00:01
All right, so we have two circuits to draw here.
00:03
I'm just going to do the prediction part because the rest of this is about a lab that you're actually going to have to perform.
00:09
So the voltage source here with two resistors in series, r1 and r2, and another circuit that has a voltage source with two resistors in parallel and r2.
00:25
Let's analyze these in turn.
00:28
First, we're assuming that our voltage source is 3 volts.
00:32
So if we put ground arbitrarily just on the bottom here, this is 3 volts here.
00:38
I'm going to label the voltages across r1 and r2 as respectively v1 and v2, but the current through them of course is the same.
00:50
So i just have one current i, and then here we have a net current i that's coming through this, but it's being split here into i1 and i2.
01:05
And i think this parallel circuit is actually going to be a little bit easier to use.
01:08
So just using ohm's law, we know that this node has 3 volts in it.
01:13
So i2 is equal to, by ohm's law, 3 volts over r1.
01:20
I1 is 3 over r1, and i2 is 3 volts over r2.
01:30
And we're told that i is equal to the sum of these by the loop law.
01:35
So that's 3 over r1 plus 3 over r2.
01:42
So that's great.
01:44
Our voltages are v1 of course is equal to 3 volts as is our v2, so that's useful.
01:56
Now the series circuit is a little bit more intricate to analyze, but not too much.
02:00
By the loop law here, we have an increase of 3 volts is equal to our decrease of 1 or equal to v1 plus v2.
02:12
Right, it's equal to v1 plus v2.
02:14
Go up 3, down v1, down v2.
02:17
We have to get down to 0 by the loop law.
02:21
And moreover, our current here by ohm's law has to be the same as v1 over r1, v1 over r1, and that also equal to v2 over r2.
02:39
We can solve for v2 here.
02:42
V2 is equal to 3 minus v1.
02:47
So v1 over r1 is equal to 3 minus v1 over r2.
02:54
We can write this here as 3 over r2 minus v1 over r2, so that we can move v1 over here.
03:09
V1 over r1 plus v1 over r2 is equal to, yeah this is 3 over r2.
03:18
Multiply both sides by r1 times r2, we get r2 v1 plus r1 v1 is equal to 3 r1 r2 over r2.
03:32
Those cancel out, so this is just 3 r1.
03:37
3 r1.
03:39
Factoring out r2 plus r1, v1 is thus equal to 3 r1 divided by r1 plus r2.
03:51
I suppose we can actually just plug in v1 here to find i.
04:04
I is equal to 3 r1 divided by r1 plus r2 divided by r1.
04:13
So we just cancel out this r1 in the numerator, we get 3 over r1 plus r2.
04:17
So here is the current through each resistor.
04:24
This is the same current, it's the same current through both of those, it's the current through the resistors individually, and the voltages across them, which is supposed to be 3 r1 over r1 plus r2, and our other voltage by symmetrical construction will be 3 times r2, or r1 plus r2...