Question

Five cells have been connected in parallel to form a battery. The emf and intenal resistances of the cells have been shown in figure. If a load resistance $R$ is connected to the battery, then

Answer 1

The internal resistance of equivalent cell is given by

$\frac{1}{r_0}=\frac{1}{r}+\frac{1}{2r}+..........+\frac{1}{16r}$

$\Rightarrow \frac{1}{r_0}=\frac{1}{r}[\frac{1}{1}+\frac{1}{2}+.....+\frac{1}{2^4}]$

Using , $S_n=\frac{a(1-r^n)}{(1-r)}$

$\frac{1}{r_0}=\frac{1}{r}\times 1\times \frac{[1-(\frac{1}{2}{)}^5]}{[1-\frac{1}{2}]}$

$\Rightarrow \frac{1}{r_0}=\frac{31}{16r}$

Clearly, $r_0$ is smaller than any value of internal resistance of cells connected in the circuit.

So, option (c) is wrong.



If we write the voltage equation for a loop containing any cell ($n^{th}$ cell) and load resistance $R$ , we get

$i_n{r}_n=E_n-IR$

$\Rightarrow i_n=\frac{E_n}{r_n}-\frac{IR}{r_n}$

From the diagram, it is clear that the value of $\frac{E_n}{r_n}=\frac{E}{r}$ for all the cells.

$\therefore i_n=\frac{E}{r}-\frac{IR}{r_n}$

Thus, $i_n$ is maximum for the cell which has largest value of $r_n$.

So, option (a) is wrong and option (b) is correct.

Equivalent enf $E_0=\frac{\frac{E}{r}+\frac{2E}{2r}+\frac{4E}{4r}+\frac{8E}{8r}+\frac{16E}{16r}}{\frac{1}{r_0}}$

$=\frac{5E}{r}\times \frac{16r}{31}=\frac{80E}{31}$

Hence, options (b) and (d) are the correct alternatives to the given question.