How to find voltage drop across each resistor for complex circuit?

Question

I don't know how to find the voltage drop across each resistor for http://i229.photobucket.com/albums/ee8/flaviozelenka/circuit.jpg Using loop analysis, I get the voltage across R_1, V_R1 is 9.41V, V_R2 = 8.58V, and V_R3 = 2.59V. However, this cannot be true as that totals to 20.58V, and only 18V are available.

Answer 1

The dodge for this is to convert each battery and its associated resistor from a Thevenin equivalent (as it is now) to a Norton equivalent.  This means replacing the voltage source in series with a resistor by a current source in parallel with one.  Once this is done, you'll have three resistors in parallel, driven by two current sources, also in parallel, and from that point the computation is trivial.  (The magnitude of the Norton current is simply I = E/R.)

Answer 2

I find redrawing the circuit helps.

Pull the 6volt battery under the 12 volt battery you will see it more clearly