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 are 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 that at the end A. which of the two faults, listed above, would be there in the circuit? Give reasons in support of your answer in each case.

Answer 1

Given: 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

To find the two possible faults in the circuit that could result if the galvanometer deflection at the end B is (i) more (ii) less than that at the end A. and which of the two faults, listed above, would be there in the circuit.

Solution:

Let point where the probe touches the wire be 'C'.

The galvanometer reading will be zero when the potential $E_{ac}$ is equal and opposite to E'.

Two important conditions for this to happen are:

1. Positive terminal of unknown emf, E' should be connected to X.
2. $E^{\prime }\le E$ (because $E_{ac}\le E_{ab}=E$)

The galvanometer reading will never be zero if either one of these conditions are not satisfied.

Now lets compare deflections at A and B for both of these 'Errors'.

1. If positive terminal of E' is connected to Y, then net potential across G will be $(E_{ac}+E^{\prime })$. Therefore, net potential and deflection will be max when $E_{ac}$ is max i.e., point B will give maximum reading.
2. If $E^{\prime }\ge E$, net potential across G will be $(E^{\prime }-E_{ac})$. This will be highest when $E_{ac}=0$, i.e., at A. Point A will give maximum reading.