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Wheatstone Bridge Experiment.
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[QUOTE="ehild, post: 4800673, member: 481"] I think you used a potentiometer to set the balance of the Bridge. You have chosen the constant resistance ( set by the decade box) closest to the theoretical value of the unknown resistance. You expect them only slightly different. The balance condition is Rx/Ro=R1/R2, where Ro is the decade resistance and R1 and R2 are the resistances of the two parts of the potentiometer. If Rx=Ro R1=R2 follows, the two parts of the potentiometer are equal, its slide contact is at the middle. So you set it at the middle initially, and then move slightly in both directions checking if the galvanometer current decreases. It can be derived that the bridge will give most accurate results (i.e. the fractional error in Rx, dRx/Rx, will be the smallest for a small change of the bridge ratio if the bridge balances near its center. That is why you set the constant resistance about equal to the unknown one. ehild [/QUOTE]
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Wheatstone Bridge Experiment.
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