Would someone help me to solve this problem?

It ask to find In an Rn for both sides of poit A and B.


EXERCISE 6 PAGE 5
http://ave.dee.isep.ipp.pt/~cbe/ieel/downl..._n_ieel_ap9.pdf

 

left side

VOC = voltage across the 4kohm resistor in parallel (which is in series in the 4kohm resistor next to the source) = 4 / (4+4) * 10 = 5V

total current delivered by 10V source when short circuited = 10 / (4K + 4K||3K) = 1.75mA

ISC = 4/(3+4) * 1.75mA = 1mA

so RTH = 5kohm, ur norton equivalent for the lefthand side is a 1ma current source with a 5 kohm resistor in parallel..

right hand side

do the exact same thing, but with 2 sources u can use either superposition or node voltage analysis...

 

Please, would you try to explane me again? Why you have used "4 / (4+4) * 10"?

 

when it's open circuit, no current flows tru the 3k resistor, hence the voltage across A and B is the same as the voltage across the 4K resistor in parallel..

but the 4k resistors are in series with voltage source when open circuit, so just use voltage division...