# Electrical Useful Equations

• ## Cable Capacity:

• For Cu Wire Current Capacity (Up to 30 Sq.mm) = 6X Size of Wire in Sq.mm
• Ex. For 2.5 Sq.mm=6×2.5=15 Amp, For 1 Sq.mm=6×1=6 Amp, For 1.5 Sq.mm=6×1.5=9 Amp
• For Cable Current Capacity = 4X Size of Cable in Sq.mm ,Ex. For 2.5 Sq.mm=4×2.5=9 Amp.
• Nomenclature for cable Rating = Uo/U
• where Uo=Phase-Ground Voltage, U=Phase-Phase Voltage, Um=Highest Permissible Voltage.
• Short Circuit Level of Cable in KA (Isc)=(0.094xCable Dia in Sq.mm)/√ Short Circuit Time (Sec)
• Cable Voltage Drop(%)=(1.732xcurrentx(RcosǾ+jsinǾ)x1.732xLength (km)x100)/(Volt(L-L)x Cable Run.
• ## Size of Cable according to Short circuit (for 11kV,3.3kV only)

• Short circuit verification is performed by using following formula:
• Cross Section area of Cable (mm2)S = I x√t / K
• Where:
• t = fault duration (S)
• I = effective short circuit current (kA)
• K = 0.094 for aluminum conductor insulated with XLPE
• Example: Fault duration(t)= 0.25sec,Fault Current (I) = 26.24 kA
• Cross Section area of Cable = 26.24 x √ (0.25) / 0.094= 139.6 sq. mm
• The selected cross sectional area is 185 sq. mm.
• ## Current Capacity of Equipment:

• 1 Phase Motor draws Current=7Amp per HP.
• 3 Phase Motor draws Current=1.25Amp per HP.
• Full Load Current of 3 Phase Motor=HPx1.5
• Full Load Current of 1 Phase Motor=HPx6
• No Load Current of 3 Phase Motor =30% of FLC
• KW Rating of Motor=HPx0.75
• Full Load Current of equipment =1.39xKVA (for 3 Phase 415Volt)
• Full Load Current of equipment =1.74xKw (for 3 Phase 415Volt)
• ## Earthing Resistance:

• Earthing Resistance for Single Pit=5Ω ,Earthing Grid=0.5Ω
• As per NEC 1985 Earthing Resistance should be <5Ω.
• Voltage between Neutral and Earth <=2 Volts
• Resistance between Neutral and Earth <=1Ω
• Creepage Distance=18 to 22mm/KV (Moderate Polluted Air) or
• Creepage Distance=25 to 33mm/KV (Highly Polluted Air)

• Minimum Bending Radius for LT Power Cable=12xDia of Cable.
• Minimum Bending Radius for HT Power Cable=20xDia of Cable.
• Minimum Bending Radius for Control Cable=10xDia of Cable.
• ## Insulation Resistance:

• Insulation Resistance Value for Rotating Machine= (KV+1) MΩ.
• Insulation Resistance Value for Motor (IS 732) = ((20xVoltage (L-L)) / (1000+ (2xKW)).
• Insulation Resistance Value for Equipment (<1KV) = Minimum 1 MΩ.
• Insulation Resistance Value for Equipment (>1KV) = KV 1 MΩ per 1KV.
• Insulation Resistance Value for Panel = 2 x KV rating of the panel.
• Min Insulation Resistance Value (Domestic) = 50 MΩ / No of Points. (All Electrical Points with Electrical fitting & Plugs). Should be less than 0.5 MΩ
• Min Insulation Resistance Value (Commercial) = 100 MΩ / No of Points. (All Electrical Points without fitting & Plugs).Should be less than 0.5 MΩ.
• Test Voltage (A.C) for Meggering = (2X Name Plate Voltage) +1000
• Test Voltage (D.C) for Meggering = (2X Name Plate Voltage).
• Submersible Pump Take 0.4 KWH of extra Energy at 1 meter drop of Water.
• ## Lighting Arrestor:

• Arrestor have Two Rating=
• (1) MCOV=Max. Continuous Line to Ground Operating Voltage.
• (2) Duty Cycle Voltage. (Duty Cycle Voltage>MCOV).
• 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).
• ## Transformer:

• Current Rating of Transformer=KVAx1.4
• Short Circuit Current of T.C /Generator= Current Rating / % Impedance
• No Load Current of Transformer=<2% of Transformer Rated current
• Capacitor Current (Ic)=KVAR / 1.732xVolt (Phase-Phase)
• Typically the local utility provides transformers rated up to 500kVA For maximum connected load of 99kW,
• Typically the local utility provides transformers rated up to 1250kVA For maximum connected load of 150kW.
• The diversity they would apply to apartments is around 60%
• Maximum HT (11kV) connected load will be around 4.5MVA per circuit.
• 4No. earth pits per transformer (2No. for body and 2No. for neutral earthing),
• Clearances, approx.1000mm around TC allow for transformer movement for replacement.
• Fault Level at TC Secondary=TC (VA) x100 / Transformer Secondary (V) x Impedance (%)
• ## Diesel Generator:

