# Thevenin problem #3

Discussion in 'Homework Help' started by gengm, Apr 12, 2010.

1. ### gengm Thread Starter New Member

Apr 7, 2010
23
0
Thanks for the help guys
I need a little more help, here are the last 2 Thevenin related problems I need help with

Here is the problem 1

I was thinking about changing the CS and the || resistor into a Thevenin Equiv. Then using node volt analysis, solve for Vx, then input the Vx back into the original circuit and then finding the Vth, then using the 1Amp rule to find Rth.

Problem 2

Would changing the Dependent current and || resistor to a Thev equiv be a good start?

Update: I've tried both my methods on both circuits and I keep getting them wrong. I solved past problems like problem 1 but some odd reason, NVA won't work this time around

Last edited: Apr 13, 2010
2. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,366
352
Rather than try to come up with an ad hoc (http://en.wikipedia.org/wiki/Ad_hoc) method for these various problems you've been posting, you would be better served by learning how to write network equations. That will always lead you to a solution.

I think if I had to recommend a method ahead of time, it would be nodal analysis. Often, once you see a particular circuit, you will recognize that another method might be quicker, but nodal analysis will (almost) always work, except in certain cases (circuits containing transformers, for example).

Since nodal analysis uses KCL, a good plan is to convert voltage sources to current sources when you can easily do this. For example, in your first circuit, you can convert the 9 Vx source and the 5 ohm resistor into a Norton equivalent and then there is only one nodal equation, which you can write by inspection.

Why don't you try this on these two problems, and post your work here.

3. ### gengm Thread Starter New Member

Apr 7, 2010
23
0

good start?. Not sure what to do with the Dependent and independent CS though.

4. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,366
352
Now try writing an equation.

gengm likes this.
5. ### gengm Thread Starter New Member

Apr 7, 2010
23
0
I got it. It came to me this morning. I just took a good 15 minutes to look at it and it came to me.
Also our exam was today, wasn't too bad, Had one easy thevenin problem, an Operational Amp, then some other Inductor and Capacitor problems.

6. ### Ghar Active Member

Mar 8, 2010
655
73
The first one has an interesting trick.

The voltage of the dependent source with the resistor is:
V = Vx + 9Vx = 10Vx

You also know that:
Vx = 5 I

Therefore V = 10*5 I = 50 I

Or equivalently it's a 50 ohm resistor...

Even simpler;

Put in a current I, find voltage:
V = 5*I + 9*(5*I) = 50*I