 Anyways, I'm going to project all the questions. So without wasting much of time, let us start solving questions. So you can see in your screen one of previous AJ questions. Please try solving this. So the hint is that you can assume it to be like two capacitors in series. The capacitor is having a different dielectric constant. So like that you can try it out. Anybody got the answer? No one. Should I solve it or should I wait for you guys? Okay. The hint is after time t, the thickness of the oil will become d by 3 minus v into t. Right? Because with velocity v, the level is decreasing. Okay, people have started telling the answer. Yes, it is option A. See, you can treat as if it is two capacitors in series. This is C1 and that is you can say that to be C2. Okay. C1 is equal to epsilon naught A divided by what? D minus x. Fine. Where x is this? All right? And C2 will be equal to K times epsilon naught A divided by x. Fine. Now, these two capacitors are connected in series. So the equivalent capacitance you have to calculate like this. 1 by C equivalent is 1 by C1 plus 1 by C2. All right? If you substitute these values, you will get option A to be correct. Fine? So like this, you have to solve the first question. Any doubts with respect to this question? Quickly tell me. Meanwhile, I'll move to the next question. Same thing. No doubts, right? Okay. Try doing this question. This came in 2011. I think these are standard questions. So if you have good practice, you should be able to solve it. All right. So all of, like most of you are getting option D and that is the correct answer. So I'll quickly talk about how to go about this question. The two microfarad capacitor is charged as shown in figure. Okay. So this two microfarad capacitor, this capacitor is connected. Okay. This one is connected to battery of, okay, battery is potential V only. Now we need to find percentage of energy stored after the switch S is turned to position 2. Now, first of all, it will store some energy and it will store some charge also. How much charge it will store? It will store a charge of 2 into 10 to the power of minus 6 times V. V is a potential difference. Okay. Now, when it is turned to switch to, okay, when you turn it to switch to the energy need not be conserved. Okay. Some energy may get lost also, but charge will be definitely conserved. Okay. So charge into microfarad should be equal charge into microfarad plus charge in 8 microfarad. Okay. This should be equal to total charge which is 2 into 10 to the power of minus 6 V. Fine. And also when you connect to microfarad parallel to 8 microfarad, their potentials are same. Okay. So since the potential is also same, I can say charge in 2 microfarad. Okay. Divided by the capacitance, this is potential across 2 microfarad. This should be equal to potential across 8 microfarad. So this divided by 8 into 10 raise to the power minus 6. Okay. Now, if you are not very comfortable writing like this, you can just write Q1 and Q2. Q1 for 2 microfarad, Q2 for 8 microfarad. Right. So like that, you will get charges in these two capacitors and once you get a charge in the capacitor for a given capacitance, you know that potential energy inside the capacitance will be equal to Q square by 2C. Okay. Like that, you can find out the total energy of 2 microfarad plus 8 microfarad. Okay. And that is energy afterwards. Energy initially was half CV square or you can just take this charge, which was initial charge, square divided by 2 microfarad divided by 2 also. So this is initial charge. So initial energy is Q square by 2 into, so basically I am doing Q square by 2C. Okay. This is initial energy and afterwards, once you get charge into microfarad and 8 microfarad, you get final energy as U1 plus U2. Okay. Now, just subtract final energy from the initial energy, you will get option D to be correct. Fine. Any doubt with respect to this question? Here is another question. There is a small correction. This line doesn't exist. Try solving this. One of the hints is that 2 microfarad and 3 microfarad, they are connected parallel to each other. So their potential will be same. Okay. We have Kushal and Tarun giving C as the answer. Tapas. Right. C is the correct answer. Okay. So basically, if you look at this circuit, you have this capacitor, the situation is like this. Okay. So, there is an isolated system. You know, what I mean here is that if you take these three plates, the next charge initially on these three plates, where is zero? Okay. So, whatever was the charge initially has to be charged finally. So, if you give 80 microcoulomb charge on the upper plate of this, then automatically the lower plate will accumulate minus 80 microcoulomb. Okay. This comes from Gauss law. Okay. Now, if this is 80 microcoulomb and let's say this value is Q1 and that is Q2. All right. So, this charge should be minus Q2 and that charge should be equal to minus Q1. Fine. So, according to conservation of charge, Q1 plus Q2 minus 80 microcoulomb, 80 into 10 minus power minus 6. When you add up all of this, this should be equal to zero because initially the charge, the sum of the charge on these three plates were zero. Okay. And also, this capacitor and this one, they both are in parallel. All right. So, Q1 divided by 2 microfarad. So, there is a 10 to the power minus 6 factor, but that will come left-hand side and right-hand side also. So, I don't need to write and then cut. So, Q2 divided by 3. Fine. So, like this, you will get Q1 and Q2. Fine. And what we need to answer here, the charge in the upper plate of 3 microfarad. So, like this, the value Q2 is the answer. Right. So, just write Q1 in terms of Q2 over here and substitute in this expression. You'll get the answer. Okay. And you'll get option C to be correct over here. In case of any doubts, please type in. Okay. All of you are getting A as the answer and that's correct over here. Right. So, see, this is the question where you have to evaluate the options. Fine. You cannot find a unique answer to this question. All you have to do is just take one by one and see which one of them will give you equivalent capacitance of 10 by 11 microfarad. Okay. For example, the arrangement in A has how many capacitance in parallel? They have 5 capacitance in parallel. So, it will be like 5 into 2 microfarad. This is in series with the 2 capacitors that are in series. The equivalent of that will be C by 2. Okay. So, this is 1 microfarad. Okay. So, it's like 10 microfarad connected series with 1 microfarad. So, the equivalent capacitance which is C1 C2 divided by C1 plus C2 when things are connected in series. So, it will be 10 into 1 divided by 10 plus 1 microfarad. Right. So, this will give you 10 by 11 microfarad. Fine. So, if the single option is correct, I will not even check for the others. I will mark option A and move ahead. Okay. Usually, when you have to evaluate the options, option A is not correct. It has to be either C or D. Like that it happens. But then, you know, at times option A may also be correct. All