Source-Load Current and Voltage Split - Electronic Circuits Step by Step

A source resistor and a load share one loop, so the same current sets both voltage drops and both power terms. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

A source resistor spends part of the source voltage

This is still a DC circuit model. The source resistor is an explicit element, so the load only receives the voltage left after the source-resistor drop.

V_s = V_{\text{source resistor}} + V_L

Each row splits voltage across source and load

Equal source and load resistances make a clean half-and-half voltage split, but the current is still computed from the total series resistance.

\begin{array}{c|c|c|c|c|c}V_s&R_s&R_L&I&V_s\text{ drop}&V_L\\12\ \text{V}&6\ \text{ohm}&6\ \text{ohm}&1\ \text{A}&6\ \text{V}&6\ \text{V}\\24\ \text{V}&4\ \text{ohm}&4\ \text{ohm}&3\ \text{A}&12\ \text{V}&12\ \text{V}\\30\ \text{V}&5\ \text{ohm}&5\ \text{ohm}&3\ \text{A}&15\ \text{V}&15\ \text{V}\\\end{array}

The same current gives useful and source-resistor watts

The load watts and source-resistor watts both come from the same loop current. There is no separate hidden current path.

\begin{array}{c|c|c|c}I&P_{\text{source}}&P_{R_s}&P_L\\1\ \text{A}&12\ \text{W}&6\ \text{W}&6\ \text{W}\\3\ \text{A}&72\ \text{W}&36\ \text{W}&36\ \text{W}\\3\ \text{A}&90\ \text{W}&45\ \text{W}&45\ \text{W}\\\end{array}

The rendered row closes the energy budget

For the rendered row, the source provides 72 watts. The source resistor and load each receive 36 watts.

24\ \text{V}\cdot3\ \text{A}=72\ \text{W},\quad 36\ \text{W}+36\ \text{W}=72\ \text{W}

Every ampere and every watt has one accounted path

The source resistor is not a vague loss term. It is a circuit element with a voltage drop, a current, and a power term that must appear in the same ledger as the load.

V_{R_s}+V_L=V_s,\quad P_{R_s}+P_L=P_{\text{source}}