Question

A potentiometer arrangement is shown in the figure. Driving cell has e.m.f. $e$ and internal resistance $r$. Resistance of potentiometer wire $\text{AB}$ is $R.\text{F}$ is the cell of e.m.f. $\frac{e}{3}$ and internal resistance $\frac{r}{3}$. Balance point $\left(J\right)$ can be obtained for all finite values of

• A. $R>\frac{r}{2}$
• B. $R<\frac{r}{2}$
• C. $R>\frac{r}{3}$
• D. $R<\frac{r}{3}$

Answer 1

The correct option is A $R>\frac{r}{2}$

Since $J$ is the balance point of the given potentiometer, so current will not flow through galvanometer.

So, current in wire $\text{AB}$ (primary circuit) will be

$I=\frac{e}{R+r}$

And potential difference across the wire will be

$V_{AB}=IR=\frac{eR}{R+r}$

To obtain a balanced point, potential drop across wire should be greater than emf of cell in the secondary circuit.

$\frac{eR}{R+r}>\frac{e}{3}$

$\therefore R>\frac{r}{2}$

Hence, option $\left(a\right)$ is correct.

Key Concept: If either the e.m.f. of the driver cell or the potential differences across the whole potentiometer wire is lesser than the e.m.f. of the experimental cell, the balance point will not be obtained.