# Расчет параметров электрической цепи - синусоидальный трехфазный ток, страница 3

∆Uc = Rac * Ic = 8* (2.624 + i27.782) = 20.992 + i222.256 B

Ua = iXLA * Ia = 6* (17.551 – i13.164) = 105.306 – i78.984 B

Ub = iXLB * Ib = 6* (-20.176 –i8.168) = -121.056 – i49.008 B

Uc = iXLC * Ic = 6* (2.624 + i27.782) = 15.744 + i166.692 B

Uab = Ua – Ub

Uab = 105.306 – i78.984

-121.056 – i49.008

226.362 – i29.976

Ubc = Ub – Uc

Ubc = -121.056 – i49.008

15.744 + i166.692

-136.8- i215.7

Uca = Uc – Ua

Uca = 15.744 + i166.692

105.306 – i78.984

-89.532 + i245.676

Проверяем баланс напряжений:

Uab + Ubc + Uca = 226.362 – i29.976

-136.8- i215.7

-89.532 + i245.676

∑ = 0 + i0 = 0

Ec - ∆Uc – Uca + ∆Ua + Ea = -109,6965 + i190

20.992 + i222.256

-89.532 + i245.676

140.408 –i105.312

219,393

∑ = 0 + i0 = 0

Eb - ∆Ub – Ubc + ∆Uc – Ec =  -109,6965 – i190

161.408 – i65.344

-136.8- i215.7

20.992 + i222.256

-109,6965 + i190

∑ = 0 + i0 = 0

Определяем напряжение смещения нейтрали нагрузки относительно нейтрали генератора.

Eb - ∆Ub – Ub – Uгн = 0

Uгн = Eb - ∆Ub – Ub = -109,6965 – i190

161.408 – i65.344

-121.056 – i49.008

∑ = 0 + i0 = 0

Действующее значение токов и напряжений:

∆Ua =

∆Ub =

∆Uc =

Ua =

Ub =

Uc =

Uab =

Ubc =

Uca =

Uгн = 0 B