thevenin analysis?

Question

how can i get the vth of a circuit if the element to be removed is a source

this is the link of the image of my circuit,the element to be removed is the 5V source
http://img301.imageshack.us/img301/5740/eeek4.png

Answer 1

first lets simplify the right side of the circuit
--> take 3k and 1k resistors in series=4k
then take this 4k in parallel with 2k resistor(not the resistor connect with 5v,the another one)
u got =1.3k
now apply tansforamtion source ;
u have 1.3k resistor connected with parallel source current ,so convert the current source to voltage source(=(2m)*(1.3k)=2.6V) which is series with 1.3k resistor.
now u have 2.6v with R(1.3k) in series in the right side
and the left side there 6V with R=3k in series also ,and between them there is the 5v and R=2k,now u can apply Thevenin .
remove the 5V and calculate the voltage between the 1.3k and 3k resistor.

(note that there is no current flowing through the 2k resistor which now connected with the open Vth,so we can ignore it when we calculate vth only)

u can use kirchof 's law to right node or left node.
lets take the left side;
-6+(3k)(I)-Vth=0--->1
we have to calculate I firse,
I=(-2.6+6)/(1.3k+3k) =0.79mA
go back to one u got v th = -3.63
to calculate Rth shorted all voltage source
Rth=2.9k

Answer 2

Just remove the voltage source as if it were a resistor and perform the Thevenin Analysis as usual. It does seem like an odd request, but it is one way determine the current in the leg with the 5 volt source without having to write and solve the loop and/or node equations directly.

I didn't attempt to do it, but by inspection the Rth is at least 2 kilo-ohm.

Answer 3

The analysis is no different in principle whatever element you remove. Looking at your circuit, I'd do the analysis in two stages, starting by replacing the current source and the three right hand resistors by their Thevenin equivalent circuit before I started fooling about removing the 5v source :-)

Answer 4

where among the resistors would you compute the thevenin voltage.