Current Divider Calculator

Related tools: Parallel Circuit Calculator, Series Circuit Calculator, and Ohm’s Law Calculator.

A current divider calculates how a known Total current splits among parallel resistance branches. It produces the estimated current in each active branch, along with the network’s total conductance, equivalent resistance, and parallel voltage.

This calculation applies to a fixed-resistance parallel circuit: a source current enters a common node, divides through two or three resistor branches, and recombines at the return node. Lower-resistance branches carry a larger share of the total current because they provide greater conductance.

For the values entered in this worksheet, the primary result is each Branch current. That value can support circuit analysis, resistor selection, branch load review, dissipation calculations performed separately, and verification of an assumed current path before a design is released.

Parallel Branch Current

Enter the current supplied to the complete parallel network in Total current (A). Then enter each branch resistance in Resistance 1 (ohm), Resistance 2 (ohm), and, when used, Resistance 3 (ohm).

A resistance value of 0 omits that branch from the calculation. Only branches with an entered resistance participate in the current-divider result.

The calculator returns:

Result	Electrical meaning
Total conductance	The sum of the conductance of all active parallel branches, in siemens (S)
Equivalent resistance	The single resistance that would draw the same total current at the calculated parallel voltage
Estimated parallel voltage	The voltage across every active branch in the ideal parallel circuit
Branch current 1	Current through Resistance 1
Branch current 2	Current through Resistance 2
Branch current 3	Current through Resistance 3, or 0 A when omitted
Branch current total	The sum of calculated branch currents; it should equal Total current within displayed rounding

All active branches have the same voltage across them. Their current differs according to resistance.

Conductance Method

The calculator uses conductance because conductances add directly in parallel circuits.

For each active branch:

\(\displaystyle G_n = \frac{1}{R_n}\)

Where:

• G_n = branch conductance in siemens
• R_n = branch resistance in ohms

The total conductance is:

\(\displaystyle G_T = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}\)

Only active branches are included. A resistance entered as 0 is omitted rather than treated as a zero-ohm branch.

Equivalent resistance is then:

\(\displaystyle R_{EQ} = \frac{1}{G_T}\)

The estimated voltage across the parallel network is:

\(\displaystyle V_P = I_T \times R_{EQ}\)

Each branch current is calculated from the common parallel voltage:

\(\displaystyle I_n = \frac{V_P}{R_n}\)

The same relationship can be written directly as a current-divider equation:

\(\displaystyle I_n = I_T \times \frac{G_n}{G_T}\)

For a two-branch circuit, this produces the familiar inverse-resistance form:

\(\displaystyle I_1 = I_T \times \frac{R_2}{R_1 + R_2}\)

A branch with half the resistance of another branch carries twice its current when both branches share the same applied voltage.

Calculation Example

Enter the following values:

Input	Value
Total current (A)	10 A
Resistance 1 (ohm)	10 ohm
Resistance 2 (ohm)	20 ohm
Resistance 3 (ohm)	0 ohm

Resistance 3 is omitted. The active branch conductances are:

\(\displaystyle G_1 = \frac{1}{10} = 0.10\text{ S}\)

\(\displaystyle G_2 = \frac{1}{20} = 0.05\text{ S}\)

\(\displaystyle G_T = 0.10 + 0.05 = 0.15\text{ S}\)

The equivalent resistance is:

\(\displaystyle R_{EQ} = \frac{1}{0.15} = 6.6667\text{ ohm}\)

With 10 A entering the parallel network, the estimated parallel voltage is:

\(\displaystyle V_P = 10 \times 6.6667 = 66.6667\text{ V}\)

The resulting branch currents are:

\(\displaystyle I_1 = \frac{66.6667}{10} = 6.6667\text{ A}\)

\(\displaystyle I_2 = \frac{66.6667}{20} = 3.3333\text{ A}\)

The calculator therefore reports:

• Total conductance: 0.15 S
• Equivalent resistance: 6.6667 ohm
• Estimated parallel voltage: 66.6667 V
• Branch current 1: 6.6667 A
• Branch current 2: 3.3333 A
• Branch current 3: 0 A
• Branch current total: 10 A

The 10-ohm branch carries two-thirds of the total current, while the 20-ohm branch carries one-third. The calculated branch currents add back to the supplied 10 A.

Design and Field Limits

The result is ideal fixed-resistance current-divider arithmetic. It does not establish conductor ampacity, AWG or kcmil selection, feeder or branch-circuit rating, overcurrent protection, terminal rating, insulation temperature rating, voltage-drop compliance, raceway fill, or an AHJ-approved installation.

Do not apply ideal resistor current division directly to parallel building conductors. Actual current sharing between parallel conductors depends on conductor material, length, size, termination resistance, routing, reactance, temperature, installed configuration, and connection quality. Those conductors must be designed and installed under the applicable electrical code requirements and equipment instructions.

The calculation also excludes resistor wattage, tolerance, temperature effects, nonlinear loads, AC impedance, conductor impedance, fault current, protective-device behavior, and manufacturer limitations. For resistor or load selection, evaluate branch power separately using the calculated branch current:

\(\displaystyle P_n = I_n^2R_n\)

For AC circuits, use impedance rather than resistance when inductance or capacitance materially affects current division.