Engine Power in Star vs Delta Connection: Understanding Through P=√3*U*I*cosφ Formula

Hello I have a question. What is the engine power in a star connection and in a delta connection? I searched the forum, but I found conflicting answers, that the power in a star connection drops by 3 times, some write that it is √3 times. Please support your statement with examples. I know that in a delta connection the current is 3 times greater, but the voltage is also √3 times greater. Let us derive our arguments from the formula: P=√3*U*I*cosφ

Assume that only resistances are important. In a star connection, the power on each phase is P=Uf*If. And for a threesome? P=√3*Uf*√3*If.

przemoo24 wrote:

...I know that in a delta connection the current is 3 times higher, but the voltage is also √3 times higher.

After delta switching, the current in each of the three motor windings is √3 times greater, as is the voltage. The phase current is three times greater and is distributed equally over the two windings in the triangle.
If you reduce the voltage on the bulb by half, the current it consumes will also be halved, and the power dissipated as the product of current and voltage will be 4 times smaller than before.

Correct. I brought this topic up because yesterday I found this post on the forum:
" the useful power of the engine (running power) will decrease by √3 times. Why? Because we assume that the "phase" (band) current of the engine will remain unchanged, i.e. at the value I/√3, where I - current when operating in triangle. Only the starting power is reduced by 3 times. The engine power will be the sum of the power on each winding (phase), i.e. 3×I/√3×230 V, and before switching 3×I/√3×√3×230 V. "

Personally, in this post I changed the term starting power to starting torque, which is a more professional term in my opinion.

I've read a bit and it seems to me that these answers should be refined. The motor power in star/delta connection depends on the available voltages. If we have, for example, a 1.1 kW, 220V/380V motor; triangle/star; 4.9A/2.8A; cosφ; then this engine will achieve its rated power rating in both configurations because:
Py=√3*380*2.8*0.77=1419W
PΔ=√3*220*4.9*0.77=1437W
(Power should be outweighed by efficiency, there will be power on the shaft)

The power will drop significantly when we want to power it not with 3*380V but with 3*220V in the star. Then all the answers that the power will drop 3 times will make sense. The same will happen if we are dealing with a 380V/660V delta/star motor and we want to power it from the home network. We can power it in a delta configuration (we have rated voltages available), but also in a star configuration with a voltage of 3*380V, but then the power will be lower.
To sum up, the engine power will change, but in the case of a star configuration, we will power it with delta voltage. If we supply power with rated parameters, the power in both configurations will be the same.


Ps. I wrote the old markings, i.e. 220/380V, so as not to mix things up too much.

przemoo24 wrote:

The motor power in star/delta connection depends on the available voltages. If we have, for example, a 1.1 kW, 220V/380V motor; triangle/star; 4.9A/2.8A; cosφ; then this engine will achieve its rated power rating in both configurations because:
Py=√3*380*2.8*0.77=1419W
PΔ=√3*220*4.9*0.77=1437W

Which should not be surprising, because in both configurations, 220V voltage is deposited on each of the three motor windings and 2.8A flows through each of them.

The question should be . How much will the engine power drop at the same voltage after star switching?

That's right, but we already have the answer, so the matter is solved.