The figure shows a circuit with three ideal batteries. Find the magnitude and direction of currents in the branches AG, BF and CD.

By Kirchhoff's laws (at point B): $\displaystyle I_{1}+I_{2}=I_{3}$ …….($\displaystyle 1$) In the closed loop AGFBA: $\displaystyle 3+2 I_{3}-6+4 I_{2}+2 I_{3}=0$ $\displaystyle I_{2}+I_{3}=\frac{3}{4}$ …….($\displaystyle 2$) From ($\displaystyle 1$): $\displaystyle 2 I_{1}+I_{2}=\frac{3}{4}$ …….($\displaystyle 3$) In closed loop BFDCB: $\displaystyle -4 I_{2}+6+2 I_{1}-6+2 I_{1}=0$ $\displaystyle I_{2}-I_{1}=0$ $\displaystyle I_{2}=I_{1}$ ……..($\displaystyle 4$) Putting in ($\displaystyle 3$): $\displaystyle I_{1}=\frac{1}{4} \mathrm{~A}$ From ($\displaystyle 4$): $\displaystyle I_{2}=\frac{1}{4} \mathrm{~A}$ From ($\displaystyle 2$): $\displaystyle I_{3}=\frac{1}{2} \mathrm{~A}$