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(B) The circuit can be analyzed using Kirchhoff's laws.Let the current in the top branch be $I_1$ (flowing upwards),the current in the middle branch b

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$-0.4$

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(B) The circuit can be analyzed using Kirchhoff's laws.Let the current in the top branch be $I_1$ (flowing upwards),the current in the middle branch be $I_3$ (flowing towards the left),and the current in the bottom branch be $I_2$ (flowing downwards).Applying Kirchhoff's Current Law $(KCL)$ at the left junction:$I_3 = I_1 + I_2$ ........... $(i)$Applying Kirchhoff's Voltage Law $(KVL)$ for the top loop $(ABCDA)$:$-30I_1 - 40I_3 + 40 = 0$Substituting $I_3 = I_1 + I_2$:$-30I_1 - 40(I_1 + I_2) + 40 = 0$$-70I_1 - 40I_2 = -40$$7I_1 + 4I_2 = 4$ ........... $(ii)$Applying $KVL$ for the bottom loop $(ADEFA)$:$-40I_2 - 40I_3 + 80 + 40 = 0$$-40I_2 - 40(I_1 + I_2) = -120$$40I_1 + 80I_2 = 120$$I_1 + 2I_2 = 3$ ........... $(iii)$Multiplying equation $(iii)$ by $2$:$2I_1 + 4I_2 = 6$ ........... $(iv)$Subtracting equation $(iv)$ from equation $(ii)$:$(7I_1 - 2I_1) + (4I_2 - 4I_2) = 4 - 6$$5I_1 = -2$$I_1 = -0.4 \, A$.

In the given circuit,the current $I_1$ is .............. $A$.

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