Question

You're writing the instruction manual for a power saw, and you have to specify the maximum permissible length for an extension cord made from 18-gauge copper wire (diameter 1.0 mm). The saw draws 7.0A and needs a minimum of 115V across its motor when the outlet supplies 120V. What do you specify for the maximum length extension cord, given that they come in 25 foot increments?
Please, help

Answer 1

and then?

Answer 2

So - you know the current through the wire, and you know the voltage drop, so you can work out the RESISTANCE of the wire

You know the wire is made of Copper - you need to find its property called RESISTIVITY

You know the diameter of the wire and you know the maximum allowable resistance.

Resistivity links resistance to Area of wire and length.

So you need a few steps, and to find the value for resistivity of Copper.

Answer 3

from V =2.5 V and I = 7.0 A, I have \(R_{wire} = \dfrac{V}{I}= 0.3571 \Omega\)
resistivity of copper is \(\rho = 1.68*10^{-8}\)
and area of the wire is \(A = (0.5*10^-3)^2*3.1415 = 7.853575*10^{-7}m^2\)
Now, put everything on formula \(R= \dfrac{\rho *l}{A}\rightarrow l = \dfrac{RA}{\rho}\) = 16.68

Answer 4

Looks good - I haven't checked the sums - but the process looks right.

(Could oyu have done it without help from here?) (Just asking - np with posting here - but you seem to have a good grasp of the subject without my guidance....)

Answer 5

no, I just follow what you suggest

Answer 6

I am new in physics, hehehe

Answer 7

OK, let me try, I will tag if I don't have the correct answer. Thanks a lot

Answer 8

gtg now - Keep up the good work!

Answer 9

have fun

Answer 10

I think that you did not convert feet to meters:
The resistance of a 25 ft (7.62 m) cord is
$$
R_{25ft}=\cfrac{\rho l}{A}=\cfrac{1.7\times10^-8 \Omega~m\times7.62~m}{7.8\times10^-7~m^2}=0.17\Omega
$$
The max voltage drop across this cord is
$$
7\times .17=1.19~Volts
$$
This means that each wire of the cord drops 1.19 volts. Where one wire goes from power to saw and the other wire goes from saw to power. Each cord has two such wires.

So the total voltage drop for a single cord is
$$
2\times1.19=2.38~volts
$$
Let x be the max number of cords allowed, then
$$
x\times2.38 \lt5~volts\\
\implies x\lt2.1
$$

So you can not have more than 2 cords.