You have a 2.0 Ω resistor, a 3.0 Ω resistor, a 6.0 Ω resistor, an...

Question

You have a 2.0 Ω resistor, a 3.0 Ω resistor, a 6.0 Ω resistor, and a 6.0 V battery. Draw a diagram of a circuit in which all three resistors are used and the battery delivers 9.0 W of power.

Accepted Answer

Hello, everyone. Let's take a look at this practice problem. Create a circuit diagram incorporating a 10 resistor, a 20 resistor, a 30 resistor, a 90 resistor, an 18 0 resistor and a 12 volt battery. Such that the power delivered by the battery is 12 watts. This is a problem where we're expected to essentially design a circuit based on the specifications using the specific tools given to us in the problem. The most important specification here is that the power delivered by the battery is 12 watts. Now recall that the formula for electric power is equal to the square of the potential difference. Essentially the voltage of the battery divided by the total equivalent resistance of the circuit. We know how much power we want delivered by the battery and we know the voltage of the battery. So what this will tell us if we algebraically rewrite this equation to solve for resistance, then we all know what equivalent resistance we want for the circuit. So we multiply both sides of the equation by the resistance and divide both sides of the equation by power. And we find that the resistance is equal to the square of the potential difference divided by the power. So that is going to be the square of volts divided by a power of 12 watts. We put that into a calculator or just do the math, we find 12 S. So the circuit we want to design must have an equivalent resistance of S. Now, this is where the problem might become a little bit of a puzzle because now we need to figure out what arrangement of resistors will give us a circuit with a 12 0 equivalent resistance. We can figure this out using the formulas we have for what happens when we have resistors that are either in series or in parallel. The equivalent resistance or resistors in series is equal to just the sum of the resistances. However, if we have resistors in parallel parallel and the equivalent resistance or rather the inverse of the equivalent resistance is equal to the sum of the inverses of all the resistances that are in parallel. So we will have to use these two formulas to figure out what arrangement we want. And note that when we have resistors that are in series, the equivalent resistance for them will always be larger than any of the individual resistances. But when we have resistors that are in parallel, their equivalent resistance will often be smaller than some of the resistances on their own. And that's relevant for this problem because you might notice that two of the resistors in particular seem to have a really high resistance. We have a 90 resistor which already takes out a majority of the 12 S we want to cover and we're also given an 18 resistor, which is more than the 12 os we want. So we need to have some way to reduce that. So just from that alone, we can tell that the 18 0 resistor will have to be in parallel with another one. So let's use some good old fashioned sh trial and error and see what happens if we choose the two largest resistances to put in parallel. And let's see what we get. So that's one divided by the 18 0 resistor plus one divided by. And then the +90 resistor, the second large resistance we have, if you put this into a calculator or do the math, then we find that this gives us an equivalent resistance of six S because if we add these two fractions up, that's equal to one divided by six S. And since, and since this is equal to one divided by the equivalent resistance, we can just invert this to find that an equivalent resistance for the parallel resistors of six S is found. And this is incredibly helpful to us because if we look at the list of resistors that the problem allows us to use the remaining three resistors that we haven't touched yet. 12 and three, all numerically just sum up to six on their own. So if we put the first three resistors in series, that gives us six S, and if we put the last two resistors in parallel, that also gives us six S. So if we add those two together, then that adds up to 12 S and we'll get the, exactly the circuit that we want. So let's do that. Let's design a circuit that has the first three resistors in series, the second two resistors in parallel and then put that parallel unit in series with the rest of it so that the two unit, the two resistances will add up. So let's begin designing this circuit. So first, we're going to have the 12 volt battery, 12 volts, then the wires coming out of it, we'll start putting some of the resistors in series. So the three ohm resistor goes in series along with the one and the two S. So I'll put the 10 resistor here and then the two resistors add here right there. And then on this side, I'll add in parallel a spot where the 90 and the 18 0 resistors can go. So this connects to some wires in parallel. We'll add one resistor here and another resistor here. So now we have two resistors in parallel and one of them will be the 18 resistor and the other will be the 90 resistor. And with that, we have drawn a circuit diagram that completely fulfills the specifications asked by the problem and that is all for this problem. I hope this video helped you out and please consider checking out some of our other tutoring videos which will give you more experience with these types of problems. That's all for now and I hope you all have a lovely day. Bye bye.

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You have a 2.0 Ω resistor, a 3.0 Ω resistor, a 6.0 Ω resistor, and a 6.0 V battery. Draw a diagram of a circuit in which all three resistors are used and the battery delivers 9.0 W of power.

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