Kirchhoff's Rule / finding unknown resistances and voltages

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Homework Statement



Find the current passing through R1 and the voltage passing through the cell to the immediate left of R1.

R1 is 2 Ω, and R2 is 6.5 Ω.

http://img255.imageshack.us/img255/9520/problem2ux3.png

Homework Equations



Kirchhoff's Rule; Ohm's Law

The Attempt at a Solution



Not sure if I'm approaching this the correct way. What I had in mind was to do some mesh analysis while considering I(1) to be 3 A, doing Kirchhoff's for the bottom half of the circuit, finding I(2), and using Ohm's to find I through that 2 Ω resistor.

Little unsure about finding the voltage of that mystery cell, though.
 
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I'd first combine R2 and the 4 Ohm resistor -- no need to keep them separate for this problem. Then ground the right side, at the - side of the 24V voltage source. You then have 2 unknown node voltages that you can write the KCL equations for, and once you solve for them, you have the solutions for the question.

BTW, instead of saying "the voltage passing through the cell to the immediate left of R1", it would be better to say "the voltage across the cell". Current passes through an element in response to the voltage placed across the element.
 
berkeman said:
I'd first combine R2 and the 4 Ohm resistor -- no need to keep them separate for this problem. Then ground the right side, at the - side of the 24V voltage source. You then have 2 unknown node voltages that you can write the KCL equations for, and once you solve for them, you have the solutions for the question.

BTW, instead of saying "the voltage passing through the cell to the immediate left of R1", it would be better to say "the voltage across the cell". Current passes through an element in response to the voltage placed across the element.

Well, that worked beautifully. I kept thinking I couldn't combine terms in the first part of the Kirchhoff's work for some reason; took me a second to realize that a voltage and the product of a current and a resistance value (Ohm's, anyone?) definitely are combinable.

My "duurrrrr" moment for the day, I suppose.

Thanks.