Changing one resistor changes how the source voltage is divided. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Start with a balanced divider

When the bottom resistance is 3 ohm, the output is 6 volts.

Vout,low=6 VV_{\text{out,low}} = 6\ \text{V}
Balanced dividerThe output node reads the checked bottom drop.12 V3 ohm3 ohm2 A+-6 V+-6 Vtopoutputbottom

Increase the bottom resistance

Change only the bottom resistance to 9 ohm.

Rbottom=9 ohmR_{\text{bottom}} = 9\ \text{ohm}
Changed dividerThe bottom resistor is larger, changing the output.12 V3 ohm9 ohm1 A+-3 V+-9 Vtopoutputbottom

The output rises

The output changes from 6 volts to 9 volts.

Vout:6 V9 VV_{\text{out}}: 6\ \text{V} \rightarrow 9\ \text{V}
Changed dividerThe bottom resistor is larger, changing the output.12 V3 ohm9 ohm1 A+-3 V+-9 Vtopoutputbottom

More bottom resistance raises the node

With the top resistor fixed, increasing the bottom resistor moves the output node closer to the source voltage. The first and last rows are the two diagrams used above.

VinRtRbIVtVout12 V3 ohm3 ohm2 A6 V6 V12 V3 ohm6 ohm43 A4 V8 V12 V3 ohm9 ohm1 A3 V9 V\begin{array}{c|c|c|c|c|c}V_{\text{in}}&R_t&R_b&I&V_t&V_{\text{out}}\\12\ \text{V}&3\ \text{ohm}&3\ \text{ohm}&2\ \text{A}&6\ \text{V}&6\ \text{V}\\12\ \text{V}&3\ \text{ohm}&6\ \text{ohm}&\tfrac{4}{3}\ \text{A}&4\ \text{V}&8\ \text{V}\\12\ \text{V}&3\ \text{ohm}&9\ \text{ohm}&1\ \text{A}&3\ \text{V}&9\ \text{V}\\\end{array}
Changed dividerThe last table row is the checked changed diagram.12 V3 ohm9 ohm1 A+-3 V+-9 Vtopoutputbottom