Calculate branch currents in parallel resistor networks using the current divider rule. Supports any number of resistors with power analysis.
The current divider rule calculates how current splits among parallel resistors. For two resistors, I₁ = I_total × R₂/(R₁ + R₂) — current through a branch is proportional to the OTHER resistor's value. For N resistors, each branch current is I_n = I_total × (R_eq/R_n), where R_eq is the parallel equivalent.
The key insight is that current favors the path of least resistance. In a parallel circuit, the lowest-resistance branch carries the most current. All branches share the same voltage V = I_total × R_eq, so current is simply the shared voltage divided by each branch resistance.
This calculator handles any number of parallel resistors, computes individual branch currents, power dissipation, percentage distribution, and equivalent resistance. It supports input as total current or source voltage, with a visual current distribution chart showing how current is shared among branches. Check the example with realistic values before reporting. That makes it practical both for textbook current-divider problems and for real circuit checks where branch heating and power sharing matter.
Manually calculating current distribution for 3+ parallel resistors requires finding the equivalent resistance, voltage, and then each branch current — a tedious process with many opportunities for arithmetic errors.
This calculator instantly handles any number of resistors, provides power dissipation for thermal analysis, and visualizes the current distribution. It is essential for circuit design, ammeter shunt sizing, and understanding parallel circuit behavior.
I_n = I_total × R_eq / R_n. For 2 resistors: I₁ = I_total × R₂/(R₁+R₂). R_eq = 1/(1/R₁ + 1/R₂ + ... + 1/Rₙ). V = I_total × R_eq. P_n = I_n² × R_n.
Result: I₁ = 0.6667 A (66.7%), I₂ = 0.3333 A (33.3%)
R_eq = 1/(1/100 + 1/200) = 66.67Ω. V = 1 × 66.67 = 66.67V. I₁ = 66.67/100 = 0.667A. I₂ = 66.67/200 = 0.333A. Or: I₁ = 1 × 200/(100+200) = 0.667A.
Starting from Kirchhoff's current law at the junction: I_total = I₁ + I₂ + ... + Iₙ. Since all branches share the same voltage V: I_total = V/R₁ + V/R₂ + ... + V/Rₙ = V × (1/R₁ + 1/R₂ + ... + 1/Rₙ) = V/R_eq.
Therefore V = I_total × R_eq, and each branch current is I_n = V/Rₙ = I_total × R_eq/Rₙ.
For the special case of two resistors: R_eq = R₁R₂/(R₁+R₂), so I₁ = I_total × R₁R₂/(R₁+R₂) / R₁ = I_total × R₂/(R₁+R₂). This is the classic two-resistor current divider formula.
**Ammeter shunts**: The most common practical application. A galvanometer (sensitive current meter, typically 50-100µA full scale) is placed in parallel with a precision low-resistance shunt. For a 10A range with a 100µA, 1kΩ movement: R_shunt = 1000 × 100µA / (10 − 100µA) ≈ 0.01Ω. Only 0.001% of the current flows through the movement.
**Current balancing**: When paralleling power supplies, MOSFETs, or LEDs, current doesn't divide equally unless impedances match. Ballast resistors force current sharing by adding resistance that dominates the natural imbalance.
For AC circuits with complex impedances (resistors, capacitors, inductors), the current divider formula becomes I_n = I_total × Z_eq/Zₙ where all quantities are complex. A parallel RC circuit with R = 1kΩ and C = 1µF at 1kHz: Z_C = −j159Ω, Z_eq = R×Z_C/(R+Z_C). The capacitor branch carries more current at higher frequencies, forming a natural high-pass/low-pass current split — the basis of crossover networks in audio speakers.
In parallel, all branches share the same voltage. By Ohm's law, I = V/R, so lower resistance means higher current. The current divider formula I₁ = I × R₂/(R₁+R₂) shows I₁ is proportional to R₂ (the other resistor), not R₁.
Voltage divider uses series resistors: V₁ = V × R₁/(R₁+R₂). Current divider uses parallel resistors: I₁ = I × R₂/(R₁+R₂). Note the "crossover" — in voltage divider, larger R gets more voltage; in current divider, larger R gets less current.
If N resistors are equal, current divides equally: each branch carries I_total/N. The equivalent resistance is R/N.
A shunt ammeter places a low-resistance shunt in parallel with the meter movement. Most current flows through the shunt (e.g., 99.9% for a 10A range with 100µA movement), protecting the delicate movement while measuring large currents.
Yes — replace R with Z (complex impedance) for AC circuits. The same formula applies: I_n = I_total × Z_eq/Z_n. The result gives complex currents with magnitude and phase.
KCL states that currents entering a node equal currents leaving. The current divider is a direct application of KCL combined with Ohm's law: I_total = I₁ + I₂ + ... + Iₙ, and all branches share the same voltage.