The calculation of the three phase circuit is listed in detail and the calculation process is added

Hehe, the score is too high, I have to come...

Let's do the easy ones first:

The total power is 10 + 10 + 10 = 30kw without suspense, because there is no neutral wire and the three resistors are all connected to the voltage of 380V (with the neutral or neutral line there is the possibility of connecting 220V voltage), so the added are rated voltage, of course, they all play the rated power.

Let's talk about the current:

The title says that the phase is current, but I understand it here as the phase.

First find phase A, phase A and phase B are connected with only a 10kW resistor.

A phase current i = p u = power voltage = 10000 380 = amps.

Then find phase C, two 10kw resistors are connected in parallel between phase b and phase c, and the equivalent power is 20kwb phase current i = p u = power voltage = 20000 380 = ampere.

Building group, building group, why do such simple questions give such a high score, what a big impact, look at which people upstairs have used many complex methods, it's strange if you're not dizzy, hehe.

My humble opinion: when you say 380 220 circuit, it means that the three-phase circuit is with a center line. This is easy to do, and the voltage is clamped at 220V without phase.

Therefore, those who say that they are connected to the AB2 phase or something are false, but in fact they are the change of star shape, the B phase is not connected to the load, the A phase has a resistor, and the C phase has two star circuits in parallel.

The current calculation is simple. r=u^2/p=

i1=220/ p=220*

i3=2i1= p=2*

Total power p=10kw

The presence or absence of a centerline is important for three-phase asymmetric circuits. If you don't have a midline in your question, you need to take one more step to calculate.

First of all, the problem of Guohaibozyz is a short circuit, but even if there is a short circuit in the middle line system, the phase voltage is still 220V, and the short circuit has no effect on the intact phase, so I do this.

For a three-phase circuit without a neutral wire, the midpoint nn must be found first'Voltage. This circuit is very special, the B phase has no resistance, and the midpoint voltage is clamped at 220 120 °V. Therefore, both resistors are subjected to a line voltage of 380V, and the power should also be rated power.

i1=p/u=10e3/380= p=10kw

i3=2* p=20kw

The total power is 30kw.

If the three-phase load is asymmetrical according to the pure resistance, the three-phase is calculated separately and then added to obtain.

Total power p=(80+100+105)*220

w do not know the nature of the load (inductive, capacitive, pure resistance?) The power factor cannot be calculated.

1. Resistance value r1 = r2 = r3 = r = u 2 p = 380 * 380 10000 = ohms;

2. A phase current ia=uab r=380, (when there is no current between CAs, ia=IAB);

3. B phase current IB=UBC (R 2)=380, (same as above, IC=IBC);

4. P total = PAB + PBC = 10 + (10 + 10) = 30KW.

In this way, the calculation is clearer, and the reason is the same as upstairs.

Note: UAB=UBC=UCA=380V is fixed and constant, and the resistance value is constant.

If there is also a resistance (load) between CAs, I am afraid that [Thevenin's theorem] will be used to analyze this circuit.

"The Sun Follows" must be a master of electricity. I don't understand what you mean, if the B phase is not connected to the load, then the B phase electromotive force through the B phase line, the load midpoint, and the neutral wire will not form a loop and short circuit? Can you explain that a little bit more?

I think it should be calculated with vectors:

Let the power supply electromotive force vector.

ea=220∠0，eb=220∠-120,ec=220∠120.

Resistance r=u*u p=380*380 10000=ohm.

A phase current vector ia=(ea-eb) r

220 [1 + 1 2 + J (root number 3 2)] r220 * root number 3 r 30

C phase current vector IC = (EC-EB) (R 2)220[-1 2 +J(root number 3 2) +1 2 +J (root number 3 2)] r

2 * root number 3 * 220 r 90

By Kirchhoff's First Law:

Phase B current vector IB=-ia-IC

220*root: 3 r, 30 - 2*root, 3*220 r, 90, root, 21*220 r

Substituting the numerical value to solve:

ia=,ib=,ic=

Because all three resistors are connected to the rated voltage, so.

Total power = 10kw * 3 = 30kw

What you say: 380 220 volts This power supply has no neutral wire, pure 3 live wires.!!

So how do you know your 380 220 volts, how do you measure them???

Mistakenly, let's elaborate on who 220 volts is for!

You can understand what you said, isn't it good to say it? It's okay to have no neutral line, but there must be a neutral point (at most a shift)! Neutral line you don't notice, you don't see!

It's right on your boat, right next to you! In addition, the name "3 resistors with a rated power of 10 kW" means that the withstand power of each resistor is 10 kW! So, you don't say the value of each resistor!

It seems that only you know the resistance value of each resistor!

At this point, you should have solved your problem! All you need to do is set the resistance value.

Think about it!! You can solve it completely, no need to ask questions!

Fill in this "quasi" neutral line, it's right under your feet!

Another: 380 220 circuit, indicating that the three-phase circuit is with a center line.

Not necessarily, please note.

Because of the above, so this problem should be unsolved for the time being! ）

Personal opinion: 1. The resistance algorithm is the same as above: r1=r2=r3=r=u 2 p=380*380 10000=

2. The current of phase A is the same as above, ia=uab r=380 3, phase c current ic=ubc (r 2)=380 4, phase b current ib=ia+ic=

The rated power is 30kW as above

On the two categories of baddies:

1. Copy-paste pie villain;

2. Plagiarism villains.

Wulecheng is the second type of villain!

