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Bridge circuit impedance

Boure

Nov 5, 2020
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Impedance of a Bridge circuit
It is usually easier to analyse a circuit once you reduce it entirely to Thevenin equivalent circuit; a source connected to an impedance in series.
Impedance in series and parallel are easily reduced to a single impedance value, however a bridge circuit is not readily reduced to a single impedance, but it is not impossible to do so.
Alternative methods exist for reducing bridge networks to a single impedance to a single impedance, such as;
  • conversion of sections of a bridge into equivalent star and delta networks
  • reducing a complex network into a Norton equivalent circuit.
Both methods require memorizing some formula but I am a firm believer in derivation from first principles. Finding the arbitrary current flow in the bridging impedance then working out current distribution using Kirchhoff current law. Ultimately, the impedance is then found by the ratio of voltage applied to current drawn from the source.
Any thoughts concerning this approach.
 

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Kabelsalat

Jul 5, 2011
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I think the most apropriate method is to convert b-y-z from a delta connection into an y connection.

It is some years ago since I did calculation on this, but I may be able to find the papers.
 

Boure

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You're right.... sections can be converted into equivalent star networks and you can find the total current drawn into the bridge.
In my opinion....trying to remove the hassle of trying to memorize the conversion formula for star to delta and vice versa.(which many people usually get mixed up especially in my case:)).
 

Ratch

Mar 10, 2013
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Impedance of a Bridge circuit
It is usually easier to analyse a circuit once you reduce it entirely to Thevenin equivalent circuit; a source connected to an impedance in series.
Impedance in series and parallel are easily reduced to a single impedance value, however a bridge circuit is not readily reduced to a single impedance, but it is not impossible to do so.
Alternative methods exist for reducing bridge networks to a single impedance to a single impedance, such as;
  • conversion of sections of a bridge into equivalent star and delta networks
  • reducing a complex network into a Norton equivalent circuit.
Both methods require memorizing some formula but I am a firm believer in derivation from first principles. Finding the arbitrary current flow in the bridging impedance then working out current distribution using Kirchhoff current law. Ultimately, the impedance is then found by the ratio of voltage applied to current drawn from the source.
Any thoughts concerning this approach.


What impedance are you talking about? The impedance across points 2-3? Or 1-4, 1-2, what? What is Zbridge? Wouldn't the diagram be more clearer if block "z' were replace with "b" and block "b" replaced with some letter from the middle of the alphabet like maybe "m"? All the words of "flow" can be dropped because current already means "charge flow". Once I know what you want, I can help you.

Ratch
 

Harald Kapp

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What impedance are you talking about?
The assignment calls for the impedance V/I which would be between points 1 and 3.
Wouldn't the diagram be more clearer if block "z' were replace with "b" and block "b" replaced with some letter from the middle of the alphabet like maybe "m"?
Not if that is how the assignment was originally phrased.

All the words of "flow" can be dropped because current already means "charge flow".
Please, please, do not start this discussion again. You've made your point clear quite a few times and while you are totally correct with regard to the physical background, "current flow" is a very common expression. Disagree as you like, but don't bring up this discussion time and again. It will confuse less experienced users more than it helps to clarify matters.
 

Ratch

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I think you want to read the attached file from my "Old Gold" collection. It sure is old, but its information is golden.
Wheatstone1.JPG
Wheatstone2.JPG
Wheatstone3.JPG
Ratch
 
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Ratch

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Thanks moderator for the edit job. I wanted to upload the whole 1947 Weston file, but the site squawked because it was too big. Lots of good info in the rest of the Weston Engineering Notes which I could not upload. Ratch
 

The Electrician

Jul 6, 2012
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Ratch, I think the final result in the image attached to post #1 is incorrect. Have you worked out the impedance seen at terminals 1 and 3, and compared your result with the result in the image?
 

Ratch

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Electrician,
I did not work out any formulas. Are you confusing me with someone else? I only submitted Weston's Lab Notes.
Ratch
 

The Electrician

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Electrician,
I did not work out any formulas. Are you confusing me with someone else? I only submitted Weston's Lab Notes.
Ratch
There is an image attached to post #1 with the problem worked out. You, like me, tend to check results, Have you worked out the solution to the impedance seen at terminals 1 and 3, and compared to the result in the image attached to post #1?
 

Ratch

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Electrician,
No, I was lazy this time. However, I would put my money on Weston Labs as being correct. That is ground that has been plowed over and over throughout many generations of electrical engineers. Those Weston Lab Notes were published in 1947 before Mathcad, Mathematica or other computer programs that could do the gritty algebra . Hats off to alll those pencil pushers who did all that work the hard way. Ratch
 

The Electrician

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Electrician,
No, I was lazy this time. However, I would put my money on Weston Labs as being correct. That is ground that has been plowed over and over throughout many generations of electrical engineers. Those Weston Lab Notes were published in 1947 before Mathcad, Mathematica or other computer programs that could do the gritty algebra . Hats off to alll those pencil pushers who did all that work the hard way. Ratch
I know you have Mathematica, so I figured you would have checked the post #1 result. :) I solved it with Mathematica, and the post #1 result appears incorrect.
 

The Electrician

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Now that I look again at the final result in the image attached to post #1, I see immediately that it's incorrect. If the resistance "b" is not there its value becomes infinite. The final expression given says that if "b" is missing (becomes infinite), the bridge resistance becomes infinite, which is obviously wrong.
 

Ratch

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Now that I look again at the final result in the image attached to post #1, I see immediately that it's incorrect. If the resistance "b" is not there its value becomes infinite. The final expression given says that if "b" is missing (becomes infinite), the bridge resistance becomes infinite, which is obviously wrong.

It no doubt is. I would compare your calculations with the Weston Lab Notes. That would be the clincher. Ratch
 
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