• # question_answer A torch battery consisting of two cells of$1.45\,\,V$and an internal resistance$0.15\Omega$, each cell sending current through the filament of the lamps having resistance$1.5\Omega$. The value of current will be A) $16.11\,\,A$ B) $1.611\,\,A$ C) $0.1611\,\,A$ D) $2.6\,\,A$

Emf of two cells$E=2\times 1.45=2.9\,\,V$ Total internal resistance of two cells                 $r=2\times 0.15=0.3\Omega$               $R=1.5\Omega$ Current, $i=\frac{E}{R+r}$                 $i=\frac{2.9}{1.5+0.3}=1.611\,\,A$