Circuit analysis with V2, R1, R2, R3: which resistor to use in second loop?

[Alpha Russ Omega]

Hello.
This is a strange problem that I ran into in my homework. I did not use one of the resistors in the second loop. I don't think this is correct and I am really confused. I have no idea how else I could set this up to solve for the needed variables.

Find Vs and Is, if V2 = 20V, R1 = 6K, R2 = 10K, and R3 = 2K.
http://img115.imageshack.us/img115/5062/ca202nrs4.gif

Here is what I did:
http://img169.imageshack.us/img169/7892/newrf2.jpg

If it is wrong; how should I set this up correctly?

 

[Astronuc]

Well for one, V_s + V₂ does not equal zero because of the presence of R₁ and the current I_s.

One should follow the conventions of Kirchoff's voltage and current laws, i.e. the conventions for assigning potential difference across a voltage source and the direction of the current.

The potential across R₂ is the same across R₃, therefore they will have different currents passing through them, but the sum must be I_s.

 

[Alpha Russ Omega]

Oh, ok. I think I'm starting to understand this.

I redrew the set-up by labeling more currents and labeling other voltages.
Did I set-up this drawing correctly?
http://img405.imageshack.us/img405/4074/new6wi9.png

Would this set-up work?
Is=I2+I3

"left loop"
Vs+I2*R2+Is*R1=0

"right loop"
-V2+V3=0

 

[Astronuc]

Yes, I believe that's correct.

One then has three equations, but one can reduce it by find the eqivalent resistance to R2||R3 and adding that to R1 = Rtot. The Vs = Is * Rtot.

Or one can solve for I3 = V2/R3, because V3 = V2.

I notice that Is and V2 are written in the opposite direction of convention. Normally current flows out of the + terminal of a voltage source (battery), and into the positive potential of the resistor. The + side of the resistor would normally be on the side closest to the + terminal of the voltage source. If one obtains a negative current or voltage, it means that it is opposite of the way it is written.

 

[Alpha Russ Omega]

Yes, I believe that's correct.

One then has three equations, but one can reduce it by find the eqivalent resistance to R2||R3 and adding that to R1 = Rtot. The Vs = Is * Rtot.

Or one can solve for I3 = V2/R3, because V3 = V2.

I notice that Is and V2 are written in the opposite direction of convention. Normally current flows out of the + terminal of a voltage source (battery), and into the positive potential of the resistor. The + side of the resistor would normally be on the side closest to the + terminal of the voltage source. If one obtains a negative current or voltage, it means that it is opposite of the way it is written.

Ah, I see. My book described that as well but the problem given already had the Is pointing into the + side of Vs so I figured I would just follow the original drawing. You are correct, the Vs did come out to be negative due to the odd sign placement.

Thank you for your help!