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In this video, we’re learning about parallel circuits.
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The circuits are parallel in contrast to series circuit, the other main type of electrical circuit.
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There are lots of great things to learn about parallel circuits.
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And as we go on, we’ll discover a helpful analogy between parallel electrical circuits and the parallel flow of water.
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So let’s get right into it.
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And to start, we’ll define what a parallel circuit is.
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Perhaps, the best way to understand a parallel circuit is to know that, in such a circuit, current follows more than one path.
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Another way to say this is to say that current splits or divides at some point in a parallel circuit.
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For example, let’s consider this circuit.
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If we follow the path of conventional current, we see that that current will travel in a clockwise direction.
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It will curl around this bend, travel through this first resistor.
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And then, it will get to this branchpoint.
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At this point, something interesting happens with the current.
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As we can see, there are two pathways to get to this part of the circuit over here, past the parallel branch.
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One pathway is to go through this upper branch, with resistor 𝑅 two.
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And one pathway is to go through the lower branch.
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Which way do you think current will go?
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It turns out that, based on the ratio of 𝑅 two to 𝑅 three, some of the current goes through the upper branch.
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And the rest of the initial current goes through the lower branch.
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This is really quite interesting because we might think that all of the current would go to whichever of these two branches has the least resistance.
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And none of the current would go to the other one.
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But that’s not what actually happens.
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Even when one of these resistance values is greater than the other, still that branch with the greater overall resistance gets some current flowing through it.
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So then, we have current travelling through the upper branch.
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And we also have some current travelling through the lower branch.
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When these branches join back up so that there’s only one path for current to flow, the current rejoins and continues on around the circuit.
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There are a couple of things we can notice about how this current travels through the branched part of this parallel circuit.
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First, let’s apply a few labels.
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Let’s say that the total current in our circuit is 𝐼.
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And then, we’ll say once the current has branched out, the current moving through the upper branch we’ll call 𝐼 sub 𝑢 and the current moving through the lower branch we’ll call 𝐼 sub 𝑙.
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It’s possible to relate these three currents mathematically.
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We can say that the total current 𝐼 is equal to the sum of the currents across the upper and lower branch.
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That makes sense if we think about our flowing water analogy to parallel circuits.
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If we had a stream flowing along and it reached a branching point, then we know that the total water that was in the branches would have to add up to equal the total water that was in the stream to start with.
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That is, the water in the branches has to come from somewhere.
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And where it comes from is farther upstream.
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In the same way, the currents flowing through the different branches of our parallel circuit have to add up to equal the total current that supply those branches.
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And by the way, this is true whether we have two branches or three branches or four or any number of branches in our parallel circuit.
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One other thing to notice about current in a parallel circuit is that when that current divides up across branches of the circuit, as we mentioned, those current values might not be the same.
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In other words, in our case 𝐼 sub 𝑢 might not equal 𝐼 sub 𝑙.
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The relative value of those currents depends on the ratio of the resistance of the branch that each is travelling through.
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Here, our flowing water analogy can be helpful.
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For the branches of this flowing stream, say that one branch has a lot of obstructions in it, lots of sticks and rocks and leaves in the way.
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But on the other hand, say that the other branch has very few obstructions.
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More or less, the water can flow unimpeded.
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Which of these two branches do you think will have more current flowing through it, the one with many obstructions or the one with few?
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What we would find is that the branch with fewer obstructions is able to handle and does handle more current flow.
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This lower branch where the water can flow without running into many obstacles might be most of the water in the stream whereas the upper branch might just pass a trickle.
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The same sort of thing happens in electrical circuits.
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If one of our parallel branches has a much greater overall resistance than the other, then that would mean that, comparatively, less current travels through that branch compared to the other branches.
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There’s still some that flows, but not nearly so much as branches with much less resistance.
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And by the way, we’ve just been arbitrarily underlining the top branch to indicate that it has more resistance.
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We haven’t actually seen that to be true yet.
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And for now, we’re gonna leave these relationships general.
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𝑅 sub two could be equal to, greater than, or less than 𝑅 sub three for our purposes.
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But anyway, if one of these branches in the electrical circuit did have more resistance than the other, then less current would travel through it than the least resistive branch.
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Very briefly, if it did happen that 𝑅 sub two was equal to 𝑅 sub three, that is, the resistances of our parallel branches were identical, then that would mean the current will divide evenly across these branches.
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In other words, 𝐼 sub 𝑢 would be equal to 𝐼 sub 𝑙.
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So that’s a bit about current in parallel circuits.
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Now let’s move on to talking about resistance in these circuits.
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We’ve already touched on this a bit with our flowing water analogy.
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But really, when it comes to parallel circuits, the burning question around resistance is what is the resistance of parallel branches in a circuit.
