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In this problem, we are going to analyze an ac circuit, specifically find a couple of unknown currents that are running through two different components.
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A capacitor, i2, and the diagram is running through a capacitor.
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And i1 is running through an inductor.
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And there is a resistor that is in series with the voltage source, which is an a .1.
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Ac source.
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Now, what's nice about this situation is that the components inside the circuit are already represented in phaser form.
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So we are going to be using phaser algebra and normal circuit rules, namely reduction into equivalent impedance in this case, not resistance.
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And some junction rules, and specifically there is one junction we'll be looking at where the current splits, in order to find those unknown currents.
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But this is the nice thing about ac circuits, is that the kirchhoff's rules for loops and junctions still apply, reduction of resistances by using parallel and series concepts still apply, but you have to use the concept of impedance.
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Instead.
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Phaser algebra, as a reminder, is algebra on imaginary or complex numbers in the complex plane.
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So a reminder that you can think about two different components of the things in the circuit.
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There are real parts of each component and imaginary parts of each component.
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And therefore components such as currents, voltages, and even impedances, have a real and imaginary part.
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They also have a magnitude and a phase.
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Okay, so the magnitude, if you have a complex number, it's just the hypotenuse in this imaginary plane, and the phase angle represents the angle that that component makes to the real axis.
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And i won't say too much about it other than the voltage source.
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What we know is it has an amplitude of 40 volts, and it has a phase of zero degrees, so it would lie completely along the real axis.
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But let's not get too much into the algebra.
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What we see is that there is a parallel branch, which i will lightly outline.
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So we have two resistances or impedances, we should call them in parallel.
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And let me be more clear and call that z parallel.
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We're going to find that in the usual way by taking the inverses of the impedances and inverting.
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That, and all this is going to be algebra on complex numbers.
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I'm not going to write in units.
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That could get a little confusing, but what we're basically doing is taking the impedance of the inductor.
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And i promised i would not put in that unit.
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We're taking the impedance of the capacitor and putting them into a usual parallel type of situation.
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And so what we're trying to do is find a z total, and then we can figure out the i total as the voltage of the source divided by z total.
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We're going to use the junction rule.
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I'll show the junction again, junction.
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That i is the sum of currents coming out of it, which is an i1 plus an i2, and we can use a parallel arrangement to separate out the i1 and the i2.
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So these are four steps that we're going to have to take in order to analyze this circuit.
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Okay, and i may not go through all the little steps in gory detail, but what we basically have is i1 times 5j.
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So voltages on the two parallel elements should be the same.
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So i1 times 5j should be equal to i2 times minus 10 j.
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Voltages in parallel are always the same is what we're saying there.
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Okay, so here's step one.
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We're finding the z parallel.
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I will jump ahead with some of the, well, i'll go ahead and show some of the steps.
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So you get a feel for that.
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That would be 10 minus 5 over 50j to the minus 1.
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And that turns out to be after you flip everything around 10j.
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So that parallel then is in series with the 8 oms.
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So our z total is 8 plus 10j.
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And now what we want to do is convert that into phaser.
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Well, actually combine it with the voltage source, which is in phaser form...