Norton equivalent Circuit: Why was the capacitor of -5j ohm neglected while calculating the Norton current between terminals a-b?

WBahn

Joined Mar 31, 2012
25,906
Without knowing what the problem was, it's hard to tell. I see two possible reasons. First, the capacitor is being treated as the load, and second, the circuit is being analyzed to find the Norton current, which is the short-circuit current between the terminals. If a short is places across the terminals, what effect does the capacitor have?
 

Thread Starter

Neophyte_2.0

Joined Aug 21, 2019
14
Yes, the circuit is being analyzed to find the Norton current. I don't understand why the capacitor isn't included in the calculation of the Norton current by current division (the second to last step in the picture) ?
This is the problem [picture (b)] and the solution:1.JPG2.JPG
 

WBahn

Joined Mar 31, 2012
25,906
Go ahead and don't ignore the capacitor and do the current division between it and a 0 Ω short and see how much of the current goes through the capacitor. Then reflect on why it is what it is and then see if the shortcut of ignoring it during the analysis makes sense. I suspect it will be a useful exercise for you because you are inquisitive enough to be asking the question in the first place.

As a complete aside, is the solution you posted from the text? If so, could you do me a favor and look in the first chapter or so of the book and see what the author says about units? I suspect that they will say something about how units are really important; most textbooks do give some kind of lip service to units even though, as demonstrated here, they don't really mean it and will just tack onto their answer whatever units they want or expect.
 

Thread Starter

Neophyte_2.0

Joined Aug 21, 2019
14
Go ahead and don't ignore the capacitor and do the current division between it and a 0 Ω short and see how much of the current goes through the capacitor. Then reflect on why it is what it is and then see if the shortcut of ignoring it during the analysis makes sense. I suspect it will be a useful exercise for you because you are inquisitive enough to be asking the question in the first place.

As a complete aside, is the solution you posted from the text? If so, could you do me a favor and look in the first chapter or so of the book and see what the author says about units? I suspect that they will say something about how units are really important; most textbooks do give some kind of lip service to units even though, as demonstrated here, they don't really mean it and will just tack onto their answer whatever units they want or expect.
ah yes, I understand now :) The current takes the path of least resistance and because the short has a resistance of 0 Ω, there will be no current through the capacitor. Thank you :D

Yes, it's part of the solution manual of the textbook (Fundamentals of Electric Circuits, Alexander; Sadiku; 6th ed).
And as you suspected, chapter 1 of the book is about the units and about how important they are. But even in exercises from this chapter they just add the units they expect/want at the end :p
 

WBahn

Joined Mar 31, 2012
25,906
Yes, it's part of the solution manual of the textbook (Fundamentals of Electric Circuits, Alexander; Sadiku; 6th ed).
And as you suspected, chapter 1 of the book is about the units and about how important they are. But even in exercises from this chapter they just add the units they expect/want at the end :p
I can't help but wonder how many millions of man-hours of effort have been wasted, not to mention millions of dollars of equipment damaged or destroyed, or the number of people injured or killed because these textbook authors, the majority of whom have little to no experience outside of academia, show by their example that tracking units is beneath them regardless of whatever pronouncements they tout in the first chapter.
 
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