Electrical Thumb Rules (Part-2)

Useful Equations:

  • For Sinusoidal Current : Form Factor = RMS Value/Average Value=1.11
  • For Sinusoidal Current : Peak Factor = Max Value/RMS Value =1.414
  • Average Value of Sinusoidal Current(Iav)=0.637xIm (Im= Max.Value)
  • RMS Value of Sinusoidal Current(Irms)=0.707xIm (Im= Max.Value)
  • A.C Current=D.C Current/0.636.
  • Phase Difference between Phase= 360/ No of Phase (1 Phase=230/1=360°,2Phase=360/2=180°)
  • Short Circuit Level of Cable in KA (Isc)=(0.094xCable Dia in Sq.mm)/√ Short Circuit Time (Sec)
  • Max.Cross Section Area of Earthing Strip(mm2) =√(Fault Current x Fault Current x Operating Time of Disconnected Device ) / K  
  • K=Material Factor, K for Cu=159, K for Alu=105, K for steel=58 , K for GI=80
  • Most Economical Voltage at given Distance=5.5x√ ((km/1.6)+(kw/100))
  • Cable Voltage Drop(%)=(1.732xcurrentx(RcosǾ+jsinǾ)x1.732xLength (km)x100)/(Volt(L-L)x Cable Run.
  • Spacing of Conductor in Transmission Line (mm) = 500 + 18x (P-P Volt) + (2x (Span in Length)/50).
  • Protection radius of Lighting Arrestor = √hx (2D-h) + (2D+L). Where h= height of L.A, D-distance of equipment (20, 40, 60 Meter), L=Vxt (V=1m/ms, t=Discharge Time).
  • Size of Lighting Arrestor= 1.5x Phase to Earth Voltage or 1.5x (System Voltage/1.732).
  • Maximum Voltage of the System= 1.1xRated Voltage (Ex. 66KV=1.1×66=72.6KV)
  • Load Factor=Average Power/Peak Power
  • If Load Factor is 1 or 100% = This is best situation for System and Consumer both.
  • If Load Factor is Low (0 or 25%) =you are paying maximum amount of KWH consumption. Load Factor may be increased by switching or use of your Electrical Application.
  • Demand Factor= Maximum Demand / Total Connected Load (Demand Factor <1)
  • Demand factor should be applied for Group Load
  • Diversity Factor= Sum of Maximum Power Demand / Maximum Demand (Demand Factor >1)
  • Diversity factor should be consider for individual Load
  • Plant Factor(Plant Capacity)= Average Load / Capacity of Plant
  • Fusing Factor=Minimum Fusing Current / Current Rating (Fusing Factor>1).
  • Voltage Variation(1 to 1.5%)= ((Average Voltage-Min Voltage)x100)/Average Voltage
  • Ex: 462V, 463V, 455V, Voltage Variation= ((460-455) x100)/455=1.1%.
  • Current Variation(10%)= ((Average Current-Min Current)x100)/Average Current
  • Ex:30A,35A,30A, Current Variation=((35-31.7)x100)/31.7=10.4%
  • Fault Level at TC Secondary=TC (VA) x100 / Transformer Secondary (V) x Impedance (%)
  • Motor Full Load Current= Kw /1.732xKVxP.FxEfficiency

About Jignesh.Parmar
Jignesh Parmar has completed M.Tech (Power System Control), B.E(Electrical) from Gujarat University. He has more than 13 years experience in Power Transmission-Power Distribution-Electrical energy theft detection-Electrical Maintenance-Electrical Projects(Planning-Designing-coordination-Execution). He is Presently associate with one of the leading business group as a Deputy Manager at Ahmedabad,India. He is Freelancer Programmer of Advance Excel and design useful Excel Sheets of Electrical Engineering as per IS,NEC,IEC,IEEE codes. He is technical Author for "Electrical Mirror" and "Electrical India" Magazines. 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.

16 Responses to Electrical Thumb Rules (Part-2)

  1. kashif hussain says:

    Short Circuit Level of Cable in KA (Isc)=(0.094xCable Dia in Sq.mm)/√ Short Circuit Time (Sec)……….what and why is 0.094?.. thanks in advance

  2. Edward Carlos Donkoh says:

    i so grateful the information,PLEASE I WILL like to know how to calculate surge circuit breaker

  3. P.Sivanpandi says:

    One small issue in our plant to find the neutral line
    all phases are giving correct voltge level with phase to neutral also if we give supply to appliance it getting off 4 days we are tried to find the problem finally we got the answer that neutral line is disconnected earth line acting neutral line
    in this case how can judge and find the neutral line fault

  4. vipulkadakia@TORRENTPOWER.COM says:

    Thanks for sharing the knowledge. Keep sending.


    Vipul Kadakia

    Manager ( Dist – NZ).

    Torrent Power Ltd


  5. Sri Nivas says:

    Very Informative………. Keep on going guys… very good job…

  6. Rajan Bhandari says:

    Sir, can you explain how A.C. =D.C./0.636 .

  7. Vrajesh Gadaria says:

    11kv system na current nu KWh,KW,MW,MWh karvu hoy to ? 66 kv system na current nu KWh,KW,MW,MWh karvu hoy to ? Same as. 132kv,220kv,400kv,765kv

  8. Vrajesh Gadaria says:

    JE Getco

  9. vamsi says:

    as a site engineer , it is great collection of formulae for me, bt regarding (spacing of conductor in transmission line how the value (500 ,18,50)in *500+18*p-pv+2* span length /50 came from?

  10. prasildas p says:

    what is thumb rule for voltage drop?
    sir ,please answer it

  11. Dear Sir,

    I have small doubt about your earth bar calculation, this is which standard because i am getting correct sq.mm that only ask. In case of any customer asking means how to i explain.

    Mohan K

  12. k.srikanth says:

    Hw is measured by dyn1 nd dyn11 vector connection in transformer. plz send it sirji…..
    Thanks sir

  13. Edwin says:

    what are all the standard fault level as per IS?

  14. ashok says:

    Dear Jignesh sir,

    In power transformers can we get value of Inductor (L) .

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