Question

The diagram below shows a potentiometer set up. On touching the jockey near to the end X
of the potentiometer wire, the galvanometer pointer deflects to left. On touching the jockey
near to end Y of the potentiometer, the galvanometer pointer again deflects to left but now by
a larger amount. Identify the fault in the circuit and explain, using appropriate equations or
otherwise, how it leads to such a one-sided deflection.

Answer 1

Answer:

Here when battery E1 is connected with opposite terminals in secondary circuit then it will show the error and we will not obtain any null deflection point.

Explanation:

Potentiometer is a device which works on the principle of null deflection point

Here when we connect the jockey with the wire then it shows deflection in galvanometer which shows current is flowing in the secondary circuit

Now if there is no current in the galvanometer then it will show zero deflection and at that position the potential difference will same on the two points of wire and on the secondary circuit

Here in this case the deflection in galvanometer is increased as we move the jockey from left to right

This may be due to wrong connection of battery in secondary circuit due to which potential difference of battery in secondary circuit is opposite to the potential difference on the wire which will increase the deflection.

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Topic : Potentiometer

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Answer 2

The positive of E1 is not connected to terminal X. In loop PGJX, E1 – VG + EXN=0 VG = E1 + EXN VG = E1 + k ℓ So, VG (or deflection) will be maximum when ℓ is maximum i.e. when jockey is touched near end Y. Also, VG (or deflection) will be minimum when ℓ is minimum i.e. when jockey is touched near end X.Read more on Sarthaks.com - https://www.sarthaks.com/299173/diagram-below-shows-potentiometer-touching-jockey-near-potentiometer-wire-galvanometer