Question

The current is drawn from a cell of emf E and internal resistance r connected to the network of resistors each of resistance r as shown in the figure. Obtain the expression.

Answer 1

$\frac{1}{R_1}=\frac{1}{r}+\frac{1}{2r}$

$\frac{1}{R_1}=\frac{3r}{2r^2}$

$R_1=\frac{2r}{3}$

Therefore$R_2=\frac{2r}{3}\left(same\right)$

$R=R_1+R_2$

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R}=\frac{3}{2r}+\frac{3}{2r}$

$R=\frac{r}{3}$

wkt $I=\frac{3E}{4r}=\frac{9E^2}{16r^2}\times \frac{4r}{3}=\frac{3E^2}{4r}$