Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.
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.
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 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...
This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register.
By continuing to use this site, you are consenting to our use of cookies.