right. So, we'll move to next question. This one. It's a huge question. I mean, you have to read a lot. So, I would have left it initially in the first go while I'm taking the exam and come back to it later on. But then, since it is problem practice, you can solve it now itself. And more than one options may be correct. Okay. D, if there are more than one options correct, okay. All right. So, let us try to solve this question. I think you are getting different, different answers. All of you. So, a parallel plate capacitor, plate area A and separation is D, charged with potential difference B. Okay. Then, battery is disconnected and afterwards, a slab of directly constant K is inserted between the plate. Okay. So, in this process, what will be constant when this happens, when you disconnect from the battery and then put a dielectric inside, what is constant in this process is potential difference constant, charges constant, or something else. What is constant? Charges constant. Okay. Okay. See. So, charge is constant over here. Guys, see these questions are not something which you might not have already seen. And it's very easy to find out all these questions. There are past year J questions that are coming from different, you know, authors. So, do not look at it. Okay. Do not refer it. Otherwise, you are harming yourself and nobody will get affected, but you only. Fine. So, whenever there's a problem practice session, make sure that you do not look at these questions beforehand. Okay. Fine. So, charge is constant between these two processes. So, charge initially, what was what when it was getting charged with potential difference V, charge was C naught times V. C naught is what? Y naught A divided by D. Okay. This was C into V is the charge. Okay. So, this much will be the charge initially as well as finally. Okay. So, B cannot be correct. Okay. So, if you say B is correct, then since K is more than one, you're saying charge is created. Okay. Charge is constant. So, A is correct. Fine. Then C and D, let's see how we can evaluate. Okay. Now, when you put a dielectric constant K, the capacitance becomes K times the initial capacitance. Right. And charge is this charges C naught times V. Okay. So, potential difference between the plate, which is defined as charge divided by capacitance. This you can write like this charge, which remains C naught V divided by new capacitance which is K times C naught. Okay. So, the new potential difference is V naught by K between the plates. Okay. And since electric field is uniform, we know that between the plates, electric field will be V divided by D only, distance between the plates. Okay. So, hence, the electric field is V naught divided by KD. Okay. So, option C is correct. All right. Now, let's say, let's see whether D is correct or not. Now, in order to find the work done, we will use that age old formula that we are using again and again. Work done is U2 plus K2 minus U1 plus K1. Okay. Now, there is no need of kinetic energy over here. Your work done is simply change in potential energy. Okay. Now, how will you find initial potential energy? See, initial potential energy you can find in different ways. You know the potential difference between the plates. You know the charge of the capacitor. So, any formula you can use, so I am using half C V square over here, right. C is epsilon naught A by D, this into V square. Okay. This is initial energy, U1. Okay. And how will you get the final energy? U2, final energy, final energy, also I can use half C V square because now I know the potential difference between the plate is V naught by K. Okay. Now, I can use Q square by 2C, whichever you may feel comfortable with. So, I can use half C V square, so I will use that half, now C is half, C is actually K times C naught. So, this into V naught by K whole square. Fine. So, just substitute the value of C naught, which is epsilon naught A by D and do U2 minus U1, you will get option D also to be correct. Fine. So, the answer is A, C and D. Okay. Any doubts with respect to this question, please type in, no doubts. Some issue with my tablet, so my handwriting is not as good as it used to be. I hope you are able to, you know, make sense of whatever is written on this, this. Here is this question, try solving this. And charge B lost, no Tarun, in order to lose the charge, see, there is no circuit over here. It's a simple one capacitor where you are putting the dielectric, okay. So, there is no question of losing charge, if charge is losing, then charge should flow, there is no circuit. The question is not clearly seen. Okay. So, I will, actually I wanted to hide the options, let me do it, now solve it. No Tarun, it can be lost to the surroundings, if surround, maybe we are treating surrounding to be non-conductor, charge cannot just flow in the air, unless of course, lightning is happening. Solve this question, all of you. Tarun, after the class, you can see there is a number in the YouTube channel. Just message me Tarun, I will add you to our J-Advance group, okay, there we do many other activities. Anyone able to solve this? You need to first draw the circuit, circuit representing this scenario, okay, then only you can solve this, 7 by 3, others 2 by 3, okay, the answer is 7 by 3. I think all of you are making silly errors, okay, let me, I have to put it here, okay, see here you can think this entire thing like this, you have 2 capacitors in series, which represents this one and that one, and what you are doing, you are dividing it into 2 half, if you see this line, this line, I am taking the just upper part of it, alright, so these 2 represents the upper part of left hand side and right hand side, and then we have one capacitor that represents the thing which is bottom, okay, so basically the equivalent of these 3, let's say this is C1, C2 and C3 will give you the final capacitance, don't confuse these C1, C2 with what is written over here, okay, so C2 is actually the final capacitance of all these 3, C1, you can write C1 as, see dielectric constant of this material is 2, this one, and for that material it is 4, so C2, sorry, C1 is K times which is 2 times epsilon naught A by D, now A is what? A is the area of the plate, now you can see that only half of the area is included in this capacitance, so it is 2 epsilon naught A by 2 divided by distance between the plate which is what? Between this and that which is D by 2, okay, so this is C1, C2, C2 can be written similarly as just only difference is the dielectric constant, the area and distance between the plate they are same, this divided by D by 2, this is C2 and C3 is dielectric constant 2 epsilon naught, area is A by 2 divided by distance between the plate, between this and that plate which is D, okay, so simply they, you know C1 and C2 they are in series, find out the equivalent of C1, C2 and then add it to C3, you will get the new capacitance which in the question is represented by C2, all right, and C1, what was C1? C1 was in air, so C1 is simply epsilon naught A divided by