Three-phase circuit multisim

[ColdKurd]

Hello out there. Can anyone show me how to construct a 3-phase circuit like this in Multisim? 
I have searched for serval topics on this forum and seen videos on youtube, but I have not been able to find anything useful. 

It would be a huge help. 

[tipa]

-так ,если правильно понял...

[GGoodwin]

Hi ColdKurd and tipa,

These should be clarified, from the sketch

• R₁ is from L₁ to neutral (N)
• R₂ is from L₃ to neutral (N), not from L₂ to neutral (N)
• R₃ is from L₁ to L₃, not from L₃ to neutral (N)

Best regards,

G. Goodwin

[tipa]

-согласен, хотя в данном случае это не принципиально. Перекоса фаз нет и все условия абсолютно идентичны...

[GGoodwin]

Specifications must be followed. The circuit that you created and what was demanded from the specification may not be the same.

The difference might be big and important. The magnitude of voltages across each of the loads are the same. However, this is a three-phase AC circuit, you cannot simply ignore that the phases on the three loads may not be the same if you follow the specifications. Moreover, the current supplied by the three phases of the source might not be the same, you can try to probe to be sure.

[tipa]

-и опять я согласен с вами. Поскольку вы тащите меня в дебри, то предупрежу вас: я слишком тесно общался с дизель-генераторами от 1 квт до 500 квт. Монтировал, настраивал, обкатывал, гонял в разных сочетаниях и под разными перегрузками, имитировал обрывы, замыкания, встречные напряжения (в линии 760 в выскакивало), обрыв нуля, когда свистопляска по фазам  начиналась, с изолированной нейтралью (к причалам атомных подлодок). То же самое с комплектными трансформаторными подстанциями 10/0, 4 кВ. Да, фазы A,B,C путать нельзя.
А вот вы ответьте мне на загадку: равносильно-ли такое подключение?(у меня не было нужного сечения кабеля, поэтому сделал замену). И чем чревата такая замена?

-а как бы вы сделали?

[tipa]

-прошу прощения, Гудвин! У нас говорят:"из-за трёх сосен леса не увидел". Задача задана именно на неравномерную нагрузку по фазам. Сделаем анализ перекосов.

[tipa]

-извиняюсь за свою невнимательность при слизывании схемы. При неравномерной нагрузке по фазам (перекос) в нулевом проводе появляется ток. А в случае обрыва нейтрали на "земле" появляется опасное для жизни напряжение (133 в)

[GGoodwin]

Hi tipa,

I am sincerely accepting your apology, pardon me too if I was not able to send the message to you much earlier. I just got busy with important matters and I can't check the updates in this forum.

Best regards,

G. Goodwin

[GGoodwin]

This portion is for any reader who might get interested in investigating or reviewing polyphase (especially, three-phase) AC circuits.

I reviewed my previous reply and found that there is a sentence which can convey an erroneous thought about the characteristics of three-phase AC circuits. Although I asserted that the load currents and the phase currents from the source will not be the same as in the case of balanced grounded load, I mistakenly stated that the magnitude of voltages across each of the loads are the same. This is only true for R₁ and R₂ but not for R₃. R₁ and R₂ are grounded while R₃ is connected across L₁ and L₃ so the voltage impressed in this third resistor/load is not equal to that of the previous two.

To find the voltage across R₃ we simply subtract the voltages from L₁ and L₃ but treating them as vectors. When doing this it is important to clarify the phase assignments for L₁, L₂, and L₃. Although it can be safe to assume that L₁ has 0° and L₃ has 240° (-120°), this is not always the case. In fact I will show how to determine the voltage across R₃ using phasors but with L₃ chosen as the convenient reference then impose the angular displacement of 120° to L₁. The phasor diagram below shows V_L1 as a blue phasor displaced at 120° and V_L3 as a green phasor used as a reference. V_R3, the voltage across R₃, is the red resultant phasor forming the longer diagonal of the rhombus in second quadrant.

By inspection, with the aid of the phase angle of V_L1, V_R3 is inclined 30° with respect to the horizontal axis. V_R3's phasor divides the rhombus into two congruent isosceles triangles, from any of these two triangles, the magnitude of V_R3 can be determined

          sin 120°

│ V_R3 │ = ────────── ∙ │ V_L │

          sin  30°

where:

V_L = V_L1 = V_L2 = V_L3

Simplifying

          sin 60°

│ V_R3 │ = ───────── ∙ │ V_L │

          sin 30°

          (√3) / 2

│ V_R3 │ = ────────── ∙ │ V_L │

            1 / 2

│ V_R3 │ = (√3) ∙ │ V_L │

So │ V_R3 │is larger than the magnitude of the line voltage by a factor of √3. From the previous phasor diagram, V_R3 is displaced 30° from V_L1 and 150° from V_L3.

Creating circuits and conducting simulation runs will help verify the results above. The circuit shown below uses three single-phase AC sources but connected to form a three-phase Y source. The magnitude is 100 V(rms) but can be easily scaled to a desired voltage. It is important to pay attention to the phase sequence, especially, when comparing to other reference materials. V₃ is at 0° while V₁ is displaced 120°. This will facilitate the comparison to the phasor diagram shown previously.

The graph of the voltages and currents is below

The other circuit below uses a single Three-Phase Y AC Source. It should be noted that it is Phase 2 that is placed at 0° here. The magnitude is 5.2173913043478260869565217391304 V(rms), which was obtained from 3 * 400/230 V.

The graph of the voltages and currents is below