Physics - Current Electricity Question with Solution | TestHub

The given circuit can be redrawn as,where, equivalent emf of the combination, $\begin{array}{rl}{E}_{\text{eq }}& =\frac{E_1{R}_1+E_2{R}_2}{R_1+R_2}\\ & =\frac{2\times 4+2\times 4}{4+4}=2\mathrm{V}\end{array}$ and $R_{eq}=\frac{R_1{R}_2}{R_1+R_2}=\frac{4\times 4}{4+4}=\frac{16}{8}=2\Omega$ Net emf of the circuit, $\begin{array}{rl}{E}^{\prime }& =E_2+E_{\mathrm{eq}}\\ & =2+2\\ & =4\mathrm{V}\end{array}$ $\therefore$ Current flowing through branch $AB$, $I=\frac{E^{\prime }}{R_{eq}}=\frac{4}{2}=2\mathrm{A}$ Clearly from figure, it is clear that direction of current is from point $A$ to $B$ clockwise is loop.