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How to convert this circuit into its Norton Equivalent?

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Farrukh

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The problem is represented in figure (a)!

I don't understand how the original 3V battery was removed and then -8V battery connected in series with 100Ohms. Then the 5V.. It's totally over my head.

Can somebody tell me how is the Norton Current Source calculated here?

I thought all the current will flow through AB, instead of passing through the resistances 100 and 220, since AB offers no resistance!! But I get the wrong answer that way.

ps: This is Problem 25, Chapter 8 of Floyd's Principles of Electric Circuits, 7th Edition. Problem 28 (I think) of 8th Edition.\]
 

It seems there are 3 voltage supply on circuit.
terminal AB current calculated with use of super position theoream.
in each step just one voltage (or current ) source is active.
and all others are disabled (short circuit for vlotage supply and open circuits for current supply).
as you write total current is sum of these currents.

For norton equivalent resistor, simply disable all supplys and we see that all three resistors would be paralleld.

Regards
Davood.
 

No, the original problem is in figure (a). There is only 1 source, the 3V source. This is the solution to the problem that I have posted, but I don't understand it. How can we remove the 3V source, and put a -8V source with the 100Ohm resistor? Can we remove sources like that? I don't think so!!
 

!!!!!!!!!!!!!!
the answer is what i said before [above].
 

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    Farrukh

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The original problem had the three batteries in the circuit simultaneously. Then superposition is used to find the output short circuit current from each and the three currents are added.
 

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    Farrukh

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Ok, it means there is an error in the book. Because it only shows one voltage source, the 3V source. It doesn't show any other voltage source. But the solution shows 3, so I guess it's an error in the book.
 

I agree with Davood.
This problem was solved using superposition (because of the existance of 3 voltage sources).
If the problem was only as shown in (a), then the answer would have been In1 only.

- AD
 

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