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That is, if we were to consider this section of the circuit as a whole, what would the equivalent or effective resistance of this part be.
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In answer to this question, there’s an addition rule for resistors arranged in parallel that we can learn.
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When it comes to this particular example circuit we have for a parallel circuit, we see we just have two resistors arranged in parallel.
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But, in general, we could have any number of parallel resistors.
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We could have 𝑛 resistors in parallel.
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Keeping this number completely general, we can say that one over the total resistance of a set of branches arranged in parallel is equal to one over the resistance of the first branch plus one over the resistance of the second branch plus dot dot dot, up to one over the resistance of the 𝑛th branch.
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Notice that this equation, which is four resistors in parallel, has a very different form than that for resistors in series.
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We can’t use the same equation for both.
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As we said, this equation for the resistance of resistors in parallel assumes a general number of resistors, 𝑛 could be any integer.
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It so happens though that, often, we encounter a circuit that has two parallel branches, no more, no fewer.
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This situation of having two resistors arranged in parallel is common enough that it’s worth simplifying this general equation for the case where 𝑛 is equal to two.
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When 𝑛 is equal to two, we then have two parallel branches.
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And one over the total resistance of those branches is equal to one over the resistance of the first branch plus one over the resistance of the second.
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And let’s say that, as a next step, we multiply both sides of this equation by 𝑅 sub 𝑡 times 𝑅 sub one times 𝑅 sub two, in other words, all three of the resistances mentioned in the equation.
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If we do this, then after cancelling out common factors that appear in numerator and denominator, we have that 𝑅 sub one times 𝑅 sub two, the resistances of the two parallel branches, are equal to 𝑅 sub 𝑡 times 𝑅 sub two plus 𝑅 sub 𝑡 times 𝑅 sub one.
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We see that, on the right-hand side, we can factor out an 𝑅 sub 𝑡 term.
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If we do that and then divide both sides of the equation by 𝑅 sub one plus 𝑅 sub two, that term on the right side cancels.
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And then, if we write out the resulting equation, we find that 𝑅 sub 𝑡, which in this case is the total equivalent resistance of the two parallel branches, is equal to the resistance of the first branch times the resistance of the second branch, divided by the sum of their resistances.
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Since circuits having exactly two parallel branches are fairly common, this relationship is worth keeping in mind although, as we know, we can always derive it from the general equation for 𝑛 resistors.
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Looking at this equation, we can see an interesting comparison between the total resistance and the individual resistors 𝑅 sub one and 𝑅 sub two.
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To see this more clearly, let’s come back over to our example circuit.
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And we’ll apply actual resistance values to our resistances of 𝑅 sub three and 𝑅 sub two.
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Let’s say that 𝑅 sub two, the resistance of the top branch, is two ohms and 𝑅 sub three, the resistance of the bottom branch, is four ohms.
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Now, here’s a question.
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What will the overall resistance of these two resistors arranged in parallel be?
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In other words, what is 𝑅 sub 𝑡, the equivalent resistance.
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According to our equation for two resistors, the total resistance here is equal to their product, two ohms times four ohms, divided by their sum, two ohms plus four ohms.
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We get eight ohms squared over six ohms.
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And a factor of ohms cancels out.
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And in its simplest form, the total equivalent resistance of these two resistors in parallel is four-thirds of an ohm.
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Notice something interesting.
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This total value is less than either of the individual values of the resistors, 𝑅 two or 𝑅 three.
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So by arranging these resistors in parallel, we’ve effectively decreased their overall resistance.
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And in fact, this is always true of resistors in parallel.
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The overall or effective or equivalent resistance of a set of parallel branches is always less than the resistance of any one of the branches.
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That’s really interesting because it means that we can decrease an overall resistance by adding more resistors.
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They just have to have smaller and smaller values.
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Okay.
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So far, we’ve talked about current and resistance in parallel circuits.
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Now, let’s talk about potential difference.
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And here, the punchline is that potential difference across parallel branches of a circuit is the same.
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That means, in the case of this circuit, that if we measured the potential difference across the upper branch, it would be the same as the potential difference across the lower branch.
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There’s a good reason why that’s the case.
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Instead of thinking of it in terms of electrical potential, let’s think for a moment in terms of gravitational potential.
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Say that we have two heights, one higher than the other.
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We’ll call them ℎ one and ℎ two.
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Let’s imagine further that we have an object that starts out at ℎ one.
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And then, it follows a particular path to get to ℎ two.
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But then, there’s another object that starts at ℎ one and takes a different path to get to ℎ two, ending up at the same point.
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And say there’s yet a third object starting at ℎ one.
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This one takes a very unusual path but then ultimately does end up at ℎ two.
