 So today we are going to revisit current electricity, fine. I know there are few topics on wave optics, that is single sheet experiment and on the wave optics, that is optical instruments that we have left. So after completing the current electricity, we will complete that also. And yeah, today we will be exclusively focusing on this chapter, current electricity. Now, whenever this chapter name comes in everybody's mind, then the first thing that will strike in their head is Ohm's law vis-a-vis IR, okay? Because that appears to be a simple thing and that you have learned in class 10 also, the numericals you have done on it and of course finding equivalent resistances where everything is connected in either series or parallel, fine? But then this chapter is much more than that because finding equal resistance and using Ohm's law, you have studied in class 10. So definitely in class 12, there has to be something beyond that, right? So don't make fool of yourself by thinking that this chapter is all about these two things only. This chapter has many other details, which I'm sure you guys have brushed up your, your notes before coming to the class, right? So many things like, for example, applying symmetry to find the resistance and all that, it's not part of theory. You learn those tricks by solving numericals. So we'll be taking three numericals with it to it, all right? I cannot tell you a single way wherein you can apply to solve all the problems in finding equal resistance. If that is possible, then everybody will score full, right? Every numerical is different. All right. So let's quickly revisit the, you know, scan through the, you know, scan through the, let's say the equations and the formulas, basic concept, and then we'll do the problem solving. So first thing is the definition of current, which is the rate at which the charge is flowing, okay? So this chapter name is current electricity. So half of the chapter is devoted to introduction to what is current, which is valid for alternating current also, all right? The other half of the chapter and wherever the battery is connected, that part of the chapter deals with direct current, where in a battery, a constant voltage source is connected. So if there is no source, and if there is a general discussion on what is current, then that is valid for the alternating current also. In fact, the Kirchhoff loop rule and junction rule is also valid for alternating current for a particular instant, right? So anyways, this is the definition of a current. So if at any place the rate of flow of the charge is this, dq by dt, then at that place the current is i, okay? For example, if you take, let's say a charge q rotating on a circle, let's say charge q is rotating in a circle with angular velocity omega, then the current, you may claim that current at every location could be different. So let's say I'm finding current only at this location over here, okay? Then what is the rate at which current is, charge is flowing at that place, which is how much charge is flowing per second? I know that after every time period t, q amount of charge is flowing. So in one second, q by t will be flowing on an average, so this is an average current, okay? And time period can be written as 2 pi by omega. So this is the current, this is the average current at this location, which is same at that location, at that location everywhere, the average current is same. So I can say that throughout an average current of this much is flowing. So be comfortable using the basic definition of the current, because if you are aiming for j ones, at least not there, the questions which are there on the face of it will look difficult, but when you look at the solution, they focus more on the basic, rather than too much of knowledge. They don't care whether you know many complicated things, they just care about whether you know the basic things clearly. Okay, so the first equation you see in this chapter in class 12 is this, current density is equal to sigma times electric field, okay? J is current density, sigma is what? Anyone? What is the name of sigma? Conductivity. Conductivity. Conductivity. Conductivity and easy electric field. So at any moment, if you see, if you see inside any material, electric field is E, whose conductivity is sigma, then the density of current at that location, where electric field is E, will be conductivity times electric field, okay? Don't worry about the battery, where is battery, what is the resistance and all that. Resistance and everything else is the outcome of this basic equation. This equation comes first, V is equal to IR comes later. So we know that the conductivity is 1 by rho, all right? And yes, the current density can be written as the current divided by area that is perpendicular to the current. From here, you can say current is equal to current density into area that is perpendicular to the current. This can also be written as, if you take area as a vector, it can be written as J.A, okay? So take a dot product of area vector with J, you'll get the current through that area. So then after that, you have current is equal to NEA into drift velocity, right? So like this, you can directly find the current at a cross section whose area is A and number of free electron, total volume is N. So like this, you can find out where drift velocity, how can I write drift velocity? Is there any equation to write in terms of electric field? Minus E square by M. Drift velocity, you don't remember E by M. M into tau. Tau, tau. Tau is the average time between two collisions, right? Tau is the average time between two collisions, fine? Now collision happens where? How collision happened to me? The free electrons are moving, where it collides? So with each other. With each other, okay? Anybody else has any other idea? Electron collides with what? In a material, okay, no one has any idea. So we have discussed it when I was teaching current electricity. So this is the material wherein the electrons are moving, but the rest of the atom will be stationary relatively. So electrons, when they are moving, they are so tiny with respect to each other, they don't collide with each other. You know, you can ignore them. The main collision that happens is electron with the crystal, sorry, electron with the lattice atoms structure. So that is a collision we're talking about, which will change the momentum, fine? Not the walls of the wire. A wall is invisible, the electron, foreign electron, things which are visible will be other electrons and the atoms, okay? So the assumption over here is that after every collision, after every collision, what happens to the velocity? After every collision, velocity becomes random. We can't predict where it will go, right? So average velocity after every collision, immediately after collision is zero, okay? All right, now comes the Ohm's law. What does Ohm's law say? Ohm's law is V is equal to IR, where R is the concept of proportionality between V and I. It is assumed that voltage and current will be proportional. That need not be true every time, okay? So it is usually true for most of the materials, but not for all the materials. So if it is not written in the question that Ohm's law is valid, you have to assume it is valid. If it is not valid, there will be some relation given between voltage and current, or it will be clearly written that it is a non-ohmic device. Semiconductor is clearly a non-ohmic device, which is where V is not proportional twice. So they need not write for semiconductors. So R is equal to rho L by A. So resistance depends on length area, as well as the property of material, which material it is, which is resistivity, fine? Now, if suppose area is not uniform, then how will you find it? For example, in this kind of scenario, how will you write resistance over here? Anyone has any idea? So you can take them in series, small length. What you can take them? Law section and then of length dx. So you are telling me that take a strip like this? Yes, sir. Why? So then these small strips are in series with one another and we can integrate. That's fine, but why you have taken a strip? I can take entire... So because uniform over that small... What is uniform? The area. That I was looking for. Area you can say is uniform in the strip because strip is very thin. So because area is uniform, I can use this formula because this formula is valid only for the uniform area. So dr is equal to rho L is dx divided by A. Area depends on x. So you can write down area in terms of x and then integrate to find the R. You can directly integrate it because all of them, all the strips, they are in series between this point and between that point, okay? But if I find, if