Hello my friends, the picture of my motor is below. How do I calculate how much current is in each individual motor wire? It is a BLDC motor.
Thank you my friends.

Kv (Motor Velocity Constant)	185 RPM/V
Kt (Motor Torque Constant)	0.0516 Nm/A
Km (Motor Constant)	0.1924 Nm/√(W)
Maximum Continuous Current*	50 A (180 s)
Maximum Continuous Power*	2755 W (180 s)
Voltage Range	22.2 V (6S LiPo) - 52.2 V (12S LiHV)
Io (@10v)	0.6 A
Rm (Wind Resistance)	0.072 Ω

 

RC Outrunner motor , current rated 50a.

 

RC Outrunner motor , current rated 50a.

RC Outrunner motor , current rated 50a.

Thank you my friend.

Are you saying that the current is in each individual motor wire is 50 Amps or is the current in each wire 50/3 = 16.67 Amps?

Thank you friends.

 

The current drawn will exist in each pair of conductor in turn.

 

Maximum Power input 2755W @ max line volts 52.5

P = √3 VL IL so ​

​

IL= P/√3 VL​

= 2755/(√3 52.5) = 30A​

 

Maximum Power input 2755W @ max line volts 52.5

P = √3 VL IL so ​

​

IL= P/√3 VL​

= 2755/(√3 52.5) = 30A​

Thank you my friend. Is it 30A per each wire or is it 30/3 = 10 A per each wire? We can see, there are 3 wires to the motor. I want to calculate the current in each individual wire.

Also, why do we have 1.73 as the multiplying constant?
Thank you friend.

 

Thank you my friend. Is it 30A per each wire or is it 30/3 = 10 A per each wire? We can see, there are 3 wires to the motor. I want to calculate the current in each individual wire.

Also, why do we have 1.73 as the multiplying constant?
Thank you friend.

IL is Line Current.. Its a 3-phase system so each wire carries part of the total at all times. Why do you need to know the individual wire current? And do you want the RMS, peak or peak-peak value?

The difficulty answering your question is we don't know how the operating voltage is spec'd. Its probably the DC input to the ESC so probably represents the peak phase voltage. Also the standard calculations assume a sine wave but with cheap hobby ESCs and motors that may not be true; it could be more like a trapezoidal waveform.

Assumig a sine wave, the RMS current per phase (line) is given by

\( I = P/(\sqrt3 V_{LL} .cos (\phi). \eta) \)​

where
P is input power in Watts
VLL is RMS line to line voltage
cos(Φ) is power factor - I'm using 1 here, but 0.9 would be typical
η is motor efficiency - I'm using 1 here, but 0.85 would be typical

Taking the 52v as peak phase voltage, then VLL is 63v RMS

So I = 2775/(√3 * 63) = 24.4A RMS = 36A peak.

Here's what it looks like graphically for VLL = 63v RMS equal to a phase (per line) voltage of 52v peak (37v RMS).