Question

${}^{\prime }{n}^{\prime }$ cells, each of emf ${}^{\prime }{e}^{\prime }$ and internal resistance ${}^{\prime }{r}^{\prime }$ are joined in series to form a row. ${}^{\prime }{m}^{\prime }$ such rows are connected in parallel to form a battery of $N=mn$ cells. This battery is connected to an external resistance ${}^{\prime }{R}^{\prime }$
Show that current ${}^{\prime }{I}^{\prime }$ flowing through the external resistance ${}^{\prime }{R}^{\prime }$ is given by:
$I=\frac{Ne}{mR+nr}$

Answer 1

$I=\frac{ne}{(\frac{nr}{m}+R)}=\frac{ne\times m}{nr+mR}=\frac{nme}{nr+mR}$
but $nm=N$
So, $I=\frac{Ne}{nr+mR}$