00:01
Here we're going to look at how a digital volt meter affects the reading of a voltage across a capacitor.
00:09
And we are going to be analyzing a circuit with an r &c in series with a battery.
00:16
The voltmeter, of course, is in parallel with the capacitor.
00:21
And for the volt meter, even though it is supposed to have ideally an infinite resistance, we will give it a finite resistance, little r, in parallel with the capacitor.
00:35
To analyze the circuit, we're going to use kirchhoff's circuit rules.
00:43
So recall that there is a junction rule.
00:48
We'll pick a particular junction, and we'll show a current through the battery that we'll call i, coming into the junction.
01:00
And coming out, we've got an i -1 through the little resistor, and we have a current, we'll call it ic, going through the capacitor, which we are going to write as dq by dt.
01:22
Okay, i -c is the time rate of change of the current, of the charge, building up on the capacitor.
01:34
Okay, so we also need two loop equations, and i'll pick the two.
01:42
I'll pick one loop.
01:44
Okay, we'll close the switch.
01:46
But one loop will start on the negative terminal of the battery and go all the way around and up through the dvm.
01:57
We'll call that loop one.
01:59
Loop 2, i think i'm just going to go inside of the capacitor dvm circuit.
02:10
So recall that we are adding up voltages to get zero.
02:14
All the voltage gains equal the voltage drops.
02:18
And so in loop 1, we gain going from right to left over the battery.
02:27
Then we lose going through r with current i.
02:33
And we again lose going through the dvm with i1.
02:40
So i've written this in a balanced way with voltage gain equals voltage lost.
02:47
Voltage loss.
02:50
For two, we basically have a parallel situation.
02:56
So i1r is equal to q over c.
03:04
Here i've just used that the voltage difference over a capacitor times c equals q.
03:13
And that gives us a voltage across the capacitor.
03:21
Okay, so now we are going to combine these, and like usual, we want to get rid of currents, but we want to do it in such a way that we wind up with the q on the capacitor.
03:33
So the first thing i'll do is i'll sub in for i from the junction equation.
03:40
And we get i1 times big r plus little r plus big r times dq by d t is equal to e and then i will simply substitute for i 1 in terms of q and that will give us e equals q over c big r plus little r over little r plus little r plus big r plus little r plus big r, dq by dt.
04:29
And this is almost ready to separate and integrate.
04:32
This is a nice differential equation that we can work with.
04:36
I usually like to play with it a bit algebraically in order to get it where it's easier to integrate.
04:45
So, of course, i'll bring over the q to the opposite side.
05:10
And i'm going to be doing some things with the r's, but.
05:16
We'll go ahead and divide out my big r.
05:19
I like to get leading terms in front of my q and my dq by dt.
05:25
The other thing i like to do is, like i said, get a leading term in front of my q.
05:35
But i see i have something that looks very much like resistors in parallel.
05:41
In other words, the r times little r over big r plus little r.
05:47
We're going to call that an r equivalent.
05:53
And i simply have minus q over rc plus e.
06:00
That's r equivalent.
06:03
E over r is equal to tq by dt.
06:10
One last step is to get the q term positive with nothing in front of it.
06:23
And so i must multiply the e term by those same things.
06:28
So i have not done anything too bizarre other than a little bit of algebra.
06:45
But at this point, we solved the equation by the technique of separate and integrate.
06:55
Separate means nothing more than switching the q and the time stuff.
07:02
So everything that joins with the q goes to the opposite side in the denominator and the dt.
07:10
Flips on over.
07:16
And what i see is that i've got something that looks like a resistance times of capacitance, normalizing the dt...