 you are visible. That is fine. You can keep your webcam on. Nobody has an issue. Hello. All right. So good evening, everyone, waiting for others to join in. So we'll wait for one minute at max and then we'll start. So can I ask you a doubt? You waited for seven days for class to happen. You could have said most of the time, I got the road just like yesterday only. Tell me. So like the sign convention you gave that Kirchhoff's law. So what do you mean by like goes from positive to negative terminal, the battery? Battery has one positive terminal and one negative terminal, right? You have seen the battery? Yes, sir. So which way is it positive or negative? I'm a little confused in the direction. Speak a bit loud. Sir, I'm confused with the direction. A battery is denoted by two lines like that. Getting it? The bigger line is positive, the smaller line is negative. Current is going. Current goes this way or that way. That doesn't matter. How are you going through it? That matters. Look at the current. If you're going from a negative battery, negative potential to positive potential, changing potential is positive. When you're traveling the loop like this, changing potential is negative. Don't look at the current when you're traveling the battery. Understood? Okay, so still couple of joining good evening. So just a few seconds will wait. Yeah, Param, I got those questions, but those questions, when I get some free time, then I'll solve it. They can't be solved just like that. You've given me some some Olympiad questions, right? So I need to get some free time to solve, but I'll get back to you. I got those questions. All right, guys. So let us start. So today, we will be continuing with this chapter, current electricity. Fine. And and we have already done substantial amount of chapter already. So very little theory is left over. Okay, so if you just talk about theory, it can get over very quickly. All right, but then the idea is not just to do theory itself. So we will devote entire four hours on this chapter itself. So I have taken a lot of different types of questions as well. So whenever, like as soon as theory gets over, we'll start solving those questions one by one. Okay. So let us start today's session. Your schools have started, right? I can see attendance is pretty thin. I think a couple of you haven't joined in yet. School started, right? Yes, today was the school. Yes, but this week, it was only half day from next week, we are full day. Okay, okay. So how it is going on school? So it's fine. Yes, sir. Like this time, the school itself created ideas for us, our teams. So like we have access to like submitting online homework and all that, like submitting the homework online. Okay. And is it like very hectic or you have in between gaps and it is manageable? So we don't have much gaps at all. Like between switching classes, we don't have, we have like five minutes gap. So then short break and lunch break is like now. Sir, but we barely do anything in a period. Like a lot of the Wi-Fi is slow and stuff. So teachers also struggle in China. Oh, and our thing, like we've been going really strong, like in each math class, we finish at least one exercise. We did like three questions in our math class. Okay, okay. It is everything is getting recorded. So see, that is fine. I think they've started just now as a time progress. They will also get used to it. And at least they have taken a pain to start the classes in school. There are many schools where classes haven't started itself. Okay, only in cities like Bangalore, Bombay, these cities online classes have begun. Okay. So it actually saves your travel time also. So, you know, it is now it is more like the quota kits itself, the way quota people used to get a lot of time, you're also having a lot of time. And you can manage, you know, the classes also. As in, are your webcams on when you're attending the classes? No, sir. All right. So let us start. So we are talking about the second part of the chapter, current electricity. So let us take a first topic. First topic today is electrical energy and power right now. So this is nothing new. You have already seen it and you know, done this in standard 10th also. Okay, so we'll be continuing the same thing. So it is well known fact that if a current passes through a resistance, if a current passes through a resistance, then temperature of the resistance, temperature of the resistance increases. Okay, as if, as if heat energy got created. Fine. Now, energy can neither be created nor can be destroyed. So from where the energy is coming from? What is the source of that? That is a very broad way of answering it. You need to tell me like, you know, in a better manner, as in more crisp manner, I can further ask you like, you know, that so what, what is electrical energy, how it is converted and things like that. So let's get into the basics of it. So when the electrons are moving in the wires, they keep colliding with atoms so that like, because of that fiction, there's heat generated. Okay. All right. So basically, resistance is what? What is resistance? Resistance increase in entropy causes randomness causing heat loss. Okay. I mean, we are not discussing all that. Okay, be confined to the discussions related to the chapter. All right. So anyways, there is actually entropy increases not there. Right. The randomness remains same. Anyways, so here, we are talking about the resistance, which is increasing its temperature. Now, resistance is what? Resistance is resistance is the quantification. What does it quantify? It quantification of what obstruction quantification of obstruction, the free electrons will have while drifting forward. Okay. So while the electrons are moving forward, whatever obstruction they are feeling, right, that obstruction is quantified in terms of resistance. Fine. Now, what is this obstruction? As you have rightly pointed out, the obstruction is basically collision, right, collision between what? Between electron and electron? Between electron and molecule of the resistors. Okay. Electrons and atoms, atoms or nucleus. Fine. Electron to electron collision is ignored because there's a lot of, you know, free space. So it is, it hardly happens. Fine. So the main source of obstruction is electron to the atom collision. Fine. And the atoms are relatively fixed. The atoms are in a form of crystal lattice. Sorry, not crystal lattice. They are, they're having some sort of structure like this. They are relatively fixed. Fine. Three dimensional structure like this. Now, when the electrons are moving, while moving can hit this, after hitting this, what will happen after every collision? After every collision, the velocity becomes random. Velocity becomes random. If it is already random, it is, you know, it remains that only, random only. But if the electron tries to drift in a particular manner, while drifting, if it collides with the nucleus, then its velocity becomes random again. Fine. And because of this randomness, on an average, on an average, electrons are drifting with a constant, constant velocity. Okay. Which is VD. Now tell me, is the electric field inside the conductor doing any work? Yes or no? Yes it is. All of you, please type in. What do you think here? Does the electric field inside the metal does work when electrons are moving, drifting? Okay. All right. So there is a force in a particular direction due to electric field. The movement, the drift is also, you know, in that line only, along that line only. So there is a displacement, net displacement of electron along the line of the electric field. So electric field does some work. Okay. Now, is there a change in kinetic energy of electron? Is there a change in kinetic energy of electron? Net net I am talking about. Does the kinetic energy of electron change or not? So it doesn't. On average. It doesn't change. It doesn't change. It remains half M into V drift square. Fine. So what happens is, if you apply work energy theorem, whatever work is done, this should be equal to change in kinetic energy plus change in potential energy plus whatever heat loss. Right. This is a source of energy. Whatever work you're doing, that work only will create these three kinds of energies. Fine. Now I know that kinetic energy change is zero. Fine. The potential energy more or less remains the same only. So whatever work is done by the electric field, everything gets converted into heat. Fine. So electric work that is done, electric work, entire electric work gets lost as heat. Fine. So electric work, write down work done due to the electric field is basically heat. And heat can be written as delta Q also. So I didn't get why change in potential would be zero. What is potential energy? Potential energy is because of shape, size, location and position. So on an average everywhere, the same number of electrons are there. The atoms are relatively, there is no change in positions if you take all of that. So but if you put in a battery that has a high potential difference, then doesn't that matter? But battery has higher potential and lower potential. But that doesn't create any potential energy. The battery, you know, battery qualifies as an agent to create electric field inside the metal. Okay. So electric field, once it is created, now that electric field pushes the electron forward and backward. All right. Now, if you have to talk about the electrostatic potential energy, electrostatic potential energy is basically charge into potential at that point. Okay. So charge doesn't remain stationary there. Charges are continuously moving. So there is no net charge anywhere. So work done, work done, electrostatic work done. What do you think it will be? Work done by the battery only. This electric work is done by the battery. How much it will be? Suppose delta Q charges flown across the battery, delta Q into V is the work done by battery. All of you understand this? Dhruv, is it clear? Yes, sir. So if there is a delta Q charge that has flown across a battery like this, the work done by the battery is delta Q into V. This is the work done. Okay. Battery pushes the charges forward while pushing. It does the work. So this is equal to delta Q, the heat generated. Okay. So delta Q, which is heat generated, is equal to, this is heat. Okay. Capital delta Q is small delta Q is the charge. So small delta Q can be written as I into delta T. Fine. This into V. So we have got rate of heat generation through a resistance is V into I. This is rate at which heat is generated. Okay. So if V is equal to IR, this can be also written as VI equals to I square R. Same thing can also be written as V square by R for the cases for which the, you know, it is an ohmic resistance. Okay. Now if this heat generate, this is what rate of heat generation, right? Rate of heat generated, dQ by dt is rate of heat generated. And what is this? I delta T into V is a work done. Right. So work done divided by delta T is V into I. So this thing is power supplied by the battery. Rate of work done by the battery is power, right? So this is power supplied. So whatever power is supplied, everything gets converted into heat finally. All right. Now this heat, let's say if you know, if I want to create a question in which I am mixing the concepts of heat transfer also, then suppose conduction is happening, then this dQ by dt can be also equal to Ka delta T by delta X. If there is a conduction happening and temperature remains, I mean, it is a steady state, then these two will be equal. All right. So these are the higher order questions. You'll not see it when you just prepare for mains only, but I'm just throwing an idea wherein, you know, you can always include a concept of heat transfer also because delta Q by delta T is heat transfer itself, rate of heat generation, which should be transferred if temperature, temperatures are equal. Okay. Any doubts here to anyone? No one has any doubts, doubtless. So it's delta X. Say it again. Delta X in this formula. Delta X. Where is delta X? So Ka delta RT by. Oh, that is length. You forgot that formula, heat transfer, conduction heat transfer