Question

Prove that:

I

f we have $n$ cells each of $emf$ $E$ Of these polarity of $m$ cells (where $n>2m$) is reversed. Then, net $emf$ in the circuit is $(n-2m)E$ and resistance of the circuit is $R$. Therefore,
$i=\frac{(n-2m)E}{R}$

Answer 1

The numbers of cells connected in one direction are $\left(n-m\right)$ and the numbers of cells connected in reverse direction are $m$.

The net potential is given as,

$V=\left(n-m\right)E-mE$

$=\left(n-2m\right)E$

Since the resistance of the circuit is $R$.

The current in the circuit is given as,

$i=\frac{\left(n-2m\right)E}{R}$