# In a delta wound 3-phase motor, how are the phase and line voltages equal?

After researching the 3 phase delta configuration, I'm confused as to how these three line-to-line and phase relationships can all be true. I have found the voltage/current relationships here , and the resistance relationship here and also in the my motor's documentation boxed in green:

After assuming Ohm's Law, I am not getting equal line and phase voltages, as the theoretical equations I found on the internet suggest. This is probably a dumb arithmetic mistake, or maybe a minute theory misunderstanding. I will lay out my problem now:

Theoretically, in a 3-phase delta winding, the relationship between the line-to-line and phase values are as follows:

On top of these relationships, I'm assuming that Ohm's Law holds for line-to-line and phase values:

When I combine these, I'm reaching the conclusion that the line and phase voltages are not equal, since the current and resistance relationships between the two configurations don't lead to a 1 = 1 statement when plugged into Ohm's Law:

Could somebody please explain to me what I am wrong?

• Ohm's law is for resistors and simple resistive circuits.
– JRE
Jun 18, 2021 at 16:01
• Impedance Z_Line = Z_Phase for delta. Where did you get R_Line = 2/3 R_phase because that is your problem. Jun 18, 2021 at 16:12
• @StainlessSteelRat The resistance relationship is gathered from the website cited in the post and also the motor documentation that I am currently using. Jun 18, 2021 at 16:41
• The concept of resistance dealing with a motor makes that site questionable. A motor winding has resistance, but this ignores the winding inductance. So unless that site talks about impedance, you are starting on a weak foundation. Jun 18, 2021 at 16:49
• @StainlessSteelRat I added an image of my motor's documentation that claims an identical relationship. When it refers to "resistance", it is referring to winding inductance. On top of this, using the impedance relationship you proposed, this just not solve the problem, since the line and phase currents are also unequal. As JRE suggests, I believe that the real issue lies in the use of Ohm's Law. Jun 18, 2021 at 16:59

Your problem is you are trying to use calculations for a motor tester testing an unconnected motor.

So looking in to two terminals for a star, tester would see 2R for a unconnected star motor. But delta windings are connected, so tester would see R || 2R for an unconnected motor.

$$R || 2R = \frac{R \times 2R} {R + 2R} = \frac{2R^2} {3R} = \frac{2} {3} R$$

So your math is flawed, because you are using info for a stand alone motor tester, which just sees motor as a resistance from the motor windings.

• This doesn't solve the issue. It just makes it so that now the answer comes out to sqrt(3) = 1 instead of (2*sqrt(3))/3 = 1. Jun 18, 2021 at 17:05
• ROFL. You are trying to extrapolate data for a stand-alone motor tester to a wired motor. Jun 18, 2021 at 17:07
• Could you please explain how you would achieve satisfying Ohm's Law given that the line current is sqrt(3) * phase current? Or why exactly is it misguided as to why using that relationship is incorrect? I'm not on this StackExchange asking questions because I'm an EE expert. Thank you. Jun 18, 2021 at 17:18
• Equating Ohm's Law to a motor does not make sense, because the load has more to do with the load torque as opposed to windings. Jun 18, 2021 at 17:29
• Yes, the torque determines the current draw. The reason I'm asking this question is because I'm trying to determine phase voltage given a torque and speed. Meaning, I can get line current via the torque vs speed graph. Also, since I'm given the winding phase resistances, I can define the entire load by calculating winding loss and back emf. However, when I go back and try to calculate line and phase voltages, since I'm using the relationships listed in the question (and provided by the motor documentation), they are not equal. Jun 18, 2021 at 17:42