Electronic – Circuits problem

There is a misconception here that needs to be cleared up quickly.

When one does KVL 'round a loop one is summing the voltages across the circuit elements in the loop.

However, you've marked node voltages on your schematic which are, by definition, the voltage from that node to the ground (datum, common, zero reference...) node.

To make it clear, the voltage you've denoted V1 is the sum of the voltages across R1 and R3. This is a consequence of KVL.

So no, you cannot meaningfully sum the node voltages and set them to zero. That would be a misapplication of KVL.

Here are some examples to consider:

By KVL, the voltage across R1 is (V1 - V2).

By KVL, the voltage across R2 is V2 (V3 = 0).

By KVL, the voltage V1 is the sum of the above voltages: V1 = (V1 - V2) + V2

Here's yet another way to look at it.

The voltage V1 is measured by placing the red lead of the voltmeter at node 1 and the black lead at the ground node.

The voltage across R1 is measured by placing the red lead at node 1 and the black lead at node 2.

The voltage across R2 is measured by placing the red lead at node 2 and the black lead at node 3 (the ground node).

By, KVL, the measured voltage across R1 plus the measured voltage across R2 must equal the measured voltage across V1.

The answer is definitely wrong. Here's a quick way to tell. If you have R=infinite, then you have 40V/300ohms = 0.133 amps if you ignore current provided by the 20V supply. That's the absolute minimum current that will flow through I. Any lowering of R from infinite will only increase the amount of current flowing through I. That means that their initial statement of I=80mA is impossible.

The only exception to that is if we do allow negative resistances as you have calculated. A negative resistance would be kind of like a voltage/current source. You're likely correct in your negative resistance calculation.