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(A) Let the potential at point $A$ be $V_A = 0 \ V$.Applying Kirchhoff's Current Law $(KCL)$ at junction $C$:The current entering from the battery bra

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(A) Let the potential at point $A$ be $V_A = 0 \ V$.Applying Kirchhoff's Current Law $(KCL)$ at junction $C$:The current entering from the battery branch is $i_1 = 1 \ A$.The current flowing through the $2 \ \Omega$ resistor is $i_3$ (downwards from $D$ to $C$).The current leaving towards $F$ is $i_2 = 2 \ A$.According to $KCL$ at junction $C$: $i_1 + i_3 = i_2$.Substituting the values: $1 \ A + i_3 = 2 \ A$,which gives $i_3 = 1 \ A$.Now,we find the potential at $D$ by traversing from $A$ to $D$ via $C$:$V_D = V_A + 1 \ V - i_3 	imes (2 \ \Omega) = 0 + 1 - (1 	imes 2) = -1 \ V$.Now,we find the potential at $B$ by traversing from $D$ to $B$:$V_B = V_D + 2 \ V = -1 + 2 = 1 \ V$.Thus,the potential at point $B$ with respect to point $A$ is $V_B - V_A = 1 - 0 = 1 \ V$.

In the circuit shown in the figure,the currents in different branches and the value of one resistor are given. Then,the potential at point $B$ with respect to point $A$ is $.......V$.

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