Answer check to challenging problem

Thread Starter

ihaveaquestion

Joined May 1, 2009
314
This is a challenging problem to me, so I'd like to make sure my thinking is correct for my way of solving:

http://img24.imageshack.us/img24/88/imguu.jpg

http://img23.imageshack.us/img23/8479/img0001ty.jpg

Given the original circuit drawn at the top (in the first link), I am asked to find the voltage vo by using Thevenin's theorem. So the strategy is to find Vth, find Rth, then use voltage division to find vo, i.e. vo = Vth(2) / Rth + 2.

I do normal nodal analysis to find Vth = V2 in the first link.

On the second link I find Rth by replacing the 2ohm resistor by a 1V voltage source, shorting the original voltage source, and do nodal analysis again. In this circuit with the excitation 1v voltage, I know v2 = 1v... so I find out what V1 is... the reason for all of this is because I want to determine what the current through my excitation voltage source is so I can determine Rth by Vs / i1 (labeled in my work). Once I find out what V1 is I do KCL at v2... I'm wondering if the directions of the currents at v2 matters... I said the current coming in from the left into V2 was v1-v2/j2... I calculate i1 (current through excitation Vs) and calculate Rth...

I finally do my voltage division with my Thevenin equiv..

what do you guys think?

Thanks in advance.
 

mik3

Joined Feb 4, 2008
4,846
Replace the capacitors and inductor with their equivalent impedance. Remove the 2R resistor and find Vth. Then replace the voltage source by a short circuit and the current source by an open circuit and find the impedance (Rth) looking into the terminals across the 2R resistor (without 2R connected).
 

Thread Starter

ihaveaquestion

Joined May 1, 2009
314
Replace the capacitors and inductor with their equivalent impedance.
This is shown in my work.

Remove the 2R resistor and find Vth. Then replace the voltage source by a short circuit and the current source by an open circuit and find the impedance (Rth) looking into the terminals across the 2R resistor (without 2R connected).
The current source is dependent, can't open circuit it. This is why I applied the 1v source in my work.
 

Thread Starter

ihaveaquestion

Joined May 1, 2009
314
As an FYI - you don't need to check all my numbers, just want confirmation that my procedure and way of thinking is correct.
 
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