Practice: What is the equivalent resistance of the following combination of resistors?

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Intro to Current | 7 mins | 0 completed | Learn |

Resistors and Ohm's Law | 14 mins | 0 completed | Learn |

Power in Circuits | 11 mins | 0 completed | Learn |

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Combining Resistors in Series & Parallel | 37 mins | 0 completed | Learn |

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!! Resistor-Capacitor Circuits |

Concept #1: Combining Resistors in Series & Parallel

**Transcript**

Hey guys. So, in this video we're going to talk about combining resistors, which is a key skill that you have to master for this chapter, let's go. Alright, so in circuit problems you're often going to be asked to collapse or combine or merge multiple resistors into a single equivalent resistor, single equivalent resistor, just one, and resistors can be connected to each other in two ways, they can be connected in series or in parallel, and this looks like this, in series you're going to have resistors sort of side by side like this, in parallel you're going to have resistors like this. Alright, and what you want to do is we want to go from a number of multiple resistors, let's say here, three, into a single equivalent resistor, same thing here, you want to turn these three resistors into just a single resistor.

Now, what makes this a series connection is that you have a direct connection between the resistors without any splits in the wire. So, imagine, if you are an electron traveling through this wire you go straight through with no forks on the rope. Alright, so and then here, the wire splits and because the wire splits you get some loops. So, let's say, we connect this to, let's connect this to a battery, to a voltage source, kind of like a battery, right? Remember, charge, charge moves out of, the current is going to flow out of the positive the larger terminal here, it's going to go this way, when he gets here, it has the choice of moving this way or this way. So, some of the charge will go one way, some of the charge go the other way so the wire splits and it forms a loop. So, one other thing that's important to note is, you have a parallel connection, whenever you have two resistors that are alone an opposite, on opposite sides, on opposite branches. So, let me write this, alone on opposite branches, or opposite sides, you can think of it that way as well. So, for example, this guy is alone in this branch, this guy is alone in this branch. So, they are alone on opposite sides or on opposite branches therefore they are in parallel alright. Now, the way you combine them is by using the equivalent resistance equation, which is different for series and parallel, and you have to memorize this one, the equivalent resistance if you're in series is just the addition of individual resistances, R1 plus R2 plus R3, for example, if this is a 1 ohm resistor and this one is 2 and this one is 3, this equivalent resistor here is going to be 6, you just add up the numbers.

Now, if you are in parallel the equation is a little more complicated, it's 1 over R equivalence equals 1 over R1 plus 1 over R2, and you keep adding on one of these fractions for each resistor, here I have three. So, I'm going to write 1 over R3, okay? And, what it means to be an equivalent resistor is that you behave the same as the original group of resistors. So, let me illustrate this, let's connect the battery here, positive side of the battery, so the current is going to flow out this way, the idea is that this group of resistors, this group of resistors behaves just as this single resistor would here, as far as the battery is concerned it is the same exact thing, the battery cannot distinguish these three resistors from a single equivalent resistor, which is why they're called equivalence, same thing here, if you combine these three resistors into a single equivalent resistor, as far as the battery is concerned, the battery sees the same exact amount of resistance, okay? One last point I want to make is that, if you combine resistors in series like we did here, the equivalent resistance, we're always going to, is always going to be higher than the individual resistances, and this should makes sense since you're adding the numbers, right? 1 plus 2 plus 3, the total number is obviously going to be higher. Now, if you have parallel connection the number is always going to be lower and that's because in this case, remember I mentioned that the current splits. So, now because the electrons have an option of going one way or the other there's effectively less resistance because they have a choice, cool? So, let's do two examples here.

