Question

Describe briefly, with the help of a circuit diagram, how a potentiometer is used to determine the internal resistance of a cell.

Answer 1

Measurement of internal resistance ofa cell using potentiometer is shown in figure. The cell of emf, E is connected across a resistance box $R$ through key $K_2$.


When $K_2$ is open, balance length is obtained at length $A{N}_1=l_1$
$E\propto l_1$
$E=\varphi l_1$ ........ (i)
When $K_2$ is closed :
Let $V$ be the terminal potential difference of cell and the balance is obtained at $A{N}_2=l_2$
$V\propto l_2$
$V=\varphi l_2$ ........ (ii)
From equations (i) and (ii), we get
$\frac{E}{V}=\frac{l_1}{l_2}$ ..... (iii)
$E=I(r+R)$
$V=IR$
$\frac{E}{V}=\frac{r+R}{R}$ ....... (iv)
From equations (iii) and (iv), we get
$\frac{R+r}{R}=\frac{l_1}{l_2}$
$\therefore r=R(\frac{l_1}{l_2}-1)$
We know $l_1,l_2$ and $R$, so we can calculate $r$.