00:01
Here we have to solve a given problem.
00:02
First we have to find the current if current density and cross -sectional area is known.
00:09
So current equals to current density types cross -sectional area and here we have to convert 2 .69 millimeters squared, 2 meters squared.
00:29
Let's do this.
00:34
So one linear millimeter equals to 10 power by negative 3 linear meters that's why 1 squared millimeter equals to 10 power by negative 3 meters squared that is 10 power by negative 6 squared meters so therefore 2 .69 millimeter squared equals to 2 .69 times 10 power by net 2 .69 millimeter squared times 10 power by negative 6 meter squared over millimeter squared.
01:24
So that's why current in question 1 equals to current density times area.
01:32
There is 6 .01 times 10 power by 6 meters squared multiplied by 2 .69 millimeter squared times 10 power by negative 6 meter squared, per millimeter squared.
01:50
We can do unit analysis.
01:54
So the resulting current will be in amperors as required.
01:58
And that equals to 6 .01 times 2 .69.
02:07
That is 16 .1 ampires.
02:12
That is yeah, that's sorry...