• # question_answer In the adjoining figure, the emf of the cell is 1.8 V and the internal resistance is $(2/3)\Omega .$ The current in $3\Omega$ resistance is: A) 0.4 A B) 0.8 A C) 0.2 A D) 0.1 A

 $8\Omega$ and $2\Omega$ are in parallel and similarly $6\Omega$ and $4\Omega$ are also in parallel combination $\therefore$Equivalent circuit is shown below 0.9 is divided in ratio$\frac{1}{3}:\frac{1}{4}:\frac{1}{6}$ ${{I}_{1}}:{{I}_{2}}:{{I}_{3}}=\frac{1}{3}:\frac{1}{4}:\frac{1}{6}$ ${{I}_{1}}=\frac{\frac{1}{3}}{\frac{1}{3}+\frac{1}{4}+\frac{1}{6}}\,\,\times \,\,I=0.9\,\,\times \,\,\frac{1}{3\times \frac{4+3+2}{12}}$$=\frac{0.9}{3}\times \frac{12}{9}=0.4A$