Voltage drop in DC circuit

[jreed]

I'm a plastics guy looking for some basic info on electricity. This is probably a very simple question:

In a DC circuit (such as in a car), how is voltage drop calculated in a specific conductor? Must the current drawn thru this specific conductor be known to calculate the voltage drop? I know the conductor dimensions and can get the resistance values for the conductor from a table.

Would appreciate any responses.

Jon [sig][/sig]

 

[jbartos]

Vd = R x I
where:
Vd voltage drop in Volts along the two conductors (plus and minus) from point A to B having length L in m (meters). The L is a length of two-conductor (plus and minus) DC circuit.
R is resistance of plus and minus conductors in Ohms added over the length L in m (meters)
I is current in Amperes flowing through the plus and minus conductors. It is the same value I since one plus is supplying the current and minus is returning it.
R=(rho) x 2L/A
A is cross-sectional area of a conductor = (PI)r**2 in mm**2 (in millimeters square)
r is radius or 2r is diameter of a conductor (usually both conductors have the same size) in mm (millimeters)
rho is resistivity of the material (cu = copper has rho = 0.0172 (Ohm mm**2)/m, al = aluminum has rho = 0.0278 (Ohm mm**2)/m)
Similarly in the inch system.
[sig][/sig]