DC voltage drop

Not to avoid the question, but what exaclty are you trying to determine. Are you trying to decide on wire size and/or length?
 
The DC voltage drop is simpler because it does not have to consider an AC power factor. On an AC circuit, if the power factor is not 1, then the effects of reactance must be considered (as well as the resistance).

For a DC circuit, a simple calculation of voltage drop would be:

VD = R X CL/1000 X A

where:

R = Resistance of wire (Ohms) per 1000 feet
CL = Circuit Length (distance out to load and back)
A = Total Circuit Load in Amperes

If your wire R value is per 100 feet, then change the 1000 divisor to 100.

You can see that the voltage drop is dependent on how heavy the circuit is loaded (not just the size of the wire and the length). This is a fact often forgotten or ignored.

For an AC single-phase circuit, for "R" in the above formua you have to substitute the effective "Z" (Impedance = Resistance + Reactance) for the circuit power factor.
 
Last edited:
newguru said:
what is the formula to calculate DC volage drop? Is it different than AC voltage drop. Thanks in advance.

E=I X R in both AC and DC circuits, if it is a purely resistive circuit. If there is reactance, in the AC circuit, then the R will be replaced with XC or XL or Z if you have a combination of resistive and reactive components.

That is the wonderful thing about the rules, once you know them.


 

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