Ka delta T by L. Remember that? Yes, I got confused by the delta. So that because that is because you haven't solved many questions on heat transfer. So make a note of it that you have to solve few questions on heat transfer. Okay. This clearly means that itself. Nothing else. So when it is a uniform cross section rod, you directly take it as L. If cross section is changing, then you take a small length, create a differential equation and then solve it. But again, that can be explained only when you do some basic problem practice. Any other doubt? Anyone has any other doubt? No doubts. Okay. So you know, why we have discussed this in particular is because when we are receiving electricity at our home, the power has been generated at a place which is far away from your home, probably hundreds of kilometers away. And it is getting transmitted through wires. So there is every chance that, you know, because of the resistive heating of the wire, the power can get lost as heat itself. All right. So later on, okay, later on focus here. So there is every chance that some energy, some energy may get lost in the transmission of the power itself. Fine. So that is the reason why you are using transformer. So there is a small story about the power transmission itself. So long back, there was this very famous scientist whose name was Thomas L. Y. Addition. Okay. He invented the bulb also and many other things he has invented. Fine. So and in modern days, you have heard about the company General Electric. It's a huge, huge company multinational and Thomas L. Y. Addition was a founder of that. So he came up with an idea of power transmission as in generate the electrical power at one location and transmit it to the multiple places, multiple homes and start charging them money for it. So that was a prime idea of, you know, General Electric when it was founded. Now, when he founded General Electric, there was only one form of current or the power that was known. It was direct current. Okay. So using the chemical reactions he used to generate the direct current and that used to get transmitted from the station to the households. The problem with, you know, this idea was multi-fold. The first problem, the first major problem was the generation of the power itself. You need a lot of chemical reactions and entire setup becomes very, very complicated if you have to generate a lot of power. And second was that whatever is a requirement of home, let's say I want two ampere current in my home, then two ampere current should be generated also if I'm using direct current, because current should be constant. Whatever current is generated, same is transmitted. So he could not find out any, you know, any ideas to cope up with this problem. He assumed that, okay, this problem will be there. And because of that, he was, you know, more into the mode of creating the power stations near to the home itself. So if Thomas Elba Additions idea would have been the only idea, then, you know, every society, for example, others palm retreat will have own power station, then some other society will have its own power station, because if power is supplied from far away places, the power get lost as heat itself, fine. But at that time, when Thomas Elba Addition was busy with, you know, creating his, making his company bigger and bigger by bringing in a lot of investment, there was this person, Nikola Tesla. So he was considered to be the, the biggest or the most famous practical scientist of all. There are many theoretical physicists, many theoretical scientists who has given theories after theory, Nikola Tesla was the one who was giving the practical application, the instruments and many other things. So when he, you know, he joined Thomas Elba Additions company as an employee. So he used to work under Thomas Elba Addition. So while working itself, he was challenging the ideas of Thomas Elba Addition. So Thomas did not like this, that every time he used to challenge and say that we can use alternating current and we can use AC motors and things like that. So by, you know, jokingly, Thomas has told Nikola that you just designed one, you know, AC motor and I'll give you some $50,000 prize money. So that point in time, $50,000 is a huge, huge amount. So to, to the Thomas's surprise, Nikola Tesla came up with the design and submitted that. So he, rather than, you know, feeling good about it, Thomas Elba Addition thought that he's challenging his authority. So he has actually fired Nikola. So Nikola was, you know, not very rich person. So when he got fired, he was very poor and he started, you know, working in Thomas's company as a labor. So he used to lay down cables for General Electric. So he, he got thrown out and then he was laying cables for General Electric. So while he was working, he used to talk about his idea about alternating current. So once you might have heard about JP Morgan, so he, JP Morgan is a great banker. So he overheard Nikola talking about his idea and he was impressed and he was like, okay, fine, I'll fund your ideas. So from that point onward, Nikola Tesla got financial support and he has made the idea of alternating current pretty big. So, but then at the same time, Thomas also was pretty big. So then the battle have begun so much so that, you know, Thomas once has electrocuted two horses, one by using alternating current, one by using direct current and used to say people that see when I electrocuted using alternating current that the horse died much earlier. So such crazy things happened and then Nikola Tesla heard about the discovery of, you know, this transformer somewhere in the Europe. So from that point onward, there is no looking back on idea of alternating current because it was proven that generating alternating current is straightforward. You just need a coil to rotate between the magnets and transmission made easy by using transformer. So transformer can increase or decrease the current and the power loss gets diminished by lot more. So because of that, then General Electric has to accept the idea of alternating current. But yes, Nikola Tesla died pretty poor and Thomas still enjoyed a lot of luxury. So this, you know, there is a long, long story behind every little thing you learn in physics. But anyways, so we were talking about transmission of electric power itself. So write down transmission, transmission of electric power. So let us say that, you know, let us say that we are generating a constant power in a powerhouse. All right. And I'm supplying a voltage V from the powerhouse at a current I. This is my power. Now, the rate of heat loss in a transmission wire is what current I square times R. Fine. So this is loss of power. Now, you know, transformers are used only for alternating current. In fact, this idea of transmission should be there in alternating power, alternating current, not in direct current chapter, but then still since it is about the power only. So we are discussing here itself. So in terms of voltage, I can write down loss of power to be equal to P square by V square times R. So you can see that the power loss is lesser. Power is lesser power loss. Sorry. So loss in power loss in power is lesser if voltage is very high. Supply voltage is very high. Now, when you generate the power, voltage may not be very high. Okay, so you have a device called transformer, which will make sure that power is constant. But when you increase the voltage, current gets decreased. So V into I is constant in a transformer. So you can, you know, suppose you're generating 400 volt only, you can make 400 volt to be equal to 40,000 volt. And then you can see the power loss will be minimized quite a bit. Okay. So lesser love lesser loss in power is achieved by increasing the voltage. Fine. But then problem is once you make it 40000 volt, 40000 volt, once it reaches to your home, your equipments are not capable to handle this much potential difference. So what you do, you have another transformer in every society, you will see there is a transformer in your society also, you can see there will be a transformer, local transformer in your society, which will be stepped down transformer. What it does, it takes 40,000 volt and again makes it whatever, whatever volt you want probably 400 volt. So like that, you can use a transformer and decrease the voltage near to your home, you do this. So when you decrease the voltage, even the current is higher now. But now the wires are very close to your home. So power loss will be lesser, but you cannot afford to have lesser voltage at a place which is hundreds of kilometers away from your home when you're getting supply. Okay. So this is a small introduction of the power and heat generation when it is getting supplied. Okay. All right. So the other formula, same formula. Okay, see there are three formulas of power, i square r, v square by r and v into i, all are correct. All are correct. Fine. But then voltage and current should be through the resistance. Whatever voltage and current is written here should be through the resistance. It may not be supply voltage. Okay. If you know voltage, if you know voltage only, then you should use v square by r. If you know current through the resistance, you can use i square by r. It is up to you. Clear Advoc? Advoc? Okay. Any other doubts to anyone? Anyone has any other doubt? No doubts? Okay. So we will now take up a couple of questions related to power. So I have one question. And now when I'm about to give a question here, what is the doubt? Tell me quick. So like if I was transmitting with alternating current, then like will be actual, like let's say I had direct current at the same voltage as alternating current, then which one will lose power faster? That you'll understand only when we are doing alternating current chapter because you know the power supplied will also continuously change because voltage is a function of time. In alternating current, current is also function of time. So even power is a function of time. Okay. So wait for that alternating current chapter. You'll understand then. Fine. Param? Yes. That's what I said. We have unnecessarily had this topic in this chapter, but then since it is there, I had to do it. Okay. So okay, let's do the basic one first. Do this. One by one will do. Okay. These are from 1981. Probably when your dads are writing the J exam. Done with the first question. What is the answer for first one? Okay. All right. So I can see many of you have got some answer. Should I solve it now? Fine. So answer for the first one is B. So, great. So let's see how to solve the first question. So current, sorry, in the circuit shown heat producing 5 ohm is given to be 10. The heat generation through 4 ohm is asked. So let's say this current is I2 and that current is I1. Now you can see between these two points, both the sets of resistance are connected, 5 ohm and 4 and 6 ohm. But if you want, you can use the loop rule also here and you can say that plus 5 I1 minus 4 I2 minus 6 I2, that is 10 I2 is equal to 0. Fine. So I1 will be equal to 2 times of I2. This is your first question. Fine. Now what is given to us is through the 5 ohm, I1 square into 5 is given as 10. You need to find I2 square into 4. This is equal to what? Let's say this is x. So if you divide it, you get I1 square by I2 square, I2 square into 5 by 4 divided by 10 is equal to 1 by x. So x will be equal to 4040 I2 square divided by 5 I1 square. Now I1 is 2 I2. So when you solve it, you'll get it as 2. Now I have not changed the units in calories because I'm taking ratios. When you take the ratio units and you get cancelled away. What about this one? Next one. It's a simple question. Equivalent resistance is what? 30 plus 30? 60? 60 parallel to 20. 60 parallel to 30. 