• # question_answer A network of resistors is connected to a 16 V battery of internal resistance of $1\,\Omega$ as shown in the figure. (i) Compute the equivalent resistance of the network. (ii) Obtain the voltage drops ${{V}_{AB}}$ and ${{V}_{CD}}.$

(i) $4\Omega$ and $4\Omega$ are in parallel combination. $\therefore$ Equivalent resistance, ${{R}_{AB}}=\frac{4}{2}=2\Omega$ Similarly, equivalent resistor of $12\,\Omega$ and $6\,\Omega$ resistor $\frac{1}{{{R}_{CD}}}=\frac{1}{12}+\frac{1}{6}\Rightarrow \frac{1}{{{R}_{CD}}}=\frac{1+2}{12}\Rightarrow {{R}_{CD}}=4\Omega$ Now, the circuit can be redrawn as Now, $2\,\Omega ,4\,\Omega$ and $1\,\Omega ,1\,\Omega$ are in series combination. $\therefore$ Equivalent resistance of the network             ${{R}_{eq}}=2\,\Omega +1\Omega +4\Omega +1\Omega =8\Omega$ (ii) Current drawn from the battery             $I=\frac{V}{R}=\frac{16}{8}=2A$ This current will flow from A to B and C to D. So, the potential difference in between AB and CD can be calculated as Now,     ${{V}_{AB}}=I{{R}_{AB}}=2\times 2=4V$ and       ${{V}_{CD}}=I{{R}_{CD}}=2\times 4=8V$