# Wire gauge

• Jason03
ACIn summary, the problem the expert is working on states that they have a power factor of .5, a real power of 3600 watts, a voltage of 240 V, and a distance between a motor and transformer of 100 feet. They are trying to find the Minimum wire gauge that could be used to power the appliance. They found that even though it is AC, they can use the RMS current. They then solved for the I which yielded 30 amps. The load will be different than 16 ohms, and they need to find the losses in the wire.f

#### Jason03

The problem I am working on states that I have a power factor of .5, a real power of 3600 watts, a voltage of 240 V, and a distance between a motor and transformer of 100 feet.

Im trying to find the Minimum wire gauge that could be used...

Do I need to find the current and than go to the AWG tables to find a match?...Since I am working with AC I am a little confused.

The problem I am working on states that I have a power factor of .5, a real power of 3600 watts, a voltage of 240 V, and a distance between a motor and transformer of 100 feet.

Im trying to find the Minimum wire gauge that could be used...

Do I need to find the current and than go to the AWG tables to find a match?...Since I am working with AC I am a little confused.

Even though it's AC, you can use the RMS current.

Ok...so I was looking at a few equations...could I use the equation for REAL POWER which is:

P = VI cos(theta)

because I have P...and I have V...and I can find cos(theta)...

my only question would be since it 240 volts...im assuming that's in RMS already so I can just plug it straight into the equation?

Yes, it is 240 V RMS.

ok so I used the formula P = VI* power factor...this allowed me to solve for I which gave me 30 amps...

So I looked up the AWG chart and found that AWG No. 10 can have a max of 30 amps...

So if that is correct...I now have to find the losses in the wire...

What is the easiest way to find the losses in the wire??

Would I need to find the resistance of the current through the wires by first using R = V^2/P...than

going to the AWG table and finding the resistance of the wire for the 100 foot length of the wire?

You're okay up to calculating I=30A

However, the load will be different than 16 ohms. For starters, you did not include the power factor, and secondly the load is not resistive. Notice you got a current that is only about 1/2 of the 30A you had calculated before(!)

While I am not an expert in this area, I know of two factors that limit wire gauge:
• overheating of the wire insulation
• too much voltage drop for long wire lengths

Given the stated 100' distance, it looks like we are dealing with the 2nd limitation. Were you given any guideline for how much voltage drop from the wiring is within tolerance?

I thought the power factor was included in the 3600 watts since its the real power...and if its not resistive what is it?...Im just using the methods that are shown in my text...but the example I am going off of is in DC...