Deriving a balanced bridge equation

[quaker]

I am having trouble finishing this problem. I am supposed to first derive an equation for V_5 in this circuit:

http://img292.imageshack.us/img292/599/picture3qy2.png

I applied a Delta-Wye transformation to get here:

http://img292.imageshack.us/img292/9672/workot8.jpg

From there I am supposed to show that the circuit is balanced when R1/R2 = R3/R4 for any value of R5. This is where I'm getting stuck. My attempt starts by setting the V5 equation equal to zero and then trying to reduce it. This isn't working very well since the V5 equation is pretty messy to begin with.

Am I going about this correctly, or is there an easier method I should use? Thanks for your help.

 

[doodle]

Yes, I think the V5 equation is messy. But there's an easier way to show the condition for a balanced bridge.

The bridge is balanced if V2 = V4 since there will be no current then across R5. This also means V1 = V3. Then, from applying Ohm's law to the 2 equations and canceling the currents, you should get R1/R2 = R3/R4 quite easily.

 

[piercas]

Hello,

Thank you for reaching out for assistance with this problem. I can understand how challenging it can be to derive equations and solve complex circuits. I will do my best to provide a response that will help you with this problem.

First, let's start by reviewing the Delta-Wye transformation, since it seems like you have already used it in your attempt to solve this problem. The Delta-Wye transformation is a technique used to simplify a circuit by transforming a Delta (Δ) configuration into a Wye (Y) configuration. In this transformation, three resistors are connected in a triangle (Delta) and are transformed into three resistors connected in a Y shape. The transformation follows the equation:

R1 = (Rab * Rac) / (Rab + Rac + Rbc)

R2 = (Rab * Rbc) / (Rab + Rac + Rbc)

R3 = (Rac * Rbc) / (Rab + Rac + Rbc)

Now, let's apply this transformation to the circuit given in the problem. We can see that the resistors R1, R2, and R3 are already connected in a Delta configuration. So, we can transform them into a Wye configuration by replacing R1, R2, and R3 with the following resistors:

R1' = (R1 * R2) / (R1 + R2 + R3)

R2' = (R1 * R3) / (R1 + R2 + R3)

R3' = (R2 * R3) / (R1 + R2 + R3)

This transformation will give us the following circuit:



Now, we can see that the circuit has two parallel branches, each with two resistors in series. This means that we can use the voltage divider rule to find the voltage across R5.

V5 = (R4 / (R3' + R4)) * (R5 / (R1' + R5)) * V4

= (R4 / (R3' + R4)) * (R5 / (R1' + R5)) * (V3' + V5)

= (R4 / (R3' + R4)) * (R5 / (R1' + R5)) * (V3 + V5)