I ask you to find equivalent resistance between this and between that, I mean, you can't do this, okay? Then I mean, it will become very complicated, in fact, nobody will ask you that. Huh, so now talking about the resistivity, not only depends on the material, it depends on the temperature also. With temperature resistivity increases. Why it increases? Because the random motion of the electron increases, more number of collision happens. Tau decreases if collision happens and drift velocity also decreases because you can see here, the drift velocity depends on the tau and everything else is anywhere fixed. Okay, so this is the resistivity variation. Alpha t is coefficient, the temperature coefficient of the resistivity, all right? Suppose I have to find that conductivity. So sigma is equal to, now you just need to take inverse. One by rho naught will be sigma naught. Just take inverse both sides, alpha t, delta t raise to power minus one. Now alpha t, delta t is so small that you can use binomial approximation and write it as sigma naught one minus alpha t, delta t. Okay, by the way, what is conductance? What is conductance? One by R. Conductance is one by R. Okay, just like conductivity is one by resistivity, conductance is one by resistance, fine? Now, after all of this, we know that Ohm's law is valid only for one resistance at a time. When you write v is equal to I R, v should be potential difference across the resistance and I should be the current through the resistance, okay? But suppose there are multiple resistances, then what you will do? Then you will try to use Ohm's law by saying that all of these resistances are equivalent to a single resistance. So there is a concept of grouping of resistance or finding equivalence. So there are few questions only on finding the equivalent resistance, nothing else, okay? So the standard way of connection series, standard way is that R equivalent is simply summation of all these resistances that are in series. In series, the current will be equal, okay? And if it is connected in parallel, you will have a one by R equivalent is equal to one by R1 plus one by R2 and so on, okay? So in parallel connection, the equivalent is less than the least, okay? So suppose if you have like one lakh Ohm resistance over here, you want to decrease this resistance less than one Ohm only. You want to connect a resistance of one Ohm in parallel to it, equivalence between one Ohm and one lakh Ohm will be less than one Ohm itself, it will be roughly one Ohm, okay? So now finding the equivalence when you do that, I mean it will not be so straightforward, right? Otherwise it will become a class 10th numerical which lakhs and lakhs of students will be able to solve. So that is as if the question is not asked and if all those who are writing the exam, everybody get the answer, that's a useless question. It doesn't create any difference. So there will be some tricky questions. So there are a few tricks which you have to apply sometimes to find out the equivalent resistance. The first one is to try to see if it is a balanced Wheatstone bridge, okay? If it is a balanced Wheatstone bridge, then what you can do, you can remove the, you can remove the middle resistance, okay? A Wheatstone bridge looks like this and you are aware of it. There a Wheatstone bridge and between these two points, you can connect a battery. So basically this middle one, across this one battery should not be connected. Across the other two ends, battery should be connected, okay? So it is R1, R2, R3, R4, R5. So if it is that R1 by R4 is equal to R2 by R3, it is balanced and you can remove resistance. It actually means that potential of this point is equal to potential of that point. So potential difference across R5 is zero. There is no current in R5. So R5 is not contributing to anything. You can remove that. Second trick that is there, most common trick is finding the symmetry. Now symmetry is something which is pure intuition based, okay? I can take example and explain it, but I cannot teach what is symmetry. Symmetry, you know, in a very layman term, symmetry means that things are similarly positioned with respect to something, okay? So that's what I can say in abstract matter. The symmetry implies that similarly placed resistance will have same current. Okay, this we will apply in some of the numerical and get into details of it. Second is that if you find point A and B for which potential difference is zero, need not be just mean strong bridge, remove the resistance between A and B. And if nothing else works, if nothing else works, in exam you leave the question. Otherwise it will eat up your lot of time, okay? Save your time. But if you are not getting any other question also, then you can do this one which will be little lengthy. So connect a battery, connect a battery between two points for which you're finding the equivalent resistance, get the current through the battery, then your R equivalent will be potential of a battery divided by current through the battery. How you get the current through the battery by applying nature of loop rule and junction rule which you already know, okay? These are the few tricks. Then not only equivalent resistance but equivalent EMF also you should be aware how to find it. So the ideal EMF has no resistance but in reality it has an internal resistance, that's E and R, fine? So if there is a current I flowing through it, A to B, like that if current is going, then potential difference between A and B is E minus IR. Current I'm assuming going from negative terminal to positive terminal, okay? So if two EMF are connected like this, they're internal resistance R1, R2, EMS, E1, E2, okay? So you will have equivalent EMF as E1 plus E2 and equivalent internal resistance as small R1 plus small R2, okay? And not only that, if let's say you have two batteries connected in parallel with an internal resistance R1, R2. Now this, you know, you need to understand internal resistance is as good as external resistance. So if there is an external resistance connected to an EMF, you can say it is like an internal resistance to the battery. So even if I tell you that there is an external resistance, you can include that as internal and then say that this is equivalent to a single EMF and a single resistance for which E equivalent divided by R equivalent is equal to E1 by R1 plus E2 by R2. And 1 by R equivalent is this. So after this finding equivalence and all, you will have Kirchhoff rule and down Kirchhoff rule. Now it is called Kirchhoff rule. It is not called Kirchhoff law because it is derived from the law. Which law? Conservation of energy, conservation of charge. The Kirchhoff rules are implication of them. So the junction rule is what according to junction rule at any junction, the sum of the current that is going into that junction should be equal to sum of the current that goes out from that junction. So what is a junction? This is an example of a junction, this point. So I1 and I2 goes in, I3 and I4 goes out. So I1 plus I2 is equal to I3 plus I4. This is what it is, simply. This is the junction rule. Then you have loop rule. So according to loop rule, algebraic sum of change in potential in any loop is zero. So basically you need to travel the loop from one point, travel the entire loop and come back to the same point. So it can go clockwise or anti-clockwise, up to you. So when you are traveling, you need to see from your previous point to the next point what is a change in potential. If the potential is decreasing, change in potential is negative, if potential is increasing, change in potential is positive. So if you want to use it like a thumb rule, what you can say that there is a resistance like this and current is like this. If you go from point A to B, change in potential is positive, which is plus IR. If you go, no, minus IR. This is minus IR. If you go from B to A, the potential increases by plus IR. So all these things, I need to take care of. Anyone has any doubts? Till now? No doubts. If there is no doubt, there is no doubt. So you know, Kirchhoff loop rule is called loop rule, but then you don't need a loop to apply a rule like that. So for example, over here, which you will be doing it, what I am interested in finding over here is the relation between VA and VB. This is a part of a bigger circuit. I am showing an entire circuit to you, but whatever is of your interest, I am showing there is a current I and I1 goes this way. So remaining current is I minus I1. This is R1, E1, R2, E2. So I want you to go from VA to VB, write the equation relating them. All of you do that. Others? You haven't written VA and VB. So what is the potential of this point? The potential of that point is