60 parallel to 30 is actually 20. Okay. So the current is equal to 2 divided by 20 which is 1 by 10 ampere. So this is the second question is your class 10th level question, isn't it? Same type of question we have done when we were there in 10th. All right. So let us take one more than all right. Only one student have answered till now. Lot of calculations, right? Should I solve or should I wait? Everyone. All right. So let us solve this question now. Heater is designed to operate with a power of 1000 watts in a 100 volt line. Okay. So the power is 1000 watt and the voltage is 100 volt. So this is what this is rated power. So you might have seen in every equipment, there will be every electrical equipment, there will be some rated power that will be written. By the way, answer is 5. Okay. So in every equipment, there will be a rated power. Now that rated power doesn't mean that it will consume the rated power only every time. Okay. Rated power is an optimum level of power and voltage. Thus equipment can operate fine. But if you change the voltage, even the consumed power will be changed. So how can I use this combination power and voltage? I can find out resistance. Fine. Because resistance won't change, even if you change the voltage, because you can see that it is connected with like this to 100 volt mains. So now the it is not operated on 100 volt the heater which is connected. The overall voltage is 100, not the heater voltage. Fine. So I know that power is V square by R, resistance of heater. Okay. Resistance R is different. So I can take it as small r. Small r is resistance of heater. From here, resistance of the heater is V square by P. How much it is? 10 ohms, right? 10 ohms. So resistance is 10 ohms for the heater. So this circuit, you know, it always helps to draw a familiar looking circuit. If you have just started learning it, never skip steps like this. Alright, so what should be the value of R so that heater operates with the power of 62.5 watts. So the power consumed by this is given as 62.5. Okay. So first, I need to find out what is the current over here. Right. So how to get that this circuit is equivalent to 10 ohms and this will be 10 R divided by 10 plus R. They're parallel like this. Okay. So the current I through it will be equal to what current I will be 100 volt divided by 10 plus 10 R divided by 10 plus R. This is the current. Fine. And this given to us I write. So what's given to us is the current in the the heaters branch. Where it is given. So it's given that heater operates at 62.5. So I in the 10 ohm branch is given. Power in 10 ohm branch is given power, not I. Yes, we can find it. It'll be 2.5. So the resistance of the heater is correct. So the resistance is given to us. Very nice. So I was approaching it in a different manner, a linear manner. So forgot about that. So I square R, which is I square times resistance of the heater will be equal to 62.5. Okay. So the, what should the value of R write correct. So current will be equal to 6.25 under root. How much it is? That's 2.5. 2.5. 2.5. 2.5 ampere is the current. Okay. So if 2.5 ampere is the current, then how much is that current? The potential difference across that will be what? 25. 25. 25 is a potential difference. Then what to do? Then potential drop across that 10 ohm separator resistance is 75. Is 75. So how does it help? Then you can find current I will be 7.5. Correct. So you can get the current through the 10 ohm because potential difference across 10 ohm is 75. So this current I, this is let's say I1, main current let's say I0, you can get by 75 divided by 10. V by R is current. This is 7.5 ampere. So the current through the resistance let's say is I2 will be equal to 7.5 minus 2.5 which is 5 ampere. Okay. So good that you have done it like this because I was trying to approach it through a brute force method. It would have been very lengthy. So all of you should understand how it is solved. So this is how you can solve it pretty neatly. Anyone has any doubts? Quickly ask. Any doubts? No doubts? Okay. One more question. No, two more questions. This one. Anyone got it already? The answer is B. So sometimes we are in a hurry and we mark wrong option. So let us see how it is done. Some heat is developed. The heat developed will be doubled if, which of them? So there's a constant voltage. So when you compare the heat you should use V2 by R formula. You should not use I2R because current will be different in two cases. So this R can be written as rho L by A. So this is V2A divided by rho L. V and rho, they will be constant in these two scenarios. Only area and length will be changed. So look at option B. If length is doubled, L will be 2L. If radius is doubled, what will be A? 4A. 4A. So it becomes two times, right? The heat developed will be doubled if this happens. Correct. All right. So just one more question then we'll get into the next concept. All right. So multiple options correct, okay? Which variable will be constant between the two segments? Current. Current will be constant because, yeah, because they're in series. So resistance of A to B is rho L divided by area that is pi into 2R whole square. So that is 1 eighth of rho L by pi R square. Resistance of B to C is rho L by 2 into pi into R square. So that is 1 by 2 times rho L by pi R square. So I can say that resistance of B to C is 4 times resistance of A to B. Okay? So voltage across AB is what? Current I into RAB and over here it is current I into RBC. So just multiply I both sides. You'll get voltage B to C is equal to 4 times voltage A to B. Okay? So A is not correct. Current density will not be equal. Current will be equal. So current per unit area will be different. Electric field will also be different. The reason is J is equal to conductivity times electric field. J is different. So electric field also be different. Only option left is this, which you can see. I square R is the power. So option B is correct. Got it? Yes sir. All right. Let's take the another topic on the chapter. Now we are getting into the electrical instruments. Fine? So we'll be talking about few electrical instruments and see how they are operated. The first instrument that we are going to learn is wheat stone bridge. So wheat stone bridge looks like this. Okay? So first all of you draw the circuit and then we'll discuss how to. So it's lagging for me. Others, is it lagging everyone? Also for me it's fine. So it's fine for me. Yes sir. I think you can rejoin. It is because of your internet speed. I have done a lot of changes to my settings. It should not lag. I'll just take a quick poll here. Tell me. Is it lagging? Is there a lag? All right. So you can see here what everyone is saying. Those who are facing lag just leave the meeting and rejoin. Okay? All right. So we are talking about the wheat stone bridge. This is how wheat stone bridge looks like. Okay? So we'll name the resistances R1, R2, R3, R4 and this is let's say R5. Okay? So what are the characteristics of wheat stone bridge? Wheat stone bridge should have a square sort of arrangement of resistance and there should be another resistance at the center like this. Okay? This is the wheat stone bridge. Is it correct? No. This is not right now the wheat stone bridge. You have to connect battery across these two. Okay? Now this is a wheat stone bridge. If you connect a battery like that, if you connect it like this, then it is not a wheat stone bridge. Fine? So this you should understand. All right. So let us analyze this circuit here. So basically wheat stone bridge is used. This is an equipment that is used to find an unknown resistance if three known resistance are there. If three known resistances are there. Okay? So now this wheat stone, this is a wheat stone bridge. And if, right now, if the current through R5 is equal to zero, then this becomes balanced wheat stone bridge. Right on balanced wheat stone bridge. So we need to find out the conditions when it is a balanced wheat stone bridge. Okay? So let's say the current I is going like this. Okay? And current I1 goes there. Current I2 goes there. What will be the current here? If it is balanced? I1. All of you understood? It will be I1 only. There is no current on the middle branch. This one will be I2. Okay? Now I want you to write down the Kirchhoff's loop equation in this loop and in that loop. There are two independent loops. Write down the equations for those two loops. Kirchhoff's loop rule. Done. So in the first loop, it can be written as minus of I1 R1 plus I2 R2 is equal to zero. So I'm getting it as I1 R1 is equal to I2 R2. Okay? In the second this thing, this other loop, I can write down it as minus of I1 R4 plus I2 R3 is equal to zero. See, this element doesn't even come R5 because there is no current. Current is zero. So zero into R5. So that will not come in. So if I modify that, it will become I1 R4 is equal to I2 R3. Now we will get a relation for it to be balanced. A relation which doesn't depend on whatever current is flowing. Okay? So I'm getting it as R1 by R4 being equal to R2 by R3. Okay? So if this condition is satisfied, then it is a balanced Wheatstone bridge. Okay? So what is this condition? R1 by R4 is equal to R2 by R3. Now R1 by R4 can be modified and can be written as R1 by R2 being equal to R4 by R3. Like that, also you can write down. Fine? So R1 by R2 is equal to R4 by R3. So any two corresponding ratio today. This by that is equal to this by that or this divided by that equal to that divided by this. Any corresponding ratios of the resistance, if they are equal, then it will be a balanced Wheatstone bridge and there is no current through R5. So if there is no current through R5, is R5 contributing anything? Is there any contribution from R5? No. So I can throw it away. So R5 is useless. So R5 doesn't contribute anything. So R2 and R3 are in series. R2 and R3 are in series. R1 and R4 are also in series. And then they are parallel to each other getting it. So Wheatstone bridge is very popular as in, you know, every time you look at some complicated circuit, the circuit can get simplified very easily. If you identify that there is a balance Wheatstone bridge in between, you can just throw away the center resistance and the circuit starts looking very, very simple. I hope this thing is clear to everyone. Now in this relation, you can see this relation. If one resistance is unknown, other three are known, you can find out the unknown resistance, isn't it? Fine. So this is the Wheatstone bridge. Now we have learned Wheatstone bridge of the resistance. Similarly, one thing I forgot to mention here is that many a times instead of R5, we connect galvanometer instead of R5. So if we connect galvanometer instead of R5, what is the purpose of that? Why we do that? If it's balanced, then it won't show deviation. So you need to make sure that it is balanced, then only you can use this relation. Otherwise, you can't use it. So you'll continuously change the resistances R1, R2, R3, R4 using rheostat or not rheostat actually, using some equipment wherein you can measure the resistance, whatever you are changing. So you continuously change it till the current through R5 or the current to galvanometer becomes 0 and then it will balance and you can use this relation. All right. So now we are going to learn about the Wheatstone bridge of the capacitor. So even Wheatstone bridge can exist due to capacitor also. Wheatstone bridge of capacitors. So instead of resistance, there will be capacitors. So draw this. This is the Wheatstone bridge made out of capacitors. So E, let's say this is C1, C2, C3, C4 and C5. Now this Wheatstone bridge is balanced. When what will happen? When if it was resistance, I say that current to the resistance should be 0. If this is a capacitor, then what should happen at C5? For it to be balanced? There won't be charge separation. So charge to it should be 0. Charged C5. Every evening it starts raining. I think it is about to rain now also. Capacitor C5 charge is 0. Okay. So charge on C5 is 0. Now if charge on C5 is 0, that capacitor is useless. Capacitor is used to store the charge and that capacitor is incapable to store any charge. So I can throw it away. So C1, C4, they are in series. C2 and C3, they will be in series and then their equivalence will be in parallel to each other if it is balanced. Okay. So we need to find out conditions, condition for it to be balanced. Fine. So let me first ask you at steady state, will there be current anywhere? Current