For the potentiometer circuit shown in the given figure, points X… - Vidyadip

Question

For the potentiometer circuit shown in the given figure, points X and Y represent the two terminals of an unknown emf E'. A student observed that when the jockey is moved from the end A to the end B of the potentiometer wire, the deflection in the galvanometer remains in the same direction.

What may be the two possible faults in the circuit that could result in this observation? If the galvanometer deflection at the end B is (i) more, (ii) less than at the end A, which of the two faults, listed above, would be there in the circuit? Give reason in support of your answer in each case.

✓

Answer

The two possible faults in the circuit may be (i) emf E′ is greater than emf E.
(ii) Terminal X of unknown emf is negative (while it should be positive).
If galvanometer deflection at end B is more than that at end A, then terminal X is negative, because in this case net current in galvanometer along AB due to both cells is additive.
If galvanometer deflection at end B is less than that at end A, then E′ > E, because net current in galvanometer due to both cells’ emfs E and E′ is subtractive.

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