Question 2.Q.50: Find the current IL in the circuit shown in Fig. Q2.50.1 us......

The circuit can be redrawn as shown in Fig. Q2.50.2 and then the parallel connected voltage sources can be converted into a single source as shown in Fig. Q2.50.3

{R}_{\mathrm{eq}}={\frac{1}{{\frac{1}{2}}+{\frac{1}{8}}}}=1.6~\Omega   ;  {E}_{\mathrm{eq}}=\left({\frac{12}{2}}+{\frac{0}{8}}\right)\times1.6=9.6V

{R}_{\mathrm{eq}2}=\frac{1}{\begin{array}{l}{{\frac{1}{2}}+{\frac{1}{3}}}\end{array}}=1.2\cdot\Omega    ;

With reference to fig Q2.50.3,