Question

The specific resistance and cross-section area of potentiometer wire is $\rho$ and $A$ respectively. If a current $I$ is passed through the wire, the potential gradient of the wire will be
(A) $\dfrac{{I\rho }}{A}$
(B) $\dfrac{I}{{\rho A}}$
(C) $\dfrac{{IA}}{\rho }$
(D) $IA\rho$

Answer 1

At null condition the current through the galvanometer is zero. In this condition the potential drop between the wire of length $l$ , is equal to the emf $E$ of the source voltage. The potential gradient of a potentiometer is a constant and it is given by, $K = \dfrac{E}{l}$ having the potential drop of $E$ in the wire length $l$ .

Complete step by step answer:
We know that the potential gradient of a potentiometer is given by, $K = \dfrac{E}{l}$ having the potential drop of $E$ in the wire length $l$ .
Now, we know from Ohm’s law we know, $V = Ir$ where, is voltage drop, is the resistance across the terminal $I$ is the current through it.
Also, we know, $r = \dfrac{{\rho l}}{A}$ where, $l$ is the length of the wire, $A$ is the cross sectional area of the wire and $\rho$ is the resistivity.
So, the voltage drop of an wire can be written as, $V = I\dfrac{{\rho l}}{A}$
So, putting the value of potential drop in the expression of potential gradient of the potentiometer will be, $K = \dfrac{{I\dfrac{{\rho l}}{A}}}{l}$
Up on simplifying we get, $K = \dfrac{{I\rho }}{A}$
Hence, the expression for potential gradient is $\dfrac{{I\rho }}{A}$ .

Note:
The balance condition of a potentiometer is acquired when the current through the galvanometer is zero. So, the net current through the galvanometer is zero. For, a cell of e.m.f $E$ with balancing length $l$ , having the potentiometer of wire length $L$ with a source of ${E_0}$ volt, the balance condition is, $E = l\dfrac{{{E_0}}}{L}$ . Or, $\dfrac{E}{l} = \dfrac{{{E_0}}}{L} = K$ . So, the potential gradient is constant for a particular potentiometer.