D, okay, so like that you can do it or you can, no just leave it, okay, this is C1, okay, this let me call is C1 dash, C2 dash and C3 dash, so that there is no confusion, all right, so you can write C1 dash as 2 times C1 also, C2 dash as 4 times C1, C3 dash as C1, so like that also you can find equivalent capacitance in terms of C1 also, okay, any doubt with respect to this question, please type in, I will move to next one, meanwhile, do this one, see at times your live streaming will accumulate the lag, so you can just check whether there is a lag, you can see the, that bar that comes below the video, right, so if it shows minus 5 seconds or minus 23 seconds, so just drag it till the end of the right hand side, all you have to do is use Kirchhoff's loop rule and use conservation of charge, energy need not be conserved, okay, as in the potential energy, some potential energy may get lost as heat, okay, so here you will see there are two isolated systems, okay, in fact there are three, let's say this is plate number one, two, three, four, five and this is six, okay, so one and two they are isolated, they are not touching anything else, okay, five and six isolated and three and four again isolated, so these are isolated and hence you can conserve charge, so whatever charge was initially in these two plates combined will be equal to final charge, right, so let's write that conservation of charge thing, so we have Q3 minus Q0, okay, this is total charge in one and two which is equal to what, whatever was the charge initially, now this plate was negative initially and the charge was equal to C times V, potential difference is 180 into 2, so minus of 2 micro farad which is minus of 2 into 10 to the power 6 into 180, okay, so this is first equation, okay, then we have Q1 minus Q0, this will be equal to charge on this plate initially, which was positive charge whose value is, this is potential difference 100 into capacitance that is 3 into 10 to the power minus 6, okay and then these bottom two plates as well, so minus Q1 minus Q3, they should be equal to the, they should be equal to the charge initially in these two plates, now see this equation which this equation will be, you know the sum of these two charges will be equal to whatever the charge initially, okay, so initially the charge was on plate 3, the charge was positive 2 into 10 to the power minus 6 into 180, okay and the charge on plate number 4 was negative and it was 100 into 3 into 10 to the power minus 6, fine, so you have actually these equations, probably one of these equations is redundant, okay, so it's like you know you just, this particular equation could be a combination of these two, all right, so you need to use the Kirchhoff's loop rule as well, so if you use Kirchhoff's loop rule then Q1 divided by that capacitance which is 3 into 10 to the power minus 6, okay, I am going from this point upwards like that, okay, then minus of Q0 divided by 2 micro farad capacitance, okay and then minus Q3 divided by 2 micro farad, this should give you 0, all right, so you have these equations and once you solve it you'll get Q1, Q2 and Q3 and then you can find out electrostatic and potential energy as well, all right, one by one you find out the energy Q2 divided by 2C and add them all, okay, any doubts on these questions, sorry on this one, any doubt please type in yes or no, this one is Q1 plus Q0 actually, let's take what is Q and what should not be, what you're asking, which capacitance, what is this charge, oh like that, see when you go from plate 4 to plate 5, potential is increasing negative to positive, so Q1 by 3, then you're going from positive to negative 6 to 1, so minus Q0 by 2, then again you're going from 2 to 3 that is positive to negative, so again potential is decreasing, or if you're confused between the initial charge and final charge, the Q1, Q3 and Q0 I have taken these charges afterwards, finally, okay, now it may happen that Q3 comes out to be a negative charge, Q3 might be equal to minus 3, all right, but then it doesn't matter, so as long as you're consistent with the formula, so you may get Q3 to be minus, then automatically this may become positive, right, but then I have assumed that it is plus Q3 and plus Q1 over here, why plate 2 is, plate 2 positively charges plate 2, Purvik I'm not saying Q3 is positive charge, Q3 is just a charge, okay, so if Q3 is a charge then minus Q3 should be on 3, Q3 can be negative charge also, it will come out when you solve the question, or you can take it as minus Q3 plus Q3, okay, then Q3 will come out to be positive if it is like that, okay, I'll move to next one, now we'll take up current electricity chapters question and these will be J mains level only, we'll not go to advanced level now, I'll put two questions, okay, first one B, no that is wrong, second one is B, B for Bombay, first one is not B, okay, let us solve the first one, so R is equal to R0, 1 plus alpha delta t, okay, so usually we take R0 at, you know, R0 as 0 degree Celsius, fine, so 1 ohm is equal to R0, 1 plus, now from 0 degree Celsius which is 273 to 300 Kelvin, so that is delta t, so 27 times alpha is this, so 2 ohm should be equal to R0, 1 plus alpha into let's say delta t is t only, okay, so just solve these two, all you have to do is just divide it, okay, when you divide these two equations and put the value of alpha, you'll get temperature, you'll get t equal to 1127 Kelvin, this is actually delta, this one is delta t, okay, so delta t is t minus 273, okay, so this is delta t, once you solve this, you'll get t as 1127, okay, so remember this thing that the resistance, this resistance, the first one you write R equal to R0, 1 plus alpha delta t, R0 is a resistance at 0 degree Celsius, okay, now I guess most of you are getting the second question's answer correct which is B, let me quickly do this, a constant voltage is applied between the two ends of the uniform metallic wire, okay, so a uniform metallic wire is there, constant voltage is applied, some heat is developed, the heat developed is doubled if which one of these should happen, now if you see here the same constant voltage will be applied, okay, so power is actually V square by R, this is what you should use to compare because V is constant between the two situations, okay, current is not constant, so unnecessarily if you bring in current also then it becomes two variables to compare, so let's write everything in terms of V only, so V square divided by R is rho L by A, why I'm writing it as rho L by A, because again rho is a constant of material, it doesn't depend on the dimension, so it is better to write it like this, so it is V square by rho multiplied by A by L, okay, so if you look at option B, if both length and radius are double then what will happen, the area will become four times because area is pi R square, so if R becomes two R, area becomes four times the earlier area, okay, and if length is two L then four by two becomes two, so that is why option B is correct over here, fine, any doubts on