Changing only the collector resistance moves load demand against the same drive limit. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Six ohms asks for only two amperes

With collector resistor 6 ohm, load demand is 2 A. Drive stays 3 A, so collector current is 2 A and output is 0 V.

Iload=2 A,3 A>2 AI_\text{load}=2\ \text{A},\quad 3\ \text{A}>2\ \text{A}
Collector load-resistance scanLoad demand and actual collector current are checked.output 0 Vdrive 3 Aload 2 Acollector 2 Asaturation

Four ohms is exactly the saturation boundary

With collector resistor 4 ohm, load demand is 3 A. Drive stays 3 A, so collector current is 3 A and output is 0 V.

Iload=3 A,3 A=3 AI_\text{load}=3\ \text{A},\quad 3\ \text{A}=3\ \text{A}
Collector load-resistance scanLoad demand and actual collector current are checked.output 0 Vdrive 3 Aload 3 Acollector 3 Asaturation

Three ohms asks for more current than beta can supply

With collector resistor 3 ohm, load demand is 4 A. Drive stays 3 A, so collector current is 3 A and output is 3 V.

Iload=4 A,3 A<4 AI_\text{load}=4\ \text{A},\quad 3\ \text{A}<4\ \text{A}
Collector load-resistance scanLoad demand and actual collector current are checked.drive 3 Aload 4 Acollector 3 Aoutput 3 Vactive