Question

Two batteries each of emf $E$ and internal resistance $r$ are connected turn by turn in series and in parallel, and are used to find current in an external resistance $R$. If the current in series is equal to that in parallel, the internal resistance of each battery is:

• A. R
• B. R/2
• C. R/3
• D. R/4

Answer 1

Answer: (A)

$I_{S}=\frac{2 E}{2 r+R}$ and $I_{P}=\frac{E}{\frac{r}{2}+R}=\frac{2 E}{r+2 R}$
If $I_{S}=I_{P}$
$\frac{2 E}{2 r+R}=\frac{2 E}{r+2 R}$
$2 r+R=r+2 R$
$r=R$