00:01
So we have some givens.
00:01
We have a current through the wire of 18 amps.
00:05
We have a voltage equaling 220 volts.
00:12
And then we have a diameter of the wire, 1 .628 millimeters.
00:18
We can say that the radius is going to be 0 .814 times 10 to the negative 3rd meters.
00:27
We can say the length is equal to 3 .5 meters as well.
00:31
So for part a, they're asking for the power.
00:33
We know the power is simply the product of the current times of voltage.
00:37
So here it's going to be 18 times 220.
00:41
And we're looking at 3 ,960 watts.
00:49
At this point, for part b, we can see the power dissipated.
00:53
And this is going to equal i squared times the resistance.
00:58
The current square times the resistance.
01:00
This is going to be i squared times the resistivity of copper, times the length of the wire divided by the cross -sectional area of the wire.
01:09
And this is going to be equal to i squared times the resistivity of the copper times the length divided by pi r squared.
01:20
Now for this we can solve.
01:23
So the power dissipated is going to be 18 squared times the resistivity of copper, 1 .68 times 10 to the negative 8 oms meters.
01:36
And then we'll have 3 .5 and then divided by pi 0 .814 times 10 to the negative third, and this will be squared.
01:51
So the power dissipated is going to be equal to approximately 18 .3 watts.
02:00
So not too much...