Physics - Current Electricity Question with Solution | TestHub

The circuit diagram can be redrawn asFor the loop $ABCDA$, $$+2-2I^{\prime }+2-2I=0$$ For the loop $ABFEA$, $$\begin{array}{r}2-2I+2-2\left(I-I^{\prime }\right)=0\\ 4-2I-2I+2I^{\prime }=0\end{array}$$ $$\begin{array}{rl}4-4I+2I^{\prime }& =0\\ 2& =2I-I^{\prime }\end{array}$$ From Eqs (i) and (in), we get $$2=2I-I^{\prime }\cdot 2=I+I^{\prime }$$ $$4=3I,I=4/3=1.33\mathrm{A}$$