What is the equivalent resistance of the following resistors? What we want to do is get four resistors into a single resistor and there are, sometimes there are multiple ways of doing this, multiple paths that you can take to get to the answer, sometimes there's only one path and what you have to sort of do is map out how are you going to go from four to one and what you want to do is look for places where you can easily combine these resistors. So, for example, one thing you might do is, look at one and start comparing it and start seeing how it's connected to every single one of these. So, is one in series or in parallel with two? Well, remember, series means that they're they have a direct connection, is there a direct connection from one to two? So, if you're, if you're a charge you're going through here and then right here the wire splits, there is no direct connection. So, one and two are not in series with each other, okay? They're not in series, what about in parallel? Well, parallel means that they're alone and opposite sides of a branch, they're not even in the same loop, this is a loop, this is a loop, there are different loops. So, they're not in parallel either. So, you cannot combine one and two, right? Now, what you can do is you can combine one and four because 1 and 4 are alone on opposite sides, on opposite branches of the loop. So, you can combine these two in parallel. So, you can combine these two in parallel and you can also combine these two in parallel and if you do that you go from having two to one and you go from having two to one so now you have a simpler circuit. So, let's do that, let's show that here, I'm going to go and redraw this, where this entire thing is going to become a single resistor, let's call this single resistor R1 and over here, I'm going to have another resistor R2, okay? Now, how could I get these two guys to be just a single resistor? Well, these guys are in series with each other, hopefully you see that right away, because they're just sitting next to each other, there's no splits on the wire, right? One flows directly to the other. So, they are in series, which means, I'm going to be able to easily combine them into a single resistor, let's call that R3, so the idea that everything, that's inside of green here, becomes just this single resistor right here. So, what I'd like to do is I like to draw the paths without doing the math, get all the way to the end. So, I know my path and then we're now actually going to calculate, okay? So, the first thing I did is I merged these two into R1. So, let's calculate R1, R1 is in parallel, R1 is the parallel connection of one and four. So, to find R1, I'm going to have to write the parallel equation, which is this one here, cool? So, it's going to be 1 over R1, which is what I'm looking for, and then 1 over plus 1 over, I like to make room to plug into variables, just so it's a little bit more organized and then we're going to plug in the numbers one and four, one and four. Now this is a fraction. So, you have to get a common denominator. Remember, and to get a common denominator between one and four you can just multiply this side by 4 and if you do that in the bottom have to do at the top and then you can multiply this side by 1 which is effectively not doing anything and do this thing up here. Now, you have 4 you have 4 over 1, I'm sorry, 4 over 4 plus 1 over 4 and because the denominator is the same I can combine them into the 5 at the top adds, and the bottom combines 5 over 4, are we done, is that the answer? No, there's one last step which is, notice that the R1 is in the denominator it's in the bottom of the left side and I have to solve for R1, one way to do this that I like is, you can just flip these two but then if you flip that on the left side, you have to flip on the right side, as well, so I'm going to get 1 over R1 and then flip, and I'm gonna get the 5 over 4 and flip and it's going to become 4 over 5. So, R1 is 4 over 5 which is 0.8 ohms, cool? We're going to do the same thing for. So, we got this guy, boom, I'm going to do the same thing to find R2, one over R2 equals 1 over parentheses plus 1 over parentheses and this is the, this is the parallel connection between the 2 and the 2 over here, in blue. So, we're going to put a 2 here, and a 2 here, the denominator is already the same. So, I'm just going to merge and say 1 plus 1 is 2 the and then this is 1. Now, remember I have to flip this, but here because you have a 1, and remember there's always an implicit one here, if you flip one over 1 you still get a 1 long story short, R2 is simply 1 ohm, okay? We're almost done, I have R1, I have R2. Now, we're just combining them in series, whenever you add something, whenever you combine resistors in series you just add their resistances. So, finally, R3 is R1 plus R2, which is 1 plus 0.8, which is 1.8 ohms, cool? That's that, that's the final answer for this one, let's do one more and what you might want to do here is maybe pause the video and at least lay out the steps of how would you combine these guys, okay?

I'm going to keep rolling here, if you, if you, hopefully you saw right away that these are in series with each other, and then these are in series with each other as well, let's color code, you can't merge anything else yet, okay? So, this is going to give me, if you'll follow the wire here, carefully you're going to go up and then this entire red thing over here is going to be replaced by, let's call this guy R1. So, that's going to be R1 and I'm actually going to make this black. So, that's R1 and then down here, the two resistors in blue are going to become R2 and then once I have these two resistors, hopefully so this is series, this is in series, hopefully you see that these guys are in parallel because they're alone on opposite sides, on opposite branches of the loop, and I can go one step further and say that these two would just combine into R3, okay? Now, let's calculate R1, R2 and then R3, R1 is the series connection. So, it's just 1 plus 2, 1 plus 3 is 4. So, I can actually just put it over here, this is 4 ohms, this is 2 and 4 which is 6 ohms. So, to find R3 I just have to merge the 4 ohms with the 6 ohms, let's do it over here, r3, 1 over R 3 is 1 over the first resistor, 1 over the second resistor, in this case, we have a 4 and a 6. So, I have a 4 and a 6, the best way to get, the easiest way to get a common denominator here is just to multiply this 4 by 6 and then put a 6 on top and then this 6 by 4 and then put this on top as well, 6 times, 4 times 6 is 24 on both of them, and on top I have 6 plus 4 which is 10. So, I got 10 over 24, are we done, is that the answer? No, be careful, you got to flip the r3. So, I'm going to flip the left and right. So, I'm going to get R3 over 1 or just R3 equals the flip of these guys, 24 divided by 10 which is just 2.4 ohms and that's the end of this one, hopefully makes sense, hopefully you got it, hopefully you think it's easy, let's keep rolling.

