Question

$n$ identical cells, each of internal resistance $\left(r\right)$ are first connected in parallel and then connected in series across a resistance $\left(R\right)$. If the current through $R$ is same in both the cases then:

• A. $R=r/2$
• B. $r=R/2$
• C. $R=r$
• D. $r=0$

Answer 1

The correct option is C $R=r$

If the emf of each cell be E, then for parallel connection, equivalent emf is

$E_{eq}=\frac{\frac{E_1}{r_1}+\frac{E_2}{r_2}+....}{\frac{1}{r_1}+\frac{1}{r_2}+....}=\frac{\frac{E}{r}+\frac{E}{r}+....}{\frac{1}{r}+\frac{1}{r}+.....}=E$

And equivalent internal resistance of the cells, $r_{eq}=\frac{r}{n}$

So net current in parallel connection is

$I_p=\frac{E_{eq}}{r_{eq}+R}=\frac{E}{\frac{r}{n}+R}$

And for series connection of cells, eqv emf , $E_{eq}=E_1+E_2+...=nE$

and equivalent internal resistance for series, $r_{eq}=r_1+r_2+....=nr$

So current ,$I_s=\frac{E_{eq}}{r_{eq}+R}=\frac{nE}{nr+R}$

Now according to question, both the currents are equal.

So, $I_p=I_s$ reduces to $\frac{E}{\frac{r}{n}+R}=\frac{nE}{nr+R}$ which gives $R=r$