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Voltage drop and cables
- Thread starter doctor1
- Start date
Welcome to the forums.
You hijacked an old thread - http://forum.allaboutcircuits.com/showthread.php?t=25955, so I placed your post here to save confusion.
You hijacked an old thread - http://forum.allaboutcircuits.com/showthread.php?t=25955, so I placed your post here to save confusion.
Any cable will have a finite amount of resistance and will therefore, just like a "proper" resistor, develop a voltage across itself whose magnitude is simply I.R where I is the current flowing through the cable and R is the resistance of that cable.
Clearly, the thicker the cable (ie: the greater its cross-sectional area) the less its resistance and the longer the cable, the more its resistance.
Thus R is proportional to L/A therefore R = ρL/A where ρ is the constant of proportionality and is known as the cable material's resistivity. For copper this turns out to be 1.72 x 10^-8 Ωm (ohm.meters Note: NOT ohms PER metre).
Therefore, for any length of a particular cable we can work out its resistance and, since, R = V/I, if the current is known we can calculate the volts drop along the cable.
Here's an example:
A 10mm^2 cross-section copper cable runs for a length of 20 metres carrying a current of 30A. What is the voltage drop across the cable?
Well, R = ρL/A but R = V/I so V = ρL.I/A
thus, V = 1.72 x 10^-8 x 20 x 30 / 10 x 10^6
(not forgetting to convert the cross-sectional area to m^2)
So V = 1.02 volts
If the system, in which we are installing the cable, is say a low voltage 12v boat installation, this would be a significant loss and would represent around a 15% power loss before the current reaches the appliance.
Also, in a boat system for example, the hull is never used as a return wire (to avoid corrosion) so the cable length that one must use in the calculation would be the total OUT and RETURN length (or 40 metres in this example) resulting in a totally unacceptable 2v loss. Hence we would have to install a far thicker cable.
You might like to calculate what cross-sectional area this would have to be to get the TOTAL voltage loss down to an acceptable 0.3v.
If you insert the cable length in metres and the cross-sectional area in mm^2 then the formula becomes (for copper cables):
V = 0.017 x L(m) x I / A(mm^2)
Clearly, the thicker the cable (ie: the greater its cross-sectional area) the less its resistance and the longer the cable, the more its resistance.
Thus R is proportional to L/A therefore R = ρL/A where ρ is the constant of proportionality and is known as the cable material's resistivity. For copper this turns out to be 1.72 x 10^-8 Ωm (ohm.meters Note: NOT ohms PER metre).
Therefore, for any length of a particular cable we can work out its resistance and, since, R = V/I, if the current is known we can calculate the volts drop along the cable.
Here's an example:
A 10mm^2 cross-section copper cable runs for a length of 20 metres carrying a current of 30A. What is the voltage drop across the cable?
Well, R = ρL/A but R = V/I so V = ρL.I/A
thus, V = 1.72 x 10^-8 x 20 x 30 / 10 x 10^6
(not forgetting to convert the cross-sectional area to m^2)
So V = 1.02 volts
If the system, in which we are installing the cable, is say a low voltage 12v boat installation, this would be a significant loss and would represent around a 15% power loss before the current reaches the appliance.
Also, in a boat system for example, the hull is never used as a return wire (to avoid corrosion) so the cable length that one must use in the calculation would be the total OUT and RETURN length (or 40 metres in this example) resulting in a totally unacceptable 2v loss. Hence we would have to install a far thicker cable.
You might like to calculate what cross-sectional area this would have to be to get the TOTAL voltage loss down to an acceptable 0.3v.
If you insert the cable length in metres and the cross-sectional area in mm^2 then the formula becomes (for copper cables):
V = 0.017 x L(m) x I / A(mm^2)
sgardner025
- Joined Nov 5, 2009
- 79
If you can get a copy of the National Electrical Code book, chapter 9 table 8 may be of interest to you.
For a complete answer to your question, the following example shows how to compute the voltage drop across a cable. This is in agreement with NEC 2008.
http://www.skm.com/support/kb/120000.pdf
http://www.skm.com/support/kb/120000.pdf
