Kelvin double bridge?

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Bluggo
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Kelvin double bridge?

Postby Bluggo » Thu Feb 26, 2009 7:16 pm UTC

I am reading about bridge circuits, and in particular about the Kelvin double bridge

Image

I think I get the principle behind it: given the standard bridge circuit, you put two additional resistances Rm and Rn, proportional to RM and RN, in the left side of the circuit.
In this way, even if Rx and Ra are small compared to the resistance of the wire, the voltmeter measures zero if and only if Ra/Rx is approximately RM/RN, and you can use this to find the value of the resistance Rx - basically, you are adding large, proportional terms to the fraction Ra/Rx in order to "drown out" small errors.

However, what I do not see is why you need the "leftmost" path from Ra to Rx, the one which connects them without passing through any resistance: unless I am misunderstanding something, that wire is connected in parallel with the rest of the circuit, so it should have no effect at all on the voltage drops across Ra and Rm (and therefore, no effect at all on the output of the indicator).

All it seems to do is getting most of the current passing through the circuit - since it is far less resistive than the other paths - but as the indicator measures voltage, this should be of no consequence.

Does it have some purpose I did not notice? Or is my understanding of how a bridge circuit works erroneous?

Thanks!
Mary Ellen Rudin wrote:Let X be a set. Call it Y.

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parallax
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Re: Kelvin double bridge?

Postby parallax » Thu Feb 26, 2009 7:22 pm UTC

My guess, and I know almost nothing about circuits, is that the leftmost path is the original circuit, so you don't exactly "need" it, but you are probably not going to remove it either.
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Bluggo
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Re: Kelvin double bridge?

Postby Bluggo » Thu Feb 26, 2009 8:58 pm UTC

After thinking about it a little more, I became convinced that the analysis I wrote earlier is completely wrong: if you remove that wire, all the current would pass through Rm and Rn, and of course this would change the voltage drops across them.
Moreover, the whole concept of "drowning the error by adding terms to the numerator and denominator" makes very little sense.

The voltage drop across the wire from Ra to Rb is exactly the one induced by the resistance of the wire, and the "right part" of the left branch divides this smallish drop - which would be the "error" in the usual bridge circuit - between Rm and Rn, proportionally to their resistances.

Since there resistances are proportional to RM and RN, you are splitting the error between the upper and the lower part of the circuit, and as the indicator shows 0 if the ratio between the resistances "above" and "under" the indicator is the same in the two branches the error disappears.

Is this new explanation correct?
Mary Ellen Rudin wrote:Let X be a set. Call it Y.


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