will not be there. Okay. At steady state, write down steady state capacitor is an open circuit. Capacitor is open circuit. So there is no question of current anywhere if we are talking about steady state. So, but charges will be there. Right. So if this one has Q1 charge, this plate has Q2 charge. This one will have minus of Q1. How much charge will be there on that plate? On this plate, what is the charge? So plus Q1. Plus Q1. What is the reason? So because conservation of charge in that system. Conservation of charge. You can see that these three plates, they are isolated from everything else. The charge on these three plates, some of the charges should be constant. They are not getting charged from anywhere. It is all induced only. So if this plate gains minus Q1, the adjacent one should gain plus Q1 and this will be minus of Q1 because it is second plate of capacitor. This will be Q2. This will be minus Q2. Similarly, this will be plus Q2. This will be minus of Q2. Okay. Now, you can use the Kirchoff loop rule in the case of capacitors also. So try writing down the equation for these two loops. Sir, if CFI is already charged before we connect the battery, how will that change things? That will not change anything. What happens is, the potential difference between these two points will become zero if it is balanced. So whatever charge will be flown away. Okay, sir. All of you written, all right. On the first loop, if I write the Kirchoff's loop rule, I can, you can see that the first of all potential drop across a capacitor is Q by C only. Right? So if I go from that plate to this plate, should I write plus Q by C or minus Q1 by C1? What should I write? Minus. Minus? No. It is like going from negative potential to positive potential. Minus charge to plus I'm going. Okay. So plus potential increases by Q1 by C. If you're going from negative plate to the positive plate. And then I'm going from positive negative. So this time negative. So minus Q2 by C2. There is no charge on C5, no question of potential difference. So this will be equal to zero. So I can write down it as Q1 by C1 being equal to Q2 by C2. This is the equation for the first loop for the second loop. I can say here Q1 by C4 minus Q2 by C3 is equal to zero. Okay, so I can write it as here Q1 by C4 equals to Q2 by C3. Fine. Now I can divide it. When I divide, I'll get condition independent of the charge. Okay. So Q1, Q1, Q2, Q2 gone. This will give me C4 by C1 equal to C3 by C2. So you can see the ratio is exactly the same way it is there in the case of resistance. So if corresponding ratio of the capacitances are equal, then it is a balanced Wheatstone bridge. Getting it fine. So if this is true, then you can throw away C5. It is useless and C2, C3 will be series. C1, C4 will be series and they'll be in parallel to each other. Any doubts? No one has any doubts? Doubtless? Okay. So there are a few variations to the Wheatstone bridge also. So write down extended Wheatstone bridge. Okay. So here it is R1, R2, R3, R4. Okay. Let's say this is R5, R6, R7, R8. Suppose this is balanced. This extended Wheatstone bridge is balanced. So through which resistance is, through what resistances the current will be zero? If it is balanced, if it is balanced, then current through R7 should be equal to current through R8 and that will be zero. There will not be any current here and there. So you can see that R2, R6 and R3 are in series. R1, R5, R4 will be in series. You can throw away R7 and R8 then. Okay. Now let's quickly analyze it. Suppose current I1 goes there. This is I2. No current goes through R7 and R8. So same current reaches there. I2 and here also I1 like that. Okay. If I write the loop equation, this loop equation if I write, all of you please write with me quickly. I'm doing it. This will be I1, R1 minus I2, R2 is equal to zero. So I1, R1 is equal to I2, R2. This is your first equation. And for the second loop, we have minus I1, R4 plus I2, R3 equal to zero. So I can say that I1, R4 is equal to I2, R3 second equation. Fine. So if I divide it, I'll get R1 by R4 to be equal to R2 by R3. Okay. So if this happens, you know, same ratio, if it is true, you can throw away R7 and R8 in case of extended Wheatstone bridge as well. Okay. So this completes the Wheatstone bridge. Anyone has any doubts? Quickly ask. Wheatstone is a person's name. Don't confuse it with the Wheatstone which you have in the villages. So over here, why don't we consider the central loop? You can consider, but without considering that itself, I have got all the equations. What is the need? No, like if you could get a relation for R5 and R6 also. You can get the relation between R5 and R6, but this relation is primary one. Whatever relations you're getting from here and here, this is the primary one. R5 and R6, you will not get relation. The reason is current over here is different from there. You get a relation between I1, R5 and I2, R6. You will not get relation only between R5 and R6, but this one is a relation between the resistance itself. Understood? Yes. Any doubt? Keep on asking doubts. Anyone has any doubts? Sir. Sir, I had a doubt. Sir, but if we do write the loop equation for the middle square part, then we get that like R2 by R1 is equal to R6 by R5 is equal to R3 by R4. Sir, so if R6 by R5 is not equal to R2 by R1, would that mean that there is some current through the R7, R8 branches? Okay. That's a very interesting thing. Here, if you write the loop equation in this loop, this one is I1 and this one is I2. So when you write it, you'll get I1, R5 to be equal to, sorry, I1, R5, yes, to be equal to I2, R6. This is what you're getting. Fine? So yes, it is very true. I have missed that this I2, R2 and then I will write like this. This one is equal to I2, R6. Then even R1 by R5 should be equal to R2 by R6, right? So yes. That's what even I was trying to say. That's what I'm writing. Yeah, yeah, that is correct. I missed that part. So R1 by R, this, there is a relation for R5, R6 also. R5 by R6 should also be equal to this by that. So all these ratios should be equal. R1 by R2 should be equal to R5 by R6 should be equal to R4 by R3. The corresponding ratio should be equal. Okay? So this is what the extended reach to reach. I'll write it down here again. R1 by R2 should be equal to R5 by R6 should be equal to R4 by R3. This should be R3. Anyways, okay, so this is the extended reach to reach. Usually extended reach to reach is not asked in the exam. But if it is asked, you can now deal with it. Okay, let's take up a question. We have done wheatstone. This one. Tell me, is it balanced? Is it balanced wheatstone bridge? It is not balanced. So is it near balanced, near to balanced? 500, 600. So it is near to balanced. So current through the gallon meter will be there, but it will be very less. So that is, yeah, that it is near to the balance. So current through the gallon meter will be very less. So using that only you have to solve it. Now you need to get the value of all the currents I1, I2 and IG. In fact, you just find out what is IG. So this is another opportunity for you to apply the loop rule and get the value. So if you get the correct value of IG in this question, you will never have to practice loop rule again. But then yes, it is not very easy. Equations are easy. Solving equation is not easy. So the third loop will be ADC and then through the battery and back to A, right? Yeah, there are three loops, three independent loops. You have to graze entire area. Have you written down the equations? Yes, sir. I'll just write the equations at least here. I'm not going to solve it. Okay. I'll just tell you the final answer because these are linear equations. I'll not teach you how to solve linear equation that you should know. Three independent loops or three equations. Be very honest with yourself. You can easily fool yourself. You know, the books are there everywhere. Don't fool yourself. All right, so these are the three equations. Anybody is getting answer? Okay. Two answers I've got till now. Both are incorrect. Yeah, Param got it. That is, yeah, but then it is approximately. I think the actual thing is zero, zero, wait, I'll type it. No, that's not correct. Yes, I know it's close to that. All right, so if you solve it, if you solve these equations properly, you'll get galonometer current to be equal to 4.8 milliampere, which is 0.0048 ampere. Okay, so this is how you do this particular question. So not all weights on weights, bridges are balanced. So sometimes you have to work hard also. Anyways, so we will take a lot of questions on Wheatstone Bridge later on. Right now, let us proceed. Let us proceed with the next topic and complete this chapter before the break so that after the break, we can focus only on problems. Write down potentiometer. Yeah, Ruchir, write down potentiometer or not potentiometer. Sorry, sorry, sorry, sorry. There is meter bridge before that. So meter bridge is a device to find one unknown resistance if only one resistance is known. If one resistance is known, other is unknown, you can get the known resistance. In case of the Wheatstone Bridge, you need to have three known and one unknown. But here you just need to have one known. Okay. Now, if you draw the circuit diagram of meter bridge, it exactly looks like a Wheatstone Bridge. So we will not draw the circuit diagram, rather we will draw the experimental setup. And you'll see that it is actually a Wheatstone Bridge itself. So all of you draw the circuit diagram. First we'll draw it and then we'll discuss what it is, how it is and why it is. The middle section of the wire is thicker. It is made up of different material. So that is I made it orange. This is key. This is a battery. Now these are elbow shaped one and this one, these are basically copper plates, copper plates having large areas. So that is why they look like this. They don't look like a wire. The reason why I'm taking large area is because I don't want this copper plate to introduce any resistance. Resistance of this copper plate I want to ignore. So that's why I've taken the thicker copper plates and arranged like that. Fine. And this point is called A. This is called B. Fine. Now is a question why it is called meter bridge. It is called meter bridge because there is a one meter scale over here. This is a one meter scale which is right below that orange wire so that you can measure the lens, any lens if you want to measure. So this is the setup and right now the upper circuit is not complete. In order to complete the upper circuit you'll connect one unknown resistance here. This is unknown resistance R and this one is known resistance S. S. R and S are the two resistances. And from the center, from the center you're connecting a wire through a galonometer and that galonometer is connected to a jockey. Okay. Have you seen the jockey? Yes. Others it's a pen like thing. So I'll just show you how does it look like. Is it like those alligator clips? No, no, no. That's like a stylus kind of thing. Ah, like stylus. This one. Have you seen this? Yes, sir. Yeah. So I had to use it like a fountain pen. So why it is like this? What you're doing is you are taking this pen and sliding on the wire. Okay. So while you're sliding on the wire, you don't want to break the circuit. You want to maintain the connection, electrical connections. So that is why it is like this. And electrical circuit is complete through the pen. The current passes through the pen and then the circuit will be complete. And so this pen is basically the jockey which is connected to the galonometer. So if there is any current through the jockey, galonometer will detect it. Okay. Now this orange wire, this orange thick wire is an alloy wire. Orange, not thick. Just fight it as orange wire. It's an alloy wire. It is, it's a metal only but then it doesn't have conductivity of copper. You don't want that. Resistance of any portion, let's say resistance of length L of the orange wire is proportional to the length of the orange wire. Okay. So while you're sliding the jockey on the wire, you're changing resistance of this side and that side. And