Parallel Current Divider - Electricity & Circuits Step by Step

A three-row current-divider scan shows branch currents before the junction total is summarized. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Current divides because branches share voltage

Both branches connect across the same 12 volt source, but each branch can have a different resistance.

I_{\text{branch}} = \frac{V}{R_{\text{branch}}}

Scan three branch pairs

Each row has the same source voltage. Lower branch resistance draws more branch current, and the total is the two branch currents added at the junction.

\begin{array}{c|c|c|c|c}R_{\text{a}}&R_{\text{b}}&I_{\text{a}}&I_{\text{b}}&I_{\text{total}}\\2\ \text{ohm}&2\ \text{ohm}&6\ \text{A}&6\ \text{A}&12\ \text{A}\\2\ \text{ohm}&4\ \text{ohm}&6\ \text{A}&3\ \text{A}&9\ \text{A}\\4\ \text{ohm}&12\ \text{ohm}&3\ \text{A}&1\ \text{A}&4\ \text{A}\\\end{array}

Compute the top branch

The top branch resistance is lower than the bottom branch in the shown case, so it draws the larger current.

I_{\text{a}} = \frac{12\ \text{V}}{2\ \text{ohm}} = 6\ \text{A}

Compute the bottom branch

The bottom branch has more resistance, so the same voltage gives a smaller current.

I_{\text{b}} = \frac{12\ \text{V}}{4\ \text{ohm}} = 3\ \text{A}

The junction total is conserved

At the split, the 6 ampere top-branch current and the 3 ampere bottom-branch current add to the 9 ampere source current.

I_{\text{total}} = 6\ \text{A} + 3\ \text{A} = \hl{9}\ \text{A}