Question

$N$ identical cells are connected to form a battery. When the terminals of the battery are joined directly (short-circuited), current $I$ flows in the circuit. To obtain the maximum value of $I$,
[Given $\sqrt{N}$ is an integer]

• A. all the cells should be joined in series
• B. all the cells should be joined in parallel
• C. two rows of $N/2$ cells each should be joined in parallel
• D. $\sqrt{N}$ rows of $\sqrt{N}$ cells each should be joined in parallel,

Answer 1

The correct option is B all the cells should be joined in parallel

Let each cell have emf $\epsilon$ and internal resistance $r$.

Assuming all cells having same polarity are in series connection, So,

${\epsilon }_{eq}=N\epsilon$
$r_{eq}=Nr$

So, $I_s=\frac{N\epsilon }{Nr}=\frac{\epsilon }{r}$

Assuming all cells having same polarity are in parallel connection, So,

${\epsilon }_{eq}=\frac{\frac{{\epsilon }_1}{r_1}+\frac{{\epsilon }_2}{r_2}+\frac{{\epsilon }_3}{r_3}+............+\frac{{\epsilon }_m}{r_n}}{\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+.............+\frac{1}{r_n}}$

Since ${\epsilon }_1={\epsilon }_2={\epsilon }_n=\epsilon$

$r_1=r_2=r_n=r$

${\epsilon }_{eq}=\epsilon$

$r_{eq}=r/N$

So, $I_p=\frac{\epsilon }{\left(r/N\right)}=\frac{N\epsilon }{r}$

Here, we can clearly see that,

$I_s<I_p$

So, parallel combination have higher value of $I$ and rest option will have in between the $I_s$ and $I_p$.