Question

The potential gradients on the potentiometer wire are ${V_1}$ and ${V_2}$ with an ideal cell and a real cell of the same e.m.f in the primary circuit then.
A) ${V_1} = {V_2}$.
B) ${V_1} > {V_2}$.
C) ${V_1} < {V_2}$.
D) ${V_1} \leqslant {V_2}$.

Answer 1

Hint:Potential difference between two points is defined as the energy required for moving the point charge from point 1 to another point 2. The potential difference is defined as the ratio of work done upon the charge which begins moving from point 1 to 2.

Complete step by step solution:
It is asked in the problem about the inequality of the potential difference of the potentiometer and e.m.f of the primary circuit. Here we must observe that the internal resistance of an ideal cell is zero but this is not the case with the real cell, the real cell will have some internal resistance and therefore there will be some drop in potential difference of the real cell. So the potential difference across the potentiometer is less than the e.m.f of the cell. The correct relation for the potential difference is equal to,
Potential difference of potentiometer ${V_1} > {V_2}$.
The correct answer for this problem is option B.
Additional information:The potentiometer is an instrument in which the resistance can be varied according to the need and as we change the resistance the electric current is also changed accordingly. The potentiometers are of two type’s rotary potentiometer and linear potentiometer. The rotary potentiometer is used for varying the supply voltage and linear potentiometer has the same working but the movement is always linear.

Note:The potential difference is the main factor which is responsible for the flow of electricity. The internal resistance is the resistance offered by the cell or battery itself and due to this internal resistance there is also some drop in the potential difference.