A group of Nidentical batteries of emfand internal resistance rmay be connected all in series (Fig. 27-80a) or all in parallel (Fig. 27-80b) and then across a resistor R. Show that both arrangements give the same current in Rif R=r.

In series arrangement, equivalent resistance is sum of all resistances. In parallel arrangement, reciprocal of equivalent resistance is sum of reciprocals of all resistances.

Formulae:

For series connection R_eq= R_A + R_B

For parallel connection$\frac{1}{R_{eq}}=\frac{1}{R_A}+\frac{1}{R_B}$$\frac{1}{R_{eq}}=\frac{1}{R_A}+\frac{1}{R_B}$

Consider the batteries are connected in series,

emf = Ne

For series connection

r_eq= r₁ + r₂ +...+r_N

If,

r₁ = r₂=.....=r_N,

Rewrite the equation:

r_eq =Nr

For the given series connections of batteries and resistor and write as:

R_eq= R+Nr
Then:
$$\begin{gathered} i_{series}=\frac{e}{R_{eq}} \\ =\frac{Ne}{R+Nr} \end{gathered}$$ …… (1)

Consider the batteries are connected in parallel:

emf = e

$\frac{1}{r_{eq}}=\frac{1}{r_1}+\frac{1}{r_2}+.....+\frac{1}{r_N}$

If r₁=r₂ =.....=r_N

For a given parallel connection of batteries and resistor, we can write

R_eq=R+r/N

We have,

…… (2)

Both currents are same when R=r,

Therefore,

$i_{parallel}=i_{series}=\frac{Ne}{(N+1)R}$

By using the formulae for series and parallel arrangement of resistances, we have found out the equivalent resistance for asked conditions.