Parallel Current Divider Calculator

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.