Practice: What is the equivalent resistance of the following combination of resistors?

Concept #2: Shortcut Equations for Resistors Parallel

**Transcript**

Hey guys. So, in this video I'm going to show you some awesome shortcuts that you can use in some cases when you have resistors in parallel and this is going to keep you from having to do a lot of work with fractions, let's check it out. Alright, so if you have two resistors in parallel there's a shortcut equation you can use. Now, remember the general equation is that the equivalent resistance in parallel is 1 over R equivalence equals to one over R one plus 1 over R2, this is, if you have two resistors, this is the general equation, okay? And, I'm going to solve for R equivalence so that you never have to do this again and you can just use this shortcut if you have two resistors, which is the most common thing you're going to get, so the first thing if you remember we have to do is, get a common denominator, and to do that you're just going to multiply R2 here and then multiply R1 here. So, R1 times R2. So, if I do the bottom I also have to do on the top, R2, and then R2 times R1 times R1, let's extend the little fraction thing. So, what happens now is, I have R1, R2 at the bottom R1, R2 at the bottom. So, I have a common denominator. So, I can write R1 times R2 and then at the top, I have R2 plus R1 or I'm going to write it R1 plus R2. So, it's in order, are we done? No, remember, we have to flip the sides. So, if I flip here, I get R equivalence over 1 or simply R equivalent equals, and then if I flip the left I got to flip the right and this is going to look, I'm sorry, this was a 2 right there, hopefully you caught that, R1 times R2 over R1 plus R2. Now, notice I'm drawing the R1 and R2 in the bottom really far apart and the reason I'm doing this is because one of the biggest problems with this equation is that people forget whether the times is on top or the plus is on top, you don't know what goes where, you might forget. The way I remember, which is silly but works, is because this is a dot, it's a tiny dot, the R1, R2 are really close together and this is a fat plus sign, which takes up a lot more space, the bottom variables are farther apart and the skinny top and fat bottom gives you sort of a triangle, okay? Super silly, but maybe, hopefully it works for you, whatever works, right? So, that's the equation. So, whenever you have two resistors in parallel you can just use that equation instead of playing with fractions, okay? Now, super important is that you cannot do this for three or more resistors. So, I actually want you to draw, right? This here, this is wrong, but I want you to write it, let's say you might want to think that you can multiply R1, R2 why not just add an R3 here, and then at the bottom here, do a plus R3? Well, this is wrong, this does not work, I want you to write it, scratch it out, and say don't, okay? Do not do this, this only works for two resistors, it does not work for more than two, okay? So, let me give you a super quick example here, let's say you have a 4 and a 6 and you want to combine them into a single resistor, you would just use this equation here and say that the equivalent resistance is 4 times 6 divided by 4 plus 6, I'm setting up the parentheses so I can put the numbers inside 4, 6, 4, 6 and this is 24 divided by 10, 2.4, right? Much faster than play with fractions. So, that's the first shortcut, the most important thing always, is that you know the general equation because this is going to work for everything but this shortcut is pretty handy as well, it's more important shortcut number 2, the shortcut number 2 is super simple, if you have resistors of the same resistance you can also use a shortcut equation. So, let's say I have a, let's make it creepy 6, 6 and 6, cool? Let's say you have something like this, what's the equivalent resistance? Well, if you write the general equation. Remember, you cannot write this equation right here, the one we just talked about because that works only for 2, but if you have this you end up with something like this, 1 over 6 plus 1 over 6 plus 1 over 6 and the denominator is already the same so you end up with 3 over 6 or 6 over 3, which is 2, the fast way to have done this, is to just say that the equivalent resistance, when they're all the same in parallel is just the same resistance, which in this case is 6, divided by the number of resistors. So, you could have just done 6 divided by 3 is 2, for example, if you have 8, 8, 8 and 8, the equivalent resistance here is super easy to calculate, it's 8, and there are 4 of them, the equivalent resistance is just a 2, cool? So, now I'm going to do an example that sort of merges all these ideas, you now know how to easily combine, things are the same, and you know how to easily combine when they're the same resistance and you know how to easily combine if there's two of them, using that first equation. Now, you can actually use those two rules to your advantage. So, if you get a question like this, this might look hairy but it's actually really simple. So, we're going to do. Notice, we have a 9 a 9 and a 9 and even though they're not right next to each other you could technically rearrange them to be right next to each other and you can say, you know what, this 9 with this 9 and this 9, because they're the same I can write the equivalent resistance of this three 9, it's just 9 over 3, I'm using this shortcut right here, there are three 9. So, it's just 9 divided by 3 which is 3, okay? You can do the same thing for the 12s, there's two twelves. So, I can say the equivalent resistance of the 12s combined are going to be 12 divided by 2 just 6. So, what I can do is all the reds become a simple, a single 3, and the and the blue becomes a 6, and now if I combine these 2, because I have two resistors that are parallel to each other, right? Let's put the little connectors this way. So, there's two resistors here that are parallel to each other, I can use the first shortcut equation, which is, you multiply at the top then you add at the bottom. Remember, the pyramid, right? It's multiply and add, so this is going to be 3 times 6, 3 plus 6, 3 times 6 of course is 18, 3 divided by, 3 plus 6 is 9, so the answer here is 2 ohms. So, notice how we're able to combine everything really, really quickly 9, 9, 9, 3, 12, 12 is 6, putting together and you get a 2. Alright, that's it for this one, let's keep going.