these two questions, please type in immediately, it is T, look at the wordings of the question, they want you to answer this resistance of the wire will be two ohm at what temperature, so they are not asking what is the difference in temperature, they are asking you final temperature, do this one, third question, option A, correct, so this is again more of theoretical question which is based on knowledge itself, don't have to solve anything, same dependence on R, what does it mean same dependence of R, it means that it should have one by R square law, okay, now potential follows one by R, it doesn't follow one by R square, okay, similarly you know C also doesn't make sense but if you just read option A, it talks about intensity and we know that intensity, intensity actually falls by one by R square law, okay, so like this you have to solve the question, we will move to the next one, see at times J means is also about speed, okay, so those who are getting the question very quickly, as in they are getting questions right, try to solve it quickly, okay, Kushali is giving one answer, others this question, option B is correct, let's see how to solve this one, heat produced in 5 ohm resistance due to the current flowing through it is 10 calorie, so let's say current flowing here is I1 and here the current that is flowing is I2, fine, now the same circuit can be visualized as this 10 ohms 6 plus 4, they are in series and this 5 ohms, okay, so if this is current I1 and that is current I2, okay, then first of all these two resistance they are in parallel, fine, so 10 times I2 should be equal to 5 times I1 or I1 will be equal to 2 times of I2, all right and one more condition is given here is that 10 calories per second is consumed by 5 ohm resistance, so we have I1 square into 5, this is power, power is 10 calories per second, now I know that this is not SI unit, but then let us try to assume that, I mean let's assume that it is in the units which has energy in terms of calorie will not unnecessarily convert in SI units because the options are also in terms of calorie, so I1 will be equal to under root 2, fine and because of that I2 will come out to be 1 by root 2, right because I1 is 2 times I2, so I2 is 1 by root 2, now we need to find out heat generated in 4 ohms, so heat generated will be I2 square which is 1 by 2 into R which is 4, so this is 2, all right, so that is why option B is correct, now let's look at question number 2, this one, okay, here the current in the circuit is asked, now you can see that these two resistance 30 ohm and 30 ohm they are in series, so this becomes 60 and then this 30 is in parallel to 60, so total resistance will be 60 into 30 divided by 60 plus 30 which is 90, okay, so it comes out to be 20 ohms, okay, so entire circuit is reduced as if 2 volt is connected to 20 ohms, right, so current will be equal to voltage divided by resistance, so 2 divided by 20 ohms or 1 by 10 amperes, so option number C is correct over here, all right, so this is how you have to solve this particular question, we'll take up other questions now, do these two, I can see only 5 or 6 students are messaging the answers, others don't be like a spectator, okay, it's not a movie that is going on, you just can't watch it with snacks in your hands, you, this is a class, okay, so work it out, otherwise you're wasting your time, first one is D, yes, metals conductivity will decrease when you cool the resistance, so resistance of the metal will decrease when you cool the metal because the random motion of electron will decrease, so they can move systematically in one direction, whereas germanium which is a semiconductor, if you give heat free electrons may get generated, so some electrons from the, you know, from the valence band may go to conduction band, so there's a chance to do that, so it will utilize to increase its conductivity or decrease its resistance, so option D for the first one, D is correct for the first one, okay, the next one, see the net resistance, so the equivalent resistance of this entire circuit is how much, how much you're getting equivalent resistance, the equivalent resistance connected across potential E is what, ignore the internal resistance, apart from that what is the equivalent resistance, you know it is 2R, see it's a Wheatstone bridge, all of you appreciate that, it is a Wheatstone bridge, okay, so you can, you know, modify this entire circuit, like this, so R plus R I'll put here, this is equal to 2R, okay, and here you have R, okay, I'll draw it in other way, suppose I draw it like this, so you have 6 ohm resistance connected across, like this, okay, this will be R, all right, this will be 2R, and then this will be 2R, and this will be 4R, okay, this is 6R, there is an internal resistance to the battery, like this, this is 4 ohms, fine, so this entire circuit can be constructed like that, fine, so if you visualize it like this, it's a balanced Wheatstone bridge, R divided by 2R is equal to 2R by 4R, so this 6R resistance can be just thrown away, fine, so now you have 3R and 6R, they are connected parallel to each other, so 3R resistance is parallel to 6R, fine, and hence the equivalent resistance is 3 into 6 divided by 3 plus 6, which will come out to be 2R, all right, now power delivered by the battery will be maximum when the equivalent resistance become equal to the internal resistance of the battery, so when 2R becomes equal to 4, the power will be maximum, or when R will become equal to 2 ohm, so this law you should remember, this is Theveny's theorem, it, this is, this you will learn in engineering, but I don't know why it is asked here, but anyways, remember this, so if equivalent resistance is equal to the internal resistance of the battery, then the power consumed by the battery will be maximum, okay, any doubt with respect to this question, I'll move to the next one meanwhile, you can ask these two, okay, first one is A, okay, in the circuit P is not equal to R, right, so if this balance Wheatstone bridge, then only that is possible, so potential at A should be equal to potential at B only, right, so in case of balance Wheatstone bridge that happens and that is why current in P should be equal to current in Q, in case of balance Wheatstone bridge, there should not be any current that should go from A to B, okay, so IP is equal to IQ and IR is equal to IG, second one, what is the answer, so when there is a steady current, the capacitor will behave like an open circuit, no current passing through the capacitor, ninth one, sorry, second one, what is the answer, C, many people are typing in C, others, anyways, so let me solve this question, so this is an open circuit because of capacitors, no current passing through this, okay, so whatever is the current that is flowing, that will flow in the outer circuit like this, okay, so you can find out the value of current by applying Kirchhoff's loop rule, so V minus IR minus I into 2R, wait, what is this 2V, another battery, how come you guys are getting answer, if you don't know