11 KVA LOAD BANK, P/N 9384L

Calculation of Inductance:

    • Input Voltage = 208 VAC L-L, 120 VAC L-N
    • Continuous Current = 30 Amps
    • Frequency = 400 Hz
    • R = V / I = 120 VAC / 30 Amps = 4 Ohms
    • R = 2 x π x F x L
    • Therefore, L = R / (2 x π x F) = 4 / (2 x 3.142 x 400) = 1.6 mH

The customer application is shown below. Three each 1.6 mH, 30 Amp Inductors will be connected in a Wye fashion to offer an 11 KVA Load Bank.

 

 

The details of the 1.6 mH, 30 Amp Inductor are shown below:

 

 

Specifications:

  • Gapped Inductor, P/N 9384L
  • L (1, 2) = 1.6 mH ± 25%, 30 Amps, 400 Hz
  • Copper windings
  • Approvals: None
  • Copper bus bars are 1/4″ x 1″ thick with one hole
  • Open core and coil: 5″H x 5.5″W x 5″D approximately
  • Estimated weight: 25 lbs.

A photo of a similar Inductor is shown for reference below:

 

 

To discuss your specific application, please call our phone number at (714) 624-4740, or send us an email at quote@lcmagnetics.com.