Need help with Determining a formula for a Resistance using a linear system for loop currents I am studying for a test, and I am stuck on a Linear Algebra problem that involves some knowledge of currents.
Here's the textbook problem:
The Wheatstone Bridge is used to determine an unknown resistance $R_3$ by adjusting a known resistance $R_1$, so that the measured current (in $r$) is zero. Determine the formula for $R_3$ using the linear system for the loop currents.
Here is the given image:

I know that the point of balance (when $r = 0$) is $R_3/R_1 = R_4/R_2$.
Thus, $R_3 = R_4R_1/R_2$.
Using Kirchoff's Current Law, I started getting a linear system.
$$I_1 R_1 - I_rr - I_3R_3 = 0$$
$$I_2  R_2 - I_4R_4+I_rr = 0$$
Which would also represent the point of balance.
Though, I don't know how to rearrange the linear system so that it equals $R_3$. Am I correct so far?
 A: The point of balance always occurs when $R_3 = R_4R_1/R_2$, no matter what the resistance $r$ is. In fact, $r$ plays essentially no role here. A derivation is given below. First let me label four points in your diagram A, B, C, and D.

Let us assume the resistance $R_1$ is adjusted so that we are at the balancing point, i.e., that $I_r = 0$. This means exactly that the voltage $V_B$ at $B$ has to be the same as the voltage $V_C$ at $C$. We know that 


*

*$V_B - V_A = I_2R_2$.

*$V_C - V_A = -I_4R_4$ (the negative comes from the choice of direction of current in our diagram).

*$V_D - V_B = I_1R_1$.

*$V_D - V_C = -I_3R_3$.


Since $V_B = V_C$, equations (1) and (2) combine to give $I_2R_2 = -I_4R_4$, and equations (3) and (4) combine to give $I_1R_1 = -I_3R_3$. These are two equations in our system. 
Another law we need to use is that the total current leaving any given point in the circuit is $0$. At point B, this law translates to $I_2 = I_1$, since by assumption $I_r = 0$. Similarly, at point C, this law gives $I_3 = I_4$. Putting this all together, we have four equations: 


*

*$I_2R_2 = -I_4R_4$.

*$I_1R_1 = -I_3R_3$.

*$I_2 = I_1$.

*$I_3 = I_4$.


Plugging equations (3) and (4) into (2) gives $I_2R_1 = -I_4R_3$. On the other hand, using equation (1) we can solve for $I_4$ to get $I_4 = -I_2R_2/R_4$. Combining these gives $$I_2R_1 = -(-I_2R_2/R_4)R_3,$$ which, after rearranging, will give $R_3 = R_1R_4/R_2$.
