I'm having trouble with this problem..

So far I've tried combinging the 2 branches and get (4-j2)/5 for the combined impedance

I found the open circuit voltage by taking all of the 20A flowing into its series circuit through the (4-j2)/5 and getting that voltage and subtracting 2Ib from it , therefore the open circuit voltage came out to be 25.3<-71.57 degrees

then when solving for Isc, i just did a voltage division and treated the controlled source at the top as a 2 ohm resistor since its in a form of R*I and its a voltage source. so i used current division to find how much of the 20a goes to the short circuit and i got that to be 6.32<-18.44. after dividing Voc/Isc, i found Zth= 4 < -53.13 and then got the norton equivalent from just drawing a diagram of the Zth and Isc in parallel. can someone verify if my steps are valid? if not, how else do I approach finding the numbers?

If someone could guide me on how to further approach this i'd REALLY appreciate it! thanks

soo yea

So far I've tried combinging the 2 branches and get (4-j2)/5 for the combined impedance

I found the open circuit voltage by taking all of the 20A flowing into its series circuit through the (4-j2)/5 and getting that voltage and subtracting 2Ib from it , therefore the open circuit voltage came out to be 25.3<-71.57 degrees

then when solving for Isc, i just did a voltage division and treated the controlled source at the top as a 2 ohm resistor since its in a form of R*I and its a voltage source. so i used current division to find how much of the 20a goes to the short circuit and i got that to be 6.32<-18.44. after dividing Voc/Isc, i found Zth= 4 < -53.13 and then got the norton equivalent from just drawing a diagram of the Zth and Isc in parallel. can someone verify if my steps are valid? if not, how else do I approach finding the numbers?

If someone could guide me on how to further approach this i'd REALLY appreciate it! thanks

soo yea

Last edited: