Explanation: In the circuit (Fig. 8F12), points A and D are at 0 volts (ground), and points B and C are at +5 volts. In a simple resistive network powered only by positive voltage and ground, all intermediate nodal voltages must fall between 0V and +5V. For "the voltage at point E" to be -1.5 Volts, E must represent a *differential voltage* across two points within the circuit, typical of a Wheatstone bridge configuration. If E represents the voltage difference between two nodes, say Vp1 and Vp2, such that E = Vp1 - Vp2, then a negative value is possible if Vp2 is greater than Vp1. For example, if Vp1 = +2.0V and Vp2 = +3.5V, then E = 2.0V - 3.5V = -1.5V. This indicates an unbalanced bridge where one potential is lower than the other. Options B, C, and D are incorrect because they represent positive voltages, which would be expected for a single node's potential referenced to ground in a simple voltage divider, but not for the specific condition that yields a negative result from a differential measurement.