Finding the voltage and the current of a ridiculous circuit.

Consider the circuit shown in the figure below. (Let R1 = 5.00 Ω, R2 = 5.00 Ω, and emf = 12.0 V.)
http://www.webassign.net/serpse8/28-p-01…

i need voltage and current across r1. i'm stuck, this goes into much more detail than anything we did in class. please help. 10pts today best answer

You should replace parallel and series resistor combinations with their equivalent resistances, eventually simplifying the circuit.

The 5 ohm in parallel with the 10 ohm is 50/15 = 3.33 ohm That is in series with R2 (5 ohm) making 8.33 ohms Those three resistors can be replaced with on 8.33 ohm resistor across R1. R1 is in parallel with th 8.33 ohm giving a net resistance of 3.12 ohms; that in turn is in series with 2 ohms so the total resistance across the battery is 5.13 ohms. The current drawn from the battery is then 12.0/5.13 = 2.34 A; the voltage drop across the 2 ohm resistor is then 4.68 v so the voltage across R1 is 12 - 4.66 = 7.32 V. Since R1 is 5.00 ohms, the current is 7.32/5.00 = 1.46 A

Let the voltage at the battery negative node be 0V. Let X be the voltage at the 2/5/5/10 node, and Y the voltage at the 5/5/10 node. Then by Kirchoff:

(x-12)/2 + x/5 + (x-y)/5 + (x-y)/10 = 0
y/5 + (y-x)/5 + (y-x)/10 = 0

Solving: x = 7.32 Volts, y = 4.39 Volts

The voltage across R1 is x, i.e., 7.32 Volts.
The current through (not across) R1 is 7.32/5 = 1.46 Amps

if R1 is equal to 5 ohms then since it is in series with the 2 ohm resister that would be 7ohms total
so volts over resistance would be (12/7 equals 1.7A @ 12V)

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