Current Divider Calculator — Formula, Examples & Parallel Circuit Current

This current divider calculator instantly finds the current through any branch in a parallel resistor network. Below is the complete theory, formulas, and worked examples.

What Is a Current Divider?

A current divider is a parallel circuit configuration where a total input current ($I_{in}$) splits between two or more branches. The key principle: more current flows through lower resistance.

This is the dual of the voltage divider (series resistors): in a voltage divider, more voltage drops across higher resistance. In a current divider, more current flows through lower resistance.

Current Divider Formula

Two Resistors in Parallel

When $I_{in}$ enters a node with two parallel resistors $R_1$ and $R_2$:

$I_1 = I_{in} \times \frac{R_2}{R_1 + R_2}$

$I_2 = I_{in} \times \frac{R_1}{R_1 + R_2}$

Notice: to find the current in branch 1, you multiply by the resistance of branch 2, and vice versa. The branch with lower resistance gets more current.

General Formula for N Branches

For any number of parallel branches, the current through branch $x$ is:

$I_x = I_{in} \times \frac{R_{eq}}{R_x}$

Where $R_{eq}$ is the total equivalent parallel resistance:

$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \cdots + \frac{1}{R_n}$

Equivalently, using conductance ($G = 1/R$):

$I_x = I_{in} \times \frac{G_x}{G_1 + G_2 + \cdots + G_n}$

Step-by-Step Example

Given: A 10 A current source feeds two parallel resistors: R1 = 4 Ω, R2 = 6 Ω.

Find: Current through each resistor.

Solution:

$I_1 = 10\,\text{A} \times \frac{6}{4 + 6} = 10 \times 0.6 = 6\,\text{A}$

$I_2 = 10\,\text{A} \times \frac{4}{4 + 6} = 10 \times 0.4 = 4\,\text{A}$

Verification (KCL): $I_1 + I_2 = 6 + 4 = 10\,\text{A}$ ✓

The branch with lower resistance (4 Ω) gets 6 A; the branch with higher resistance (6 Ω) gets 4 A.

Three-Branch Example

Given: $I_{in} = 12\,\text{A}$, R1 = 2 Ω, R2 = 3 Ω, R3 = 6 Ω.

Step 1 — Find $R_{eq}$:

$\frac{1}{R_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 0.5 + 0.333 + 0.167 = 1\,\text{S}$

$R_{eq} = 1\,\Omega$

Step 2 — Find each branch current:

$I_1 = 12 \times \frac{1}{2} = 6\,\text{A}$

$I_2 = 12 \times \frac{1}{3} = 4\,\text{A}$

$I_3 = 12 \times \frac{1}{6} = 2\,\text{A}$

Check: $6 + 4 + 2 = 12\,\text{A}$ ✓

Current Divider vs Voltage Divider

Frequently Asked Questions

Why does more current flow through the lower resistance branch?

Both branches share the same voltage across their terminals (Kirchhoff's voltage law). By Ohm's Law ($I = V/R$), a lower resistance means higher current for the same voltage. The total input current distributes in inverse proportion to each branch's resistance.

Does the current divider rule apply to AC circuits?

Yes, using impedances ($Z$) instead of resistances. Replace $R$ with complex impedance $Z = R + jX$, and the same formulas apply using complex current phasors. The calculator assumes DC (resistances only).

What happens if one branch is a short circuit (0 Ω)?

All current flows through the short circuit — the 0 Ω branch has zero voltage across it, leaving zero voltage across all other branches and therefore zero current in them. This is why short circuits are dangerous: the full current concentrates in one path.

Can I use this formula with current sources in parallel?

If you have multiple current sources in parallel, their currents simply add (or subtract if opposing polarity) to give $I_{total}$. Then apply the current divider formula with the combined total current.