VA. How much VA? Minus I1. Minus I1. Then potential of that point minus E1. Potential of that point minus I1, R2. This point plus E2 is equal to VB. So like this, you can travel from any point in a circuit. All you have to make sure that if you are going across the battery from negative to positive terminal, you can use plus of EMF from positive to negative, negative of the EMF. And for the resistance and current we have already discussed. Then we have the instruments. This chapter has become more and more important over the years because number of questions have increased. How do you determine sign up for potential difference across capacitor and inductor? So if it is a capacitor, the potential difference is Q by C. So if you go from positive plate to negative plate, will you write plus Q by C or minus Q by C? Minus Q by C. Minus Q by C. Similarly negative to plus plate plus Q by C. So that you have to see from which plate you are going to what plate. For inductor, what you will do? So L into Y. EMF for inductor is how much? Potential difference across inductor. Minus L Di by T. Minus L Di by T you will write every time. And you should go in the direction of current. Got it here? Best would be to find out a numerical practice yourself because learning these things as a theory is just useless. So now you know, now you find out some numericals only on that and then solve it. That is how you guys should prepare going forward. You should not be blind as in, okay, I have this book. Let me solve one by one all the questions. No, like for example here has doubt that what is the how to write a potential difference between capacitor and inductor. Now I told her, so now she should go back, find out questions only on that particular concept and solve it. So like that you should target one concept at a time. The first instrument is the what is that Wheatstone bridge. This is a Wheatstone bridge which is used to find out the unknown resistance R1 R2 R3 R4. In the middle section you usually have a gallium meter. Gallium meter itself will have some resistance. So you choose resistance in such a way that gallium meter has zero current and then at that moment R1 by R4 is equal to R2 by R3. The experimental version the setup for the Wheatstone bridge is called the meter bridge which looks like this. All of these things should be handy to you like any time if I ask you to draw the meter bridge, potentiometer, circuit diagram of that it should be there in your head you should be able to draw it. These are the copper plates then you have an alloy wire like that and a battery within these two point you have one resistance then these two you have another resistance R and S and then you have a gallium meter which is connected by a jockey like that which you can slide on this alloy wire and we know very well that the resistance is proportional to length. So if this alloy wire is uniform the resistance of this portion is proportional to length of it let's say L. Total length is 1 meter or 100 centimeter so the other length is 100 minus L. So it looks like a Wheatstone bridge wherein this is R, this is S this one is proportional to L R3 is proportional to 100 minus L so if I write if I convert this is a circuit diagram actually for this I can say R1 by R2 is equal to R4 by R3 it is same the same way R by S is equal to L divided by 100 minus L the resistance is proportional to length so if you use a meter bridge you just need to know one of the resistance you can find out the other one okay over here you need to know three resistance to find the fourth one fine this is the meter bridge this one is Wheatstone alright next one is potentiometer the last thing what does potentiometer is used for everyone calculating the EMF internal resistance of any given battery correct so it can be used to find the EMF and the internal resistance so for J kind of exam or especially the advanced level remembering formula I mean of course you should not forget FORGE-VAL to master an explanation what I am trying to say is that the EMF and the meter bridge formula all that doesn't help what helps is to analyze the entire setup suppose the experimental setup of potentiometer is this if I ask you to draw the let's say what the circuit diagram you should you should be able to draw it quickly so let me draw the experimental setup then you draw the circuit diagram so this is how it is there in your book the blue line is the alloy wire I want to use alloy wire because the resistance I can say depends on the length it has higher resistance than the connecting wires like copper so from here from this point I have two EMFs I have a keyboard then you have a galvanometer G which you can slide on it and wherever galvanometer shows a zero deflection you can stop fine so suppose this is E1 this is E2 connecting these two this battery is driver cell this is called the driver cell so can you draw the experimentals sorry the circuit diagram for this all of you draw the circuit diagram let me know once it is done then all of you you have the driver battery and driver cell that's enough this is R1 R2 what is R1 R2 where it is there in the experimental setup where it is there going to go from where to where A B what is the resistance R1 is between A and B R2 is between B and B all of you clear yes sir okay so now if it is let's say the balanced or you can say galvanometer shows zero deflection then whatever current comes from here there is no current in this branch because zero deflection in galvanometer so entire thing goes there like that so this current is always fixed so if I use the Kirchoff loop rule over here I can say that even is equal to I times R1 where current is V divided by R1 plus R2 which will remain later on also so if you divide even in E2 you will get it as equal to L1 by I2 because R1 R1 will change R1 is this length resistance L1 when you connect E2 this L1 will become L2 so this R1 changes so initially this I R1 later on it will be I R1 dash so they are proportional to length so that is how it is okay so we will see numerical also further on this let us solve numericals now enough of theory two different answers already I have so see you guys are getting different different answer so definitely only one answer is correct you guys are making serious okay give me 30 seconds I just take it I am just writing formulas I mean it is a direct formula substitution you continue then all of you now okay I cannot give you so much time for every little question right I will proceed 1.7 divided by number of three electrons point volume 8.5 10 square 28 E 1.6 10 so minus 90 area is a d square d is going to take 1 1.02 is roughly 1 okay so 1 into 1 divided by 4 pi square by 4 and it is in mm so that 10 is 4 minus 6 so you will get around 1.5 into 10 is 4 minus 3 meter per second this is what is given in the book calculation I am not going to teach how to do it is sad that you guys are not getting the correct answer this year J mains and J advance both were calculation intensive if you are not good at calculation it will be very very difficult alright okay so if sodium iron goes this way chlorine iron will go in opposite direction in the same direction go in the opposite direction this is the direction of Cn minus this will be Na plus so their currents added or subtracted their individual currents what do you think so how do we know they go in opposite direction electrify will always be in one direction inside the solution it can't be both directions okay yes sir they will be getting added up because the opposite direction of the negative charge will be the current so that is why so the total current will be 6.5 into 10 is power 16 plus 4.2 into 10 is power 16 into charge of one electron because that is a total charge that divided by amount of time it takes to flow that is one second okay so now you can solve and get the answer so this is around 1.7 you can see 6.5 plus 4.2 is roughly 10 1.6 slightly more than 10 so 1.7 so 1.7 into 10 is power this will be minus 2 MPS many of you got the wrong answer just get the expression in terms of small r and capital R okay small r and capital R just get the expression j is equal to some constant into r square constant is 3 to transfer it r is r do not need to calculate current density is not constant right what do you do small thin ring when you integrate then the rings ring at a distance of r thickness dr when I am asking something better you type in first when I say you can speak please speak because others also doing it so tell me please type in how much will be the current through this ring no current okay j at that place is equal to c into r square now the ring is so thin that you can say r is practically remaining same for the ring