Practice: What is the equivalent resistance of the following combination of resistors?

Example #1: Weird Arrangement (Re-Drawing Resistors)

**Transcript**

Hey guys. So, in this video I want to go over an example of a network of resistors that looks scarier than it really is. So, here we want to find the equivalent resistor of these guys, and it looks really hairy but I want to show you how you can make it look more familiar to solve this question, okay? And the key thing we're going to do here is, we're going to move the wires around so that they look more familiar. So, for example, the 5 and the 4 are sort of an angle which is unusual and it doesn't, it makes it harder to sort of visualize what's really going on here. So, the first thing I'm going to do is, I'm going to try to make this for vertical and I can actually just move this point here, where the three of them, three wires touch, I can move this point right here, and then imagine if you have a 4 and then you're grabbing the bottom, right? You're grabbing the bottom and just doing this, okay? So, if I do that, let's redraw, I'm going to get a 2 at the top, a 3 here and then this red point right here is now going to be right here. So, that I get a 4 like this and then there's still a 5 here, let's leave that alone, will take this slowly, and now this hopefully looks more familiar, the other thing we can do is. Notice that this is a single branch right here, that's all on this green line, there are no points where the wire splits. Now, obviously the wire splits here, on the green dots, splits on the red dot but within those two dots nothing splits, what that means is that you can actually move the 3 and just redraw the 3 over here and is exactly the same thing, functions just the same. So, now we get something a little cleaner and I'm going to draw, again, I'm going really slowly because I want to make sure you fully understand this, okay? And now it looks like this, and hopefully, now, this looks super familiar, hopefully you'd see that this is a branch with a single resistor in it, and this is a branch here, also with the single resistor on it, or in it, and because you have two resistors on opposite sides, they're alone on opposite sides of this loop here, they are in parallel. So, these two guys are in parallel. So, I can combine them, okay? Usually I would keep going but just for the sake of just getting this over with, let's just combine these two real quick and because I have two resistors in parallel I can use the shortcut equation, the equivalent resistance is going to be. Remember, the pyramid, it's times on top and plus on the bottom, okay? So, it's going to be 4 times 3 divided by 4 plus 3, this is 12 divided by 7 which is 11.7, okay? So, this entire thing can be redrawn as a this whole thing here, actually you can think of it as all of this, right? Can be redrawn, there's a 2, there's a 5 and instead of having a 4, and then 3, I'm just going to have a single 1.7 right here, and now this is a little simpler because I have 3 instead of 2. Now, we're about this other 5 here, I fixed the four made it straight, let's make the 5 straight, and the 5 is like this. So, I'm going to move the top a little bit and then I'm going to move the bottom a little bit. So, really slowly here, the 5 will be moved over here and this part of the 5 is going to be moved over here. So, that it forms sort of a straight line, and this is going to look like this, got a 2 then you got a 5 straight down and then there's still the 7 over here. Now, the 7 is sort of like this, right? If you curve the, the 7 is kind of like straight like this, the 1.7 rather, but I can extend this wire and make it look like this, okay? All these are equivalent, you just have to be careful to redraw them correctly, otherwise you made up a new circuit and obviously it's going to be wrong, okay? Again here, hopefully you see that you have a branch with a single resistor and then here you have a branch with a single resistor and because you have two resistors there alone on their branches on their opposite sides, they are also in parallel, and once again I can use the parallel shortcut equation, because I have two resistors, the equivalent resistance is going to be 5 times 1.7 divided by 5 plus 1.7, okay? And, if you do this in a calculator, you get 1.3, 1.3, I want to quickly remind you of something that we talked about earlier, which is, whenever you combine things in parallel the total resistance is smaller or lower than all of the resistances. So, it's going to be lower than the 5 and it's getting lower than the 1.7. Notice that when I merge them I got a 1.3 lower than 1.7, it makes sense, you can use that to validate that you're probably correct. So, finally here, I can draw the 2 and in this entire thing here gets replaced so instead of being two of them there's maybe just a single 1.3 and these are the points that we have there. Now, notice that between these two guys there's a direct connection between them, there's no forks between them, which means that they are in series, and finally series resistors are just added. So, 2 plus 1.3 is 3.3, so the equivalent resistance of that big old mess is just 3.3 ohms. So, please do know that you can take the Liberty to move some wires around to make things little bit more familiar to you, different people see things a little bit differently and they prefer their resistors to be organized differently, in the book might throw you, book or professor might throw you some questions that are drawn in a weird way to see, if you really know what's going on, take some liberty to draw them, but please make sure that you do it correctly, cool? That's it for this one, let's keep going.

