00:01
Okay, so for this problem, we have a resistor and inductor connected, and we want to know at what rate is the battery transferring energy to the inductor.
00:09
At this particular time, t equals 1 .61, tau, l.
00:18
Oh, is that tau l? yeah, tau inductor, tau l.
00:22
Okay, so here we go.
00:25
Let's do some derivatives.
00:26
So we were trying to evaluate d -u -d -t at t equals 1 .61 tau -l.
00:37
And then so that means we need to evaluate the derivative of the energy of an inductor.
00:46
So it's 1 half l, i squared.
00:51
And then we can plug in our formula for the current in a charging inductor, which is this v over r times 1 minus e to the minus t over tau.
01:06
So i'm basically i'm just going to plug this in here.
01:11
And then, oh yeah, we're giving some other given to all.
01:14
I'll write those down.
01:14
But anyway, i'm going to plug in this eye here and then evaluate do udt at t equals 1 .6 tau -l.
01:21
And then before i do that, let me just go ahead and write down the other givens.
01:24
So the emf, oh, i'll use voltage here.
01:38
So it's 12 .0 volts and r is 20 oms.
01:51
Do they actually give you the inductance? oh no, but luckily you don't need it.
01:59
Okay, because it's not really luck.
02:03
Okay, so anyway, so let's go ahead and evaluate u as a function of i.
02:09
So i can say that u is equal to one half l times this whole thing squared, so then that's b squared over r squared.
02:24
And then i'll just go ahead and expand it.
02:26
So 1 minus 2e to the minus t over tau, l, plus e to the minus 2 tau over, t or tau.
02:42
Okay, and let me get that other parenthesis.
02:45
Okay, wonderful...