Voltage Drop
Ohm's law and calculating voltage drop
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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 77oF - 121oF), K = 12 (temp 122oF - 167oF)
- Solid Aluminum, K = 18 (temp 77oF - 121oF), K = 20 (temp 122oF - 167oF)
- Stranded Copper, K = 11 (temp 77oF - 121oF), K = 12 (temp 122oF - 167oF)
- Stranded Aluminum, K = 19 (temp 77oF - 121oF), K = 20 (temp 122oF - 167oF)
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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Related Topics
- Electrical - Amps and electrical wiring, AWG - wire gauge, electrical formulas, motors and units
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