00:01
So in this problem, we have a circuit composing of four resistors, and we are given the corresponding values of our voltage source and our resistances.
00:10
And we are asked to find the potential difference across each resistor.
00:15
So the first thing we're going to do is write down oms law, v equals i .r.
00:22
And let's represent our oms law and apply our oms law for each potential difference across the resistors.
00:28
So let's start with our v1.
00:32
V1 will be equal to i1 times r1, v2 will be equal to i2 times r2, our v3 will be equal to i3 times r3, and last but not least, our v4 will be equal to i4 times r4.
00:57
So let's see how we can simplify the corresponding potential differences.
01:05
And so if we look at our diagram back up here, we notice that r1 and r2 are in series.
01:11
Let's go ahead and make that note over here.
01:14
Say r1, r2 in series.
01:21
And if we take a look at r3 and r4, well, those are in parallel.
01:26
So let's go ahead and make that note.
01:31
3, 4, n.
01:40
So how can we use this to, i guess, simplify and make our lives a bit easier when solving for potential differences? well, if we have two resistors in series, in this case r1 and r2, then we can say that i1 and i2 will be equal to each other, because there, that is a property of resistors in, or the current going through resistors in series.
02:04
So we can say v1 is still equal to i1 times r1, but now for our v2, we can say that's equal to the current going through resistance or resistor 1 times r2.
02:18
And how about these are potential differences v3 and v4? well, we know that for two resistors in parallel, the potential difference will be equal to each other across those two resistors.
02:34
And that's being a property of resistors in parallel.
02:38
So we can say that v3, v3 is equal to v4, or in other words, it's gonna go like that over here, v3 is equal to v4, or in other words, i3 times r3 is able to i4 times r4.
03:01
So now we have expressions to help us solve for potential differences across each resistor.
03:06
So let's try to solve, so we know the resistance values of all resistors, so now we need to solve for the corresponding currents.
03:14
So let's solve for i -1 first.
03:17
So let's see, find i -1.
03:20
First thing we're going to do is get the equivalent resistance values, and we can do this the reason, and i guess when we do this, we're essentially going to create one simple circuit with a resistance in which we summed all our resistors.
03:37
And so how do we get our equivalent resistance? well, we'll get our diagram.
03:44
We have r1 and r2 in series.
03:46
So using the property of adding two resistors in series, we're simply going to take the sum.
03:54
And now how do we add two resistors in parallel? well, we notice that we have r3 and r4 in parallel.
04:02
And if you recall, our equation for, i'll call this r3, 4.
04:10
So in parallel, if we want to simplify our resistors, r3 and 4 into 1, we write down the expression that r34 is equal to 1 over r3 plus 1 over r4.
04:25
And so when we solve for r34, we get that r34 is equal to 1 over.
04:39
We get that this is equal to, i'm just going to do this up.
04:45
So our equivalent resistance for which i'll call should be r34 is 1 over 1 over r3 plus 1 over r4.
04:59
So we're going to fill that in here to get our equivalent resistance.
05:03
So 1 over 3 plus 1 over.
05:10
And so when we calculate this equivalent resistance, we're going to get rq is equal to r1, which is given us 24...