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(B) First,find the current $I$ in the circuit using Kirchhoff's voltage law. The net electromotive force is $36 \ V - 12 \ V = 24 \ V$ and the total r

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$-2.4$

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(B) First,find the current $I$ in the circuit using Kirchhoff's voltage law. The net electromotive force is $36 \ V - 12 \ V = 24 \ V$ and the total resistance is $3 \ \Omega + 2 \ \Omega = 5 \ \Omega$.$I = \frac{24 \ V}{5 \ \Omega} = 4.8 \ A$.The current flows from the $36 \ V$ battery towards the $12 \ V$ battery,i.e.,from $B$ to $A$ through the upper branch.Starting from point $B$ with potential $V_B = 24 \ V$,we move towards $A$:$V_A = V_B - 12 \ V - I 	imes 3 \ \Omega$$V_A = 24 \ V - 12 \ V - (4.8 \ A 	imes 3 \ \Omega)$$V_A = 12 \ V - 14.4 \ V = -2.4 \ V$.

In the given circuit,if the potential at point $B$ is $24 \ V$,the potential at point $A$ is: (in $V$)

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