## anonymous one year ago Logic in the following circuit I will post it

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1. anonymous

2. anonymous

|dw:1444493174253:dw|

3. anonymous

@ganeshie8 @Lyrae

4. amistre64

the only thing that comes to mind for me is superposition. remove all but one of the power supply, and compute the parts. do this for each power supply and then add the results for each part. not too sure if its applicable tho, its the only thing that comes to mind tho.

5. Lyrae

$$I_{ab} = I_1+I_2$$ is (probably) because of Kirchhoff's current law, i.e. the current entering any junction is equal to the current leaving that junction. |dw:1444511469312:dw| If we analyse this circuit with KCL we have two incoming currents in junction a, $$I_1$$ and $$I_2$$. The only way for these currents to flow given they "choose" the path of least resistance is towards b. In a more complex circuit you might have to calculate the currents (and their direction) by using both KCL and Kirchhoff's voltage law to create and solve a set of linear equations. A small guess is also that understating Norton and Thévenin's equivalents might be helpful for that picture in your first post.

6. Lyrae

|dw:1444512913493:dw|

|dw:1444531879314:dw|Not for sure what logic is supposed to be demonstrated here, just the fact that the various voltmeters, and ammeters are working as analysis would show.

8. anonymous

I mean: it doesn't make sense for being zero voltage and drew some current. for H bridge when zero voltage = zero current, here another point of view is the following |dw:1444544265619:dw| node analysis $\frac{ b-v }{ R } + \frac{ b-v }{ R } +\frac{ b-a }{ 0 }=\frac{ b-v }{ R } + \frac{ b-v }{ R }+\frac{ 0 }{ 0 }$ then 0/0 = 2i !!

9. anonymous

-2i*