even if it is having some small thickness so the current di through that ring and we see r square into 2 pi r dr okay which you can integrate and get the total current 0 to i and you go from small r to capital R so it will be 2 pi c by 4 r square into r r q dr so it will be r to the power 4 minus small r to the power 4 pi c by 2 clear everyone okay let's move ahead if you have doubts you can stop me okay no way cross that's not correct all of it right alright i can see many of you answer something alright so if i take a cross section over here and if i have to find out the current through this current through this cross section like the definition of current dq by dt dq is lambda times dl that divide by dt lambda times dl by dt is lambda times velocity which is lambda is 2 micro coulomb per meter velocity is 4 velocity is 2 total answer will be 4 into 10 is power minus 6 amperes it's a direct formula first part what is the answer get the expression okay do not you don't need to substitute the values otherwise you become a lot of calculations are there first part you can calculate current is given to us so charge of one electron into dn by dt this is equal to current that is 32 into 10 is power minus 3 so from here dn by dt rate at which electrons are passing that divide by charge of one electron this is dn by dt second how will you get the electric field of the surface of the beam okay for the beam you can find out the velocity by equating half mass of a proton into velocity square is equal to e times v potential so from here you get velocity of the beam so from velocity of beam how will you get the electric field can i use this formula all of you lambda derivative 2 pi h lambda r can i use this this formula you remember for an infinite wire can i treat it like an infinite wire beam of current so but the charges are moving here right you are saying charges so what let it be what is the problem how does it matter all matters is that the there has to be a charge if it moves it will create magnetic field also but for electric field movement doesn't matter okay so can i convert this formula in terms of velocity and current because lambda is not given to us velocity is given to us current is given to us can i convert this formula can i write lambda as current by velocity why so the previous question was so take it as lambda into dl so what we can do is that current which is dq by dt can be written as lambda into velocity so from here lambda is current divided by velocity so you can substitute the value of lambda to be current divided by velocity and get the electric field right the third part we can do after the class also it is integral a dot dr okay now let me go to the equivalent resistance type question once we are over with whatever i am giving we can go one by one and see whatever questions are left hmm do this here ruptures ruptures vikas got something raaz di got something raaz di check your answer again advaite check your answer again fine so now let us do this so you can see that there are two connecting wires over here one connecting wire is this connective wire is this, other connective wire is that. So this entire thing, entire red color thing, is it having the same potential? I can shrink the size of this red color to be a point because there is no difference. All the points having the same potential, the same thing. So this entire thing get transformed to a circuit like that, like this. Now you can do it, you are on 2, 2, 2. So this is 2 by 3 plus 2 by 3. Equal resistance is 4 by 3 ohms. So this thing got something. Is it calculate what it is? But you are writing calculate how much it is. Give me the value, not the fraction. Decimal is 1 decimal. So I want to connect these 2 wires in series and so that they are not temperature dependent, the overall resistance should remain fixed which is R1 plus R2 should remain equal to 20. R1 is the earlier resistance 1 plus alpha carbon delta T plus the earlier resistance of 2, 1 plus alpha of the iron delta T. This is equal to 20. So R10 plus R20 initial resistances plus R10 alpha C plus R20 alpha iron that into delta T should be equal to 20. Now if this is to be independent of the change in temperature and the coefficient of change in temperature should be 0. So R10 alpha C plus R20 alpha iron should be equal to 0. So because R10 and R20 they are positive quantity, one of them alpha C and alpha iron should be negative and that is how it is. And another equation is R10 plus R20 should be equal to 20. It should remain 20 no matter what temperature it is. Solve them, you will get one of the resistance to be 11, 18.18 ohms, other one is 1.82 ohms. Mayul is it clear? Yes sir. Why is something so dull? Long time office? Solving right Mayul? Solve this all of you. Okay because got something? No because that is not correct. All of you tried? Nobody got the answer till now. So what means sir? What means sir? Coming something you guys have any exam? Come on test. So tomorrow CET test is there? Yeah yeah tomorrow for HSR it is there. So next Saturday also we have it there? Yes so next Saturday that goes. Tomorrow you have school also is it? Yes sir. Like two classes. Okay, explain the CET also. All the exams are important. Then soon we will then start the mock test for BITs will start, BIT will start, CET, mains, advance, all the kinds of exams that are there. Most of it will be doing it. But CET is a very different pattern. Focus is more on speed. And in CET you know physics test is separate. Like after physics test is done, there is 15-20 minutes break, then chemistry test, then break, then math test. It is like that in CET. All right so I can see it's only one or two of you got the answer. Can you see here one thing that this red path, this entire red path is at the same potential. Can you see that? So if it is at the entire same potential, what is the potential difference between A and D? Zero. Zero. So is this one of any use? No sir. No sir. Then just throw it away. So now the modified structure is this. And also this point is O. Let's say A to O is a single point. Okay. All of you got this one, one, two, two, two. One and one and parallel, two and two are parallel. So it will become two and two parallel is one. One and one parallel is one and one parallel is half. Then half is then connected with two like this. Okay. So half and two is in series. So that is five by two. Five by two in parallel with one. So two by five plus one is one by r equivalent. So r equivalent is five by seven ohms. Understood all of you? See anything kind of parallel to one, it answer has to be less than one. Equal relation should be less than the least. Type in all of you understood. Okay. Many of you are getting the same answer. So here you can see a Wheatstone bridge or not? Yes sir. So this resistance can be removed. Can you solve this without using Wheatstone bridge? I mean once this resistance is removed, it is straightforward. But suppose you don't want to use Wheatstone bridge concept. Sir by symmetry. Symmetry to lagak, it will come out like a resistance. Yes sir. No. So basically I connected like a battery along P and R for instance. And then current will split equally between those three thingies. The three. So let's say this is i1, i2, i3. Then what do you say? Sir all three will be equal right? Why? No. Because oh. What will be this current? This current should be what? This, this, this. It should be i1, this should be i2. Symmetry is to determine what will be the current. So for current this symmetry. So if current i1, i2, i3 goes from here because of symmetry, same current should come to this side. Getting it? So if this i1 goes there, same current i1 goes there. So what is the current over here? Zero. Zero. So without using Wheatstone bridge also you can do. You can eliminate this. Okay. Once you eliminate these two series, these two series parallel and then the certain parallel. So answer with two ohms. Getting it? Okay. Let's proceed. Okay. Different, different answers people are getting. Can I remove this one ohm? Type in. Can I remove this one ohm or not? All of you, can I remove this because of symmetry? Can I remove that or not? No. We cannot remove it. Okay. Symmetry will tell you what will be the magnitude of current. It will not. Symmetry will not enable you to throw away a resistance like that. Okay. So symmetry will tell you that if this is i1 and this is i2 then let me use different color. This is i1, this is i2. Then what will be the current over here? Same thing. i1, i2. i1 and from here i2. So symmetry will tell me that if it is over here i3. Okay. Then what will be the current over there? i3. And how much current over there? So we can assume something. i2 minus i3. Okay. All of it here. So like this you can apply the junction rule as well as symmetry to arrange this. Now do you think that the joint here matters? Whatever current comes from here goes forward. Whatever current is coming from here it goes up like that. So you can remove this joint actually. Okay. Once you remove that joint it becomes like this. Okay. Then you have, this is 1. It is 2. 