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 (Iav) = 0.637 xIm (Im = Max. Value)

· RMS Value of Sinusoidal Current (Irms) = 0.707 xIm (Im = Max. Value)

· 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 (Isc) = (0.094 x CableDia 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 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 Lighting Arrestor = √h x (2D-h) + (2D+L).

Whereh= height of L.A, D-distance of equipment (20, 40, 60 Meter), L=V x t (V=1m/ms, t=Discharge Time).

· Size of Lighting 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 bestsituation for System and Consumer both.

· If Load Factor is Low (0 or 25%) = you are payingmaximum 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 appliedfor Group Load

· Diversity Factor =

Sum of Maximum Power Demand / Maximum Demand (Demand Factor >1)

Diversity factorshould 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: 30A,35A,30A, Current Variation = ((35-31.7) x 100) /31.7 = 10.4%

· 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

· 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) = 0.637 x

· RMS Value of Sinusoidal Current (Irms) = 0.707 x

· 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 (Isc) = (0.094 x Cable

· Max

K = Material Factor, K for Cu = 159, K for Al = 105, K for steel = 58

· Most Economical Voltage at given Distance = 5.5 x √ ((km/1.6) + (

· 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

· Protection radius of Lighting Arrestor = √h x (2D-h) + (2D+L).

Where

· Size of Lighting Arrestor = 1.5x Phase to Earth Voltage or 1.5 x (System Voltage/1.732).

· Maximum Voltage of the System = 1.1

· Load Factor = Average Power/Peak Power

· If Load Factor is 1 or 100% = This is best

· If Load Factor is Low (0 or 25%) = you are paying

· Demand Factor = Maximum Demand / Total Connected Load (Demand Factor <1)

· Demand factor should be applied

· Diversity Factor =

Sum of Maximum Power Demand / Maximum Demand (Demand Factor >1)

Diversity factor

· 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

Ex: 462V, 463V, 455V, Voltage Variation= ((460 – 455) x 100

· Current Variation (10%) = ((Average Current – Min Current) x 100

Ex: 30A,35A,30A, Current Variation = ((35-31.7) x 100

· Fault Level at TC Secondary

= TC (VA) x 100 / Transformer Secondary (V) x Impedance (%)

Motor Full Load Current =

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