• Diesel Generator Set Produces=3.87 Units (KWH) in 1 Litter of Diesel.
• Requirement Area of Diesel Generator = for 25KW to 48KW=56 Sq.meter, 100KW=65 Sq.meter.
• DG less than or equal to 1000kVA must be in a canopy.
• DG greater 1000kVA can either be in a canopy or skid mounted in an acoustically treated room
• DG noise levels to be less than 75dBA @ 1meter.
• DG fuel storage tanks should be a maximum of 990 Litter per unit Storage tanks above this level will trigger more stringent explosion protection provision.
• ## Current Transformer:

• Nomenclature of CT:
• Ratio: input / output current ratio
• Burden (VA): total burden including pilot wires. (2.5, 5, 10, 15 and 30VA.)
• Class: Accuracy required for operation (Metering: 0.2, 0.5, 1 or 3, Protection: 5, 10, 15, 20, 30).
• Accuracy Limit Factor:
• Nomenclature of CT: Ratio, VA Burden, Accuracy Class, Accuracy Limit Factor.Example: 1600/5, 15VA 5P10  (Ratio: 1600/5, Burden: 15VA, Accuracy Class: 5P, ALF: 10)
• As per IEEE Metering CT: 0.3B0.1 rated Metering CT is accu­rate to 0.3 percent if the connected secondary burden if imped­ance does not exceed 0.1 ohms.
• As per IEEE Relaying (Protection) CT: 2.5C100 Relaying CT is accurate within 2.5 percent if the secondary burden is less than 1.0 ohm (100 volts/100A).
• ## Others:

• 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°)
• Most Economical Voltage at given Distance=5.5x√ ((km/1.6)+(kw/100))
• Maximum Voltage of the System= 1.1xRated Voltage (Ex. 66KV=1.1×66=72.6KV)
• Spacing of Conductor in Transmission Line (mm) = 500 + 18x (P-P Volt) + (2x (Span in Length)/50).
• 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%
• Motor Full Load Current= Kw /1.732xKVxP.FxEfficiency
 Quick Electrical Calculation 1HP=0.746KW Star Connection 1KW=1.36HP Line Voltage=√3 Phase Voltage 1Watt=0.846 Kla/Hr Line Current=Phase Current 1Watt=3.41 BTU/Hr Delta Connection 1KWH=3.6 MJ Line Voltage=Phase Voltage 1Cal=4.186 J Line Current=√3 Phase Current 1Tone= 3530 BTU 85 Sq.ft Floor Area=1200 BTU 1Kcal=4186 Joule 1KWH=860 Kcal 1Cal=4.183 Joule

## Cable Coding (IS 1554):( A2XFY / FRLS / FRPVC / FRLA / PILC)

 A Aluminium 2X XLPE F Flat Armoured W Wire Armoured Y Outer PVC Insulation Sheath W Steel Round Wire WW Steel double round wire Armoured YY Steel double Strip Armoured FR Fire Retardation LS Low Smoke LA Low Acid Gas Emission WA Non Magnetic round wire Armoured FA Non Magnetic Flat wire Armoured FF Double Steel Round Wire Armoured

Jignesh Parmar has completed M.Tech (Power System Control), B.E(Electrical). He is member of Institution of Engineers (MIE) and CEng,India. Membership No:M-1473586.He has more than 16 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”,”Smart Energy”, “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.

### 70 Responses to Electrical Useful Equations

Very useful information
Thanks

2. MASILAMANI.V says:

excellent for this words

3. Chisi says:

Man, you are the best..I never knew most of these rules.
Long live Jignesh!

4. HEENA says:

3 PHASE MOTOR DRAWS CURRENTS – 2.5 AMP/hp.if it is 230 v
and 1.5A/HP If it is 440 volt .

5. SUBRAMANIAN.P says:

Really the topics is very useful to me

6. arvind says:

VERY USEFUL AND HANDY INFO

7. arvind says:

very imp info

8. shaileshkumar.patel@Linde-LE.com says:

Dear Jignesh
I am regularly getting updateds from this websites. It is very useful.

As per Indian standrad : Emergency standby duty generator ( for
example : 1250 KVA,415 V) can be loaded for full load (i.e 1250 KVA-
constant or vairiable load ) in case of
main power source outgae for 3 days. If any supporting IS number is
In case Alternate relevant international standard is clarifying this

With best regards

Shailesh G Patel

it is very useful for electrical engineer and thank you very much Mr jignesh

10. Suresh says:

Very useful in short. Thanks

11. boy sandra says:

12. Praveen Malviya says:

Sir please help me… My question is, what is the suitable stater and breaker for 400kw,1440rpm induction motor…

13. engelect says:

Thanks for this rule of thumbs. It is really usefull for short assesment.

14. Murali.k says:

Thanks yery useful for all electrical engineers in the world

15. malebogo moupi says:

hie sir help me here’a single phase load consists of 1. 12kw of lighting & heating at unity pf.2. 8kw of motors at 0.8pf lagging & 3. 10kva of motors at 0.7 pf lagging.calculate (a)kwt (b)kvart (c)kvat (d)overall pf and the total supply current at 240v.my main wory is how to come with 30.3total kva.13.1total kvar &overall pf pliz help!