what is this 2V, the 2V battery is, there is a 2V battery that is connected like this over here, fine, so V minus IR minus I into 2R minus, that is correct, I am here but which one is positive polarity, how do you know, anyways, leave it, so V minus IR minus I into 2R minus 2V is equal to 0, fine, so you will get minus of V is equal to 3 times IR, okay, so current you are getting it as minus V by 3R, so you are getting negative current, it means that the direction of flow of current is like this and not what we have considered, fine, so this much current will be flowing V by 3R, okay, now if this is the current flowing, I need to find out potential drop across the capacitor, now potential drop across the capacitor, if it is suppose, you know, if it is like, if this is positive plate and if this is negative plate, let's say the potential drop is, you know, V1, okay, so I can use Kirchhoff's loop rule in the upper loop and we can write it like, you know, V minus IR, I am going from this point like that, so V minus IR, then I am going from negative to positive, so plus V1, okay, and then I am jumping from positive to negative, so minus V again will be equal to 0, fine, so this will give me V1 to be equal to I into R, fine, I is V by 3R, so V by 3R into R, so V by 3 is the answer, option C, okay, so like this you have to solve this question, should I move to the next one or you have any doubts, type in quickly, so every now and then you have to move the cursor of the video on the right extreme because the delay may get accumulated, so it may show minus two minutes, it means that you are two minutes behind, okay, so we will move to the next one, these two questions, solve it, the first one, what is the answer? If current is independent of resistance R6, it means that it is a balanced Weedstone bridge, there is no current in R6, so it does not matter what you connect there, nothing will get affected and if it is balanced Weedstone bridge, then what should be the condition for the balanced Weedstone bridge, so always make sure you look Weedstone bridge like this, okay, the Weedstone bridge should look like this structure, so if you think this as Weedstone bridge and say that this is R6, fine, then this will be R1, this will be R2, okay, and then R1 and R3 is this and this is R4, fine, so like this you have to visualize the Weedstone bridge, you cannot just assume any other structure as Weedstone bridge, getting it and from here you have R5 connected, now you can see that R1 by R3, if this is equal to R2 by R4, it is balanced, okay, or R1, R4 equals to R2, R3, fine, so hence option C is correct over here. Second one, now in second question, you have two balanced Weedstone bridge, right, so you can remove just, which one will you remove, will you remove small R or capital 2R, which one will you remove, so the middle resistance of the Weedstone bridge is capital 2R or small R, right, it is capital 2R, so if you look at the upper one as in this and this one, you can visualize it like this, okay, then you can say that this is 2R, this is 2R, this is R and this is R, so this is 2R, okay, so this one is the middle branch of the Weedstone bridge, so you can just remove 2R, so this 1, 2R is gone, then the second one is also connected like a Weedstone bridge only, fine, so like this it is connected, so this is 2R, now you can draw the same structure, same the way you have drawn here for the lower one, again this 2R will be gone, so this is gone and that is gone, so it's simply 2R and 2R, the upper one, they are in series, R and R in series and 2R and 2R in series, right, so you have 4R parallel with 2R and that is parallel with 4R, right, so just find out the equivalent resistance of these three parallel resistances, okay, so for this question, option number, option A is correct for this one, fine, any doubts, please type in, meanwhile I'll go to the next one, okay, let's do this question, then we'll take a small break, okay, anyone got the answer, okay, okay, first of all you should remember one thing that the power of the bulb is given as a rating, okay, for a particular voltage, fine, so if it is a 100 watt bulb that doesn't mean that, you know, it will always consume 100 watt bulb or it will always consume 100 watt power, so if it is a 100 watt bulb that power might be consumed at some voltage, let us say that standard voltage is V0, okay, so for bulb B1, I can say that V0 square by R1, where R1 the resistance of the bulb, this is 100, okay, so from here I will get the value of R1 to be equal to V0 square by 100, now why I am finding the value of R1, I am finding value of R1 because resistance will not change, okay, but the power may change depending on how it is connected, fine, similarly I will get the value of R2 for bulb B2, I can get R2 as V0 square divided by 60, this is R2 and R3 will be equal to V0 square divided by 60 again, okay, now they are connected to 250 volt as shown in the figure and W1, W2, W3 are the output powers of bulb B1, B2 and B3 respectively, so we need to compare these powers, all right, now 250 volt is connected across bulb B3, so for bulb B3, W3 is the power, right, W3 is V square which is what, 250, 250 square divided by R, R3 is V0 square by 60, okay, so this is W3 and see here B1 and B2, they are together connected to 250 volt supply, so what you can do is you will get the value of current after finding the resistance, so R equivalent for 1 and 2 will be equal to R1 plus R2 where R2 is this and R1 is that, okay, once you get R1 plus R2 as R12, you will get the value of current to be equal to 250 divided by R1 plus R2, now why I am finding current because current is something which is constant for B1 and B2, so power consumed in bulb 1 will be what W1, this you can write down as I square R1, okay and power consumed in bulb 2 can be written as I square R2, right, so once you get these expression in terms of V0, you can compare it afterwards, so R1 you can substitute from here and value of R2 can substitute from here and then you can compare, all right, so you will get option number is D, option number D to be correct, fine, so we will take a small, okay, Amogh we are asking something, the voltage across R1, R2, R3 are different, how can we write V square, yes Amogh voltage are different, but this when I write V square by R1, see the power, please understand that this is written as 100 watt power, this is 60 watt and this is 60 watt for B2 and B3, okay and 100 watt for B1 and then it is asking you to compare output powers of B1, B2, B3, so powers are directly given, why can't you just compare it, okay, now these powers when they are given, they are given against some standard voltage, okay, that standard voltage I am taking it as V0, at some other voltage like for example 250 volt you are applying, the power will be something else, okay, so I am assuming standard voltage to be V0 for all three bulbs and then using V0, I am first finding out R1, R2 and R3, then afterwards I am finding how much power these bulbs are consuming if you