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(NONE) Let the potential at the bottom node be $0 \, V$. The potential at the node above the $2 \, V$ battery is $2 \, V$. Let the potential at the no

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$1$

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(NONE) Let the potential at the bottom node be $0 \, V$. The potential at the node above the $2 \, V$ battery is $2 \, V$. Let the potential at the node between the $1 \, \Omega$ resistors be $y \, V$ and the potential at the left node be $x \, V$.Applying Kirchhoff's Current Law $(KCL)$ at node $A$ (left node):$\frac{x-2}{2} + \frac{x-y-5}{1} + \frac{x-0}{2} = 0$$x - 2 + 2x - 2y - 10 + x = 0$$4x - 2y = 12 \implies 2x - y = 6 \quad (1)$Applying $KCL$ at node $B$ (middle node):$\frac{y-x+5}{1} + \frac{y-2}{1} + \frac{y-0}{1} = 0$$y - x + 5 + y - 2 + y = 0$$3y - x = -3 \implies x = 3y + 3 \quad (2)$Substituting $(2)$ into $(1)$:$2(3y + 3) - y = 6$$6y + 6 - y = 6 \implies 5y = 0 \implies y = 0 \, V$Substituting $y=0$ into $(2)$:$x = 3(0) + 3 = 3 \, V$The current $I_1$ flows through the $2 \, V$ battery branch. At the bottom node $D$ (potential $0 \, V$), the current $I_1$ is the sum of currents entering from the left branch and the middle branch:$I_1 = \frac{x-0}{2} + \frac{y-0}{1} = \frac{3-0}{2} + \frac{0-0}{1} = 1.5 \, A$.

In the following circuit, the magnitude of current $I_1$ is $............A$.

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