Ignorant people on the whole floor, what are you shouting, if you don't know, you don't know, don't you be legally responsible? There are two configuration methods for lighting transformers on board: the first is two core transformers, one working and one standby; The second is composed of three single-phase transformers, which are connected under normal conditions and can be connected into an open triangle connection when faulty, that is, V V connection.

The landlord's problem is the problem of V connection, which is explained in the second chapter of Shi Jichang's "Marine Electrical Equipment and Systems" transformer, but the calculation of the book only discusses the efficiency of the transformer, and does not distinguish the concept of three-phase power and single-phase power.

I don't know if the power in the "3-phase current and total power" referred to by the landlord refers to the "total power" consumed by the resistor or the "total power" provided by the power supply. Let's explain each of them:

First of all, it should be noted that the "3-phase current" is no longer a symmetrical current, and the current of phase AB and phase BC can be calculated separately (the value of the total rated power of the resistor divided by the line voltage) to obtain the effective value of phase A current (calculated by the resistance connected to phase AB, i.e., 10kW 380) and the effective value of phase C current (calculated by the resistance connected to phase BC, i.e., 20kW 380, and phase B current is the phasor sum of phase A and phase C currents. If the current of phase A is , the current of phase C. By adding phasors, the phase b current is:

The phase current is the phasor sum of the two phase currents A and C. I don't need to write out the calculation process of phasor sum). That is, the phase current (equal to the line current) of the ABC three-phase is:

Phase A, phase B, phase C.

The total power consumed by the resistors (which is single-phase power) is equal to the sum of the rated power of the three resistors, i.e. 30 kW.

The electrical power provided by the power supply (which is three-phase power) is equal to the sum of the single-phase power calculated by the three phases, which is the sum of the phase voltage (220V) multiplied by the three-phase current. Phase A is 220, phase B is 220, and phase C is 220. The total power provided by the three-phase power supply is (the actual phase voltage should be, and 220V is still taken for the convenience of calculation).

p=。Finally, it should be noted that the power grid on board is very short, the voltage drop of the line is very small, and the actual line voltage of the three phases should be more than 380V, and if you want to calculate it accurately, it should be carried out with the actual measured value.

<> solution: UCB (phasor) = 380 30°V, according to the phasor diagram, then:

UBC (phasor) = 380 210°. IBC (phasor) = UBC (phasor) Z = 380 210° 10 30° = 38 180° (A).

According to the relationship between the line current and the phase current, so: ib (phasor) = (38 3) (180°-30°) = 38 3 150°(a).

According to the characteristics of the three-symmetrical circuit: ia (phasor) = 38 3 (150 ° + 120 °) = 38 3 270 ° = 38 3 -90 ° (a).

IC (phasor) = 38 3 (150°-120°) = 38 3 30° (A).

The maximum reduction and minimum depression are divided by the average number of three phases, that is, Yanbiwan (390 - 365) 365 + 380 + 390) 3 = =

<> solution: z1 is connected, load voltage = power line voltage, so load phase current: i1'=380/|z1|=380/|j10|=38（a）。

a1 is the line current of the load, so: i1 = 3i1'=√3×28=38√3（a）。

The current of the resistor r is: IR = phase voltage r = (380 3) 20 220 20 = 11 (a).

Since Z2 is Y-connected and is a three-symmetrical load, the neutral current is 0. According to KCl, then the reading of A2 is the current of the resistor R: I2 = 11A.

The active power of the z1 part, since the z1 part is a pure inductive load, so: p1 = 0.

The active power of the resistor r: pr = phase voltage r = (380 3) 20 = 7220 3 (w).

The phase current of the z2 part is: i2'= Phase voltage |z2|=（380/√3）/|3-j4|=(380 3) 5=76 3(a), so the active power of the z2 part is: p2=3 (i2'²×3）=3×（76/√3）²×3=17328（w）。

The first "3" is 3 single-phase power; The second "3" is: re(z2)=3( ).

Therefore, the total active power of the whole load is: p=pr+p2=7220 3+17328=.

Answer]: B three-phase short-circuit current calculation uses standard unit value calculation, only two benchmarks need to be selected, and voltage and capacity are often selected.

Summary. Hello, this is Xiaodu, I'm glad to have your question.

However, Xiaodu's knowledge content is limited, and he only knows.

Delta line voltage = phase voltage = 380V star line voltage = root number 3 * phase voltage, so phase voltage = 220V

Three 10 equal dry plum appreciation resistors, do triangle connection voltage of 380V in the three circuits, no old disturbance to find the line current RMS, phase current RMS.

Hello, this is Xiaodu, very high Lu Lu Tang Xing can your questions. However, the knowledge content of Xiaodu is limited, and I only know that the voltage of the triangle vertical connection line = phase voltage = 380V, the voltage of the star connection line = the root number 3 * the phase companion voltage, so the phase voltage = 220V

Xiaodu is not sure if it is right or not.,If it doesn't help you.,I'm really sorry.,Please understand.。