Power loss in connectors and wires

[Bergenheimer]

Hello,

I'm working on a project at which a large amount of current is being passed through a low voltage system (48 Volts). In one part of the system, there are connectors rated at 12 Volts in which the 48 volt DC current is passed through. How much power loss might I be suffering here?

Also, I know there is a calculable voltage drop across a wire with a certain gauge over a certain length. However, how to I calculate how much current is lost over a similar distance?

Thanks a lot in advance for some advice in this matter!



Bergenheimer

 

[phinds]

I don't know why a connector would be rated in volts. Doesn't seem to make any sense. I mean, if you actually HAVE 12 volts over the connector that means it isn't plugged in and if it IS plugged in, you surely are not going to have 12 volts across it.

"how much current is lost over a similar distance?" DEFINITELY doesn't make any sense. Current loss due to circuit resistance will be a loss throughout the entire circuit loop, not in some specific length of wire.

 

[berkeman]

Hello,

I'm working on a project at which a large amount of current is being passed through a low voltage system (48 Volts). In one part of the system, there are connectors rated at 12 Volts in which the 48 volt DC current is passed through. How much power loss might I be suffering here?

Also, I know there is a calculable voltage drop across a wire with a certain gauge over a certain length. However, how to I calculate how much current is lost over a similar distance?

Thanks a lot in advance for some advice in this matter!



Bergenheimer

Connectors have a voltage rating just to be sure that there are no arcs between the conductors. If you apply 120Vrms to a 12V rated connector, there is a good chance that you would get arcing. But applying 48V to a 12V rated connector should generally not be a problem.

And as phinds says, you are confusing voltage and current. The current will be the same at all parts of the complete circuit, and you will have voltage drops along the complete circuit depending on the resistances along the way.

You calculate the voltage drop across a section of wire by the resistance (get it from wire tables online) and the current, using V=IR. You calculate the power lost in a section of wire by the voltage drop (from above) and the current, using P=VI.

 

[Bergenheimer]

Connectors have a voltage rating just to be sure that there are no arcs between the conductors. If you apply 120Vrms to a 12V rated connector, there is a good chance that you would get arcing. But applying 48V to a 12V rated connector should generally not be a problem.

And as phinds says, you are confusing voltage and current. The current will be the same at all parts of the complete circuit, and you will have voltage drops along the complete circuit depending on the resistances along the way.

You calculate the voltage drop across a section of wire by the resistance (get it from wire tables online) and the current, using V=IR. You calculate the power lost in a section of wire by the voltage drop (from above) and the current, using P=VI.

Ok, this all makes sense to me. So the power lost over a wire with a length x is P = (Vin - Vout)*I, where Vin is the input voltage at the beginning of the wire and Vout is the voltage at the end of the wire determined by resistance R of a wire of gauge y and length x. The current through the wire is constant as given by rules of circuit analysis but the voltage drop changes depending on the length of the wire. Can i think of a wire as a resistor with a resistance that varies depending on its length, and therefore has varying voltage drops at either end?

 

[berkeman]

Ok, this all makes sense to me. So the power lost over a wire with a length x is P = (Vin - Vout)*I, where Vin is the input voltage at the beginning of the wire and Vout is the voltage at the end of the wire determined by resistance R of a wire of gauge y and length x. The current through the wire is constant as given by rules of circuit analysis but the voltage drop changes depending on the length of the wire. Can i think of a wire as a resistor with a resistance that varies depending on its length, and therefore has varying voltage drops at either end?

Yes, that all sounds correct. I use a small book that has wire tables in it when I do these calculations, but there are many online like this:

http://www.powerstream.com/Wire_Size.htm

.

 

[Bergenheimer]

Yes, that all sounds correct. I use a small book that has wire tables in it when I do these calculations, but there are many online like this:

http://www.powerstream.com/Wire_Size.htm

.

Ok, thanks guys for clearing up my minor confusion on these topics, this will help moving forward in my designs

 

[Bergenheimer]

The equation I'm finding around the internet is voltage drop Vd = (2*L*R*I)/1000, where L is the length of the wire, R is the resistance and I is the current moving through the wire. This is consistent with the logic above except it is twice as much. I'm assuming this is to account for a typical two wire system a DC or single phase system would run off of.

Now my question is, how much of a voltage drop is my load actually seeing? Is it the number in the equation above or is it half as much (the voltage drop across a single wire, which would be the positive wire in a DC system). This understanding is important in determining the efficiency of my system.

 

[berkeman]

The equation I'm finding around the internet is voltage drop Vd = (2*L*R*I)/1000, where L is the length of the wire, R is the resistance and I is the current moving through the wire. This is consistent with the logic above except it is twice as much. I'm assuming this is to account for a typical two wire system a DC or single phase system would run off of.

Now my question is, how much of a voltage drop is my load actually seeing? Is it the number in the equation above or is it half as much (the voltage drop across a single wire, which would be the positive wire in a DC system). This understanding is important in determining the efficiency of my system.

The factor or 2L probably just accounts for the round trip resistance. Your load will see the source voltage minus the voltage drop across the wires going out and coming back. Just think of each of the two wires as a resistor.

 

[GrahamN-UK]

The equation I'm finding around the internet is voltage drop Vd = (2*L*R*I)/1000, where L is the length of the wire, R is the resistance and I is the current moving through the wire.

L is the length of the wire in some units and R is the resistance of a single conductor in ohms per 1000 of the same units as you are measuring L in. You need to choose whichever units are used in the tables of wire resistance you refer to.

The tables linked to by berkeman list resistance in ohms per 1000 feet (sometimes written kft) and in ohms per 1000m (km) so you can measure L in either feet or metres as long as you use the corresponding feet or metre resistance columns.

Sometimes tables will list wire resistance per foot or metre, in which case you can omit the division by 1000 in your formula.

 

[jim hardy]

Re that table

rule of thumb for back of envelope calculations:

#10 is one ohm per thousand feet.
Resistance doubles (or halves) for each three wire sizes.
Observe #13 is 2 ohms per thousand, #7 is one half ohm per thousand.

So resistance changes by ratio cube root of two, 1.26 , for each step in wire size

Those simple numbers are handy to remember when estimating in a country that uses feet.

old jim