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what is the equation for amperage loss of a cable, depending of the width and the length of the cable?

2006-11-27 01:55:58 · 5 answers · asked by fabric1601 2 in Science & Mathematics Engineering

I mean Voltage loss as well

2006-11-27 02:07:49 · update #1

5 answers

First of all the current (Ampere) in the cable will not be lost, it will be the same along the cable. The losses will be in the voltage according to cable resistance. The equation is :

dV=L P / X S U

dV=Voltage drop from sending to receiving side (Volt).
L=cable length in (meter)
P=Power in (watt)
X=conductivity (56 for copper & 34 for Alumineum)
S=cross sectional area of cable (Sq.mm)
U=rated voltage in (Volt)

2006-11-27 02:56:35 · answer #1 · answered by z1scop2004 1 · 0 0

There is no loss of amperage in a cable. The current that goes in one end of a cable is the current that comes out the other end. One of the fundamental laws of electricity is that the sum of currents going into a node (in this case, the cable) is zero. Thus, the positive current going into a cable is exactly equal to the negative current coming out the other end.

There is, however, a loss of voltage across the length of the cable that is in proportion to the resistance of the cable. This is in accordance with Ohm's law:

Voltage (volts) = Current (amperes) x Resistance (ohms)

The resistance of the cable depends on the conductivity of the material (usually annealed copper), the cross-sectional area of the cable, and the length of the cable, according to the formula:

Resistance = Length / (Area x Conductivity)

In SI (metric) units, the conductivity of annealed copper is about 58E6/(meter-ohms). Thus a 100 foot length of 12 gauge electrical wire, with a cross-sectional area of 0.00513 square inches, has a resistance of about 0.16 ohms. With a current of 20 amperes flowing through it, this wire will have a voltage drop of 3.2 volts from end to end. Since an electrical circuit always requires a return path, the total voltage drop for a 100 foot 12 gauge extension cord with a 20 ampere load will be twice 3.2 volts, or 6.4 volts.

2006-11-27 05:53:14 · answer #2 · answered by Tech Dude 5 · 0 0

ohms law I=V/R or V=IxR or R=V/I

the resistance r = (row x l)/A where row(symbol) = resistivity of material a constant (although this can vary with temperature).

simply put, the longer the conductor the more resistance, the more area (A) the less resistance.

amperes is the unit of current so your question should have read what is the equation for current loss. but this doesnt make sense as there is no current loss. but there is power loss. P=VxI this is mainly dependant on the voltage dropped over the cable and over the load.

2006-11-27 07:24:30 · answer #3 · answered by Mark G 2 · 0 0

There is no "amperage loss" in a cable. Amperes are measures of current and current is a flow of electrons. Electrons flowing through a wire do no decrease in numbers. Could you be thinking of voltage drop or the maximum current that a wire can safely handle before it gets overheated ?

2006-11-27 02:00:52 · answer #4 · answered by Gene 7 · 0 0

hmmmm - do you perhaps mean, how do i calculate the resistance of the cable, as ampage will depend on the potentil applied using the old V=IR formula.

well that depends on the type of material the cable is made of as all materials have a different resistivity. eg for copper it is 1.7 × 10-8 ohms per meter

the formula is R = p x l / A
where p is the resistivity, l=length in meters and A = cross sectional area in meters squared.

longer l then more resistance
thicker the cable = more xsection = lower resistance

2006-11-27 02:12:35 · answer #5 · answered by alatoruk 5 · 0 0

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