Engineering Finding the constraint equation of a circuit with a dependent voltage

[The Electrician]

I question the polarity of the current controlled voltage source.
It is shown inverted in the original diagram.

Exactly what are you getting at when you say "I question the polarity of the current controlled voltage source."? That source is indeed shown inverted in the diagram as you assert, so what is your point?

 

[Baluncore]

It is OK. The 10 V source is sinking conventional current, so Id is negative.
Multiply by 20 and invert the CCVS to get +64 V, which sources the current to be sunk by the 10 V.

 

[BvU]

I question the polarity of the current controlled voltage source.
It is shown inverted in the original diagram.

Only means a negative answer instead of a positive

 

[Baluncore]

Only means a negative answer instead of a positive

Not quite.
I think if you reverse the CCVS polarity it produces 6.80851 V, and Vo becomes +7.65957 V.
The 10.0 V becomes a positive current source, so the CCVS is also positive.

 

[BvU]

My turn to be confused: if $V_2=+64$ V, the CCVS delivers $-$64 V and pumps 3.8 A towards point V2. The $-$64 is 20 times the $-$3.2 A $i_\Delta$.

If the CCVS sets $V_2=+ 6.80851$ V, the dial must read $-6.80851$ V which is minus 20 times the then $-$0.34 A $i_\Delta$.

 

[The Electrician]

My turn to be confused: if $V_2=+64$ V, the CCVS delivers $-$64 V and pumps 3.8 A towards point V2. The $-$64 is 20 times the $-$3.2 A $i_\Delta$.

If the CCVS sets $V_2=+ 6.80851$ V, the dial must read $-6.80851$ V which is minus 20 times the then $-$0.34 A $i_\Delta$.

Your first short paragraph above describes the situation as shown in the schematic of post #1.

The second short paragraph describes what would happen if the CCVS control law were changed from 20 IΔ to -20 IΔ, or if the orientation of the CCVS were flipped upside down with no change to the control law.

That's what Baluncore is referring to when he says "reverse the CCVS polarity".