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## Homework Statement:

- Derivation of error due to finite load attached to potentiometer

## Relevant Equations:

- (Included in main body, so that it formats correctly)

x = fraction of potentiometer connected to load

V

If R

If R

Therefore error is x.V

Trying to break the equation down further, I did the following :

However, the final equation, is supposed to be (R

Do you just ignore the denominator, or am I missing a step?

I appreciate any help or advice

V

_{p}in parrallel with V_{L}= x/(R_{p}/R_{L}.x.(1-x) + 1)If R

_{L}= infinite, then R_{o}= x and V_{o}= x.VsIf R

_{L}= finite, then R_{o}= x/(R_{p}/R_{L}.x.(1-x) + 1) and V_{o}= x.V_{s}/(R_{p}/R_{L}.x.(1-x) + 1)Therefore error is x.V

_{s}- x.V_{s}/(R_{p}/R_{L}.x.(1-x) + 1)Trying to break the equation down further, I did the following :

However, the final equation, is supposed to be (R

_{p}/R_{L}).V_{s}.(x^{2}-x^{3})Do you just ignore the denominator, or am I missing a step?

I appreciate any help or advice