Power capacitor current formula, such as 11 / root 3-334-1 capacitor, 11 / root 3 is the rated voltage, 334kvar, the meter measurement is 26.9 micro method current formula 11 divided by root 3 About 6.35, the unit is kV

Power capacitor current formula, such as 11 / root 3-334-1 capacitor, 11 / root 3 is the rated voltage, 334kvar, the meter measurement is 26.9 micro method current formula 11 divided by root 3 About 6.35, the unit is kV


Calculate the rated current of each phase:
I=Qc/1.732U=334/1.732×6.35≈30(A)



What standard is the calculation formula of motor three phase current unbalance?


I don't know if I want this formula. The mobile phone doesn't have a special symbol. I use this instead! This is the formula for calculating the frequency of rotor current & # 163; 2 = S & # 163; 1 = (N1-N) / N1? 1 & # 163; 1 = stator winding plus power frequency (Hz) s = slip



A three phase four wire circuit problem!
In the three-phase four wire circuit, the three-phase load is connected into a star, the known power line voltage is 380V, the load voltage RA = 11 Ω, Rb = RC = 22 Ω, find 1, the neutral line is disconnected, the phase current and line current of phase a short circuit; 2, the line current and line current of phase a short circuit when the neutral line is disconnected


1. When the neutral line is disconnected, phase A is short circuited. At this time, UA is directly applied to the midpoint of the star load. The voltage on RB and RC is 380V. IB = ic = 380 / 22 = 17.27a. IA is the vector sum of IB and ic = √ 3 * 17.27 = 29.92a. The phase current of star connection is equal to the line current. 2. When the neutral line is disconnected, phase A is disconnected



Three phase four wire power, current 5, ask how much electricity an hour?


First of all, you should know that conversion of 1 degree = 1kW / h is 1000W / h
According to Ohm's law, P = IU = 380 * 5 = 1900w
The power of such an electric appliance is 1.9kw, that is, 1.9kwh per hour