Wheatstone bridge in Linear Algebra

[Gunner1412]

Homework Statement

I remember in physics class that we saw this set up.The ratio of r1/r4=r2/r3 is the same. There is no use for the current to head through R5 if it's the same energy on both sides. Now in linear I have no idea how to use this.

Homework Equations

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The Attempt at a Solution

I tried re-arranging Kirchhoff's laws, but it didn't give me enough variables to solve...

Please help guys, I really can't figure this one out.

 

[quantumdude]

We'll need to fix our notation in order to understand each other. Start by labeling nodes in the circuit as follows.

Node a connects [itex]R_1[/itex] and [itex]R_2[/itex].
Node b connects [itex]R_1[/itex], [itex]R_5[/itex], and [itex]R_4[/itex]
Node c connects [itex]R_2[/itex], [itex]R_5[/itex], and [itex]R_3[/itex].
Node d connects [itex]R_3[/itex] and [itex]R_4[/itex].

Let the voltages at these nodes be [itex]v_a[/itex], [itex]v_b[/itex], [itex]v_c[/itex], and [itex]v_d[/itex]. Since only potential differences are physically meaningful you can set any one of these equal to zero. Just so our notations match, let [itex]v_d[/itex]=0.

Now you have to assume a direction for the currents through the resistors. Let the currents flow from top to bottom. Recalling that currents flow from higher potentials to lower ones, you should be able to write down expressions for the currents through the resistors in terms of the node voltages and resistances. Then you need to do the following.

1.) Apply KCL. There are only 2 nodes at which you can apply KCL, because you don't know the current flowing through the battery. So it should be easy to see where to do it. This will give you 2 equations, which is all you need.
2.) If the current [itex]i_5[/itex] through resistor [itex]R_5[/itex] is zero, then which two node voltages must be equal? This will eliminate one of the unknowns.

At this point you should be able to derive the result.