it may happen that at some point when appropriate resistance are found, the galonometer current becomes zero, nichrome. It is made up of nichrome. It can be made up of any other alloy also. There is no foundation as such. It doesn't matter actually. Fine. So now this is one meter scale. We measure length in centimeter. So this is 100 centimeter. So from A to D, A to D, if length is L, what will length from D to B? 100 minus L. Getting it. Now tell me, are you able to see a Wheatstone bridge here? A to D is one resistance. D to B is another resistance. Everyone please type in, are you able to see the Wheatstone bridge? So there is a Wheatstone bridge and if the galonometer has zero current, then it will be balanced Wheatstone bridge. Balance Wheatstone bridge means what? R by S should be equal to resistance of A to D. Resistance of A to D is what? Rho times rho L by A. It's a uniform wire. So resistivity and the area of cross section will be same. The resistance of D to B is this. This can be written as L divided by 100 minus L. So without knowing what is the exact resistance of A to D, I can get an expression. R by S is equal to length of A to D divided by 100 minus L. 100 minus L because it is one meter. That's why it is called meter bridge. Clear to everyone? Now here, if you can calculate the length, if S is known, you can find R using this equation. Okay? Let's take a question on it, then it will be more clear. Oh, I'll just copy this. Rajdeep, is it clear? Yes, sir. Okay, so this is the Wheatstone bridge. Why copy it here? Because the question refers to it. Solve this question. Once you're done, type in your answer. No, that's not correct. Yes. Correct. All right. So the answer I'll write down first. The value of R is 6.86 ohms and the value of S is 13.5 ohms. Okay, fine. So let us see how to go about this particular question. Null point is found at 33.7. Null point means the gallium meter is at that point where it is balanced. So if L is 33.7, then R by S should be equal to 33.7 divided by 66.3, right? This is the first equation. If a resistance of 12 ohm is connected parallel with S, so 12 ohm is connected here. The null point happens at 51.9. L becomes 51.9. So equivalence of 12 and S is connected, right? So I can say R divided by equivalence of 12 and S is 12 S divided by 12 plus S. They are parallel. This is equal to 51.9 divided by 48.1. Okay, this is your equation 2. You have to solve these equations and you'll get these answers. Everyone understood? Any doubts? Prabhupada, is it clear? So S over here is the unknown resistance. See, listen here. In experiment, yes, S will be unknown and R is known. But when I'm creating a problem based out of the meter bridge, I don't care which is unknown, which is known. I'm using a setup of meter bridge and I'm doing whatever I want. Okay, so you should be able to freely analyze it. Do not constrain your thought process like, oh, I have learned the derivation that S should be known, R should be unknown and things like that. All that is just good to know knowledge. It doesn't matter which one is known, which one is unknown, if you are having basics intact. R by S should be equal to this. Sometime S is given, maybe R is unknown, it doesn't matter. Any other doubt? No doubts. Okay, let's proceed forward. And the next concept is potentiometer. Potentiometer is the last device we are learning. As in also it is the last topic of the chapter. All right, this is the last topic of the chapter. So, but the thing is that potentiometer is very important and many times students do not learn it systematically the way it should be learned. And if a question comes from this chapter, there's a high chance that it will be either on potentiometer or on meter bridge. These are at the end of the chapter on potentiometer. So potentiometer is a device which is used to calculate EMF, EMF of the cell and its internal resistance. Okay, what voltmeter does? What voltmeter does? If you connect voltmeter across the battery, will it measure EMF? No cell, it will measure the voltage difference between the two. It will measure the voltage difference across the battery. It will not measure the EMF. Okay, so in order to measure the EMF, you can use potentiometer. Let's see how the experimental setup is. First, we will quietly draw the experimental setup. After that, we will see how it is working. Okay, let's try it. Lag is there. I'm not even speaking. How can you understand there's a lag or not? So when you're drawing, it's coming like two, three seconds after. How do you know when I have drawn it? So your mouse is going with that button. After some time, then the images. That is fine. Okay. It'll go away. It's not a lag. When I'm speaking and drawing is coming, maybe five, for three, four seconds later, that is lag. That is troubling. Is that happening? Now I'm drawing a line from galonometer and I'm joining here like this. Is there a lag? It's still not coming. Yeah, now it's okay. So it's fine for me. It's fine for everybody. Yes, that's fine. You can rejoin. Those who are feeling the lag, rejoin. Leave the meeting and rejoin quickly. I'm just drawing the diagram only. You can leave and rejoin quickly. So this is the setup This is point A. This is B, battery source B. This is a rheostat. Resistance is R. Okay. All of you have drawn, right? Now I'll tell you what it is. See, there is this battery B. This battery is called driver cell. B is driver cell. This is the name of it. Okay, drivers or driver cell. It drives the setup. Okay. From point A to, let's say, point C. From A to C, there's an alloy wire. Okay. And I've folded it. Do you know what is the reason we have folded the alloy wire like that? I do not have, you know, such big space. So I folded it. Fine. So when you fold it, you can in the same space, you can accommodate larger length. That is the only reason. All right. I can see at point A, we have two batteries, even an E2. All right. So what this experiment lets you do is that if even is known, you can find out E2. One known EMF, other unknown one you can find out. Now, how do we conduct this experiment? See, these two batteries, the battery which is driving and these two EMFs, these two are opposing each other. There's a positive battery connected with the positive of the driver cell. Fine. Both of them. And both these cells have been ended at the key box. One and two are the points where the batteries points have been ended. And three is a point through which gallium meter is connected. Now, if I connect two and three, now the circuit is completed for E2. Even is in open circuit. Even doesn't do anything. But two and three are connected. So A, E2, 2, 3, G, N1 like this, a loop gets completed for E2. So current will flow through E2 if it is allowed. And gallium meter is connected here and you are using a jockey to move over this folded alloy wire. N1 is a neutral point. What is a neutral point? Neutral point is a point when current through the gallium meter becomes zero. Fine. So you found it as N1 for E1. And for E2, you got it as N2. So when you're connecting E2, your neutral point is here. So when you bring jockey here, current through gallium meter is zero. But when you connect E1, you have to move the jockey here for current to become zero. Okay, this is what is happening. All of that we'll again learn. But step number one is drawing this circuit diagram corresponding to that. This is an experimental setup. Okay, just like meter bridge is experimental setup of Wheatstone bridge. Wheatstone bridge is a circuit diagram of meter bridge. Can you draw this circuit diagram for this potential meter experimental setup? All of you. So why is the alloy wire folded? Because I want larger length. Yes, sir. Have you tried drawing the circuit diagram for this? Okay, now all of you focus here. Very important thing. See, this is how circuit diagram is. The battery B and the resistance R looks similar in the circuit also. Now this is point A here. Then there is a resistance corresponding to A to N1 length. This length from here to here, one resistance will be there. From N1 to N2, another resistance is there. Okay, so A to N1 is, let's say, resistance R1. Okay, this is R1. And N1 to N2 is, let's say, resistance R2. This is the outer circuit. And then you have EMF connected. Let's say, I have connected E1. Fine. And E1, E1 has internal resistance also. So you, if you want, you can put as internal resistance R. And there is a galonometer connected. This is galonometer like this. Okay. And yes, there is one more battery like that. This is E2. How many of you got this correct? This circuit diagram, anybody able to do this? Have you understood this? Everyone should understand how this is circuit diagram for the... So E1 E2 part, I don't understand. Which part? How do you put E1 E2 where you put that? What is your confusion? Tell me. So like, someone's like, I'll just look at it, okay? From this point A, this is point A. There are two EMF, E1 E2. Okay. Yeah. On this circuit, I have actually connected E1 here, like this. Oh, okay, sir. Then I understood. Okay. So it is balanced. All right. This is balanced at N1. Balance means what? Current to the galonometer is equal to zero. Okay. So if current to the galonometer is zero, then tell me if current over here is I0, what will be the current over here? How much it will be? I0 only. Minus. Minus only. Over here, there is no current because galonometer current is zero, which is part of this loop only. In that loop, zero current. This is what it means. The same current I0 will flow in the outer circuit. So can I say that... I can apply the loop rule there. V minus I0 times R1 minus I0 times R2 minus I0 times R is equal to zero. So I can get the value of current to be equal to V divided by R1 plus R2 plus R. Now, can I say that if galonometer current is zero, both the cases, right? When I connect E1, I got galonometer current zero. When I connect E2, then also I got galonometer current zero. In both the cases, I'll have same current flowing in the main loop. Is it correct? Can I say that? The reason is R1 plus R2. This is what? Total resistance of wire. So from N1, you can go to N2. Then R1 will be different, R2 will be different, but R1 plus R2 will not be different. R1 plus R2 will be same, which is total resistance of the wire. Because of that, the current I0 in both the cases will be same, whether you go the balance point as N1 or you get balance point as N2. The reason is R1 plus R2, the summation is fixed. Indigely, they can change. All of you understood this? Please type in quickly, waiting for your comments. So could you just repeat it once like what we just said? What I said was in the denominator, we have R1 plus R2 plus R, which is a constant. Because R1 plus R2 is a constant and R is anyway constant. Yes. So current is a constant, voltage is same in both the cases N1 or you go to N2. So this is when E1 is connected. When E1 is connected, this one will be the current and when E2 is connected, then you move the jockey to N2. When you move the jockey to N2, R1 will be different, R2 will be different, but sum of R1 plus R2 will not be different. It will be same as initial R1 plus R2. Sir, but won't the two of them be in parallel? What do you mean? Oh, yeah, I know. I am connecting one by one, not together. Yes, sir. First I am connecting one and three, then I am removing one with three and connecting two and three, one by one. Yes, sir. Okay. So now can you write down the Kirchhoff's loop rule in this loop, all of you? Yes, sir. What do you get? E1 minus, are you getting E0 is equal to I0, R1, all of you? Internal resistance is not coming in the equation because the current is zero there. Right? You get this. Okay. Now you are connecting E2. When you connect E2, R1 becomes R1 dash, R2 becomes R2 dash. Then if you connect E2, instead of E1, E2 will come. So E2 will be equal to I0, R1 dash, because R1 will be different now. All of you agree? I0 will be same, R1 will be different because I have slide the jockey to point N1. So R1 will be now the resistance associated with this entire length from here to here. Earlier it was still here, the length, the resistance of that length. Now the length has been changed. But the current I0 won't change. I0 will remain that same. That is a trick in this experiment. Current I0 will not change. So I can say that E1 by E2 is equal to R1 by R1 dash, which is equal to, R1 is proposed to length, so L1 by L2. L1 is the initial length of A from N1 and L2 is the length of A from N2. Everyone understood any doubt? So why have we ignored the internal resistance? Can I speak a bit loud? Tell me. So why have we ignored the internal resistance while doing the loop rule? Internal resistance will not even come because current is 0. So 0 into small r will be 0 only. Yes. If you can write down like E1 minus I0, R1 then 0 into small r is equal to 0. 