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Concept of voltage drop

by birdbybird
Tags: voltage drop
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birdbybird
#1
Feb19-11, 07:29 AM
P: 3
So this is the question...

In the direct circuit diagram below (please see the attached file), Resistors 1 and 3 have fixed resistances (R1 and R3, respectively), which are known. Resistor 2 is a variable (or adjustable) resistor, and the resistance of Resistor 4 is unknown.

Show that if Resistor 2 is adjusted until the ammeter shown registers no current through its branch, then

R1R4 = R2R3

(This circuit arrangement is known as a Wheatstone bridge.)


and the solution says...
blahblahblah..and then
"Since no current flows between Points a and b, there must be no potential difference between a and b; they're at the same potential. (so i was like, ok i understand that) So the voltage drop from a to c must equal the voltage drop from b o c. <-- this is the part i don't get... why does having the same potential mean that they both have equal voltage drop?

I would be every so pleased if someone could explain this to me..
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Naty1
#2
Feb19-11, 08:25 AM
P: 5,632
That's a wheatstone bridge....look it up in Wikipedia they must explain it...

http://en.wikipedia.org/wiki/Wheatstone_bridge

"why does having the same potential mean that they both have equal voltage drop?"

two different nmes for the same thing...
dexterbla
#3
Feb19-11, 08:29 AM
P: 4
a and b are the same node if there is no current flowing across them..,so the potential drop is same

birdbybird
#4
Feb20-11, 12:55 AM
P: 3
Concept of voltage drop

Thank you Naty1 and dexterbala...
I read the wikipedia explanation for Wheastone bridge, but I still don't understand this part...

It says on wikipedia that (http://en.wikipedia.org/wiki/Wheatstone_bridge)
"Kirchhoff's second rule is used for finding the voltage in the loops ABD and BCD
I3R3 - IgRg - I1R1 = 0
IxRx -I2R2 +IgRg = 0"

...and I'm even more confused. If the current flows through the resister against the direction of the circuit, the voltage increases IR. if it flows through the resister in the same direction as the circuit flow the voltage drops by IR. but in the wikipedia it seems like it's used in the opposite way...?

Oh man. I am really really puzzled.
Can anyone explain why the wikipedia did it this way...?



Then, Kirchhoff's second rule is used for finding the voltage in the loops ABD and BCD:



The bridge is balanced and Ig = 0, so the second set of equations can be rewritten as:
birdbybird
#5
Feb20-11, 12:57 AM
P: 3
Quote Quote by dexterbla View Post
a and b are the same node if there is no current flowing across them..,so the potential drop is same
Thank you for your reply but could you explain what you mean by the same node? I know that there is no current flow across them so they are at the same voltage. But does that mean that when the two currents come together at node c, they both must have had the same voltage drop?
Mark.R
#6
Feb20-11, 05:53 AM
P: 3
You agree that b and c are at the same voltage?

i.e. b = c


Voltage drop is the difference between voltages. i.e. from c to a it is V(c)-V(a).

But since V(c) = V(b) (the voltage at c is the same as the voltage at b), then V(c)-V(a) = V(b)-V(a) by simple substitution.




Voltage is related to the potential energy. If 'a' and 'b' have the same potential energy, then the difference between each of them and a third potential energy is the same. If that makes sense :S

In an analogue with gravity on Earth, voltage is like the 'height' of a hill. If points c and b are the same height, then the difference in height (the 'potential drop') from each of them to a point 'a' is the same for both 'c' and 'b'.


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