0:00:00.025,0:00:03.170
Now, we want to solve this circuit for vx.

0:00:03.170,0:00:07.790
Let's do it this particular way where[br]we're going to use a current divider and

0:00:07.790,0:00:11.110
let's find the current right here,[br]let's call that I1 and

0:00:11.110,0:00:14.710
let's find this current right here,[br]let's call that Ix.

0:00:14.710,0:00:17.920
We can say that we can find the voltage[br]because we know the resistance.

0:00:17.920,0:00:21.710
So if we do this to the current, we'd be[br]able to find the voltage using Ohm's law.

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This is the current divider.

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When we have a current divider,[br]the easiest way to handle that is to take

0:00:26.970,0:00:31.520
our source and divide it through,[br]let's say, R1 and R2.

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The current I1 and[br]I2 is going to be found this way.

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Let's take our R and convert it to G,[br]which is 1 over R1, and

0:00:41.153,0:00:45.013
let's convert this one to G2,[br]which is 1 over R2.

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Then we know that I1 is[br]equal to G1 over G1 + G2,

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times our original source current.

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And I2 is equal to G2 over G1 + G2,[br]also times our source.

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So let's do that over here.

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I can see that if I combine these two[br]in series, that would be 12 Ohms,

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that's what I'll be using in that arm.

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So if I wanted to have G1,[br]that would be one-third and

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then converting this to G2,[br]that would be 1 divided by one-twelfth.

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So I1 is going to be 10[br]Amps times G1 over G1 + G2.

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I2 is going to be 10 Amps[br]times G2 over G1 + G2.

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Do the math to work out[br]what the current will be.

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Then because I now know I2, oops,[br]I guess I called it x2, but

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there's I2, we can say that vx is going to[br]be = I2 times the resistance which is 4.

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Mm-hm, let's work out[br]the polarity on this.

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We can see that vx is[br]defined from plus to minus.

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Let's make sure that's right.

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My current is defined in this[br]direction from plus to minus, and so

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because these two are matching,[br]I have the correct polarity.

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If I had defined my current[br]the other direction,

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then I would have had a minus[br]instead of a plus here.

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Okay, great,[br]that's one way of doing this problem.

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Let's go do it another way.

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Let's say that I have the current,[br]So here's 3, 4, and 8.

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And this is the voltage[br]that I want to find.

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This is 10 amps.

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One thing that we can do is convert[br]this thing into a voltage source and

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then be able to do a voltage divider.

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So to convert this to a voltage source, I[br]simply say, what is the voltage across it?

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It's going to be 10 amps times 3.

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So I'm going to have a voltage[br]source plus to minus vs of 10 times

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3 which is 30 volts.

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And then my original current this 3,[br]that's right here,

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is going to end up being in[br]series instead of in parallel.

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So just 3 in parallel, sorry,[br]3 in parallel became 3 in series.

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So these two things are equal,[br]then I have my 4 Ohm and

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my 8 Ohm resistance, 4 and 8.

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And what I want to find is vx here.

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Remember when I have[br]a voltage divider like so and

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I have a series of three resistances,[br]R1, R2, and R3.

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Vs =, sorry V1 like this = vs times

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R1 divided by R1 + R2 + R3,

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similar for V2, V3, and V4.

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So finding xx now, I can just say that is[br]equal to vs times 4, divided by 3 + 4 + 8.

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And by vs is 30 volts.

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So again, 30 times 4 times 3 + 4 + 8.

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So that's the second way of[br]being able to solve the circuit.