As shown in figure "C" of the illustration,...

Question 1 of 270705eccefa4bacbc34ba4d483b8

As shown in figure "C" of the illustration, due to the effect of distributed capacitance, what would be the voltage of bus phases "A" and "B" with respect to the common equipment grounding conductor (hull ground)? See illustration EL-0126.

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Question 1 of 27070

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🔍 Key Concepts

• Distributed capacitance between each phase conductor and the hull (equipment ground) in an ungrounded 480 VAC, 3‑phase system • Difference between line-to-line voltage (V_LL) and line-to-ground voltage (V_LG) in a 480 VAC system • What happens to V_LG on the unfaulted phases when one phase becomes solidly grounded (single line-to-ground fault)

💭 Think About

• First, for a 480 VAC 3‑phase system, determine the normal phase (line) voltage to ground using the relationship between V_LL and V_LG. • In figure C, phase C is faulted to ground. With that phase now effectively at 0 V to ground, think about how the voltage triangle (phasor diagram) shifts and what new voltage each of the remaining two phases (A and B) will have to ground. • Ask yourself: does the presence of the ground fault on phase C make the other phases closer to ground potential, or does it push their potential farther away from ground?

✅ Before You Answer

• Confirm the value of V_LG = V_LL / √3 for a 480 VAC, 3‑phase system before the fault occurs. • Sketch or visualize a phasor diagram of a 3‑phase system and then force one phase to be at 0° and 0 V with respect to ground; see what the magnitude of the other two phasors to ground becomes. • Make sure you’re choosing between the given options based on whether A and B are at their normal line-to-ground value or at the full line-to-line value relative to the hull.