Question

Two cells of EMF IV, 2V and internal resistance $2Omega$ and $1Omega$ respectively are connected in (i) series (ii) parallel. What should be the external resistance in the circuit so that circuit so that the current through the resistance be the same in that two cases. ?

Answer 1

given data $E_1=IV$ E_2 = 2V $

$R_1=2n{R}_2=1n$

$\to$ if connected in series

effective emt = $E_1+E_2=1+223v$

effective internal resistance = $R_1+R_2=2+1=3n$

current through circuit = $\frac{V}{R}=\frac{3}{3+r}$ ........1

where r in internal resistance

$\to$ it connected in parellel

effective emt = 2V

effective resistance = $R_1||R_2=\frac{1}{2}+\frac{1}{1}=\frac{2}{3}r$

current through external resistance = $\frac{2}{\frac{2}{3}+r}$

in two case current in same

$\therefore \frac{3}{3+r}=\frac{2}{\frac{2}{3}+r}\to (\frac{2}{3}+r)3=6+2r$

$2+3r=6+2r$

$4=4n$

$\therefore$ external resistance to be connect in

$r=4n$