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WheatStone bridge Answered

Recently i studied the wheatstone bridge in me physics book in class but i could not understand it properly. I asked the teacher about my confusion but he also dont know. i have posted the picture of topic(wheatstone bridge). My question is ;

1.Why current I1 is following through point A to D in loop ADBA. 2.Why current I2 is following through point D to C in loop DCBD.

My confusion is that why current are following through these two paths. As current always flow through level of a higher potential to a level of lower potential but here it seems opposite.

Discussions

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steveastrouk

2 months ago

Applyimg Kirchoff's laws to the solution of circuit theory questions can be confusing. The currents are virtual, a mathematical fiction. which sum to the measured currents experience in the circuit.

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iceng

2 months ago

Isn't electrical current magical..

The W bridge solves all of these complicated equations every moment you can get the galvo-motor to stand at no deflection (at rest ! :-)

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Jack A Lopez

2 months ago

Looking at this again, I think the negative signs are there, in equations 13.23 and 13.24, because the author is imagining voltage drops, around each loop.

For loop I1, starting at point A. Then D. Then B. Then A.

VA =?
VD = VA - (I1-I3)*R3
VB = VD - (I1-I2)*RG = VA - (I1-I3)*R3 - (I1-I2)*RG
VA = VB - I1*R1 = VA - (I1-I3)*R3 - (I1-I2)*RG - I1*R1
Subtract VA from both sides
0 = -(I1-I3)*R3 - (I1-I2)*RG - I1*R1

For loop I2, starting at point D. Then C. Then B. Then D.

VD =?
VC = VD - (I2-I3)*R4
VB = VC - I2*R2 = VD - (I2-I3)*R4 - I2*R2
VD = VB - (I2-I1)*RG = VD - (I2-I3)*R4 - I2*R2 - (I2-I1)*RG
Subtract VD from both sides
0 = - (I2-I3)*R4 - I2*R2 - (I2-I1)*RG

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rickharris

2 months ago

The current (electrons) will flow from negative to positive. and, assuming all the resistors are the same value will go equally across the top and bottom of the bridge. A to B to C and A to D to C. this is a balanced circuit.

If one resistor changes value then more or less current will flow from A to C through that changed resistor. This unbalances the cct and current will try to flow through the low resistance meter one way or the other thus making the meter effectively reads the change in resistance.

(Click on diagram to see it bigger)

https://www.electronics-tutorials.ws/blog/wheatsto...

WHEATSTONE.jpg
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rickharrisrickharris

Reply 2 months ago

I should add that the current flows through the meter because that will be the lower resistance pathway.

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Jack A Lopez

2 months ago

Before explaining the math, I am going to maybe spoil the surprise, regarding the Wheatstone bridge.

Specifically, I claim the Wheatstone bridge, in the condition where the galavanometer current is zero, is essentially just two voltage dividers, adjusted so they are dividing by the same ratio.

And this is suggested by the result

(R1/R2) = (R3/R4)

Moreover I claim the practical significance of the Wheatstone bridge, is it can be used to measure resistance, even with an imperfect voltmeter, or rather a galvanometer with non-infinite resistance. Because in the balanced condition, it does not matter how big, or small, RG is, because exactly zero current is flowing through RG in that condition.

Getting back to the question you're actually asking about, the loop currents, {I1, I2, I3}. These are chosen arbitrarily. These currents are an assumption, and for the sake of convenience, they are all imagined to be circling the same way, namely, counterclockwise.

Or in he author's words:

We consider loops ADBA, DCBA and ADCA and assume anticlockwise loop currents I1, I2, I3, through the loops respectively.


I admit the signs are kind of funny. The opposite of what I would expect them to be.

I mean, like, for adding up the voltage changes around current loop I1, I would be tempted to start at point A. Then consider B. Then D. Then A.

VA =?
[VB is higher than VA by I1*R1]
VB = R1*I1 +VA
[VD is higher than VB by (I1-I2)*RG]
VD = (I1-I2)*RG+VB = (I1-I2)*RG+I1*R1+VA
[VA is higher than VD by (I1-I3)*R3]
VA = (I1-I3)*R3+VD = (I1-I3)*R3+(I1-I2)*RG+I1*R1+VA

Subtract VA from both sides. Get:

0 = (I1-I3)*R3 + (I1-I2)*RG + I1*R1

And that is exactly the same as equation (13.23), except with both sides multiplied by -1.

For loop I2, I would start at point B

VB=?
VC = I2*R2+VB
VD = (I2-I3)*R4+VC = (I2-I3)*R4+I2*R2+VB
VB = (I2-I1)*RG+VD = (I2-I1)*RG+(I2-I3)*R4+I2*R2+VB

Subtract VB from both sides. Get:

0 = (I2-I1)*RG+(I2-I3)*R4+I2*R2

And that is equation (13.24), except with both sides multiplied by -1.