Practice: If every resistor below has resistance R, what is the equivalent resistance of the combination, in terms of R?

0 of 6 completed

Concept #1: Combining Resistors in Series & Parallel

Practice #1: Find Equivalent Resistance #1

Concept #2: Shortcut Equations for Resistors Parallel

Practice #2: Find Equivalent Resistance #2

Example #1: Weird Arrangement (Re-Drawing Resistors)

Practice #3: Equivalent Resistance with Variables (Literal S...

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5. RXY = 21 Ω
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In the circuit shown in the figure, all the resistors are rated at a maximum power of 1.20 W.What is the maximum emf that the battery can have without burning up any of the resistors?

When a bulb is added in parallel to a circuit with a single bulb, the resistance of the circuit: A. remains the same B. decreases C. increases D. triples E. doubles

Consider the network of four resistors shown in the diagram, where R1 = 2.00? , R2 = 5.00? , R3 = 1.00? , and R4 = 7.00? . The resistors are connected to a constant voltage of magnitude V. (Figure 1)A. Find the equivalent resistance RA of the resistor network.B. Two resistors of resistance R5 = 3.00? and R6 = 3.00? are added to the network, and an additional resistor of resistance R7 = 3.00? is connected by a switch, as shown in the diagram..(Figure 2) Find the equivalent resistance RB of the new resistor network when the switch is open.C. Find the equivalent resistance RC of the resistor network described in Part B when the switch is closed.

Use the model for electric current to rank the networks shown below in order according to resistance. Explain your reasoning.

1. Find the current through resistor a) in the figure. (Figure 1) Express your answer to two significant figures and include the appropriate units.2. Find the potential difference across resistor a) in the figure.3. Find the current through resistor b) in the figure.4. Find the potential difference across resistor b) in the figure.5. Find the current through resistor c) in the figure.6. Find the potential difference across resistor c) in the figure.7. Find the current through resistor d) in the figure.8. Find the potential difference across resistor d) in the figure 2.

Consider the circuit shown in (Figure 1) . Suppose that ε = 25V. What is the current through the battery when the switch is closed?

In the circuit at right, what is the potential at each of the noted points?

For the circuit shown in the figure, find the current through resistor R3 =6.0 Ω (the top one) and the potential difference across resistor R3=6.0 Ω

Consider the circuit shown in (Figure 1) . Suppose that ε = 25V. What is the current through the battery when the switch is open?

