00:01
In this problem, on the topic of alternating current circuits, we're told that a 5 micro -farrid capacitor is in series with a 4 -oom resistor and is charged with a 9 -volt battery for a long time by closing the switch to position a in the figure.
00:16
Now, the capacitor is then discharged through the inductor by flipping the switch to position b at t is equal to zero.
00:23
The inductance of the inductor is 40 millie henries, and we want to find the maximum current in the inductor as well as, the first time at which the current is at its maximum.
00:36
So the maximum current through the inductor i max is given by omega -0 times the maximum charge q -max, and that's the maximum charge in the capacitor.
00:54
And we know that omega -0 is equal to 1 over the square root of l times c.
01:03
Now the fully charged capacitor has a maximum charge q max of c times v.
01:12
And so therefore the maximum current imax is equal to c times v over the square root of lc, which is the square root of c over l times v and if we put our values in this is the square root of the capacitance 5 times 10 to the minus 6 ferrets divided by the inductance of 40 times 10 to the minus 3 henries times the voltage 9 volts...