Finding Thevenin voltage across R2 in circuit with R1-R5

[phyznut]

I'm trying to find the current and voltage across R2 using thenenin.
http://img209.imageshack.us/img209/4606/thv15ph.png
I've managed to find Rth, which is equal to (R1 + R4)||(R3 + R5). That comes out to about 2.9 ohms. I need help with finding the Vth (Vab) though.
http://img208.imageshack.us/img208/5002/thv28nc.png
Can someone please help me. What are the tricks involved here?

 

[lightgrav]

Your second picture IS the trick ... with I_2 = 0, what is V_ab ?
That is, with I_2 = 0 , all the current thru R_1 goes thru R_3 , R_4 , and R_5 .

 

[phyznut]

Your second picture IS the trick ... with I_2 = 0, what is V_ab ?
That is, with I_2 = 0 , all the current thru R_1 goes thru R_3 , R_4 , and R_5 .

Ok then so with R2 removed, the circuit is now series. That makes the current the same for the remaining resistors.

So if I_2 is 0, then the voltage between that point is also zero? That doesn't seem right though. I still don't get it, please give me another hint.

 

[phyznut]

Can no one help.

Would I have to use the voltage divider? How do i use it?

 

[ranger]

Yes you could use the voltage divider formula. With R2 removed, you can think of $$V_{AB}$$ as the voltage drops across $$R_3 and R_5$$

Look at it in this way. When you remove $$R_2$$, the circuit now becomes a series circuit. Look at where point A and B are. If you where to take a voltmeter and put the probes across points A and B. Your voltmeter would be giving you to combined voltage of R3 and R5.