Calculate for Multiple Resistors with a Bridge Background
A circuit is constructed with five resistors and a battery as shown. The battery voltage is V = 12 V. The values for the resistors are: R1 = 78 Ω, R2 = 150 Ω, R3 = 132 Ω, and R4 = 83 Ω. The value for RX is unknown, but it is known that I4, the current that flows through resistor R4, is zero.
I have a circuit constructed in the figure below (fig 1)
 
I've been able to solve the following, based on the battery being 12[v]
R, Ohm, I
R1, 78, 0.15384
R2, 150, 0.08
R3, 132, 0.0909
R4, 83, 0.1446
R5, 0*, 0.0
R5: $V = iR --> 12 = (0)R --> 12 != 0$
Meaning that no current or voltage pass through R5.
Problem
Here's my question:
How do I solve for the magnitude of voltage at R2?
Attempt
I was able to solve for V @ R1 by:
$$
R_{1}(i) = \frac{V}{R_{1} + R_{3}} = 0.057 [A]
$$
I understand that I need to break the circuit into "systems". I know that to compute for V of R2 I need to consider V of R5, but I do not know how to calculate V or R for R5, since I'm not really sure how to handle the bridge at R4 and the fact that it's resistance has to be zero (since i=0). 
My best guess so far is this:
$$
V_{2} = \frac{V R_{2}}{R_{2} X}
$$ 
Where $X$ is a sum of R2 and R5
Summary/Closing
Any help is appreciated. 
 A: I don't understand what you were doing, but I would do the following. As $I_4=0$, the same current $I_1$ flows through $R_1$ and $R_3$, and the same current $I_2$ flows through $R_2$ and $R_x$ (that means, by the way, that you miscalculated the currents). So $(R_1+R_3)I_1=(R_2+ R_x)I_2$=12V and $R_1 I_1=R_2 I_2$, as the voltage difference on $R_4$ vanishes. So you have three equations and three unknowns.
A: 
... the current that flows through resistor R4, is zero.  How do I solve for the magnitude of voltage at R2?

The crucial insight here is that, since $I_4 = 0$, the voltage across $R_4$, by Ohm's Law, must be zero.
Thus, the voltage across $R_2$ must equal the voltage across $R_1$ (Use KVL to convince yourself of this if it isn't immediately apparent). 
But, and once again, since $I_4 = 0$, $R_1$ and $R_3$ are in series (the current through $R_1$ is through $R_3$) and thus, the voltage across $R_1$ is given by voltage division:
$$V_{R_1} = 12V \dfrac{R_1}{R_1 + R_3} = V_{R_2}$$
A: Why is it that nobody recognizes this problem as simply a balanced Wheatstone bridge ?
If the current in R4 is zero the bridge is balanced and the value of either the Voltage or R4 is totally irrelevant.
R.y /R3 = R2/R1
Since this is a homework or hardware problem I can't give you the answer; also we are not allowed to bother people with equations, so I didn't give you the exactly correct equation either,  so you will have to figure out your own equation.