connect 250 volts, okay let us start, here is the next question, you are able to hear me, right, okay, what is the answer for this one, which can be used to verify the ohm's law, what is ohm's law, V is proportional to I, see when you plot that voltage versus current graph, okay, so you need to vary the voltage or vary the current, then only you will get different different points to plot V and I, okay, now here the voltage source is fixed, okay, so the way you are varying the voltage and current is by changing the resistance of this, this resistance is what you are changing, okay, so because of that the voltage is getting changed, okay, so you can see that the potential difference across these two points, sorry, potential difference across these two points if you track, okay, and also track current in that loop, right, then you will be able to get corresponding voltage and current values, for example if you look at here, okay, if you look at here you are measuring current in this branch and you are finding potential difference in other branch, okay, you should be doing in just one loop itself potential difference and current, that is why option A has to be correct, okay, all of you clear about it and also voltmeter cannot be connected in series, okay, so definitely C and D they are out of question, fine, so it has to be either A or B, now in A you have voltage and current measured in one loop itself, fine, we will go to the next question, do this one, okay, sign is saying option A, others see you can you know understand one thing here is that if you just talk about equivalent resistance between two points, suppose between this point and that point, this and that, all of the circuits will be same if you just consider equivalent resistance between those two points, so you can say that this is R equivalent and these two points and current in all the circuits will be I only, okay, so power dissipation will be equal to I square into R equivalent, right, and R equivalent is highest in first and lowest in the third, right, so any option which has one as highest and three as lowest will be the correct answer, so here I will just mark option A without thinking much and move ahead, fine, you don't need to break your head exactly getting the answer, just evaluate the option, mark the correct answer and move to the next one, okay, let us move to the new question 56, this is one, these two questions, try solving them, even I am not 100% sure that's how it will be in j-ment, sometimes they will just coin a term that you don't even know and probably the solution is independent of the definition of post office box, it is already shown, right, in the figure, basically post office box is used to measure the unknown resistance, so in that experiment to find out the external resistance, you want to find out the full resistance, right, so you connect between A and D, so option B is correct for the first one, so this one, for the second question, six identical resistance are connected as shown in figure, the equivalent resistance will be, now it is not clearly written between which two points equivalent resistance is, okay, even the options, in options it is clear between which point they are asking, so you can evaluate, all right, so between P and R, between P and R you can find out the equivalent resistance, so when you find out between P and R, this R and R they are in parallel, then they are in series with R, okay, then between Q and R, these three resistance are in parallel, then that is in series with R and then again parallel with these two, so I think this is straightforward, the second one, I will just tell you the final answer for the second question, it is C, is maximum between P and Q, okay, now here you can, you know, apply some logic and get the answer quickly, for example, you know that equivalent resistance of two parallel resistance will be less than the least, okay, so if you know that the equivalent resistance of these two resistance will be less than R, which will be R by 2 and this will be R by 3, this will be R by 3 and this will be R by 2 and this is R, fine, so naturally if you add up R by 2 and R by 3 and then connect in parallel with R, then there is a chance that you will get a higher resistance than adding up these two and connecting parallel to R by 3, if you connect parallel to R by 3, your equivalent resistance will be less than R by 3, fine, if you connect parallel to R by 2, equivalent resistance will be less than R by 2, but if you connect parallel to R and your equivalent resistance is coming out to be greater than R by 2, if your R equivalent is coming greater than R by 2, then this will be the option that is correct between P and Q, okay, so like that you can evaluate first between P and Q if it is more than R by 2 and get the correct answer quickly, okay, right, let us move to next question. Solve these two questions on your right hand side of the screen, let's call it as first question, second, third and this as fourth, third one you are saying option A is correct, that's the answer because you know two ohm resistance is not forming any loop with any of the, I mean it's an open circuit, two ohm is not a part of any loop, so that is why the current through two ohm should be zero because if current is flowing it has to be inside a loop, so that is why zero current is through two ohm, what about the fourth question, see charging of capacitor, I am not sure whether you guys remember charging of capacitor charged at any point in time is Cv naught 1 minus e to the power minus T by RC, okay, let me solve this now, this is Q, alright, so the potential difference across the capacitor, first of all let me draw the diagram because any physics question you should not solve without the diagram, so you have 12 volt, you have this 2.5 mega ohms and this one is 4 micro farad capacitor, use the green one, so potential across the capacitor is Q divided by C, okay, which is V naught 1 minus e to the power minus T by RC, okay and potential across resistance is current into resistance I into R, now I is what dQ by dt, so dQ by dt times R is the potential difference across the resistance, now dQ by dt is what dQ by dt will be V naught, sorry Cv naught then differentiation of e to the power minus T by RC divided by RC e to the power minus T by RC, okay and C and C will get cancelled off, so this into R also, so R also goes away, so potential difference across resistance is V naught e to the power minus T by RC, fine, now we need to find out when the potential difference across the capacitor is three times the potential difference across the resistance, this should be equal to three times V naught e to the power minus T by RC, fine, so you can cancel out V naught, okay and then you will get 1 is equal to 4 times e to the power minus T by RC, all right and then you just take log both sides now you will get 2 log 2 is equal to T by RC, fine, so T will come out to be 2 RC ln 2, okay so I hope you have done like this and you will get option A to be correct, okay, any