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Home / Technical Articles / An example of calculating transformer size and voltage drop due to starting of large motor

Calculate voltage drop

Let’s calculate voltage drop in transformer 1000KVA, 11/0.480 kV, impedance 5.75% due to starting of  300 kW, 460V, 0.8 power factor, motor code D (kva/hp). Motor starts 2 times per hour and the allowable voltage drop at transformer secondary terminal is 10%.

Medium-voltage motor starting transformer
Medium-voltage motor starting transformer (man. J. Schneider Elektrotechnik; photo credit: DirectIndustry)

Calculation can be checked by using this MS Excel Spreadsheet dedicated especially to this kind of problem.

Ok, let’s get into the calculations…

Motor current / Torque

Motor full load current = (Kw x 1000) / (1.732 x Volt (L-L) x P.F

  • Motor full load current = 300 × 1000 / 1.732 x 460 x 0.8 = 471 Amp.
  • Motor locked rotor current = Multiplier x Motor full load current

Locked rotor current (Kva/Hp)

Motor CodeMinMax
  • Min. motor locked rotor current (L1) = 4.10 × 471 = 1930 Amp
  • Max. motor locked rotor current (L2) = 4.50 × 471 = 2118 Amp
  • Motor inrush Kva at Starting (Irsm) = Volt x locked rotor current x Full load current x 1.732 / 1000
  • Motor inrush Kva at Starting (Irsm) = 460 x 2118 x 471 x 1.732 / 1000 = 1688 kVA


  • Transformer full load current = kVA / (1.732 x Volt)
  • Transformer full load current = 1000 / (1.73 2× 480) = 1203 Amp.
  • Short circuit current at TC secondary (Isc) = Transformer full load current / Impedance
  • Short circuit current at TC secondary = 1203 / 5.75 = 20919 Amp
  • Maximum kVA of TC at rated Short circuit current (Q1) = (Volt x Isc x 1.732) / 1000
  • Maximum kVA of TC at rated Short circuit current (Q1) = 480 x 20919 x 1.732 / 1000 = 17391 kVA
  • Voltage drop at transformer secondary due to Motor Inrush (Vd) = (Irsm) / Q1
  • Voltage drop at transformer secondary due to Motor inrush (Vd) = 1688 / 17391 = 10%
  • Voltage drop at Transformer secondary is 10% which is within permissible limit.
  • Motor full load current ≤ 65% of Transformer full load current
  • 471 Amp ≤ 65% x 1203 Amp = 471 Amp ≤ 781 Amp
Here voltage drop is within limit and Motor full load current ≤ TC full load current.

Size of Transformer is Adequate.

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Jignesh Parmar

Jignesh Parmar has completed M.Tech (Power System Control), B.E (Electrical). He is member of Institution of Engineers (MIE), India. He has more than 20 years experience in transmission & distribution-energy theft detection and maintenance electrical projects.


  1. Ifran
    Sep 07, 2022

    Short circuit current at TC secondary = 1203 / 5.75 = 20919 Amp.
    Short circuit current at TC secondary = 12038×100 / 5.75 = 20919 Amp.

    which is correct?

  2. Nagendra Mathur
    Jun 24, 2022

    Should we add Motor Starting KVA in Denominator to the transformer Short Circuit KVA ie the % drop = Motor Starting KVA / ( Transformer Short circuit KVA _ Motor Starting KVA

  3. Gousemydeen
    Jan 29, 2022

    In our industry, We have 4 nos of 260 KW motor, which are connected with 1600 KVA transformer. Is this correct design? Please clarify me. We are facing voltage drop upto 48 V

  4. sonam T
    Apr 13, 2021

    hello, if sir could help me out to find the size of transformer for a given load.

  5. Sougata Mukherjee
    Apr 09, 2019

    kindly mention IS or IEC standard from which voltage drop formula used for transformer sizing has been furnished.

    Feb 15, 2019

    Hello sir,

    We have to purchase new transfirmer 33 kv/ 440 volt,i know the LOAD IN KW,how to calculate how much kva transfomer we need …what is formula

    • Ahmed Sheikh
      May 27, 2019

      You need to use an application.

  7. Md. Arifuzzaman
    Feb 11, 2019

    if we start 1000 kva generator( prime power-800 kw, output voltage-440 v) and we want to produce 11 KV voltage with a tranformer(step up) to the load of generator.
    what rating s of x-er to be connected.
    if 1000 kva rating of x-er , the generator tripped?
    or if uses 1500/2000 kva x-er is it safe?

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