00:01
Okay, in this problem we have an 8 volt battery and a 2 -oom resistor and a .2 ferad capacitor.
00:10
And there's a switch, capacitor is uncharged, and then at time equals 0, the switch is closed.
00:17
So what is the current right when the switch is closed, and then what is the current half a second after the switch is closed? okay, so for this kind of problem, when a capacity, the capacitor is initially uncharged.
00:38
Right when the switch is closed, it's almost as if the capacitor's not there, right? when it has no charge on it, it gets, the way it was explained to me was it's very thirsty, so it drinks up all the charge that comes in.
00:52
But the more that it drinks, the more charge that it gets inside it, the more the, the more it slows down, right? so as it gets more and more charge put into it, it starts to slow down.
01:05
And take in less and less.
01:09
So as time goes on, you expect the current to get smaller and smaller.
01:17
We can do a kirchhoff's voltage loop here, and say, we'll see the current is going clockwise, right? so the current is flowing this way.
01:31
So our loop will look like v minus, and then the voltage across the capacitor is the charge divided by the capacitance so minus q over c and then minus i r right for the resistor okay so solving for v we get v equals oh that equals zero sorry v equals i r plus q over c well current is just the time rate of change of charge right it's dq dt okay so now you have yourself a nice differential equation...