Account for current at a parallel junction

Identify shared nodes before adding resistances

In the parallel drawing, both resistors connect between the same two nodes. Trace each route separately: each starts at the source’s positive node and ends at its negative node. Both components therefore have the full source voltage across them. Their currents need not be equal.

At 6 V, R1 = 100 Ω carries 60 mA, while R2 = 200 Ω carries 30 mA. The source delivers their sum, 90 mA. The return wire carries the same total current back to the source. Current splits at one junction and recombines at the other; it is not used up on the way.

Equivalent resistance from the branch currents

Each branch follows I = V/R. Adding currents gives I source = V/R1 + V/R2. Defining I source = V/R equivalent gives 1/R equivalent = 1/R1 + 1/R2. For two resistors this becomes R equivalent = R1 R2/(R1 + R2).

For 100 Ω and 200 Ω, the equivalent is 100 × 200/300 = 66.667 Ω, smaller than either branch resistance. The extra path allows more total current at the same voltage. Do not add branch resistances as if they were consecutive obstacles along one path.

Change one branch and inspect the other

Keep source voltage at 6 V. Increase R2 from 200 to 400 Ω. Predict which quantities change, then capture both settings.

R1 remains at 60 mA. R2 falls from 30 to 15 mA, and source current falls from 90 to 75 mA. The equivalent resistance rises from 66.667 to 80 Ω. Each branch still has a 6 V drop. The unchanged R1 current depends on this ideal source maintaining its terminal voltage; a real source with internal resistance may not maintain exactly that voltage as load changes.

Now inspect the series comparison at the same settings. Changing R2 there changes the current through R1 as well, because the components share one path. Compare connections, not only component values.

Topology matters more than page position

Two resistors drawn next to each other are not necessarily parallel. They must share both endpoint nodes. Two drawn in a row are not necessarily series if their shared node also has another branch. The browser shows only the two specified networks; it is not a solver for arbitrary wiring.

The source at 0 V remains connected and all branch currents are zero. Removing a branch would be a different topology; shorting a source with an ideal wire would fall outside this finite-resistance model. No open or short circuit is silently represented by the resistance slider.

Transfer to a new pair

Connect 200 Ω and 300 Ω across an ideal 6 V source. Predict each branch current, their sum and equivalent resistance. Explain which branch carries more current and why.

Check the parallel transfer

Branch currents are 6/200 = 0.030 A and 6/300 = 0.020 A, or 30 and 20 mA. Total current is 50 mA. Equivalent resistance is 6/0.050 = 120 Ω. The smaller resistance carries more current because both branches have the same voltage.

OpenStax: resistors in series and parallel develops equivalent resistance. These selected comparisons and transfer values are original synthetic exercises; a physical circuit requires measured component values and source behavior.