2 and 1 is in parallel. 2 and 1 in parallel becomes how much? 2 by 3. 2 by 3. 2 by 3 then plus 2. It is 3 to the 6, 8. 3 by 8. 3 by 3. No. What am I doing? 3 to the 6. 8 by 3. 8 by 3 parallel with 2. Right. So 3 by 8 plus 1 by 2. 8, 3, 4. 8 by 7. Oh. It is the answer. Okay. Okay. Different different answers I am getting. Okay. Shall we proceed? All of you? Yes sir. So if this is current i, this is current i, how much will be the current over here and over there? Did we say? Yes sir. This current will it be equal to i or something else? Middle one. Something else. Something else. So this is i1 but due to symmetry that is also i1 only. Okay. So over here you can see that current from here will go like that only. This joint is immaterial. Okay. Similarly, current from here will go like that. The joint, through the joint whatever current coming from left hand side goes forward. So this joint you can just remove. It is not required. And tell me one more thing. Is the potential of point A and the potential of point B it is same or different? Why? Because potential drop here is IR, here also IR. Okay. The potential of A is equal to potential of B and potential of C is equal to potential of D. So what does it mean? It means that it is as if there is a connecting wire between them. Okay. These two point you can connect them. They are not connected but you can treat as if they are connected because connecting automatically means potential is same and why I am connecting so that I can use series parallel connection concepts which I know already. Okay. So the entire thing will look like this. This is the middle section. It will be untouched 2R, 2R then you have this R then this is your R this and then if I touch B and A and DNC the consequence of that will be this. All of you are able to understand this? Everyone? The circuit everyone understood quickly. This is what you guys have done. Who did like this? Anyone who did like this? Sir, would there be one more R connected? Sorry, the above one. Okay. This is also R. Again this point A and point B, they have the same potential so I have connected them. That's what I did earlier. Sir, instead of doing like this we can do the other way also. See there are many ways you can do this but answer should not be defined. So let's see what is the answer for this. Within these two points you have to find the equivalent resistance. So R and R if the R by 2, this entire thing is what? R by 2, this entire thing is R. R by 2 is parallel to R which is how much? 2 by R, 1 by R, R by 3 is it? The girls were actually 2R in the diagram. The ones which were connected 2 resistors. Oh, this one 2R? Yes, sir. 2R to R, 4R then 4R. 4R parallel with 4R becomes 2R and this R and R becomes R by 2. So R by 2, 2 by R plus 1 by 2R is 1 by R equivalent for these bottom ones. So it will be 2R. So 4 plus 1. Can you do it yourself rather than waiting for me and copy down? Try doing it yourself. 2R by 5 is this. Then 2R by 5 is parallel with R by 2. This is 2R plus 4, 2R by 9. Sir, I think you already included those, sir. What? Sir, we already did for all of those. The 2 by R was both of them. Oh yeah, sorry. So this is 2R by 5, all of this is 2R by 5. This is R by 2, this is R by 2. So 2R by 5 plus R by 2, R by 2 series becomes R. So it will be 5, 2R plus 5R. So 7R by 5 is parallel to 4R. So 5 by 7R plus 1 by 4R. 4 is a 28. So 28 by 27R is the answer. All right, we will proceed now. This is an infinite ladder kind of question. Okay. Yes, most of you got it correct. Chaitanya, check your answer. Chaitanya, check what you did. No. Do you all understand that equivalent resistance between A and B is same as equivalent resistance between A dash and B dash, right? So I can say that the circuit is like this. One ohm, it is connected with equivalent resistance. Equal resistance is let's say R. Fine. So between A and B is also equivalent resistance is R. It is infinite ladder. So it doesn't matter if you remove 1, it will be still R only. That is what makes it infinite. So 2R divided by 2 plus R plus 1 is the total resistance. This should be equal to R. Okay. So 2R plus 2 plus R is equal to 2R plus R square. So R square minus R minus 2 is equal to 0. So you can clearly see that R equal to 2 satisfies this equation. So R equal to 2 ohms. Okay. I think many of you got this. I'll proceed. Others, nobody else hint is that can I write it like a single battery with a single resistance like that? This entire thing, is it equivalent to this? What is equivalent EMF over here? Anybody got this way like this? You can use the parallel connection batteries E1 by R1 plus E2 by R2. Other one is reverse battery connected in a reverse manner. So minus of E3 by R3 is equal to E equivalent divided by R equivalent. Where in 1 by R is equal to 1 by R1 plus 1 by R2 plus 1 by R3. With resistance, we don't write plus minus. Sorry, we don't write minus. It's always plus. So we can first find out equivalent resistance that is equal to how much R1 10, 20, 30. 10, 20, 30. Then E1 by R1 just substitute here the values and small r is this. You'll get E. Okay. Now once you have the value of E, what is the potential difference between A and B? All of you tell me. E minus IR. What is I? Sir, E by R1. No. Is there be a current? It's an open circuit. Will there be a current? No, sir. The answer is E only. There cannot be any current. Understood? Sir, but if you write it like a loop and you solve it when I was getting some current. Current is you are looking at internally to this. Sorry, what I'm saying is that if this entire thing, entire setup is like a battery, which if you connect to the resistance, then there'll be a current. It's like a black box. You're looking inside the black box. Okay. So that is like internal force kind of thing. Okay. Many of you got it. Others. So you have two plates between them. You are putting material. Distance between the plate is small d. Conductivity goes from sigma to two sigma. Linearly. Now, if sigma would have been a constant number, I would have directly written the resistance as, resistance is what? Rho L by N, which we can directly written as one by sigma. L, you can take it as D divided by A. Like this, you could have written. But sigma is not a constant. So which sigma will you write? Sigma or two sigma or in between? So because of that, you're going to assume, you should assume a thin strip like this inside the thin strip. You can say sigma is uniform because the strip thickness is DX, very, very less. So let's say it is at a distance of X. So what is the sigma at a distance of X inside this strip? How much it will be? Sigma into one plus X by B. Sigma into one plus X by D. This is how linear variation will be there. Okay. So I mean, whatever formula you get should be linear in nature. And when you put X equal to zero, it should give you sigma. When you put X equal to D, it should give you two sigma. How you arrive to that? Ruchir? Okay. So I mean, if it is really a problem to anyone, what you can do is you can plot a graph between sigma and X. Okay. And sigma at X equals zero is sigma and X equal to D is two sigma like that. So you need to find the equation of the straight line. All right. This is sigma. This is two sigma. So that is what it is. Or if you do a lot of such questions where linear variation is there, you figure it out automatically. So we can just assume the function to be some AX plus B. Then substitute the values and find A and B. There are many ways. I'm just telling you the coordinate geometry way because it's a linear variation. So you can do it by using algebra also directly. So we have resistance of this strip is let's say DR. DR, I can write it as one by sigma, which is sigma resistance of X, which is this, instead of L, I have to take DX and A is fixed, luckily. So you'll get DR to be equal to one by sigma A DX divided by one plus X by D. Okay. And you can see all the strips are connected parallel. So resistances are connected, sorry, not parallel in series. So they're in series. So just need to integrate this to get the total resistance. Okay. So there will be a log term also again. Clear to everyone. Type it. Is it clear? X will go from zero to D. Others, is it clear? Should I take your names? One more question before the break. It's a part of a circuit. Don't think that it's an open circuit. Current should