lizzy
Member
The calculation results to problems 3-17 & 3-18 seem incorrect. I think the solution should be as follows:
Problem 3-17:
IN = - (IA + IB + IC)
= - (54.8 <(0 – 36.9⁰) + 0.25<(120⁰-5⁰) + 96<(240⁰-36.8⁰))
= -(-44.52 - j70.49)
= 44.52 + j70.49
= 83.37 <57.73⁰
Problem 3-18:
IN = - (IB + IC)
= - (0.25<(120⁰-5⁰) + 96<(240⁰-36.8⁰))
= -(-88.34 – j37.59)
= 88.34 + j37.59
= 96 <23.05⁰
Also why has a phase sequence of CBA been assumed in the above two problems?
Any suggestions?
Problem 3-17:
IN = - (IA + IB + IC)
= - (54.8 <(0 – 36.9⁰) + 0.25<(120⁰-5⁰) + 96<(240⁰-36.8⁰))
= -(-44.52 - j70.49)
= 44.52 + j70.49
= 83.37 <57.73⁰
Problem 3-18:
IN = - (IB + IC)
= - (0.25<(120⁰-5⁰) + 96<(240⁰-36.8⁰))
= -(-88.34 – j37.59)
= 88.34 + j37.59
= 96 <23.05⁰
Also why has a phase sequence of CBA been assumed in the above two problems?
Any suggestions?