0 into small r becomes 0 because gallon meter has zero current. Let's see. There are two things. First is the first trick is that the current in the outer circuit, this circuit remains unchanged if gallon meter current is 0, no matter what EMF you connect. That is the first trick. Second trick is because gallon meter current is 0, the potential difference across the internal resistance is also 0. So internal resistance will not create any potential difference. So whatever potential difference you measure, there will be EMF only. These are the two main tricks of this experiment. Any other doubts to anyone? So how do we calculate the internal resistance now? Coming to that. All right. So next topic is the last topic of the chapter calculating the internal resistance using potentiometer. Write down internal resistance using potentiometer. All right. So we will draw the setup again. So all of you please draw it. It's totally worth it. Okay. I have drawn at least 100 times then also I draw it systematically. So you guys should also do that. Now you have only one EMF because you're trying to find the internal resistance of that EMF only. You don't need two batteries. The upper one is a resistance box. Have you seen resistance box? Yes. How it is different from rheostat? So you don't slide this thing you take out? You take out the resistance of that pin or something. Yeah. That is the mechanism. But how it is different operationally? So you only get natural numbers as the resistance. The difference in rheostat, you never know what is a resistance. Okay. But in the resistance box, while you're changing it, you will know what is the resistance. Okay. It is marked. So what are the two lines that you draw? Like the white ones that you've drawn here? That is just to increase the aesthetic of the diagram. Nothing else. Fine. So we have two neutral points again. The first neutral point is when only EMF E is connected. Second neutral point is when the resistance box is also connected in the circuit. Okay. So can you quickly draw the circuit diagram corresponding to this experimental setup? Do the circuit diagram. Let me know once you're done with the circuit diagram. So what is N1 and N2 okay? N1 is a neutral point when only E is connected. N2 is next neutral point when resistance box is also connected. There is a key here. When you close the key, resistance box and EMF will be parallel. N1 is when you don't have resistance box. You have removed the key. Neutral points are points which represent the situation when current through gallium meter is zero. So N1 is a point when current through gallium meter is zero when only EMF is connected. N2 is a point when current through the gallium meter is zero, when EMF and resistance box both are connected. Clear Mehul? Yes, sir. All right. So let us quickly draw the circuit diagram for this. This is the external circuit Rv. This is R1. This is R2. Okay. Then you have an EMF with internal resistance small r. Fine. And you have a gallium meter. And then you have a resistance box of resistance. Let's say R, this is R. Let's take it as R0 and this as R. The resistance box resistance is R, which is known to us. This is internal resistance small r. This is EMF. Okay. Right now the switch is open. The resistance box is not connected. So this is a circuit diagram. How many of you got this correct? Circuit diagram? Anyone got this correct? Yes. All right. Great. So now this is a point which is neutral point. Neutral point represents that there is no current in gallium meter. So how much current will be there at neutral point? What is the current over here? It will be zero. There is no current. Right? Correct. It will be zero. So the current in the outer circuit on that circuit, let's say the current is I0. How to get I0? Same way. It will be V divided by R1 plus R2 plus R0. So this is the current when only EMF is connected. Now when I connect this also resistance box also, then R1 will change. It will become R1 dash. R2 will become R2 dash and new current outside if there is no current in the gallium meter. It will be R1 dash plus R2 dash plus R0. So since R1 plus R2 is a total resistance of the wire, this is equal to R1 dash plus R2 dash also. So some of the two resistance is the total resistance of the wire, which is constant. So current in both the cases are equal. This is very much clear to us. Now if resistance box is not connected, then if you write the loop equation in that loop, what you will be writing here? It will be I0 R1 minus E is equal to zero. There is no current in the internal resistance. So E is equal to I0 R1. All of you understood till now? Anyone has any doubt? This is probably the trickiest of all topics. So I am doing it very slowly. In case any doubts, quickly ask. Type in if you have any doubts. Type in yes or no at least. Case number two, this is when the resistance box is not connected. Now you are connecting the resistance box. Now tell me, will there be a current in the internal resistance or not? Will there be a current? Everyone, will there be a current here or not? So there will be, yeah. Because from that loop, there is no current. But there is another loop which is formed. This one. Let's say this current is I0, small i let's say. So that current small i will be equal to emf divided by R plus R. Isn't it? This current through this and that is zero. So that current, this one can be written as E divided by small r plus capital R. Got it? Now in this loop, this one I am talking about, can you write the Kirchoff's loop equation for this loop? All of you do this and tell me once you are done. Now you will see internal resistance will come in the equation. Got it? It will be I0 times R1 minus E plus small i times small r is equal to zero. This is the second equation. Clear? So now why does the internal resistance come? Because now there is current through internal resistance. Potential difference will be there. Current into resistance. Earlier current was zero through small r. So zero into small r was there earlier which was zero. The sole purpose of connecting the resistance box is to have a current through the small r. Otherwise small r will never come in the equation. A secondary loop is getting created so that a current appears. Clear Rajdeep? Yes sir. So we can say that E minus small r, by the way the R1 will be changed now. I will be R1 dash let us say because the length is changed. You have gone to N2 from N1. Got it? So this will be equal to I0 R1 dash. So I will just write down these equations on the other page equal to I0 R1. All of you please write down again I0 R1 is equal to E and I0 R1 dash is equal to E minus I times this. Isn't it? Yes sir. Yeah that is what it is. So I can divide left hand side and right hand side because I don't want I0 in the equation. So I0 is gone. R1 by R1 dash is the ratio of lengths. So see it will be slightly different calculation. L1 by L2 is R1 by R1 dash. This is equal to E divided by E minus what is small i? Small i we have calculated is this divided by r plus r. So E divided by r plus r times this. So even E is gone from the expression. So L1 by L2 is equal to 1 divided by capital R divided by small r plus capital R. You can just rearrange the term you will get it. This r divided by small r plus capital R just rearrange it. From here r plus capital R is equal to L1 by L2 times r. So small r is r times 1 minus L1 by L2. This is how you calculate the internal resistance using potentiometer. And probably the trickiest topic of the entire chapter okay and it is at the end which people ignore quite often. So finally tell me is it clear everyone? I am waiting for everybody's response. Go through it no hurry let me know is it clear? Sir so when the resistance box is connected at that point the galvanometer will show deflection? No no in both the cases galvanometer doesn't show the deflection. You are moving galvanometer to another place when you are connecting resistance box. It goes to N2. N2 that is why R1 becomes R1 dash. Understood understood. Rajdeep understood? Yes sir. Rajdeep understood okay. Pranav is it clear? Yes sir. Pranav. Yes sir. All right so I think almost everybody has. So we will take a small break now. Sir it should be L1 by L2 minus 1. Oh yeah sir. Yes sir. Good this is the small resistance. And sir I had another doubt. So to find internal resistance we can find like another way to do it. We can find EMF using potentiometer and then we can connect it to some other circuit. Find i yeah using ammeter and voltmeter. Correct why not? You can do all that. Sir it seems easier than this that's why. No but it is more accurate here. Ammeter will have its own resistance. Voltmeter also will not be ideal. Okay when there is no current that is where the external conditions will not play much. You know experimentally this is accurate. So when we are conducting experiment accuracy is more important than simplicity. Okay sir. Any other doubts anyone? No doubts then we will take a break. Right now it is around 6.30. We will meet around 6.45 and we'll be solving only and only questions and discussing. There are many questions I have shortlisted. So come back in time. This is your break. We'll meet on the other side of the break. Bye. All right so I guess everyone is there. We can start taking up the numericals. Let's take it one by one. Few questions you might have seen earlier. Do this. What is the answer? It's straight forward. The answer is A. Zero resistance. Why zero resistance? Because by this two ohm is not part of any loop. Okay so until this two ohm is part of loop current will not flow through it. If current goes through it like this there has to be a current that is coming back and completing the loop that is not possible. So current through two ohm is zero. You don't need to write any equation as such. This one is it a balanced extended Wheatstone bridge. Yes sir. Yes sir. It is. Yes sir. So now do it. All four. A, B, C, D. All right so you can see that it is balanced Wheatstone bridge because two by four is equal to two by four is equal to two by four. So you can remove one ohm. These two one ohms you can remove. Okay so if you remove it it's gone. Current through PQ is zero. That's correct. I1 is what this is the I1. You can get the equivalent resistance two plus two plus two six connected parallel to 12. What is the equivalent resistance? 