not be there. It is not a full circuit. It's a network. Sir, are the resistances equal like same resistance? Oh, sorry, I have to write the, this is not complete. I have to put the values. Sorry about that. This is eight volts, half ohms. This is 10 ohms. This is two ohms. Sorry about that. Okay, Ruchir Singh got something. Others? Do you all understand that I1 plus I2 plus I3 is equal to zero? Yes. So now I have to write I1, I2, I3 in terms of V and everything else. For example, let's take the bottom one. So we'll have zero minus I2 into half minus eight is equal to V. So I will get I2 to be equal to minus of two times eight plus V. This is your I2. Then for this one, you have 20 minus I1 into two minus 10 is equal to V. So I1 can be written as 10 minus V by two. Similarly, you can do it for the other one also. We have minus 15 minus 10 I3 minus 6 is equal to V. So I3 is equal to, everyone is getting different answer because there is calculation involved. So I'm not going to calculate and tell you what is the correct answer. So do it properly the first time and get the correct answer. This minus V by 10. This is your third equation. Okay. So just substitute the value of I1, I2 and I3 over here, get the value of V. Anybody did like this? Yes, sir. Okay. So all three who said yes, you got different different answers. Let me tell you the correct answer. Just hold a second. Calculation is very important guys. I can't stress it enough. The correct answer is minus 44 by 7 volt. So if you are really serious about your calibration skills during the break, please try this and get the correct answer. Everything is in front of you written. You just need to do the calculation. If you are serious, you will do it. Fine. So we will meet after the break now. So we'll meet at 6.23 p.m. Fine. So let's continue. Okay, Ruchir. Current you got I1 and I2, this 10 ampere going upwards. So over here, 10 is equal to I1 plus I2 and other could be the loop equation for this. You can write it as minus 20 minus plus because I2 is coming down. So plus 4 times I2 plus 3 times I2 minus 4 I1 minus 10 is equal to 0. So you get 7 I2 minus 4 I1 is equal to 30. Second equation. So solving this, you'll get 40 by 11 as I1 and 70 by 11 as I2. Okay. Okay. Tell me which battery is supplying energy, which one is absorbing. So we have to find E also. E also. E is going to come. You apply Kitchoff loop rule over here. You get the E. You can write that E then minus 2 times into 10 then minus 4 times I1 minus 10 is equal to 0. From here you get value of E. Which one is supplying and which one is absorbing? 10 volt. Is it supplying or absorbing? That's absorbing. How do you say that absorbing or supplying? So we can say by direction of current through the cell. So it is getting charged. So it is absorbing energy. Even this is absorbing energy. No, no, no. Sorry. 20 volt is supplying. This is negative terminals. 20 volt is supplying. 10 volt is absorbing and even E is supplying. You don't need to find out the exact values. Exact value if you get the current, it will be current into potential. That is the power that is absorbed or supplied. Potential of all three points will be equal or not. Potential of A, B and C. What do you think? Yes. All of you. A and C will be different. What? What? Say it again. A and B will be different. A and B will be different. Why? Because there is a resistance across AC. A and C you are saying different, right? Yes, AC will be different. No, but you can go like this. Like this you can go. From A to C. Even the resistance is there, current will be 0. Hold on. You don't need to comment. So A to B, same potential or not? Yes, sir. B to C is same potential? Yes, sir. So A and C are same potential? Yes, sir. Okay. Now tell me between A and B is current 0 since potential difference is 0 between A and B? So current is 0 or not? All of you. Yes, Ruchad. Sir, there is no resistance. I think it wouldn't matter. Ruchad Singh, what do you think? Yes, sir. Same thing. Because... What same thing? Sir, there can be current but potential will anyways be 0 because resistance between A and B is 0. Why current be there even if potential is 0? I don't know. You need to solve the equation. Why? Potential difference is 0. So according to Ohm's law current should be 0, isn't it? Sir, but here resistance is 0. So I can either be 0 or normal 0. Correct. So resistance is also 0. So even if potential difference is 0, because resistance is also 0, current can be there. It is an indeterminate form, V by R. Both are 0. 0 by 0 it is. But when you look at A to C, there is a finite resistance. A3 has a finite resistance this much. And if it is a finite resistance, denominator is not 0 but V is 0. So it is 0 by some number which will be 0. So A3 current will be 0 because it has a finite resistance. So you can remove A3 like this. It has gone. And I should all of you now do it everyone the values of current. Solve it completely. Again, I can just put this thing. I, I1, I minus I1. Write down the two loop equations, two variables I and I1. How much will be the current here in terms of I and I1? There should be how much? I1. So A1 reads I1. And this is 2 reads I. So you need to find I and I1. Those are the answers. Get the answer. Two equations like this you have to write and solve I1 and I1. Ruchir Parek got something. Others? Okay. Others? Nobody else? The answer is 82 divided by 27 ampere and 34 divided by 27 ampere. Ohms? Because ohms look like good. Current go. All right. Let's proceed. Sure. One more chance. If you have made cellular get this one. Many times, they just test your calculations. That's all. They are not testing your concepts. Anyone close to the answer? Ohm is done. Can I write the final answer? 30 seconds. Okay. Nobody is even close. Fine. The answer is 5.1 watts. Fine. So you can do your calculations and see. So I just wanted to highlight where we are in terms of calculation. We'll come back to all these questions. Let us start the instrument. So I can see. Ruchir Parek, what is your answer? Get the check. Check again what you did. No. 3 ohms is not correct. We'll do this. So suppose you have this gallium meter. Gallium meter works on the current, right? So there is a maximum current. There is a current for which it shows the full deflection. Okay. So deflection of a moving call gallium meter falls from 60 division to 12 division with a shunt of 12 ohm connected. So basically, if there is a current of i that goes through the gallium meter, then that current is proportional to how many divisions? 60 division. Okay. So let's call it as i only. But when we connect a shunt resistance of 12 ohm. All right. Same current will split into 2. This would be i minus i1 and this is i1. Okay. So now current i1 is proportional to 12 division. This current that goes in. But from the network, i is coming in. If you connect 12, some current will bypass the gallium meter. So basically we have i divided by i1 is equal to 60 by 12. That is 5. So i is 5 times i1. Okay. And what else we have? i minus i1 times 12 is equal to i1 times the resistance of gallium meter. Let's say r. Okay. So we can put i is equal to 5 i1 over here. So it will become 4 i1 into 12. They are equal to i1 into r. So r is equal to 48 ohms like this. All of you understood? So does shunt resistance parallel to the whatever device we have? Say it again, shunt? Does shunt mean we'll have the resistor connected in parallel to the device always? Shunt will have current in parallel to device always. What does it mean? So like if there's a shunt resistance, will we always connect it parallel? Yes. Yes. Shunt, they go. The meaning of shunt is this. Shunted is a verb. Shunting is a verb. Okay. So shunting means that you are decreasing the resistance drastically. You can decrease the resistance by only connecting a small resistance parallel to a already big resistance. Okay. So by definition is that shunt is connected parallel. Here? Yes. Thank you. But if I write that there is a shunt resistance which I first connected parallel then later on my connected series, then yes, I can write like that. But just like, you know, if I don't write anything, I just need to do gravity is 9.8, but I can say take g as 10. So similarly with shunt also. If I don't write anything, it is parallel. There is a figure also. Okay. Let me draw this. It will be a gallon meter like that. R1, R2, R3. Do this. This is the voltmeter. As in, there is a voltmeter connected like this. The range is one volt. A voltmeter connected like that. The range is 10. A voltmeter connected like this. The range is 100. Okay. Good. See, here we have, this current gives the full deflection to the gallon meter. Okay. The potential difference between these two can at max be one volt. So I can say that 10 is power minus 3. 