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You are given bulbs (2 type having resistance of 50Q2 and 2 type B having resistance of 100Q). You need to connect them in such way that one type B bulb will be the...

Question

You are given bulbs (2 type having resistance of 50Q2 and 2 type B having resistance of 100Q). You need to connect them in such way that one type B bulb will be the dimmest and one type A bulb will be the brightest: The other bulbs (one type A and one type B) will have intermediate brightness, i.e_ more than the dimmest but less than the brightest: Draw the circuit diagram to show the connections

You are given bulbs (2 type having resistance of 50Q2 and 2 type B having resistance of 100Q). You need to connect them in such way that one type B bulb will be the dimmest and one type A bulb will be the brightest: The other bulbs (one type A and one type B) will have intermediate brightness, i.e_ more than the dimmest but less than the brightest: Draw the circuit diagram to show the connections



Answers

Two $\quad$ lightbulbs, one of resistance $R_{1}$ and the other of resistance $R_{2}$, are connected to a battery (a) in parallel and (b) in series. Which bulb is brighter in each case if $R_{1}=R_{2}$ ? How is your answer different if $R_{1}>R_{2} ?$

So we'll be using party with risk over our formula so we can see that the resistance off the 60 year old lightbulbs is this could over our where the voting is under 20 village is 60 worldwide balls three times out to be 2 40 home similar. The resistance of the 100 armed light ball comes out to 340. Or now if we take the ratio of the powers be won over P do when they're connected in cities that current flowing through, a lot of them are the same. So that begins our brother, where art ill, which is 2 40 over 1 40 equals well over seven. So power dissipated on Bob ones is 12/7 times power dissipated in about two. Now in parallel continuation party with respect over are so because in palace configuration will trajectories boot up tonight but at the same power is proportional to one over R. So the lower resistance bottle will go fighter

In this problem, we have given power of first ball piven is equal to 60 watt. Power of second ball pit is equal to 100 word and B is equal to World at the socket. 1 20. What we all know that physical to be square by Earth. Or simply I can write R. Is equal to P square by P. So here the resistance for possible as 60 is equal to 1 20 B square by 60 watt, which is equal to 2 40 mps. Um Also I can like our 100 is equal to 1 20 P. Holy square by 100 watt, which is equal to 1 40. Um Now I will go forward and solve it further. So solving part B. Since we all know that P is equal to I square. Uh and I can and I can also write the larger resistance dissipates more power and sign. So it will be brighter. So I can write The 60 watt bulb, right, will be signer or you can right right there It is because 240 Oh means greater than 1 40 homes. Now, for part C, I can ride the ball with the bulb with is smaller. Resistance is smaller, message tense, mm. Dc pates the most the move five would. Therefore, I can write 100 watt bulb Science brighter, so this is the answer for part C.

Ask for the given dude. Oh, here The Cubans are identical. Having the same distance are without put shortages. Delta Bi. Now let's start with step, eh? And to find the answer with respect to a swan. Now, in this case, double, they're having the same resistance, so we'll just write it down. Says, do Bubs are having same a distance, and they are connected right here in series. So argument, Lee, the same current flows so right down the same current froze through them. So the potential difference what he means the same it grows, but yeah, and B also we can see that to bugs are equal in brightness because of that. So at the same time, we can add something that is more or power distribution is given as, say, participation is equal to I screwed up so therefore couldn't and registers. Now these two bad amigos in case one are seen, Dad, for this power dissipation, it means the same This we can conclude that both bugs have seem current, huh, Bagger and even equal brightness in case one. This should be though answer now that's moving the next beach and try to find and super step be here. We are going to consider case too. Where bulbs are connecting batter. So right first here to Bob's are in battle in now would add more their resistance and potential difference. All right, seem we write more here. So these two birds, these bulbs d'oh equally? No, Since it grows the true Bubbs, it's same current two. The boats remains the same since they have same traditions. Now let's talk about the power distribution. Also, power dissipation across the bulbs remains the scene now destination that power dissipation is equal to I square multiplied by our should us. Finally, we can see that both bulbs both the bulbs I have same couldn't power and even equal brightness in case too. This is the answer. Now let's move to step three and write down and surf or step See, that's captured governed in each case now. So here, if you consider the case one and we're right against you, Cigar and I s is equal to Delta. We upon two games are because bulbs are connecting cities and here began mation. It does say I'd be one is equal to I P too. Is Delta v upon our so we can see that I p one is this current and I bread was this current since warded across both of them remains the same and they're having the same the distance. We can say that simply using home slow but Delta bi upon our in case too. Now let's write down something to explain So and this would be very important we can mation God didn't flows through each of this show. Specifically in case too is devil do that off in case one we need to understand this because in case to a distance is half of the geese one so current will be doubled. Nix signboard barber supplied is equal to is basically I square market power. Better distance. So what we can say that does increase do power supplied to each bug is four times to the power to each bug in case one. Now here the boat having more power definitely has more brightness, So no, but in case too. Okay, so we're writing the bulge in case too with the deal brighter then that off in case one, this is what weekend Right now let's go to page for to write down the answer first MD Now here we get nation directly about case one encased. You sold her first, right? About case one. So, in case one, since the bugs are connected and see the sword right down says the bulbs here, uh, being series, the same gun flows through them. So if any, we're right here off the bugs. So if any of the bulbs is field, God didn't cannot feel two The part because our fridge that either by I could not glow. This is what you need to understand. Since the same current flow straight. This could be the answer for the case. One. We're right on the next page for gays to soo in case, too, which is a battery circuit. We're right here, since the bulbs are connected in battle in if the pot through which couldn't flows. Two. The but A is broken then but a they're not go, but but be. But glow continuously says it's spot Couldn't it means the scene that's a legitimately God didn't two but be remains the same and pretty close with same brightness. Even if the bulb is not working

Hi the given problem E M. F of the battery. He's given us a silent and resistance of each bowl, the three bulbs each having a resistance of our. Now in the first part of the problem, The three Bulbs are joined in serious combination. So in serious combination, net resistance R. S will be given by R plus R plus R means this is three times of art. So power from the battery power deliver from the battery will be given by PS in sequel to E square museum if you square divided by net resistance. So this becomes the answer for the first part of the problem. Now, in the second part of the problem, in parallel combination, Net resistance here will be given by our by three. So now the power delivered by the battery will be given by again, E M F square divided by net resistance, which is our by three. So here, this time it will become three times of absalon is square divided by our So this is the answer for the second part of the problem. And in the 3rd part of the problem, we have to find which combination will be giving the brightest bulb glowing. So it is clear that the burl will be glowing brightest in parallel combination, as the power given by the battery is maximum in parallel combination. Thank you.


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