Answer by periblepsis for Calculating Vab Without Reducing the Circuit Using Source Transformation

You write that the following are not to be used:

• no circuit reduction
• no mesh (KVL) analysis
• no nodal (KCL) analysis

You instead decided to use source transformation, with the idea that this doesn't involve circuit reduction. But it inevitably does, because that's the point of performing source transformation -- to allow circuit reduction.

What remains, in my mind, is superposition, where you alternatively turn off all sources except one, analyzing each, and then summing them back at the end to get the final result.

I think you can argue that this doesn't perform circuit reduction. It leaves everything in place. It just sets their values to different ones:

1. Leave the current source alone and set the two voltage sources to \$0\:\text{V}\$. This leaves all the components in the circuit, as before.
2. Leave the upper-right voltage source alone and set the lower-left \$12\:\text{V}\$ voltage source to \$0\:\text{V}\$ and set the current source to \$0\:\text{A}\$. This leaves all the components in the circuit, as before.
3. Leave the lower-left voltage source alone and set the upper-right \$12\:\text{V}\$ voltage source to \$0\:\text{V}\$ and set the current source to \$0\:\text{A}\$. This leaves all the components in the circuit, as before.

If you perform each of these and then analyze each -- and you can do this easily by inspection to create the expression to calculate a value -- and then sum up each result then you will get the right answer.

Take these in order:

1. By inspection, I see that there is \$2\:\text{mA}\$ driving into \$\left(1\:\text{k}\Omega+3\:\text{k}\Omega\right)\:\vert\vert\: 1\:\text{k}\Omega\:\vert\vert\: 2\:\text{k}\Omega\$ and that this sets the voltage at the node where the current source points into. Multiplying those provides the voltage at that node, which is then divided down by the voltage divider caused by the \$1\:\text{k}\Omega\$ and \$3\:\text{k}\Omega\$ resistors (where node \$a\$ is located.)

   Compute \$\frac{1\:\text{k}\Omega}{1\:\text{k}\Omega+3\:\text{k}\Omega}\cdot 2\:\text{mA}\cdot \left[\left(1\:\text{k}\Omega+3\:\text{k}\Omega\right)\:\vert\vert\: 1\:\text{k}\Omega\:\vert\vert\: 2\:\text{k}\Omega\right]=\frac27\:\text{V}\$.

2. By inspection, I see a simple series loop consisting of the \$12\:\text{V}\$ voltage source and three resistances that are added together in series: \$1\:\text{k}\Omega\$ resistor, \$3\:\text{k}\Omega\$ resistor, and \$1\:\text{k}\Omega\:\vert\vert\: 2\:\text{k}\Omega\$. The loop current is then \$\frac{12\:\text{V}}{1\:\text{k}\Omega+3\:\text{k}\Omega+\left(1\:\text{k}\Omega\:\vert\vert\: 2\:\text{k}\Omega\right)}\$. This is multiplied by the \$1\:\text{k}\Omega\$ resistor to get the voltage at \$a\$.

   Compute \$1\:\text{k}\Omega\cdot\frac{12\:\text{V}}{1\:\text{k}\Omega+3\:\text{k}\Omega+\left(1\:\text{k}\Omega\:\vert\vert\: 2\:\text{k}\Omega\right)}=\frac{18}7\:\text{V}\$.

3. By inspection, I first swap the series pair of \$-12\:\text{V}\$ and the \$2\:\text{k}\Omega\$ resistor, so that I have a \$-12\:\text{V}\$ source relative to ground on that branch, now. From here, I see \$2\:\text{k}\Omega\$ in series with \$\left(1\:\text{k}\Omega+3\:\text{k}\Omega\right)\:\vert\vert\: 1\:\text{k}\Omega\$, to ground. This sets up a current of \$\frac{-12\:\text{V}}{2\:\text{k}\Omega+\left[\left(1\:\text{k}\Omega+3\:\text{k}\Omega\right)\:\vert\vert\: 1\:\text{k}\Omega\right]}\$. This current creates a voltage drop across the \$2\:\text{k}\Omega\$ resistor. The remainder voltage is then \$-12\:\text{V}-2\:\text{k}\Omega\cdot \frac{-12\:\text{V}}{2\:\text{k}\Omega+\left[\left(1\:\text{k}\Omega+3\:\text{k}\Omega\right)\:\vert\vert\: 1\:\text{k}\Omega\right]}\$, which is then applied to the voltage divider caused by the \$1\:\text{k}\Omega\$ and \$3\:\text{k}\Omega\$ resistors.

   Compute \$\frac{1\:\text{k}\Omega}{1\:\text{k}\Omega+3\:\text{k}\Omega}\cdot\left\{-12\:\text{V}-2\:\text{k}\Omega\cdot \frac{-12\:\text{V}}{2\:\text{k}\Omega+\left[\left(1\:\text{k}\Omega+3\:\text{k}\Omega\right)\:\vert\vert\: 1\:\text{k}\Omega\right]}\right\}=-\frac67\:\text{V}\$.

Adding these up gives \$\frac27\:\text{V}+\frac{18}7\:\text{V}-\frac67\:\text{V}=2\:\text{V}\$.

This is confirmed by simulation:

I am not taking your classes and I've no idea if the above is something that's allowed. But from here, today, it looks to me that this would avoid circuit reductions, mesh, and nodal analysis per the requirements given to you.

You also could just re-arrange two parts:

Those two voltage sources are now in opposition to each other. I've labeled their center node as \$X\$. Clearly, this tells you that the other two directly related nodes are equal to each other and are \$X+12\:\text{V}\$. From that, you know that \$\frac{X}{1\:\text{k}\Omega}+\frac{X+12\:\text{V}}{2\:\text{k}\Omega}+\frac{X+12\:\text{V}}{3\:\text{k}\Omega+1\:\text{k}\Omega}=2\:\text{mA}\$. (That's KCL-ish, so perhaps forbidden.) But the solution to that will be \$X=-4\:\text{V}\$ and so \$X+12\:\text{V}=8\:\text{V}\$. And \$\frac{1\:\text{k}\Omega}{3\:\text{k}\Omega+1\:\text{k}\Omega}\cdot 8\:\text{V}=2\:\text{V}\$. So that's another way to at least double-check your results, even if it can't actually be used due to the rules.