This is an extra credit question from my homework. Can someone spot a trick here or anything? My instructor likes to throw curveballs at us sometimes. How do I go about finding the voltage from A to B?

 

First find voltage at VA (voltage drop across R4) next find voltage at VB (voltage across R6) and final VAB = VA - VB.

EDIT
Also notice that R4/(R3+R4) = R6/(R5+R6) (voltage divider) so from there we can conclude that Vab = ??

 

Thanks for the help. I came up with 200mV.

 

Thanks for the help. I came up with 200mV.

How ?? This is wrong see my additional note.

 

Thanks for the help. I came up with 200mV.

No. It is a trick question. Google Wheatstone Bridge.

 

How ?? This is wrong see my additional note.

Here are my calculations. Would you mind showing me where I went wrong?

 

Disregard that "or 1.5V." It's not related.

 

Here are my calculations. Would you mind showing me where I went wrong?

Yep, (R5 + R6) are connected in parallel with (R3+R4). As I says early please notice that this two voltage divider (R3,R4) and (R5,R6) are the same. They have the same voltage ratio R4/(R3+R4) = R6/(R5+R6).
https://en.wikipedia.org/wiki/Voltage_divider#Resistive_divider

 

This is an extra credit question from my homework. Can someone spot a trick here or anything? My instructor likes to throw curveballs at us sometimes. How do I go about finding the voltage from A to B? View attachment 103474

Just remember that Vxy is short for Vx - Vy.

So find the voltage at Node A (Va), Find the voltage at Node B (Vb). Then compute Vab = Va - Vb.

 

Just remember that Vxy is short for Vx - Vy.

So find the voltage at Node A (Va), Find the voltage at Node B (Vb). Then compute Vab = Va - Vb.

In this particular case, if you look at the values closely, you can answer the question by inspection. But, if not, just analyze the circuit as described above and you will get the answer. Once you do, then ask yourself if, in hindsight, there are things you could have spotted that would have led you to the same result with less work.

 

Is the answer 0v? I calculated 72.2V across R4 and across R6. Another calculation came up with 79.38V across both branches (R3 and R5 included). Either way it suggests The voltage across the branches are the same which is how it should behave according to what was suggested earlier about the ratios (not to mention the nature of parallel circuits in general). I hope I'm not missing something.

 

You are missing the fact that the answer can be determined by inspection.
You don't need to do any calculations, just observe the the voltage division ratio of R3 and R4 is the same as R5 and R6 so the voltage difference is 0V.
If you did the calculations and came up with some small voltage due to rounding errors in your calculation, some profs will take points off your (not quite correct) answer because you didn't observe that it should be 0V.

 

Thanks for the help everyone.

 

Hello,

This is not a trick question...balanced bridges are a fact of nature...it happens. In practice there may be a small difference, but in theory we can theorize that there is absolutely zero difference. Another way of putting this is that a balanced bridge is "unobservable".

If you have never dealt with resistive bridges however this may be new to you, so you just have to carefully calculate the two voltages and then subtract. Sometimes polarity matters, so Vab=Va-Vb and Vba=Vb-Va, although there may be cases where they are both the same (ideally). I dont want to give out too much just yet though.