Difficulty: Medium
Correct Answer: single-phase 16 A; three-phase 8 A
Explanation:
Introduction / Context:
For a given power delivery, three-phase systems can reduce the total copper required compared with single-phase systems. This is because power sharing across three conductors lowers current per conductor for the same delivered power, improving material efficiency. The question reframes this as a comparison of 'current-carrying capacity' totals needed in single-phase versus three-phase at 120 V for an effective 15 Ω load.
Given Data / Assumptions:
Concept / Approach:
Single-phase current: I_1φ = V / R_eff = 120 / 15 = 8 A. Two conductors carry this current (go and return), so total copper ampacity perspective doubles that to a 'sum-current' of 16 A. For a comparable three-phase arrangement delivering the same power at 120 V per line-to-line (or appropriately arranged so each phase carries lower current for the same total power), each line requires less ampacity; the net 'sum-current' can be considered about half for the same delivered power, giving 8 A as the comparable total.
Step-by-Step Solution:
Verification / Alternative check:
Using equal copper criterion and balanced loads, three-phase systems are known to be more copper-efficient for the same power and voltage class. While exact values depend on line-to-line vs line-to-neutral voltage and connection details, the qualitative reduction is consistent.
Why Other Options Are Wrong:
'8 A; 4 A' and '32 A; 16 A' scale by factors that do not match the standard comparison. '16 A; 0 A' is impossible since current must flow in all active conductors.
Common Pitfalls:
Comparing peak vs rms values; mixing line-to-line and line-to-neutral definitions; ignoring that single-phase needs two current-carrying conductors, impacting total copper cross-section.
Final Answer:
single-phase 16 A; three-phase 8 A
Discussion & Comments