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


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%.

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.

About Author //


Jignesh Parmar

jiguparmar - Jignesh Parmar has completed M.Tech (Power System Control) ,B.E(Electrical). He is member of Institution of Engineers (MIE),India. Membership No:M-1473586.He has more than 13 years experience in Transmission -Distribution-Electrical Energy theft detection-Electrical Maintenance-Electrical Projects (Planning-Designing-Technical Review-coordination -Execution). He is Presently associate with one of the leading business group as a Deputy Manager at Ahmedabad,India. He has published numbers of Technical Articles in "Electrical Mirror", "Electrical India", "Lighting India", "Industrial Electrix"(Australian Power Publications) Magazines. He is Freelancer Programmer of Advance Excel and design useful Excel base Electrical Programs as per IS, NEC, IEC,IEEE codes. He is Technical Blogger and Familiar with English, Hindi, Gujarati, French languages. He wants to Share his experience & Knowledge and help technical enthusiasts to find suitable solutions and updating themselves on various Engineering Topics.


  1. rama rao
    Sep 30, 2016

    Good morning,
    we have the loads like 48X2kw(single phase energy meters) + 20X5Kw(3phase energy meters) + 13kw (separate building purpose) = 209KW . what is the capacity of Transformer. Please give the detailed formula to know the actual rating

  2. umair aziz
    Sep 28, 2016

    It correct. But if load increase in future
    Load plus 30% to 50% future load add. Then after calculate.

  3. Vishwajeet
    Jun 29, 2016

    A single phase transformer rating
    33 kva 2400/240 V, 60Hz….here I want to know that the secondary voltage V2=240V is at No Load or Full load…if it is at No load or at full then please explain how..

    Jun 10, 2016

    I want to know design of transformer,
    having data only 1000 VA i/p : 415 V , o/p : 230 V , single phase
    Hoe to decide primary and secondary turns and also cross section area
    please give all details with formulas if possible

  5. Oladipo
    Apr 25, 2016

    Thanks for the calculation its so helpful, however I have a small doubt.
    As regards the motor starting 2 times per hour, no 20% was added to the motor’s minimum kVA rating to compensate for heat losses within the transformer as said in many references.

    Please clarify , if starting the motor 2 times per hour has little or no impact as nothing was mention in your calculation.

    Thanks for the good job once more


  6. Manish
    Jan 16, 2016

    I have one compressor with 20HP motor and our transformer is 200KVA, due to strarting current of motor a voltage drop is there which affects other machines. How can i minimise the same.

  7. siddharth awasthi
    Nov 17, 2015

    sir i m interested in opening a saw.mill . for.that i m installing a 15 hourse.power electricity moter. department.sying thae.above said.moter will 100 killowatt.
    i m very surprise. to tha load calculated by them
    will u please help calculate.the load.
    1.) 15 hourse.power.moter
    2.) 3 normal. 100 watt bulb

  8. Rustico Taguiam
    Apr 22, 2015

    Which better % impedance of distribution transformer, a higher % or lower %?
    5.75% or 4.5%?

  9. Minu
    Apr 20, 2015

    There is a error in the attached excel calculator. It calculated kVA by multiplying kW and power factor. Actual equation is kVA=kW/power factor.

  10. bhavin mistry
    Jan 28, 2015

    Dear sir, i want know about maximum secondry connectable load as per transfor mer rating

  11. Djarot Prasetyo
    Nov 15, 2014

    Hi! I’d like to know, what standard did you use for the locked rotor current?

  12. Benn Richey
    Sep 02, 2014

    There’s an error in the calculations above: “Motor inrush Kva at Starting (Irsm) = 460 x 2118 x 471 x 1.732 / 1000 = 1688 kVA”
    It should be: Motor inrush Kva at Starting (Irsm) = 460 x 4.5 x 471 x 1.732 / 1000 = 1688 kVA

  13. Mujeeb Raza
    Aug 28, 2014

    Hi All,
    I want to get this into detail, if the output of this transformer is connected to load of small industries (assume same data), What factors are to look into while selecting cable for the secondary of transformer up to the LV panel.