72 by 18. Four ohms. So yes I1 is three. Then I2 is what? Where is I2? This current I2. So 12 volt connected across two plus two plus two six. So yes I2 is two. The potential of S is less than Q. How to find that? Potential of S and potential of Q. See this potential is same right? This potential is let's say V. So potential of P is V like this. Potential of S they ask me right? Potential of S is V minus potential drop across two which is I2 that is two into two. I2 is two minus potential drop across further two. Two into two. So potential of S is V minus eight volts. Potential of Q is V minus. I'm starting from here now going that way. V minus that current. This current is what? I1 was three and this was two. So this is one. V minus four. So definitely VS is greater than VQ. No sorry VS is less than VQ. That's our option C is correct. Okay. So we can also say VP is equal to VQ right? Because current in PQ is zero. So potential of P is equal to potential of Q. Potential of S is different. Yes so then we can also say that potential decreases in direction of current. So in P, S. You're telling me another way to solve it. Yes another way. But this is also fine. He's saying potential of P is equal to potential of Q. When I'm going further down potential will decrease only. So that is why VS is less than VP and both are equal. So VS is less than VQ also. So that is also nice. Fine. I'm moving on to the next one. No one got the answer. Okay. Ruchir got it. Multiple correct may be possible. Okay. I'll do it now. Yeah Ruchir I don't remember the option. Correct. Okay. Let me solve it. All of you attempted right? Should I solve or should I wait? So one minute sir. You need to write Kirchhoff's loop rules. There are two loops. Just do that. Okay. So there are two loops this one and that one. Right. So let's say current I is going like this current in the resistance R2 is zero. So same current will go there also. I right. So on first loop I can say that V1 minus IR1 is equal to zero. Isn't it? You write the loop equation you get that. Okay. On the second loop on the second loop you'll get V2 minus IR3 is equal to zero. V2 minus IR3 is equal to zero. Fine. So from here you'll get V1 is equal to IR1. V2 is equal to IR3. So V1 by V2 should be equal to R1 by R3. Okay. Fine. Stop typing in multiple messages when I am telling something. I can't keep checking your messages while talking. Okay. So while I'm teaching something do not type multiple messages. So all of you understood this equation. How it comes. I've got a relation independent of current. So now this should be valid if this is valid then current through R2 will be zero because that we have already incorporated in writing these two. So V1 by V2 should be equal to R1 by R3. So V1 by V2 over here is 1 by 2. Here is 2 by 1. This is D is correct because V1 by V2 is 1 by 2 and R1 by R3 is also 1 by 2. So A and B also correct. It doesn't matter what is R2. So then why don't we consider the current supplied by V2 here? See it is not current supplied by V2 or V1. I is the net current. Okay. This is not supplied by anything. This is the net current. Now you can say that current both voltages are anyway along the same direction only. It is like one person is applying force this way. F1 other is applying like this. F2 total force will be F1 minus F2. Okay. Both are in the same direction. What you are seeing is net effect when you are drawing current IE. This is not the current from any battery. Although you have started drawing IE from here, but this is the net current. Anyone got the answer? Okay. Varun got some answer. Others? It's very simple. Two resistance are first connected. Two EMS are connected in series. Then they are connected parallel. Use the equivalence concept. Okay. So should I solve? I am solving it. Case number one, two, but this is straightforward guys. Case number one, they are connected in series. Across resistance are like this. Case number two, they are connected in parallel across a resistance are. Interference is one. This is unknown and EMF is not given. Let's consider it as E. E is the EMF. This is R. This is R. This is capital R. So if I say this is equal to, this is same as a single circuit with single EMF and single internal resistance, then this will be two times E. This will be small R. Sorry, two R. This will be capital R. Fine. Now, if I say this is equivalent to this circuit, then what is the value of EMF and small R? Everyone, small R, how will you find equivalent of small R? One by R equivalent is one by R plus one by R. So from here, R equivalent is R by two. Look at the topic where we have talked about batteries connected in parallel. This is R by two. What about this EMF? Everyone? So is it two E again? Yeah, I think it's two E. So the problem is E by R equivalent is E. E one by R one plus E two by R two. So you haven't bothered to put your own. You're just guessing it. No sir, like I did it. You need to write it down. So E is equal to R equivalent R by two into two E by R. This is E only. E equivalent is E. Fine. So J one is 2.25 times J two. What is J one and J two? Rate of heat produced. So rate of heat produced in case number one is J one. So J one, I can write it as E square, which is V square is two E square divided by two R plus capital R. This is your J one. Two small R is one only. So you could have substituted the value of R to be one. I can just put it as two plus R. This J one, J two is E square divided by R by, again, R is one small R one by two plus R. Okay, just substitute the thing in the equation. So two E square divided by two plus R. J one is 2.25 times E square by half plus R. So from here, you'll get the value of R to be four ohms. Everyone understood? You can see this is, you can, you know, sometimes we learn concept only to, you know, only to be able to derive the formula. That's all. So when you're learning the equivalent EMF, don't think that you're learning it just to derive the formula in the board exam. Fine. You have to use it also. And this is the chance you can see direct usage of the formula is there. I don't think there is any, there should be any difficulty in solving this question. Direct application of formulas. These are like most difficult questions on this chapter, which are actually, you know, turning out to be very simple ones. So there are not many varieties in this chapter. You just have to practice 30-40 questions, and you're done with this chapter. Are you able to see a Wheatstone bridge here? Yes, so the 10 is the middle. Are you able to see a Wheatstone bridge? Yes. How many? Two. There are two Wheatstone bridge. Two bridges are there, and both are balanced also. The first Wheatstone bridge, where it is? It's around the A2M resistor. Yeah, that one, two. This one. This is a Wheatstone bridge. Yeah. This is a Wheatstone bridge. So from these two points, things are connected. And this balance also, one by two is equal to two by four. So you can remove eight ohm. So this is gone. Once you remove two ohm, sorry, eight ohms, you can see one and two series that is three ohm connected parallel to six ohm, which is actually two ohm. Now, instead of this entire thing, I could have just written two ohm there. So again, this is a Wheatstone bridge. Six by two is equal to 12 by four. Right? So I can remove this as well. So what this turns out to be? 6.5 volt. This is four like this. So you can see that two and four series, six and 12, they are series. So six and 18 in parallel. So how much it is? 12 into six divided by weight. Six into 18 divided by 24. Okay, so nine by two. Nine by two is 4.5. 4.5 plus two. 6.5 ohm is the equivalent resistance connected across 6.5 volt. So the answer is one ampere. All of you have understood or not? This is an important question. All of you must be clear about it. If you are not, please tell me. You can type in your doubts or you can message or you can speak also. Sir, I didn't understand why did you remove the 10 ohm resistor. Everyone understood? I'll go to the next. Sir, are you able to hear me? Hello. Agansha, have you understood? Hello, sir. Akshit. Oh, you guys are muted. Sorry, sorry. Now you can speak. No doubts, right? So I'm going to the... Sir, may hole as it also. Yes, please, please. Speak may hole. Sir, I didn't understand why did you remove the 10 ohm resistor. 10 ohm? Okay. Yeah. See, this one, this one you understood, right? 8 ohm, how I removed. Yes, sir. Dad, you understood and have you understood that equivalent of these is 2? Yes, sir. 1 plus 2 in series 3 and 6 parallel. So it becomes 2. Okay, dad, you have understood. So this circuit becomes something like this. 6.5 volt. This is 2 ohms. This is 6. This is 2. This is 12. This is 4. And in between there is this 10. It's a bridge and is it balanced, may hole? Yes, sir. So if it is balanced, I remove 10. Oh. Got it? Yes. Yes, I got it. All right. So let's move to the next, this one. Are you able to see two resistances? How are they connected? Parallel connection. Okay, we have one answer. No. All of you understood, right? There are two, the resistances are connected in parallel. There are two resistances connected in parallel. All right. So is there any doubt in this understanding that these two resistances are connected in parallel resistance of aluminum and iron? They are parallel. No doubts. So R1 is equal to resistance of aluminum is resistivity of aluminum into area of aluminum. Area of aluminum is what? 7 square minus 2 square into 10 ratio power minus 6. Oh, sorry, this will be in the denominator. Rho L by A, right? So Rho into L 50 mm divided by the area that is 7 square minus 2 square into 10 is power minus 6. This is resistance of the aluminum. Okay, resistivity of aluminum is given. So yes, this particular question has some calculations. So this is a good example for you to understand that in community exams, the calculations sometimes can be not pretty intensive. So Rho of iron into length still the same divided by the area that is 2 square into 10 is power minus 6. Fine. So equivalent resistance is this they are connected in parallel. So R aluminum into R iron divided by R aluminum plus R iron. So you can substitute and then you'll get the answer which is C. Paramp you're there. Paramp left, is it? Sir, his electricity is going. Anyone has any doubt? See, if I tell you that there is an aluminum rod like this and there is an iron rod like that. And if I connect them like that, then you'll understand that they are parallel. Yes, sir. Okay, now how this is different from that scenario? Just that one is inside the other, that is the only difference. Other than that, it is same. Yes, sir. Right? So that is why these two are in parallel. The potential of this phase is same as, so entire phase has the same potential. The potential of iron and aluminum this side is same, potential of iron and aluminum that side is same. So that is why they're connected parallel. Like that also you can understand. Yes, sir. Basically they're like equi-potential surfaces, right? Yeah, you can, that is what. But then don't make it so complicated to make you understand, right? So keep it simple. Try to analyze it the first way I told you. Here, there is an important concept that power delivered by the battery, derivation we are not doing it. This is Theveny's theorem. Power delivered by the battery is maximum when equivalent resistance, equivalent resistance across battery is equal to its internal resistance. Now do it. Everyone, this is a balanced wish-torn bridge or not? It's a balanced wish-torn bridge. All of you agree? R and R2 are, R and R2 are. Okay. So this connection I can modify it and this is how it will look like. You know, wish-torn bridge is a favorite topic of the examiner. It's a balanced wish-torn bridge. Internal resistance here is 4 ohm. So you can remove 6R and then R and 2R series 3R parallel with 6R. So it becomes 2R. Okay? 2R should be equal to 4. So R should be equal to 2. Sir, could you repeat the name of this theorem? Theveny's. Okay. Thank you, sir. You don't need to, you know, read that. It is, it comes in the engineering first year. I am a little surprised why they have asked from that particular concept. All right, let us do some questions on capacitor with resistance. Solve this. We are talking about intermediate state here. Should I solve? You have to write Kirchhoff loop rules equation. Two minutes given. I'll just mark currents so that we all follow the same. This is not a steady state. So I am assuming current to the capacitor and at any moment charge on the capacitor is Q. Should I solve now? Okay. All of you focus here. So let us say this is loop 1 and this is your loop 2. Okay? So for loop 1, I can write V minus IR1 minus Q by C is equal to 0. Okay. For the, for the second loop, I can write down. See, when I am writing minus Q by C, I am assuming this is minus Q. The upper plate is plus Q because I am writing minus. So it means I am going up positive negative plate. Now for the other loop, I am going from negative to positive. So plus Q by C minus I minus I1 times R2 equal to 0. Did you get these two equations? Yes