16. vamsi says:

as a site engineer this info is very helpfull to me ..

17. Rasika says:

what are electrodynamic stresses?

18. pankaj says:

yes

19. Almoutaz Safwan says:

It’s useful info. thanks

20. Dharmendra Kumar says:

THIS INFORMATION IS VERY USEFUL FOR EVERY ENGINEER AND ELECTRICAL MAINT. PERSON.
THANKS JIGNESH.
_ DHARMENDRA KUMAR

21. jayesh says:

so greatful job you had done.

Jay jay garvi gujarat

22. chandresh says:

thank u sir…………………………….

23. Nilesh Prajapati says:

Thank you sir

24. Steve says:

thank you for all your hard work in putting this together….Greetings from West Texas

25. surender says:

in what case voltage reading between ground and neutral up to 230 at distribution transformer L T side .The transformer ratio is 33 kv / .415 kv .

26. Ravindra Kumar says:

thank you very much sir………

All the three phase are generate in the power plant…Sir plz tell me about the neutral generation…

28. Sumit Sharma says:

Dear Sir
i am working in residential building my one flat load is 15 kw so how to calculate diversity factor.

regards

Sumit Sharma

29. Bhavin Patel says:

sir
thanks for sharing such a information

30. K.Gopinath says:

31. Alwin Paul says:

Dear Mr. Jignesh

■1 Phase Motor draws Current=7Amp per HP.
■3 Phase Motor draws Current=1.25Amp per HP

Can you explain the above information, please?

32. Paramasivan Tsp says:

Dear Mr. Jignesh,

Kindly use reference and standards. This ‘ll help us to improve the knowledge about standards and it ‘ll also induce to search about the standards.

-Paramasivan Tsp (My FB id)

• Normally I always indeed to indicate Standard and mention for the same..But where it is not possible,it is not mentioned..
as in the case of “THUMB RULE”

33. karan says:

34. prakash parmar says:

Dear sir,
I m gating regular mail Its a very use full information from for me

Thank you

Dear
Which rule have you followed for cable current carrying capacity?

36. ikoh says:

thank you for all the information!!!

37. kashyap raval says:

Dear Sir,
Thanks a lot for such technical information which will help to solve day to day workings.
It is reallya most valuable and sound as well as perfect calculations ..
Please be in touch by mail,
I shall be grateful if send me such an electrical related informations,calculations etvc.

38. Dear sir,

Thanks for your lot of Technical information.

One our customer in UK.Their asking for Below ground earthing calculation used BS (british standard).But I have IEEE80 format Ex sheet. Their not accepting IEEE80. Hence.Hence Kindly send the BS format for below ground earthing.

This is very useful for Us.

Regards
V.vasantha kumar.

39. Jose says:

Dear Sir,

can you prepare thumb rule based on DEWA regulations.

40. Mohanty says:

Gud information for public knowladge.. Thanks lot.

41. In domestic wiring, if neutral line is shotred with housing earth line, What problem we can face? Whether it will effect on KWH meter reading? Is there any chances of getting my electronic equipments damage?

42. GK says:

Excellent information. LIVE LONG DEAR ENGR JIGNESH. GK

43. parag says:

WHAT IS 10X RULE IN VOLTAGE DIVIDER CIRCUIT

Sir
its wonderful knowledge is here
i m very thankful to u . Jagnesh ji

45. Sourabh Singh says:

can we use elcb in horizontal mounting?

46. taufik says:

Thank you sir..!

What is the minimum cable size that I can use for AL cable in building distribution. Please suggest me the reference standard also.

47. Sanjay says:

Thumb rule for line loss for 300 square mm cable

48. ronak dave says:

Cu wire current carrying capacity is wrong. If you take 25 sq.mm cable then according to you sir it should be 25*6 = 150 amp. But in the catlogue of polycab its showing 108 amp. How there is so much of difference?

49. dipak says:

What is this 6

very very important knowledge for us ,thank you sir.

51. sivakumar A says:

Hello sir….

How to make the this kind thumb rule based on the reference document???
sir i want know about detail manner each thumb rule or otherwise you mention standard number…sir… example u use copper current rating calculation=6* size of sq.mm…why it’s used 6….???

52. Swapnendu Sil says:

32 amp industrial socket what cable or wire we need? what about cu or aluminium cable ?

53. Chakravarthi says:

For cable current capacity example calculation is wrong

54. rughnath prajapat says:

very useful information thanks

55. Sunil kumar rout says:

Very usefully

56. mohamed says:

very useful Thanks

57. yogesh nagare says:

NICE SIR

58. Vinod Punia says:

useful sir

59. Anish says:

Very useful thanks

60. zakeer says:

one sqmt = how much lux sir

61. Brillants technical information for all registered electrical engineer.

62. Md. Kazi Noman Arefin says:

Need to know the rules for single phase induction motor starting current calculation

63. santosh wagh says:

very useful article

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