Thevenin and Norton Circuits

I'm having difficulty understanding the topic that is currently being covered in my Linear Circuits course. I have a couple questions due on Monday and was wondering if I could get some help solving them. The problems in question are 3.69 and 3.70 in attachment #2. Attachment #1 is my attempt at problem 3.69 (It uses problem 3.43 in attachment #3). I'm assuming it isnt right since the Vth / Rth isnt equal to the Inorton. So I was wondering if maybe someone could help me spot where I screwed up. With problem 3.70, I'm mostly having trouble understanding the wording, and I think if I can get help with 3.69, I should be able to solve 3.70 without a problem. What I'd like to know is, for 3.70, would I need to calculate the nodal voltage above R1 and then recalculate that voltage when R3 is added into the total load? Also, Rs is not included in the Rth equaivalent, correct?

 

Hello jnbfive,

For 3.69 ,while finding the thevenin's resistance
you have to consider R1 and R4 as parallel resistances and similarly R2||R5.Now these two combinations are in series.I think you have taken it as R1 series with R2 and R4 in series with R4 .

similarly if you want to find the current considering the sources independantly
say VS1 first ,you can find the current as VS1/(R1+R4). you need not include R5 and R2.Similarly for VS2 you need not include R4 and R1.

Once you find the current you can easily find the Vth or open circuit voltage.

 

Hey, thanks for the reply! Thanks for pointing out the Rth. I realize that I shouldnt have taken those in series first due to the node. I do have some more question though. I did what you said, (I think at least), and I used a mesh analysis, (i1 and i2) to calculate the current, and I wound up with i1 = 75 A and i2 = 56.25 A (i1 is the bottom right mesh). Then I tried voltage divider and wound up with the same difference, 18.75, in volts . Now I know Rth =Vth/In, so obviously that isnt right. Any suggestions on the mistakes I'm making? I didnt use a third mesh since no current can travel when you remove R3 (I think thats right?). Then for question .70, is Rs considered part of the source and therefore, not included in the Rth calculation?