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Consider the two circuits shown below. All bulbs and batteries are identical.EWith the switch in the open position as shown; how is the resistance of circuit #1 wit...

Question

Consider the two circuits shown below. All bulbs and batteries are identical.EWith the switch in the open position as shown; how is the resistance of circuit #1 with respect to the resistance of circuit #2? Explain your answer.b. If the switch is closed, what will happen to the brightness of bulb A. Explain your answer_What will happen to the brightness of bulb B and C when the switch is closed? Explain your answer

Consider the two circuits shown below. All bulbs and batteries are identical. E With the switch in the open position as shown; how is the resistance of circuit #1 with respect to the resistance of circuit #2? Explain your answer. b. If the switch is closed, what will happen to the brightness of bulb A. Explain your answer_ What will happen to the brightness of bulb B and C when the switch is closed? Explain your answer



Answers

A circuit consists of three identical lamps, each of resistance $R,$ connected to a battery as in Figure P 18.53. (a) Calculate an expression for the equivalent resistance of the circuit when the switch is open. Repeat the calculation when the switch is closed. (b) Write an expression for the power supplied by the battery when the switch is open. Repeat the calculation when the switch is closed. (c) Using the results already obtained, explain what happens to the brightness of the lamps when the switch is closed.

Okay, So for this problem, we're being asked to find an expression for the currents. Each of the circuits first circuit looks like this. We have a battery. And then instead of drawing the my boat, we're gonna represent it with a resistor, and then here closes with the battery. So this is the positive end of the battery. Here's a negative end of a battery. And then here's the Here's a light bulb. It's We're gonna call it our Okay, So the the current running through this circuit, we're gonna use homes law for that. And homes, that tells us that B is equal to IR. Now, in this case, if you go around the loop, we're gonna start here as your potential. We go through the loop here, so we gain some potential, and then this is that this is the lecture potential off this battery, and then we're gonna go ahead and lose the potential across the resistor resistor. There's gonna b minus ir, and this is gonna be equal to zero. So here we solve for the current, which is gonna be I or is gonna be equal to the potential across the battery. So then we divide both sides by our and then this is gonna be our current through lightbulb A. We're gonna label that with a subscript Sunday here, and this is gonna be equal to the absolute over our Okay, so this is the current running through this first circuit. Okay, so let's move all this to the side here and now let's work on the second circuit. Okay? So for the second circuit, we're gonna have to resistors in Siris. So we have a battery. As before, we're gonna have one lightbulb represented by this resistor there. We're gonna have the other my bold represented by this resistor. Okay. And the resistor of each light bulb is also our So register one is gonna be our or this This would be be disturb e. And this would be resistor sea or light bulb. See, which is also represented by this are here. So electric potential here along the battery is also this absolute. And here it's positive. And then here it's negative. So the currents gonna flow clockwise in this direction. It's gonna feel clockwise. Okay, So what we're gonna do now is again We'll start the zero point we gain some potential. So here we are. We gain some potential, and then we lose potential along this first. Well, these potential on the first light bulb, and then we're gonna lose potential. Well, okay, so we're gonna lose potential along the second life books. So how do we represent that? Whoa. It's gonna be a drop of potential. So minus ir and then minus i argha And this is gonna come out to be zero. By the time we drop potential twice, we're back at zero. It's gonna be zero. And then again, we're gonna solve four I. So we combine like terms. It was gonna be Absalon minus two ir. It's gonna be equal to zero. And then we move this to the right hand side. So absolute is gonna be equal to to ir, and then we saw for I. So this is the current that flows through resistor or libel B and light bulb. See? So we're gonna call that current. I said be this is exactly the same current that's gonna flow through light bulb. See, And it's gonna be Absalon over tour. Okay, So here noticed that the current running through lightbulbs B and C is half up off light bulb. So which was gonna be brighter? Well, this one here light bulb is gonna be brighter because it has the higher current. So high current means more that trans air flowing through this light bulb, and it's going to cause it to shine brighter. Here we have the current flowing through this combination of resistors gonna be half the current look that that float through a So the light bulbs are gonna be less bright. B is gonna be equally break to see. But each of these each one individually is gonna be less bright than a Okay, So if we want to figure out how, how much less bright it's gonna be, we're gonna have to bring in power. Okay, so let's go to the second board here for the power decimated through the through the light bulbs. We're gonna use the phone formula. We have power is gonna be equal. So I ve and we're gonna recall from homes law that be is equal to IR. And then we could go ahead and plug this into this expression. So as a substitute for V, this is gonna be equal to high squared or okay. So if we want to know how bright the light bulb A is gonna shine we saw for the power dissipated through life obey. So it's gonna be the current through like Kobe squared $10. So remember, from the previous white board, the current through Lipobay is absolute. Over are So let's go ahead and put that in here and the Absalon over or and this quantity is gonna be squared and then multiplied by our. So this is gonna be equal to Absalon. Squared times are over our because this denominator is squared and then one of them gets canceled out from this numerator here. Okay, so this is the power going through light bulb. Now, let's do like will be the power running through that. I noticed that the power, anything I will be is also gonna be equal to that powering through help. See, there you go, because they have the same current. And here power depends on the currents. So here we have the power running through Robo bee is gonna be its current squared times, the resistance. So if we go back to the previous board, we see that the current through libel B is Absalon over to our. So then we put that in here, absolute over to our and then we square this quantity, and then we're gonna multiply it by our. So when we square this notice that this is gonna be equal to excellent squared over four. So the power running through libel B and C, right? It's 1/4 of that running through a So that means light bulb A shines four times as bright as light bulbs would be and see. Okay, so if you wanted to write this at the end, we would put the power running through light will be is gonna be equal to the power running through light bulb. See? And that power is gonna be 1/4 of that running through light bulb A. It's gonna be 1/4 of the power running through. So they're gonna be Ah, fourth. They're gonna have the fourth of the brightness of Libel bay. Okay, this concludes those three questions about the two circuits