doubt guys please type in immediately I will move to next, meanwhile any doubts it's okay to have doubts not okay to not to ask, okay first one B, please try out the second one, meanwhile I will solve the first one for those who want me to solve it, I will gas is filled with filled in a closed rigid thermodynamic container, so coil is there 100 ohm is the resistance carrying a current of 1 ampere for 5 minutes, change in the internal energy, now all the energy is supposed to utilize to heat up the gas and since it is the closed container the volume is not changing, so whatever heat you give will be used as change in internal energy and this will be equal to I square R which is power into time, okay I is 1 so 1 square is again 1 only I square R into T, T is what 5 minutes so 5 into 60, so we will have 3 into 10 ratio power 4 joules, okay that means option B is correct, second one, okay V AB is equal to V BC that's not correct, this is not correct, why because the current in AB will be equal to current in BC or not, suppose resistance of AB is let us say R1 and resistance of BC let's say that is R2, okay so both of them they are connected in series, so the current will be same, same current will be flowing, okay but current per unit area which is current density that is different, fine, so even this is not correct and since resistance are different so I into R1 which is potential difference need not be equal to I into R2 because resistance are different so even 1 is not correct, then power across BC is 4 times the power across AB, now the area of cross section of AB is what pi into R square which is pi into 2R square, so that is 4 times of pi R square, this is area of AB, fine and area of BC is simply pi R square only, fine and length is same L by 2 and L by 2 only, okay so since current is same power in AB is I square into R which is rho A by L, now I know that area of, I am using S for area, so let me use that only, so rho into area of AB into L, now area of AB is 4 times area of BC, okay so I will write it like this 4 times I square to rho area of BC divided by L which is you know small L by 2 only for both, now this entire thing is resistance of BC and when you multiply I square with it this will give you power consumed in BC, so that is why power AB is equal to 4 times power BC, okay so that is how, this is exactly how the question is asked in J, so this is just copy pasted from the let us say from the question paper itself, so please expect it to be like this only, this came in one of the J exams, I think 2006 it came, all right let us move to next question, any doubts on this one, okay AB is 2 times VBC even this is not correct because resistance of AB is one fourth of the resistance of AB is one fourth, okay I think I made a silly error, this is L by SAB, okay so the formula itself I have taken it wrong, so I will correct myself first, rho L by 2 by area of AB, now area of AB is 4 times area of BC, so I square rho L by 2 divided by 4 times area of BC, fine so power of AB is equal to one fourth of I square times rho L by 2 by area of BC, now this is resistance of BC, this into current of BC will give me the power consumed in BC, so 1 by 4 power consumed in BC, so power consumed in BC is 4 times the power consumed in AB, okay so like that you have to do this question, all right I will move to the next one, these two this let us say is first and this is second, okay none of you getting it, the question number one refers to Wheatstone bridge, right so you can quickly draw the Wheatstone bridge, this is the wire by the time you prepare for board exams you will be well versed with all of these circuits, there is galvanometer connected point over here, okay now let us say this is two ohms, the resistance two ohms is connected by one gap and the unknown resistance, let us say unknown resistance is R, fine length is one meter and the unknown resistance greater than two ohms, all right now when it is interchanged the point shifts by 20 centimeter, now one thing you should remember is that the length corresponding to the higher resistance will be the length that is corresponding to higher resistance will be more, okay so this length will be more, why it is so because you need to maintain the ratio suppose from A to B and this is B to C, if A to B if resistance is let us say R1 and B to C the resistance is R2, I need to maintain the ratio for balance Wheatstone bridge that is 2 by R1 should be giving me R divided by R2, fine so if capital R is more than two ohms then R2 should be more than R1 also just so that the ratios are maintained, right hence since resistance is proportional to length this length B to C is more than length A to B, okay so when it says that when you interchange the point shifts by 20 centimeter so when R comes over here the galvanometer will move this side, fine so this length will increase by 20 centimeter, fine it is not given exactly it is increasing or decreasing you need to infer it, okay so right now let us say length is L and this length will be 100 minus L then, okay so you can write 2 by R to be equal to L divided by 100 minus L, okay and when you interchange the point will shift right hand side because R is higher so you can write R divided by R divided by 2 to be equal to L plus 20 divided by 80 minus L, right so now you have L and R as two unknowns and you can solve it from these two equations, so for this particular question option number 25, A is correct, option A is the answer, second one A is not correct, any other answer for the second one, second one anyone, okay let me solve this question, no I will take this one this will better, all right so you need to find once which is closed how much charge will flow from y to x, okay now when charge is flowing from y to x which plates charge will be getting affected these two plates charges will be getting affected, right so if let us say charges initially let me draw it slightly bigger so if this is charge q1 and this is charge q2 these are initial charges q1 not and q2 not, okay and finally let us say charges are q1 and q2 so if q1 not plus q2 not is not equal to q1 plus q2 then it means that charges will flow, right okay if the sum of these two charges both initially and finally are same then charges will not flow, now we need to find out what is the charge in these two capacitors before and after, okay now when the switch is open, when the switch is open these three micro farad and six micro farad they are open circuits so current will just go like this and it will you know travel like that, right so you can say that current will be equal to nine which is nine volts divided by net resistance is 6 plus 3 again 9 so you have one ampere current, okay one ampere is a current now the potential difference between these two points of the two capacitor will be what 1 into 9 which is equivalent resistance, right so potential difference, okay I don't need to even find out the current I since 9 volt is connected across these two ends so I know that potential difference is 9 volt