Question

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.

Short answer

The value of the series and the parallel current is $i_{parallel}=i_{series}=\frac{N_e}{(N+1)R}$.

Step-by-step solution

Determine the concept of series and parallel combination of resistance 

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}$

Derive the formula for the current as follows:

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

$$\begin{gathered} \frac{1}{r_{eq}}=\frac{N}{r} \\ {r}_{eq}=\frac{r}{N} \end{gathered}$$

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

R_eq=R+r/N

We have,

$$\begin{gathered} i_{parallel}=\frac{e}{R_{eq}} \\ =\frac{e}{R+{\displaystyle \frac{r}{N}}} \\ =\frac{Ne}{r+RN} \end{gathered}$$

…… (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.