Question

A potentiometer wire AB having length L and resistance $12r$ is joined to a cell D of emf $\varepsilon$and internal resistance r. A cell C having emf $\varepsilon /2$ and internal resistance $3r$ is connected. The length AJ at which the galvanometer was shown in fig. shows no deflection is:

1. $\dfrac{5}{{12}}L\\\$
2. $\dfrac{{11}}{{24}}L\\\$
3. $\dfrac{{11}}{{12}}L\\\$
4. $\dfrac{{13}}{{24}}L$

Answer 1

The above problem can be resolved using the concept and the fundamentals of the potentiometer. In this problem, we are supposed to find the distance of a null point, from either end of the wire. For this, the basic approach is to calculate the potential gradient. Then after calculating the potential gradient of wire, one can go with the calculation of emf after reaching the null point. Then, at last, by making appropriate substitution of the numerically calculated values, we can obtain the necessary result of the null point distance.

Complete step by step answer:
Given:
The length of wire is, $L$.
The resistance of wire is, ${R_1} = 12r$.
The emf of cell D is, ${E_1} = \varepsilon$.
The internal resistance of cell D is, ${R_2} = r$.
The emf of cell C is, ${E_2} = \dfrac{\varepsilon }{2}$.
The internal resistance of cell C is, ${R_3} = 3r$.
We know the expression for the potential gradient as,
$PG = I \times \dfrac{{{R_1}}}{L}$
Here, I is the magnitude of current through AB and its value is, $\varepsilon /\left( {{R_1} + {R_2}} \right)$.
Solve by substituting the values as,

PG = I \times \dfrac{{{R_1}}}{L}\\\ PG = \left( {\dfrac{\varepsilon }{{{R_1} + {R_2}}}} \right) \times \dfrac{{{R_1}}}{L}\\\ PG = \left( {\dfrac{\varepsilon }{{12r + r}}} \right) \times \dfrac{{12r}}{L}\\\ PG = \dfrac{{12}}{{13}}\left( {\dfrac{\varepsilon }{L}} \right) \end{array}$$ Let x be the length of AJ, then the emf of cell C is, $${E_2} = PG \times x$$ Solve by substituting the values as, $$\begin{array}{l} {E_2} = PG \times x\\\ \left( {\dfrac{\varepsilon }{2}} \right) = \dfrac{{12}}{{13}}\left( {\dfrac{\varepsilon }{L}} \right) \times x\\\ x = \dfrac{{13}}{{24}}L \end{array}$$ **Therefore, the length of AJ is $$\dfrac{{13}}{{24}}L$$ and option (4) is the correct answer. Note: ** To solve the given problem, it is important to remember the concept and application of the potentiometer. The potentiometer is utilised for calculating the electromotive force (EMF) of any desired cell. Moreover, it is also required to remember the fundamentals concepts utilising the terms like null point and internal resistance.