only and then I know that these two resistance they are these two capacitance they are connected in series so since three micro farad and six micro farad they are connected in series hence charge on three will be equal to charge on six and will be equal to charge on the equivalent of three and six, all right so now the equivalent of three and six is what equivalent of three and six is three into six divided by three plus six they are in series so this is how you find equivalent capacitance which is two micro farad, okay so the charge on both of them will be two micro farad into potential difference which is nine volt so nine into two that is 18 micro coulombs this is the charge on three micro farad and six micro farad so q 1 naught plus q 2 naught okay first of all you should know that this will be the negative plate and this will be the positive plate okay so charges will be I mean this the total charge will be equal to zero okay on both of them initially because this one will be minus this one will be minus 18 micro coulomb this will be plus 18 micro coulomb in fact you could have got it initially itself without even solving it you know saying that okay charge will get accumulated on this plate this one and it will get taken from that plate and these two plate will have equivalent opposite charges these two have equivalent opposite and since they are connected in series they will have same charges so that is why some of this charge plus that charge will be equal to zero initially now finally how much is the charge when the switch is closed the situation changes a bit okay white changes you can just draw the diagram and visualize it I am sure you have not drawn any diagram for this question but I can just tell you once or twice this is a diagram 9 volt 3 ohm and 6 ohm once the switch is closed fine the situation is a little different now the potential across 3 volt is the potential across 3 ohm resistance and potential across 6 micro farad is same as potential across 6 ohm resistance by getting it the potential across 3 micro farad capacitance will be current into 3 current is 1 ampere so this is 3 volts and potential across 6 micro farad capacitor will be equal to 6 volt fine current will still be 1 ampere only because through the capacitor current won't be flowing so it will be like that only but when you close the switch potential difference across the capacitors will be different now okay now you will get charge across 3 micro farad as c times v which is 3 into 3 that is 9 micro coulomb okay and across 6 you will get 6 into 6 that is 36 micro coulomb fine now just take a look at the the polarity of the plate so this will be positive this will be negative and then this will be negative and this will be positive so basically total charge on this plate and that plate will be equal to 36 minus 9 so total will be 36 minus 9 okay this comes out to be 27 micro coulomb right so now total charge is this much initially it was 0 so this much charge should have been flown any doubt guys on this question please type in yes or no as in if you have doubt you can directly type in the doubt itself fine let us let us take the final two questions for today and just pick no this is easy see last year there were two questions from current electricity and both were from the instruments so you can expect that this year probably questions may not be from the instruments but that's not the rule it may still come okay okay this is also straightforward only you can see that this chapter not many difficult questions can be made if there is a difficult question that will be in a some in some instruments maybe from galanometer weakstone bridge meter break something like that it may be there otherwise it this chapter is straightforward do these two questions that's it for today once we are done with these two we can close the session all of you done done trying this question see the actual length will be the length which is measured plus end correction so if the end correction is negative it would have been written minus 1 centimeter so it when it says 52 you take length as 52 plus 1 and if it is saying 48 you take 48 plus 2 like that so for example this length this length you should take as the end corrections at end a is 1 centimeter right so this you take 52 plus 1 so that is 53 okay and this length this length what is showing is 100 minus 52 which is 48 so what you take here as 48 plus 2 because end correction at B is 2 so this will come out to be 50 okay don't worry if 53 plus 50 is greater than 100 that's fine these are corrections fine now now you can write down the equation x divided by 53 equals to 10 divided by 50 right so x divided by this length is equal to 10 divided by that length or you can write x divided by 10 is equal to 53 by 50 okay so once you do that you'll get x to be equal to 10.6 ohm fine so first one the option B is correct and second one I guess is just sheer I mean it's brute force you just have to solve it you get the answer like for example 2 4 and 2 they are in series so total resistance is 8 ohm now 8 and 8 is 4 now again 2 4 2 again 8 8 and 8 again 4 so this entire thing can be looked at like this so this is 3 2 and this is 4 I hope you can appreciate 2 plus 4 plus 2 8 8 and 8 parallel 4 now again 2 4 2 8 8 and 8 parallel again 4 so like that so here the current flowing through 3 ohm resistance will be equal to 9 divided by 4 plus 3 7 and 2 9 so 1 ampere fine so 1 ampere is the current now through 3 ohm it is 1 ampere so A and B are not correct so we need to check at 4 ampere what is the at 4 ohm what is the current all right so we know this current which is 1 ampere and let us assume that here you have you know here you have 1 by 2 why it is 1 by 2 ampere 1 by 2 and 1 by 2 there because once you look between these two points the equivalent of all of this is 8 only so here also 8 and equivalent of 1 2 3 4 5 6 these 6 resistances also 8 so after this point the current will equally split okay so this is 1 by 2 and here is let us say unknown is i similarly here it is i here it is 1 by 2 and this is 1 so I can use Kirchhoff's loop rule in the bigger loop so I will be able to write like this 9 minus 3 into 1 okay then minus 2 into half okay minus this current i and the sum of these three resistances 8 so I can directly write 8 into i all right minus then 2 into i by 2 2 into 1 by 2 sorry and minus 2 into 1 okay this should come out to be 0 all right so once you solve this particular equation you will get option which one is option d to be correct d for del e okay now now you can solve this particular question quickly like this or you can simply write down three or four equations and solve it and you'll still get the same answer if you don't make any silly errors okay so that's it for today we will meet again probably on Monday with with new chapter revision and I'll announce it before the class so Tarun if you want to be informed about the classes or anything else just message me on that number which is on the youtube channel number I'll reply bye