Voltage Drop

Ohm's law and calculating voltage drop

Voltage drop can be calculated using Ohm's law like

E = R I         (1)

where

E = voltage drop (Volts)

R = electrical resistance (Ohms)

I = current (Amps)

Example - Voltage Drop

Voltage drop in a 100 ft power line:

• 2 x 100 ft 
• #10 copper
• electrical resistance 1.02 Ohm/1000 ft
• current 10 Amps

can be calculated as

 E = 2 (0.00102 Ohm/foot) (100 ft) (10 Amps) 

    = 2.04 Volts

Circular mils and Voltage drop

Voltage drop can also be calculated using mils like

E = K P L I / A         (2)

where

K = specific resistivity (Ohm - circular mils/foot)

P = phase constant = 2 (single phase) = 1.732 (three phase=

L =  wire length (ft)

A = wire area (circular mils)

Specific resistivity various wires

• Solid Copper, K = 11 (temp 77^oF - 121^oF), K = 12 (temp 122^oF - 167^oF)
• Solid Aluminum, K = 18 (temp 77^oF - 121^oF), K = 20 (temp 122^oF - 167^oF)
• Stranded Copper, K = 11 (temp 77^oF - 121^oF), K = 12 (temp 122^oF - 167^oF)
• Stranded Aluminum, K = 19 (temp 77^oF - 121^oF), K = 20 (temp 122^oF - 167^oF)

Example - Specific resistivity and Voltage drop

With values from the example above the voltage drop can be calculated as

E = (11 Ohm - circular mils/foot) 2 (100 ft) (10 Amps) / (10400 mils)

    = 2.11 Volt

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