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All three of these objects, we can see, have moved through the same height difference and therefore have experienced the same change in gravitational potential.
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We could think of these different pathways that we’ve drawn in as different parallel branches of an electrical circuit.
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Even though the paths are different, because their start and end point differences are the same, that means their overall gravitational potential difference is the same.
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And it’s the same way with electrical potential in our parallel circuit.
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Say we were to measure the electrical potential difference between here and here in our circuit, in other words, across the parallel part of the circuit.
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Whatever pathways electrical current can follow to go from the start point to the end point, we know that the potential difference across those paths must be the same.
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And this fact is very helpful to us when we’re trying to solve for different quantities in parallel electric circuits.
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Here’s an example.
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Let’s say that our cell in this case is providing an overall potential difference of 10 volts.
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And let’s further say that our resistor 𝑅 sub one is equal to five ohms.
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Knowing all that, what we wonder is the current running through the lower branch of our parallel section, 𝐼 sub 𝑙.
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Let’s solve for that current now, using what we’ve learned so far about parallel circuits.
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As a general solution strategy, here’s what we’ll do.
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First, we’ll work to solve for the overall current in the circuit capital 𝐼.
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We’ll do that by applying Ohm’s law to this circuit.
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Then, using that law, we’ll calculate just how much voltage is dropped over the resistor 𝑅 one.
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Once we know that, we will be able to solve for how much voltage must drop over the remaining portion of the circuit.
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And once that’s solved for, we’ll be able to use Ohm’s law once more to solve for 𝐼 sub 𝑙, the current through the lower branch.
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So let’s get started.
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And as we mentioned, we’ll start by using Ohm’s law to solve for the overall current in the circuit.
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To solve for that current, we’ll need to know the overall potential difference as well as the overall resistance.
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We know the overall potential difference 10 volts.
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But we don’t yet know the overall resistance.
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To solve for that, we’ll need to add these two resistors which are in parallel, the two-ohms and the four-ohm resistors, and then add that equivalent resistance to 𝑅 sub one.
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To add these two parallel resistors, the four-ohm and the two-ohm, we can recall our rule for two resistors arranged in parallel that their equivalent resistance is equal to their product divided by their sum.
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And recall that, in our case, we’ve already solved for this equivalent resistance.
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It’s four-thirds of an ohm.
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To get the total resistance of the circuit then, we’ll add this value, four-thirds of an ohm, to five ohms, the resistor 𝑅 one, which is arranged in series with the parallel portion of the circuit.
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When we add those resistances according to the series addition rule, we find it’s equal to nineteen-thirds of an ohm.
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That’s the total equivalent resistance of all three resistors in our circuit.
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Now, to find the overall current in our circuit, we’ll divide the overall voltage by the overall resistance, 10 volts divided by nineteen-thirds of an ohm.
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That comes out to thirty nineteenths of an amp.
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Now we know the total current in the circuit.
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Let’s figure out the potential difference across our resistor 𝑅 sub one.
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According to Ohm’s law, that’ll equal the current running through this resistor, thirty nineteenths of an ampere, multiplied by its resistance, five ohms.
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Written as a fraction, this comes out to 150 over 19 volts.
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Notice, by the way, that, as a decimal, this is approximately equal to seven and a half volts.
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In other words, it’s the majority of our overall potential difference, 10 volts.
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Our next step is to solve for the potential difference across the parallel section of our circuit.
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And we’ll do it by subtracting this value, one hundred and fifty nineteenths of a volt, from our overall potential difference of 10 volts.
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We’ll call that potential difference 𝑉 sub 𝑝.
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And it’s equal to 190 over 19 volts, that’s 10 volts, minus 150 over 19 volts.
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And this comes out to 40 over 19 volts.
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This means that each of the two branches in our parallel circuit experiences a potential difference of forty nineteenths of a volt.
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Using Ohm’s law one more time, we can say that the current in the lower branch is equal to the potential difference across that branch divided by its resistance.
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When we calculate that fraction, we find a result of ten nineteenths of an ampere.
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That’s how much current is flowing through the lower branch of our parallel circuit.
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Let’s now summarize what we’ve learned so far about parallel circuits.
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For starters in this lesson, we saw that parallel circuits have more than one path for current to follow.
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We also saw that current divides up across the various paths offered to it.
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We also saw that, for 𝑛 resistors arranged in parallel, one over their total equivalent resistance is equal to one over the resistance of the first resistor plus one over the resistance of the second resistor plus all the way up to one over the resistance of the 𝑛th resistor.
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We also saw that when we have exactly two resistors in parallel, their total resistance is equal to their product divided by their sum.
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And finally, we saw that the potential difference across parallel branches of a circuit is the same.