Question

Figure shows a long potentiometer wire AB having a constant potential gradient. The null points for the two primary cells of emfs ε₁ and ε₂ connected in the manner shown are obtained at a distance of l₁ = 120 cm and l₂ = 300 cm from the end A. Determine (i) ε₁/ε₂ and (ii) position of null point for the cell ε₁ only.

Answer 1

(i)  Let x be the resistance per unit length of the potentiometer wire and I be the constant current flowing through it. Then from the figure, we have:

\[\epsilon_1 - \epsilon_2 = \left( 120x \right)I . . . (1)\] and

\[\epsilon_1 + \epsilon_2 = \left( 300x \right)I . . . (2)\]

Dividing both equations, we get:

 

\[\frac{\epsilon_1 - \epsilon_2}{\epsilon_1 + \epsilon_2} = \frac{120}{300}\]

\[ \Rightarrow 180 \epsilon_1 = 420 \epsilon_2 \]

\[ \Rightarrow \frac{\epsilon_1}{\epsilon_2} = \frac{7}{3}\]

(ii) Now adding equations (1) and (2), we get:

\[2 \epsilon_1 = \left( 420x \right)I\]

\[ \Rightarrow \epsilon_1 = \left( 210x \right)I\]

Comparing with

\[\epsilon = \left( Lx \right)I\]we get:
length of balancing point, L = 210 cm for cell 1