once you substitute V = I_total · (R₁‖R₂).
Does the rule work for impedances in AC? It does. Swap R for the complex impedance Z and keep going. You'll need complex arithmetic to track magnitudes and phases, but the relationship I₁ = I_total · Z₂/(Z₁+Z₂) stays intact.
What if one branch is a short circuit? Every bit of current takes the short (R = 0), and the parallel resistor ends up with none. That is precisely what the formula gives you as R₁ → 0.
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The results provided by this tool are for general informational and educational purposes only and do not constitute professional, financial, medical, legal, tax or accounting advice. Always confirm important decisions with a qualified professional and official sources.