50 ohm is a resistance of gallon meter into R1. This should not exceed 1. So I can write it as 1. Then I have this same current. 50 plus R1 plus R2 should not exceed 10 volt. 10 volt is a limit this voltmeter can measure. Then this same current should be equal to 100. So from the first one, you get R1, 950 ohms. Second one will give you R2, 9000 ohms. Third one will give you R3. It is 9 into 10 is power 4 ohms. Okay. So like this you solve this particular question. Okay. It proceed. Is it 60 meters or centimeters? It has to be centimeter. Total length is one meter. Yes. Okay. Yes, that's correct. Good. So finally, I mean, good that you are not using direct formula for potential meter. Because if I generally say a potentiometer, then that formula straight away comes in your head. Even by E2 is even to L1 by L2. Then you start finding what is L1, what is L2. So don't learn potentiometer like that. Okay. So you have a circuit diagram of potentiometer like this. Like this. You have R1, R2. And with this, you are connecting a resistance of R. So you have not only R1, R2, you have R also, third resistance. R1 and R2 are the resistance of the wire in which R1 plus R2 is 10. Okay. Apart from that, you are connecting a third resistance R. It is connected in series with this. EMF, this voltage is given to us. It is, see, it is written that it's a potentiometer wire. It is not written that it is a potentiometer. Okay. A potentiometer wire is written. This connected in series with this cell of EMF three volt negligible integrations, source of EMF 10 millivolts is balanced for 10 centimeter potentiometer wire. So a battery like this is balanced. So when I connect a battery like that, 10 millivolts, that is 10 minus 2 volts. So it is balanced. So this is the current I. So there is no current over here. Entire current flows through it like that. So we know that R1 plus R2 is equal to 10. This is the first thing we know. And do I know what is, what is R1 divided by R2? 6 by 4. 60 by 40. So it is 6 by 4. Fine. So from here, R1 is 6 ohm, R2 is 4 ohms. Okay. So we got R1, R2. So the value of current can be written as 3 divided by R1 plus R2 plus R, that is 10 plus R. Okay. And I, if I apply Kirchhoff loop rule over there, then 10 is for minus 2 volt should be equal to current 3 divided by 10 plus R into R1, which is 6. Fine. So from here, if you rearrange the terms, you will get the answer. Clear to everyone? 1790 ohm. All of you, is it clear? Type in quickly and move forward then. Why Roche? No, no, that is not correct. A, G, F. But Roche, you don't know the current. Current here, do you know? How can you go like that? Anyways, we'll solve it. We have got something. See, the final answer what I have, I'm looking at that. I haven't solved it. I will solve it. I'm just looking at the final answer right now. According to final answer, none of you got the correct answer till now. Yeah, Vikas got for the current. I think it's a simple question. Why you're not getting it? Don't understand. You might be thinking that it is a difficult question. That is why you're not able to solve it. You yourself are making it complicated in your head. So this is loop 1. Loop 2 is above. Let's say current is i over here. This one is, let's say i1. That one is i minus i1. So I can say 12 minus 3i1 plus 6 is equal to 0. So i1 is 18 by 3, 6 amperes is current i1. Okay, but that doesn't solve my purpose. So I'll use the upper loop also. So that will be minus 6 plus 3 times i1, then plus 8 minus 10 times i minus i1 is equal to 0. Okay, minus 6 plus 6 is 18 plus 8 minus 10 i plus 10 into i1 is 6. This is equal to 0. So i becomes how much from here? From here i is what? 18 plus 8, 20. You're not able to do this also? From here you get 8 amperes. Okay, so this is 8. We have i1 and i, both. So emitter reading is 8. Now how will you get potential difference between f and g? Descend that, how will you get? You can go across f, h, g or gf1. So go across vg, then plus 3 times i1, which is 6, then plus 8 is equal to vf. Okay. In fact, I don't need to find i to find out vg minus vf if I go in that path. So vf minus vg is equal to 8 plus 18, that is 26 volt. So that is how you saw this. All of you clear? Good current flow through gf also sir. No, it's an infinite resistance. If it is not an ideal voltmeter, then you have to consider it as a resistance. Otherwise, if it is written, it is ideal or if it is not written anything, then you assume voltmeter is ideal, infinite resistance like an open circuit and emitter, if it is ideal, it has zero resistance like a connecting wire. Otherwise, the resistances will be mentioned. I hope I have opened your mind a little bit with respect to this chapter. It's not just Kirchhoff law and Ohm's law. I'll open it a bit more. Isn't it direct? It is reworded in such a manner that you start thinking something special is there, but it's a direct formula substitution. I'll do it now. All right. So current, emitter, voltmeter, this is voltmeter, this is emitter. So ideal voltmeter, ideal emitter. You're assuming, right? No, no, no. Emitter has resistance. 10. Voltmeter has this resistance. And what is this? R? R is not given to us. So you have to find oh yeah. So you have to find. Okay. All right. So what is given to us voltmeter reading? So potential difference across voltmeter is 99.5 volts and the resistance of the voltmeter is 99.5 ohms. So I can get the current through the voltmeter, right? Which is how much current, why do I have to get current? I have to find out. So voltmeter reading is 99.5. Emitter reading, how will you find out? What? Sir, the voltage across ammeter will be 0.5 volts. Voltage across ammeter will be incorrect because the voltmeter has an voltage of 19.5. So remaining voltage will be seen over here. Okay. Good. So that information we can use to find out the current through the emitter which is the main current I. So that current will be equal to voltages 0.5 and the emitter resistance is, is it given? Yes, sir, it's 10. 10. So the main current is how much? 0.05 ampere. So you can get the answer now, voltmeter reading divided by emitter reading. It will be around, it will be, answer given is 1990 ohms. Sorry, ohms. What is the actual value of resistance? Sir, actually the old connection was on the correct, sir. The ammeter. So I did that, I got 1990. Ammeter on the top and voltmeter. The emitter is connected to the parallel, right? So that's what you did, but then if you do that, you get the correct answer. That's what I was doing. I'm not comfortable with this. They have connected emitter parallel and voltmeter here. Who does that? So this becomes, you know, oh, this is not an ideal. Oh, they're not ideal. This thing devices. Okay. So I get the entire thing is upside down. Okay. Let's read this question again. Voltmeter has a resistance of this. Emitter has 10. I shown to calibrate this one. Voltmeter reading is 99.5. So the emitter, the voltage across emitter is 0.5. And the resistance is 10. So the current through the emitter is 0.5 divided by 10. 0.05 amperes is this current. And we have that current also current through the voltmeter, because voltmeter reading is given to us and voltmeter resistance is also there. So this main current is equal to 99.5 divided by 99.5. That is 0.1. So if this is 0.1, which is splits into 0.05 here, 0.05 will go there also. And these two are connected in parallel. The potential differences are equal. So 0.05 into emitter resistance, which is 10, should be equal to 0.05 into R. So R should be equal to 10 ohms. Okay. So like this, this is the way they have connected. We will proceed to the next question now. Okay. This also looks weird. Let's do this. You have to, you know, whenever some simple question is simple chapter is there. Make some effort to find out some weird looking question or different type of question and then practice those. Don't solve those regular type of question, which you anyway know how to solve. That will not help. Yes, I'm muted. I'm not saying anything. Get the answer. Okay. All of you tried. Sir, could you give two more minutes? Okay. Okay, Ruchalpade got something. Others? No, that's not correct. So let us do this. First of all, we'll write Kirchoff loop equation for the first one, then for the second one. So we have 10 minus 7 I1. See, after some time, temperature will become constant and resistance will also become stable. So I am solving for that condition. Temperature has stopped changing. Temperature has remained constant now. So 10 minus this minus of I1 minus I2 times R plus 5 minus 3 times I1 is equal to 0. This will give me 15 minus 10 I1 minus I1 minus I2 times R is equal to 0. Second