In this problem, we have two different scenarios. We have this interview here on the left, which is able to connected to a single light bulb, and on the right we have a voltage attracted to light bulbs. All the light bulbs are identical in terms of the resistance, and the voltages is sort of perfect. There's no internal resistance, and they're both equivalent. So the first question asks us to find expressions for the current in each of these cases. So let's look at the first case here in blue, the voltage using homes LA we have voltage is equal to current times resistance. So if I want to find the current, that's going to be equal to the voltage divided by the resistance in this case, my voltages e and my resistance is just are over here on the red case we have I is equal to V over R. However, our voltage is still E but are are in this case is to our because there's two resistors right here on. So that is our That's our expression for the to, um, currents. To make it a little bit more clear, we label them I A so the current to the 1st 1 is the MF, divided by r and I b is equal to i c on those air equal to the voltage divided by to our so part B says, How does the brightness of be compared to see Well, um, the brightness, um of be and see are equal. And the reason why is because they're both receiving the same amount of current Um however, in part C, it asks to compare the brightness of being see to the brightness of a well, if we look at these two currents right here, this one and this one, which one of these is larger? Well, it's clearly this one, Um, because there's a two here, so that's whatever current this is. This is the same divided by two, so it's always gonna be half a much so that means there's more current flow through a. That means there's more brightness, so the brightness of a is going to be greater than the brightness of B, which is equal to the brightness of C

This question covered the concept of home slaw. So for the first circuit, when the switch is open, The current passing through R. one and R. two. Let's name it as Ivan. Okay. and for the second case the current passing through R. one and R. two. Let's name it as let's throw a loop for each case. So from one floor below four World- Ivan into our one that is to home -11 into two. Home That must be close to zero. Or from this we get Ivan is equals to one M. Pierre. For the second case from home soup law for world minus, I turned to minus I do into that most because 20 And from this we get I do is equal to one mp here. So in the in both of the cases the current passing through the resistance R one is one ampere. Therefore the brightness of our one will be same in both cases. And the correct option is option E.

Send this problem. We have a circuit where each resistor represents a resistor for a light bulb and these light bulbs, their intensities. They are governed by the equation for power which is given adds current times the voltage. So as you increase current the increasing power and therefore increase intensity. In these light bulbs. I noticed two things that they are parallel to each other so each resister or each mike bulb will have the same voltage V. From the battery. And also notice that there's a switch at the top when this switch is open. Like it is now current will not flow through the last resistor. So that means no current is flowing through that light bulb so there's no power. So it's turned off basically on its closed. Then you'll have currents flowing through it so then the intensity will increase. So we'll look at today when the switch is closed. What happens to the intensity of the master's sister or light bulb? 3? So again when it's open, no current is flowing through it. So it has zero current. You need zero power meaning zero intensity. So closing the current allows are closing the switch allows the current to flow Through that last light bulb. So that means it will have a non-0 current. So its intensity will increase. Mhm. And now looking at part B. What happens to the intensity for It was just the one light bulb one and Michael to our two. So going back to our circuit, we see that R. One and R. Two are parallel so both of them will have the same voltage V. So if we use only slot for each resistor east light bulb we have voltage equals current times resistance. Whenever solving for the current we have current equals voltage divided by the resistance because they are parallel. They will both have the same voltage. So that means when the switch is closed. R. one and R. three. Yeah we'll have the same voltage across them. And if they have the same resist or right then that means they're current. Who remained the same for both of them. All right. So that means our one the current stay the same. So that means power stays the same and the intensity of light bulb say the same and the same goes for R. Three or light bulb. Three. Yeah. Mhm. Okay. Yeah.


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