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Using Potentiometer to find balancing emf.

by harjyot
Tags: balancing, potentiometer
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harjyot
#1
Mar1-14, 01:52 PM
P: 42
According to my book, if we connect a high resistance in series to the secondary cell (whose emf is to be balanced) , connect this whole apparatus to a potentiometer with an auxiliary Emf and find the null point . It's written that even after removing the Hugh resistance the null point comes down to be the same. How's that ? Doesn't the High resistance Infact contribute to increase the effective balancing length? Or do we assume that the potential gradient of the potentiometer far exceeds the potential drop created by the resistance?
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CWatters
#2
Mar1-14, 05:51 PM
P: 3,106
I would have been nice to have a circuit showing what you mean. My guess is that when in balance there is no voltage drop across the "high resistance". So no current flowing through it which means it can be replaced with a short circuit without anything changing.
sophiecentaur
#3
Mar2-14, 11:51 AM
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Quote Quote by harjyot View Post
According to my book, if we connect a high resistance in series to the secondary cell (whose emf is to be balanced) , connect this whole apparatus to a potentiometer with an auxiliary Emf and find the null point . It's written that even after removing the Hugh resistance the null point comes down to be the same. How's that ? Doesn't the High resistance Infact contribute to increase the effective balancing length? Or do we assume that the potential gradient of the potentiometer far exceeds the potential drop created by the resistance?
When the potentiometer is balanced, there is no current through the series 'protection' resistor so there is no voltage drop across it and the reference cell voltage also appears on the slider of the pot.


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