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  • 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.
    Thanks in advance. ? 

    ELv3-a4412652-4e2b-49ce-9712-0c24ee277862.jpg

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  • lal

    Member • Jan 9, 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.





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  • Abdelrahman Rabah

    Member • Jan 10, 2019

    #-Link-Snipped-# 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 ..*

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  • lal

    Member • Jan 10, 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. 


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  • Abdelrahman Rabah

    Member • Jan 10, 2019

    #-Link-Snipped-#  that quite helped a lot .. thanks :D <3  

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