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

Hint: 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.