00:01
So here after the first switch has been closed a long time, we have a current i equaling 250 volts divided by 500 ums, and this is going to be equal to 0 .5 .00 amps.
00:27
Now, when s1 is open, when this first switch is opened and the second switch is closed, this means that a current decay rl circuit is produced and this is all from the diagram all from the diagram of the circuit so we can say that let's get a new workbook and say that the function of current as a function of time is going to be equal to i initial times xp to the negative resistance the negative resistance multiplied by time divided by the inductance.
01:20
So here, we know that for part a, i not, the initial current, is going to be equal to the max current.
01:34
So this would be 0 .5 .00 amps.
01:38
For part b, we want to find the time constant.
01:41
So tau, time constant, is going to be equal to the inductance divided by the resistance.
01:46
So this would be 300 times 10 to the negative third henry's and then divided by 500 ums.
01:59
And this is giving us 6 .00 times 10 to the negative fourth second...