• # question_answer Two cells ${{E}_{1}}$ and ${{E}_{2}}$connected as shown in figure have an emf 5 V and 9 V and internal resistance of $0.3\,\Omega ,$ and $1.2\,\Omega$ respectively. Calculate (i) the value of current flowing through If resistance of $3\,\Omega .$ (ii) power dissipated in the resistance of $6\,\Omega .$

(i) Here net emf of circuit $E={{E}_{2}}-{{E}_{1}}=9-5=4\,V$ and total resistance of the circuit $R=\frac{6\times 3}{6+3}+4.5+0.3+1.2=8\,\Omega$ $\therefore$ Main circuit current $I=E/R=4V/8\,\Omega =0.5\,A$ If current flowing through $3\,\Omega$ resistance be ${{I}_{1}},$ then current flowing through $6\,\Omega$ resistance will be $(0.5-{{I}_{1}})$ and hence             $3{{I}_{1}}=6\times (0.5-{{I}_{1}})$ $\Rightarrow$   ${{I}_{1}}=0.33\,A$ (ii) Now, current through $6\,\Omega$ resistance is             $I{{'}_{1}}=0.5-{{I}_{1}}=0.5-0.33=0.17\,A$ $\therefore$ Power dissipated in $6\,\Omega$ resistance is             $P={{(I{{'}_{1}})}^{2}}\times 6={{(0.17)}^{2}}\times 6=0.173\,W$