Three identical resistors have an equivalent resistance of 30 Ω when connected in parallel. What is their equivalent resistance when connected in series?

Find the equivalent resistance between points A and B for the group of resistors shown in the figure(Figure 1). Suppose R1 = 78 Ω , R2 = 57 Ω , and R3 = 55 Ω .

Which one of the following is a correct statement for a number of resistors connected in series or parallel?a) The flow of current is different through resistors connected in a series circuit.b) The total resistance in a parallel circuit decreases as more resistors are added.c) The total resistance in a series circuit decreases as more resistors are added.d) The voltage is different across resistors connected in a parallel circuit.e) None of the above statements is correct.

Consider the circuit shown in (Figure 1). Suppose that R = 70 Ω. What is the equivalent resistance between points a and b?

Two resistors in SeriesConsider the setup shown of two ideal ohmic resistors. (Figure 1).Resistor 1 has resistance R1, and resistor 2 has resistance R2. They are connected in series with a constant voltage of magnitude V. When the two resistors are connected in this way, they form a system equivalent to a single resistor of resistance R, as shown in the next diagram. (Figure 2).Part AWhat is the effective resistance R of the two-resistor system?Express the effective resistance in terms of R1 and R2.

What is the net resistance of the circuit connected to the battery in (Figure 1)?Express your answer in terms of R.

Suppose that R = 70 Ω.What is the equivalent resistance between the points a and b in Figure 1? Express your answer with the appropriate units.Rab = ____

What is the equivalent resistance of each group of resistors shown in the figure below? In the figure, R1 = 2.0 ?, R2 = 3.5?, R3 = 7.5 ?, and R4 = 2.0 ?.(a) ______ ?(b) ______ ?

For the set-up shown, find the equivalent resistance between points A and B.

(a) Determine the equivalent resistance of the circuit shown in the figure. Express your answer to three significant figures and include the appropriate units.(b) Determine the voltage across 820 Ω resistor. Express your answer to two significant figures and include the appropriate units.(c) Determine the voltage across 960 Ω resistor. Express your answer to two significant figures and include the appropriate units.(d) Determine the voltage across 680 Ω resistor. Express your answer to two significant figures and include the appropriate units.￼

For the set-up shown, find the equivalent resistance between points A and B.

What is the value of resistor R in the figure below, in which I = 8 A and ΔV = 34 V?

What is the equivalent resistance of group (a) of resistors shown in the figure?What is the equivalent resistance of group (b) of resistors shown in the figure?Express your answers using two significant figures.

What is the equivalent resistance between points a and b in the figure? (Figure 1) Please show work.

(Figure 1) shows three identical bulbs in a circuit. What happens to the brightness of bulb A if you replace bulb B with a short circuit?A. Bulb A's brightness does not change.B. Bulb A goes out.C. Bulb A gets dimmer.D. Bulb A gets brighter.

Find the equivalent resistance of the combination of resistors shown in the figure.

What is the value of resistor R in the figure below?A) 4.0 ΩB) 12 Ω C) 36 ΩD) 72 ΩE) 96 Ω

In the figure, these three resistors are connected to a voltage source so that R2 = 5.75 Ω and R3 = 11.5 Ω are in parallel with one another and that combination is in series with R1 = 1.75 Ω. (a) Calculate the power being dissipated by the third resistor P3, in watts.(b) Find the total power supplied by the source, in watts.

Both batteries in the figure are identical and all light bulbs are the same. Rank in order, from brightest to least bright, the brightness of bulbs a to c.

Find the equivalent resistance between points A and B in the drawing (R1 = 4.00, R2 = 8.00).

It's possible to estimate the percentage of fat in the body by measuring the resistance of the upper leg rather than the upper arm; the calculation is similar. A person's leg measures 40 cm between the knee and the hip, with an average leg diameter (ignoring bone and other poorly conducting tissue) of 12 cm. A potential difference of 0.78V causes a current of 1.6 mA. What are the fractions of muscle and fat in the leg? Express your answer using two significant figures separated by a comma.

You have a collection of six 2.1 kΩ resistors. What is the smallest resistance you can make by combining them? Express your answer with the appropriate units.

You have a collection of six 1.0 kΩ resistors.What is the smallest resistance you can make by combining them?

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