In order to measure the internal resistance r₁ of a cell of emf E, a meter bridge of wire resistance R₀ = 50Ω, a resistance R₀/2, another cell of emf E/2 (internal resistance r ) and a galvanometer G are used in a circuit, as shown in the figure. If the null point is found at l = 72 cm, then the value of r₁ =___ Ω

i in primary circuit $=\frac{\mathrm{E}}{{\mathrm{R}}_0+\frac{{\mathrm{R}}_0}{2}+{\mathrm{r}}_1}$

P.D. between the points where the secondary cell is connected is

$\begin{array}{l}=\mathrm{i}\left(\frac{{\mathrm{R}}_0}{2}+\frac{28{\mathrm{R}}_0}{100}\right)\therefore \frac{\mathrm{E}}{2}=\frac{\mathrm{E}}{\left({\mathrm{r}}_1+\frac{\left.3{\mathrm{R}}_0\right)}{2}\right)}\left(\frac{78{\mathrm{R}}_0}{100}\right)\\ {\mathrm{r}}_1+\frac{3{\mathrm{R}}_0}{2}=\frac{156{\mathrm{R}}_0}{100}\\ {\mathrm{r}}_1=\frac{156\times 50}{1002}-\frac{3}{2}\times 5625\\ {\mathrm{r}}_1=78-75\\ {\mathrm{r}}_1=3\end{array}$

II method :

$$\begin{gathered} \mathrm{balancing} \mathrm{point}=\frac{72\times 50}{100}=36\mathrm{\Omega } \\ \mathrm{Remaining} \mathrm{length}=50-36=14\mathrm{\Omega } \\ {\mathrm{I}}_1 \& {\mathrm{I}}_2 \mathrm{are} \mathrm{currents} \mathrm{in} \mathrm{primary} \mathrm{and} \mathrm{secondary} \mathrm{circuits} \\ {\mathrm{I}}_1=\frac{\mathrm{E}/2}{{\mathrm{r}}_1+36}=\frac{\mathrm{E}}{2{\mathrm{r}}_1+72} \\ {\mathrm{I}}_2=\frac{\mathrm{E}/2}{39} \\ {\mathrm{I}}_1={\mathrm{I}}_2 \mathrm{we} \mathrm{get} \\ {\mathrm{r}}_1=3\mathrm{\Omega } \end{gathered}$$