Parallel Circuits

Branch Currents Add

conservation at a junction

At a junction, the total current equals the sum of the branch currents. Charge flow is conserved.

electricity A 12 V source across 6 ohm and 3 ohm branches gives branch currents of 2 A and 4 A, which add to 6 A at the junction.

Example

At a junction, the total current equals the sum of the branch currents. Charge flow is conserved. 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 splits at the junction

The total current reaches the left bus, then splits between the two resistor branches. The branch currents depend on each branch resistance.

Itotal=IRa+IRbI_{\text{total}} = I_{\text{Ra}} + I_{\text{Rb}}

At the same voltage, smaller branch resistance draws more current

Every branch gets 12 volts. A smaller branch resistance draws a larger current from that same voltage.

VRI12 V12 ohm1 A12 V6 ohm2 A12 V3 ohm4 A\begin{array}{c|c|c}V & R & I \\12\ \text{V} & 12\ \text{ohm} & 1\ \text{A} \\12\ \text{V} & 6\ \text{ohm} & 2\ \text{A} \\12\ \text{V} & 3\ \text{ohm} & 4\ \text{A}\end{array}
Parallel branches share the same busesA battery feeds two distinct resistor branches connected between the same left and right buses.12 VA6 A6 ohm3 ohm6 A2 A4 A

Compute each branch current

The upper branch carries 2 amperes and the lower branch carries 4 amperes.

branchVRIRa12 V6 ohm2 ARb12 V3 ohm4 A\begin{array}{c|c|c|c}\text{branch} & V & R & I \\\text{Ra} & 12\ \text{V} & 6\ \text{ohm} & 2\ \text{A} \\\text{Rb} & 12\ \text{V} & 3\ \text{ohm} & 4\ \text{A}\end{array}
Parallel branches share the same busesA battery feeds two distinct resistor branches connected between the same left and right buses.12 VA6 A6 ohm3 ohm6 A2 A4 A

The junction total is the branch-current sum

At the split, charge flow is conserved. Scan three possible branch pairs: the total current is always the two branch currents added.

IRaIRbItotal1 A2 A3 A2 A4 A6 A3 A6 A9 A\begin{array}{c|c|c}I_{\text{Ra}} & I_{\text{Rb}} & I_{\text{total}} \\1\ \text{A} & 2\ \text{A} & 3\ \text{A} \\2\ \text{A} & 4\ \text{A} & 6\ \text{A} \\3\ \text{A} & 6\ \text{A} & 9\ \text{A}\end{array}
Parallel branches share the same busesA battery feeds two distinct resistor branches connected between the same left and right buses.12 VA6 A6 ohm3 ohm6 A2 A4 A

Branch currents add to the total

At the junction, the 2 ampere upper-branch current plus the 4 ampere lower-branch current gives the 6 ampere total. The same values are drawn on the current arrows.

Itotal=2 A+4 A=6 AI_{\text{total}} = 2\ \text{A} + 4\ \text{A} = \hl{6}\ \text{A}
Parallel branches share the same busesA battery feeds two distinct resistor branches connected between the same left and right buses.12 VA6 A6 ohm3 ohm6 A2 A4 A