Abdelrahman Rabah
Electronics and Communication
09 Jan 2019

# Something confuses me an exam ..

As u can see in the image below .. I just don't get it .. in the fig 1, the polarity in both A and B networks were the same, that is why there was no current flowing .. but in fig 2, he reversed the connections .. so that means that he also reversed the polarity, in another meaning, the + in network a ----> the - in network b and vise versa .. am I mistaken ? or right ? *There is no clear solution for me in this question so I will be asking the one who answered a lot of questions.*
edit : network A and B only have 1 Current source + 1 Resistor
Another question, is reversing the current source direction affects the polarity ?
Sorry for being talkative, but I'm really confused and the exam is around the corners.

lal

Branch Unspecified
10 Jan 2019

Why don't you start by represeting those networks as two equivalent Norton circuits and equate? Let us know what relations you come up with.

For instance, the first connection setup would give you something like,

Ina x Rna = Inb x Rnb = 10V

*Pardon the naming. By Ina, I meant equivalent current of network A and Rna is equivalent resistance of network A.

Abdelrahman Rabah

Electronics and Communication
10 Jan 2019

@lal  I actually did all of those .. and i represented the 1st case with this way
Ra*Ia = Rb*Ib = 10
but the 2nd case is the one confuses me .. to shorten all of these talks .. if I changed the current source direction, would it change the polarity between the R ? *I did a lot of searches on google, but nothing came handy ..*

lal

Branch Unspecified
10 Jan 2019

Yes, it would, and note that it is true only if you keep the reference of measurement same in both cases.

If A and B are the terminals of a resistor, and a current is flowing from A to B, V(AB) will read positive. Now if the current direction is reversed, V(AB) would read negative. Note that terminal B is reference in both the cases.

To start with, then, consider network B to be some impedanc, Zb, and network A to be driving that load. You could work out Ia and Ra then without much chaos.

Abdelrahman Rabah

Electronics and Communication
11 Jan 2019

@lal   that quite helped a lot .. thanks :D <3