Question

With the help of a circuit diagram describe the method to find the internal resistance of a cell using potentiometer.

Answer 1

Figure shows the arrangement for the measurement for measuring the internal resistance of a battery. Let the emf of the battery be $E_1$ and its internal resistance be $r$. A known resistance $R$ is connected across the battery together with the plug key $K$. The plug key $K$ is opened and the balance point $P$ is searched on the wire $AB$ so that there is no deflection in the galvanometer.

As the key is open, there is no current through the battery and the potential difference across the terminals of the battery is the same as the emf $E_1$ of the battery. If $A{Y}_1=l$, we have

$E_1=\frac{l}{L}\times V_o$ .......................(i)

With the symbols having their usual meanings.

Now the key is closed and the new balance point $Y_2$ is searched. There is a current

$i=\frac{E1}{(R+r)}$ through the battery.

The potential between the terminals of the battery is then,

$iR=\frac{E_1\times R}{(R+r)}$

If $A{Y}_2=m$,

$\frac{E_1\times R}{(R+r)}=\left(m/L\right)\times V_o$.........(ii)

Dividing (ii) by (i),

$\Rightarrow r=\frac{R\times (l-m)}{m}$