Continued from first part: Electrical Thumb Rules You MUST Follow (Part 1)

## Useful Electrical 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 (I**) = 0.637 x Im (Im = Max.Value)_{av}**RMS Value of Sinusoidal Current (I**0.707 x Im (Im = Max.Value)_{rms}) =**A.C Current =**D.C Current/0.636.**Phase Difference between Phase =**360/ No of Phase (1 Phase=230/1=360°, 2 Phase=360/2=180°)**Short Circuit Level of Cable in KA (I**0.094 x Cable Dia in Sq.mm) /√ Short Circuit Time (Sec)_{sc}) =

(**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 Al = 105, K for steel = 58 , K for GI = 80**Most Economical Voltage at given Distance =**5.5 x √ ((km/1.6) + (kw/100))**Cable Voltage Drop (%) =**1.732 x current x (RcosǾ+jsinǾ) x 1.732 x Length (km) x 100) / (Volt(L-L) x Cable Run.

(**Spacing of Conductor in Transmission Line (mm) =**500 + 18 x (P – P Volt) + (2 x (Span in Length)/50).**Protection radius of Lighnting****Arrestor**= √h x (2D-h) + (2D+L).

Where h= height of L.A, D-distance of equipment (20, 40, 60 Meter), L=V x t (V=1m/ms, t=Discharge Time).**Size of Lightning Arrestor =**1.5x Phase to Earth Voltage or 1.5 x (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) x 100)/Average Voltage

**Ex:**462V, 463V, 455V, Voltage Variation= ((460 – 455) x 100)/455 =**1.1%**.**Current Variation (10%) =**((Average Current – Min Current) x 100)/Average Current

**Ex:****Fault Level at TC Secondary**= TC (VA) x 100 / Transformer Secondary (V) x Impedance (%)

**Motor Full Load Current =**Kw /1.732 x KV x P.F x Efficiency

PARAG NIKUMBH

will reply once sure

aakifnouman

A.C Current = D.C Current/0.636

what is 0.636??

sumitfloyd

A.C Current = D.C Current/0.636.

Please clarify this relation on how do you get it?