# Norton's Theorem Question (with dependent sources)

Discussion in 'Homework Help' started by inkosi, Dec 10, 2013.

1. ### inkosi Thread Starter Member

Dec 10, 2013
35
0
Hello friends,

I have trouble solving the attached question. How exactly do you solve this? I did it with nodal analysis and got 2.18 V which I think is correct but the question specifically says solve it using Norton's Theorem. I know you have to find Voc and Isc, but what is Voc? How do you find it? And what is Isc? Is it the current flowing to the 4k resistor on the right hand side?

Help would be appreciated.

File size:
24.6 KB
Views:
41
2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
First you need to find Vo voltage, using whatever method you like or use.
And Vo = Voc
Next you need to short Vo terminals and find short circuit current Isc.

3. ### inkosi Thread Starter Member

Dec 10, 2013
35
0
I'm a bit stuck finding Isc. Here are the nodal equations I made:

-6-Vo / 6k = (Vo / 2k) + (-3 / k) + (V1 / 4k)

And Vo / 2k = 2kIx and V1 - Vo = 2kIx. I get Isc as 3/7 mA, but apparently the answer is -1mA.

What have I done wrong?

Feb 17, 2009
3,990
1,115
Isc = Vb/R3

File size:
22.4 KB
Views:
26
5. ### WBahn Moderator

Mar 31, 2012
18,086
4,917
I'm a bit unsure what is being by "solve this"? In other words, I see several things that the problem may be asking for and am not sure which it actually is.

Are you JUST being asked to find Vo for this circuit as given? If that's the case, I don't see how Norton's Theorem really helps you that much. You really can't walk across doing source transformations and transforming the one that you can do (on the left hand side) really doesn't buy you a whole lot.

I'm just guessing here, but I think what you might be expected to do is to treat the 4kΩ across the output as a load and then find the Norton equivalent circuit for everything else and then apply that equivalent circuit the to 4kΩ resistor and solve for Vo. But that's just a guess and I can think of a few other things that would be just as reasonable.