equation, it will be minus of 4 I2 minus of 10 I2, sorry, 6. I was thinking, I will add 6 into 10. This plus I1 minus I2 times R, then it will be minus 5 is equal to 0. Okay. So we will get over here, we'll get 5 plus 10 I2 minus I1 minus I2 times R, this is equal to 0. Now, power is what power dissipated in the resistance is I1 minus I2 square times R, right? So I don't need to find out I1 and I2 individually. I can find out I1 minus I2, right? So this is the first equation, second equation. If you add them up, you'll get 15 plus 9, we put minus 15 minus 10 times I1 minus I2, then minus of 2 times I1 minus I2 times R is equal to 0, right? So from here, you get I1 minus I2 is equal to 15 divided by 10 plus 2R. This is I1 minus I2. So the power dissipated in this resistance is I1 minus I2 whole square this square times R. So when you then maximize this, dp by dr should be equal to 0, okay? Clear? So when you maximize it, you will get the resistance as 5 ohms. Clear to everyone? Chaitanya or Chaitanya did like this? Yes, sir. Right. Do the next part of the question. But we, it will never reach 5 ohms. Why do not reach 5 ohms? So the temperature is, oh, increasing or decreasing? No, temperature will decrease. Oh, yes, sir. The minimum value is 5. Room temperature will be calculated. Find out, b part. Okay, Ruchay got it. That's known as, Ruchay, how you did it? Explain. Sir, I knew that the resistance is 5 ohms and the temperature, use the temperature coefficient. So 15 plus half into delta T should be equal to 5. Delta T comes as negative, delta T is negative 20. Hmm. So initial temperature was 50. So it will be 30 now. Right? So the answer for the second is 30 degrees Celsius. All of you understood? Type in c part. Okay. Ruchay, Ruchay got something. Not correct. It follows Newton's law of cooling. So rate of change of temperature is equal to the constant times EO temperature minus surrounding temperature. What is the surrounding temperature? 20. Right? This is the law. I put it like this. DT divided by T minus 20 is equal to minus of KDT. You will go from 50 degrees Celsius to 30. Okay. And time is 0 to T. 50 to 20 you will go, right? Because power, time after which power dissipation will be maximum. It will be 30 only because delta T will be 30. Power dissipation will be maximum. Yeah, yeah, sorry, 30. 30. From here you get the answer. Okay. So here in the terms you get the answer. Yeah. Which unit will we get the answer? Time, they're asking, right? Time will be in seconds. Why is that now? Sir, because I didn't give the units, so I was confused like whether we had to multiply by 60 and divide it. Again, now it is usual units. Okay, let me see what else we can do. Okay, let's see. That's all. This one, though, all of you should get the answer. Why are you not getting it? It is very strange, not all of you getting the answer. So I'm assuming that those who are not answering it are wasting their time unnecessarily. So this is 400 ohm, this is 800 ohm and there is an emitter of 10 ohm. The current is 6 divided by 400 plus 800 plus 10. That is 6 divided by 1200 10. You can say that resistance of emitter is very less. So it is 6 by 1200 only. Similarly, a voltmeter is used to measure potential difference was 400 ohm, what will be the reading of the voltmeter. Now you can see the voltmeter resistance is very high. So you can assume it like an ideal voltmeter. So across this, the potential difference will be whatever current you have, approximate current, ignoring the resistance of the emitter into 400. If you assume voltmeter to be near ideal and emitter to be near ideal. Okay, very simple question. Okay, Vikas, Ruchak, what else? This also apparently is a very simple question. You have 10 ohms, it's internal resistance, 8 volts, it's internal resistance. These two batteries connected to 24 ohms, batteries 10 volt, 20 ohms, 200 percent, 8 and 16 ohms, there's 24 ohms. So only simple questions are remaining, looks like. So current is equal to total potential difference 18 divided by 20 plus 16 plus 24. Vikas, now yeah, 18 divided by 50, right? 60 sir. 60. 60 ampere. Hmm. Potential difference across 8 volt battery. So we have to find out potential difference between these two points. Okay, that will be 8 minus IR 16 into 18 by 16. This is what is the terminal potential difference across it. That is 3.2 volts. 3.2, this comes as 3.2, right? Yes, sir. Yes, sir. Okay. Okay, these are simple questions. Okay, first part looks like everybody got it. The current will be maximum. Current will be maximum over here, right? Over here, current is, so this one will consume the maximum power, right? i square r. So i square r should be equal to 50. So it will give me 5, 4, 20. i will come out to be half. All of you are able to understand, i will be half because of which it will be i by 4 here and i by 4 there. So potential difference between a and b will be half times 200 plus 1 by 4 times 200. All of you understand this? Type in. I am going from here to here, like this. So how much it is? This is 100 plus 50, 150 volts. Okay. Second part, find the equivalent resistance and i square r is the answer. Equivalent resistance is 200 plus. This is 100. 300 is the equivalent resistance to i square, which is 1 by 4 times 300. That is 75 volts. Okay. All of you clear, right? Okay, looks like you are finding it very easy. All of you able to solve it? Let me find out one last question. This one. I will just write it down. So first I will draw the circuit. All of you draw with me. There are 4 ohms. 4 ohms. So the wire a to b is a meter bridge. Meter bridge whose radius changes from r to 2r. It is not a uniform cross section. It is a wire which is like this. Understood? a to b, wire is like that from left to right end. You need to find out where should the free end of galvanometer should be connected on a to b so that deflection in galvanometer is 0. We need to make, we need to find out how much is this distance for deflection in galvanometer to be 0. x is what? For current in galvanometer to be 0, as in balanced midstone bridge. Total length a to b is 1 meter. As it is written, it's a meter bridge. What will be the radius at a distance x? 1 plus x by r. x by 1. So it will become x only. 1 plus x. So let's say that deflection is happening at a distance over here x. So this portion has a resistance r1. That portion has resistance r2. So what should be the case? r1 by r2 should be 4 by 4. 1? Oh, it is 4 and 4. That should be 1. It should be better if it would have been different. So 4 by 4. It should be 1 is to 1. And anything else? Can we say about the resistance? Okay, resistance. No. Okay. So you can take a thin strip like this. So the resistance of the thin strip dr is equal to what? Rho into dx divided by area. Area is pi r square. So pi r square 1 plus x square. So then you integrate to find out the resistance rho by pi r square into dx divided by 1 plus x square. So you integrate this from 0 to let's say x. And then you equate this integral to rho by pi r square same integral dx by 1 plus x square from x to 1. Understood all of you? These two guys still are equal. If it would have been like 2 by 3, then 2 times of this should be equal to 3 times of that integral wise. So this gets cancelled and then you integrate and find out. You can integrate and find out. The value of x will come out to be 1 by 3 meters. That is the answer. Okay. So that's it for this chapter. I have seen by the way some very weird kind of questions in which a jockey which is sliding on the potentiometer wire, they say that it is moving with a speed which depends on the sine function like sine omega t plus something. And you have to find out. I don't know exactly what they're trying to find out. But then there could be questions on just how this jockey is moving. Its velocity could be function of time function of x could be some weird function of time and x. So could be anything. So the bottom line is that if you know the basics, if you know that you can use Kirchhoff loop rule at any moment and it is balanced then the ratio of the resistances should be equal from both sides without forcing yourself to directly substitute the values in the equation, then you'll be able to solve all those kind of questions. So I hope I have opened your mind a little bit with respect to this chapter. It's not as straight forward as people think it to be. There can be some tricky question made out of it. So you can spend maybe 1 or 2 hours or 2, 3 hours more on this chapter finding out some tricky, difficult questions. And then we can say that this chapter is over. All right. So that's it from my side. We will meet next week with new chapter. Bye. Thank you, sir. Thank you, sir. Thank you, sir.