Output R of Wheatstone bridge

Thread Starter

Peytonator

Joined Jun 30, 2008
105
I already did ... but it hasn't helped too much.

Am I correct in saying that if Ro is the thevinin equivalent resistance, then replacing E or Vin with a short gives Ro = R3||Rx ?
 

beenthere

Joined Apr 20, 2004
15,819
The concept of output resistance and a resistive bridge doesn't connect for me. The utility of the bridge is that current through the galvanometer depends sensitively on the ratio R1 - R2 and R3 - Rx. When R1 = R3 and R2 = Rx, the galvanometer reads 0 because no potential difference exists to push current through it. Therefore, any change in Rx will result in an indication above or below 0 on the galvanometer.

Rt for the sum of R1 & R2 will remain fixed. Rt for the sum of R3 and Rx will vary with Rx. That changes the potential at the junction of R3 & Rx.
 

The Electrician

Joined Oct 9, 2007
2,971
Hi there,

Say I have a transducer Rx in a wheatstone bridge, as in this diagram. How do I calculate the output Resistance of the Wheatstone bridge?

Thanks :)
It's not clear what is meant by "output resistance". Is it the Thevenin equivalent seen at the Rx terminals? If that's what is wanted, then you would have to know the resistance of the galvanometer to calculate it.

On the other hand, perhaps it's the Thevenin resistance seen by the galvanometer. In that case, if we assume that a resistor of value Rx is connected to the Rx terminals, then it can be calculated.

Which is it?
 

Thread Starter

Peytonator

Joined Jun 30, 2008
105
It's that which is seen by the galvanometer. Perhaps I should have rather just put two terminals as V0 instead. I think the answer is actually...

Rt or Ro = R1||R2 + R3||R4

Right?
 
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