Question

Five cells each of internal resistance $0.2\Omega$ and emf $2V$ are connected in series with a resistance of $4\Omega$.The current through the external resistance is

• A. $4A$
• B. $2A$
• C. $1A$
• D. $0.5A$

Answer 1

Answer: (B)

$2A$

Since the five batteries and their internal resistances are connected in series to one another, the net emf will be $nE$ and the internal resistance is $nr$.

Current through external resistance
$i=\frac{nE}{nr+R}=\frac{5\times 2}{5\times 0.2+4}$
$=2A$