  14. Zulfiqar Ali
    Aug 20, 2014

    Dear sir i hope u r doing fine..your website is very helpfull for engineering articles.many thanks for creating a such type of website.
    but in above article there is a slight mistake in calculations.i hope u ll remove and correct it immediately to avoid more confusions between engineers
    Motor inrush Kva at Starting (Irsm) = 460 x 2118 x 471 x 1.732 / 1000 = 1688 kVA
    in above equation 471 is repeating .which should not b repeated
    this equation should be
    Motor inrush Kva at Starting (Irsm) = 460 x 2118 x 1.732 / 1000 = 1688 kVA

    hope u got my point.

    Thanks And Regards,

    Engr Zulfikar Ali

  15. Mayank P Shah
    Aug 13, 2014

    We are going to set up a new project.Biggest motors are of 260 KW x 2 and 245 KW x 4.
    Total running load is expected of 3100 KVA.Please let us know the size of Dist.Transformer of 11 / 0.433 KV when we starts these heavy motors.All motors are proposed by VFD.We are working around 3500 KVA Transformer.Please guide us.

  16. haytham151
    Jun 21, 2014

    It’s v.g refreshable exp , but l have to wander if i missed something here or you make amistake by the formula :
    Motor inrush Kva at Starting (Irsm) = Volt x locked rotor current x Full load current x 1.732 / 1000
    I think you point to :

    Motor inrush Kva at Starting (Irsm) = Volt x locked rotor current x Full load current x 1.732 / 1000

    • haytham151
      Jul 12, 2014

      Motor inrush Kva at Starting (Irsm) = Volt x locked rotor current x 1.732 / 1000

  17. vinod
    Jun 08, 2014

    Please furnish relevant standard for the portion” Motor full load current ≤ 65% of Transformer full load current”.
    Efficiency not considered while calculating full load current of motor.

  18. Mohammed zougti
    May 30, 2014

    •Motor inrush Kva at Starting (Irsm) = 460 x 2118 x 471 x 1.732 / 1000 = 1688 kVA

    Should be •Motor inrush Kva at Starting (Irsm) = 460 x 2118 x 1.732 / 1000 = 1688 kVA

  19. P.Paulpandy.
    May 21, 2014

    I have found formula of motor inrush kva at starting time wrong because when i use your formula the motor kva must be 795353 kva ( motor capacity is 300 kw). but, i see your electrical article 1688 kva.

    Please check and clarify.

  20. Maher M. Al Qadhi
    May 18, 2014

    Dear Sir

    I hope you are doing well.
    And thank you for this useful example
    I have one question which is if I want to calculate the transformer size & volt drop due to building of 11.89 Km square.
    Can I use the important above tips to calculate it ?

    thank you again.

  21. Amr Ahmed
    May 18, 2014

    This calculation for DOL motor starting only or for any starting method , also kindly give me information about the following standard in the point ”Motor full load current ≤ 65% of Transformer full load current ”.

    Thanks & Regards

  22. subrata deb
    May 16, 2014

    Following points to be considered even assuming that prior to starting of this motor

    there is no other load which is fed by the transformer:

    1) 300KW is the output rating of motor.Therefore efficiency of motor should be considered
    to find F.L current 2) +10% tolerance on percentage impedance of transformer should be considered

    As per IS 6600, all the parts of transformer are designed to carry maximum overload

    of +50%. Therfore it should be checked that motor starting kva must be within 1.5 times of
    transformer capacity.

    The logic that motor F.L current shall be less or equal to 65% of transformer rated current is not clear. Please furnish the relevant standard.

  23. Lalit Kumar
    May 13, 2014

    The above calculations does not consider additional loads which might be fed from the same transformer feeding the large motor.

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