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(D) Given that the current $I = 1 	ext{ A}$ flows from $A$ to $C$. The potential at point $B$ is $V_B = 0 	ext{ V}$.For the segment $AB$,the potenti

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$+1.5 	ext{ V}, -0.5 	ext{ V}$

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(D) Given that the current $I = 1 	ext{ A}$ flows from $A$ to $C$. The potential at point $B$ is $V_B = 0 	ext{ V}$.For the segment $AB$,the potential difference is $V_A - V_B = I 	imes R_{AB} = 1 	ext{ A} 	imes 1.5 	ext{ } \Omega = 1.5 	ext{ V}$.Since $V_B = 0 	ext{ V}$,we have $V_A = 1.5 	ext{ V}$.For the segment $BC$,we apply Kirchhoff's Voltage Law $(KVL)$ from $B$ to $C$:$V_B - I 	imes R_{BC} - E = V_C$Here,$V_B = 0 	ext{ V}$,$I = 1 	ext{ A}$,$R_{BC} = 2.5 	ext{ } \Omega$,and the battery $E = 2 	ext{ V}$ is oriented such that the potential drops across the resistor and the battery.$0 - (1 	ext{ A} 	imes 2.5 	ext{ } \Omega) - 2 	ext{ V} = V_C$$V_C = -2.5 	ext{ V} - 2 	ext{ V} = -4.5 	ext{ V}$.Wait,re-evaluating the circuit diagram: The battery is connected such that the current flows from the positive terminal to the negative terminal,or vice versa. Looking at the diagram,the current flows from $B$ through $2.5 	ext{ } \Omega$ and then through the $2 	ext{ V}$ battery. The potential at $C$ is $V_C = V_B - I 	imes R_{BC} + E$ (if moving from negative to positive terminal) or $V_C = V_B - I 	imes R_{BC} - E$ (if moving from positive to negative terminal).Based on the standard convention for such problems,$V_C = V_B - I 	imes R_{BC} + E = 0 - (1 	imes 2.5) + 2 = -0.5 	ext{ V}$.Thus,$V_A = 1.5 	ext{ V}$ and $V_C = -0.5 	ext{ V}$.

In the given part of a circuit,the potential at point $B$ is zero. Then the potentials at $A$ and $C$ respectively are

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