sir. Okay. Now how to write Q in terms of current or current in terms of Q? How to write? DQ by DQ. DQ by DQ is what? No, I1 equal to DQ by DQ. I1 is equal to DQ by DT. Fine. So if there is I1 in the equation, I can write it as DQ by DT because I1 is flowing across the capacitor. So I1 is equal to DQ by DT. I is not equal to DQ by DT. Fine. So this is the main equation. These two are anyway loop equation which everybody will be able to write. So I need to write I in terms of I1 there. So I can get IR2 is equal to Q by C minus I1 R2 or I can be written as Q by R2 times C minus I1. So this I will substitute over there and I will get V minus Q by R2C minus I1 times R1. This is J advanced question. So a little bit of calculation is involved like that. So shouldn't it be plus I1 R2 in the top part, the yellow? This one? Yeah. That's in direction of current. I'm going like this, no? Same direction the current is. So this will be equal to V minus I'll be taking Q by C constant. So we'll be getting R1 by R2. I'm combining this and that plus I1 R1 is equal to zero. Okay. So I'll just do till here. Now you have to solve it like a differential equation. Write I1 as DQ by DT and then you'll get a differential equation. Getting it? I think it is plus I1 R2, sir. See, whatever it is, these calculation errors you can manage yourself. But have you understood how we are solving it? Anyone? So one doubt, sir. So will DQ by DT be negative? Speak a bit louder. So will the DQ by DT be negative, like DQ, negative DQ? No, no, it will not be negative because charge is building up on the capacitor. Charge is coming. It is not discharging of capacitor is not creating current. Charge. The current is creating charge, not otherwise. Okay, sir. Yes. Thank you, sir. Okay. Somebody found some sillier somewhere. What was it? Is there any sillier in the calculation? Yes, sir. I R2 becomes QC plus I1 R2. Where? Which line? Sir, third from the bottom. This one? The underline. This one. I1, I R2 plus. So this will be plus. If this is plus, this will be minus. All right. So it will become a differential equation and then you'll be able to solve it. Let's move ahead. Which one? What should we do here? Yes, sir. You have to find which temperature is greater. This is true or false type. Is it true or false? The statement. Okay. So I know that with temperature, resistance should increase. Right. So if resistance is increased, same value of voltage should give me lesser value of current. Okay. So that is true for T2. T2 is greater than T1. Higher temperature means higher resistance for metals. Today is simple. You can see such simple questions are asked. There's one. Yeah, there are questions which uses symmetry, but today I don't have that. But you know, there are not many such kind of questions. You have plane of symmetry and depending on symmetry, you can identify a few things. I'll just try to see whether I can create one or two questions. Solve this one. So is it multiple correct or one option? Multiple. Okay. All objective questions are multiple. Assume that. Done. All of you. So it looks like a Wheatstone bridge, closing switch doesn't affect anything. So hence, the current through S is zero. So current through R should be equal to current through G. Isn't it? Whatever current here flows, same current there also flows. So A is definitely correct. Current through P should be equal to current through Q. There is no current over here, whether it is closed or not. But that is not in the option. So only A option is correct. You can see how simple it is. The symmetry. Let's see whether we can try this. This is let's say two ohms. This is four ohm. This is five. This is four. And this is two ohm. Is this balanced Wheatstone bridge? No. Is there some sort of symmetry? Yeah. Yes, sir. Yeah. Yes. Can you represent current using the symmetry? Try using it. I want you to write, let's say this is current I. This one is, let's say I1. This one is I2. Now I want you to write down current in every branch. Let me know once you're done. So potential difference? Dad, you don't need. I'm not asking you to find out the value of current. Just represent it as I1, I2, I3 like that. Yes. Done. Yes, sir. Okay. So this is I2. What will be the current over here? I2 minus I3. No, it will be I1 only because of symmetry. Oh, this one is I2 only. So this one will be I1 minus I1. So you use a symmetry. Okay. Did we find the equivalent resistance of the cube last class? No, sir. Okay. What? What? Somebody said yes. No, sir. I said no. He said no. So, you know, you one, you have understood how we have used symmetry here, right? So it makes calculation very easy. You have only two variables, I1 and I2. So you can find it out. But when you use symmetry, you will, you will understand that writing this loop and that loop, these two loop equation will be same, exactly same. So one of them will be redundant now. You have to write for this and that one. Okay. So you need to write an equation involving voltage, right? Otherwise, how can current be independent of voltage? Anyways, so here is the next question based on symmetry. All of you draw the cube, you may not be able to draw as neatly as what I am trying. The dotted line means it is behind. Okay. So every link represent a resistance of one ohm. Every link represent a resistance of one ohm. You need to find out equivalent resistance between this point and that point. So diagonally opposite points between A and B. Tell me what is the equivalent resistance? You want to try it out first? Try it for a minute. It's all about symmetry here. Okay. So what you do is you connect a voltage like this. Do you know, by the way, this is from your NCRT book. It's a solved exercise. Let's say this is voltage V. Then if let's say current is I, can I say equivalent resistance is equal to V by I? Can I say that because then this circuit will be, if R equivalent is the equivalent resistance, this circuit is this only? So V should be equal to IR. So R equivalent should be equal to V by I, right? Getting it? Okay. Now, if this is I, how much would be the current over there? I by 3. These three directions are exactly same. Okay. There is no difference among these three directions. So current will equally divide due to symmetry. Everywhere it will be similar. Current through here will be what? When it is going? This one will be what? I by 3. So this is also I by 3. That is also I by 3. Okay. Okay. This current is what? I by 6. Yeah. I by 6. This is I by 6. I by 6. This one is I by 6. So you can see that beautifully the symmetry works. Everywhere you know exactly what ratio the current will flow. And you can see junction rule is also valid. I by 6 from here. I by 6 from here becomes I by 3. So I've got all the current ratios. If I write down the Kirchhoff's loop rule here along this loop, I'll be able to write it as V minus I by 3 times R. Let's say R is the resistance of this side, minus I by 6 times R. This one minus I by 3 times R is equal to 0. Okay. So I can say that V minus 2 by 3 plus 1 by 6 times I R is equal to 0. So 2 by 3 plus 1 by 6 is what? 4 plus 5 by 6. So V is equal to 5 R by 6 times I. So V by I is what? 5 R by 6. So equivalent resistance is 5 R by 6 where R is a resistance of one of the sides. All of you understood how we used symmetry here to find the equivalent resistance. So which which two faces did you take for the loop? The opposite faces. Say it again. So for the loop equation, which two faces did you take? Faces? Not faces. Loop doesn't include faces. It represents one line only. This is this is the loop. Okay. Okay. The white color loop is the loop I have to write because I have to include V in the equation. Otherwise, I have to write I. Yes. Yes. Any other doubt? You can include any loop by the way. Your wish, but we should be there. We should come. You need to write V by I. No other doubt. We can take few more questions. Let's continue with those questions. Dhruv understood? Yes, sir. Little bit of electrostatics is mixed. Okay. No, that's not correct. All right. So I'll solve it now. Case number one is when switch is open. The circuit looks like this. What I can say about the charges on the capacitor. Anyone they are equal. So let's say this has Q charge. This also has Q three micro farad, six micro farad. And let's talk about charges on the plate only. So this has plus Q minus Q plus Q and minus Q. They're asking how much charge will flow through this wire, right? So I just need to take care of some of the charges on this plate and that plate because if charges flow from here that charge on these two plates will change. So some of these two charges is zero right now. So I don't need to find out Q unnecessarily, right? So do not blindly start finding the charge. So right now net charge is zero in these two plates. Now I'll find out the final charge when I close the switch. Okay. So when you close the switch, this circuit looks like this. Tell me is everyone clear how this is? This is what it is. Anyone clear about it? Okay. Now tell me what I can say about the capacitor voltage or charges or anything. Can I say anything? So voltage across the capacitor is same as across the resistor. Correct. So you can see that the capacitors are connected along parallel to three and six respectively. So potential drop across three is same as potential drop across six. Potential over six is same as that. Getting it. So the current is what at steady state, what is the current that is flowing in? Anyone? What is this current? Steady state current is what? So one, all of you clear? It'll be one because current will not be there through these two capacitors. There are open circuit at the steady state. So three and six are in series effectively. So the current is equal to nine by nine, which is one ampere. Okay. So potential difference across three microfarad is three into one, three volts. Potential difference across six microfarad is six into one, six volt. Okay. So charge on three is q equal to Cv. Right. So nine microcoulomb charge on three, charge on six is six into six, 36 microcoulomb. Now charges are different. So I just have to track these two plates, isn't it? These two plates. So let me see what it is. So this is plus nine. This is minus nine. This one now has plus 36. This one has minus 36. The sum of these two charges is what? 36 minus nine? 25, sorry, 27. 27 microcoulomb should come from this wire and then it will sit on these two plates. Earlier the total charge was zero. This plus that charge was zero. Now these two plates are acquiring charge from this wire, which is 36 minus nine. So 27 microcoulomb should flow from the wire and hence option C is correct. Okay. So all of you, any doubts from anywhere on the chapter? We are done with the chapter. Anyone has any doubts from anywhere? So what arrangement of capacitors and resistors would give me a Wheatstone bridge kind of thing? Like in this question, if I had to do it with these capacitors and resistors, how would I arrange them to have no current through the switch? There will not be any current through the switch any time. Okay. The charges will flow and current immediately will become zero. This is not a, this will never become a Wheatstone bridge, first of all. So is there an arrangement that will never make current flow at any point? This is the arrangement. Current will not flow. Charges will just arrange itself and then that's all. But is there an arrangement that wouldn't require the charges to move at all? Yeah, for that the sum of the charges on these two plates should become zero later on also. But right now it is not becoming zero. So you just change the resistance such a manner that both the two charges are equal. Change the resistance. This one, rather than having six ohm, you have it such that it will multiplication of the potential drop and that become nine microgram. But why are you doing that unnecessarily? But then yes, this is what you have to arrange it that way. So the charges on these two are equal later also. So I have a doubt. I got disconnected. So can you just repeat how you got the potential across the capacitors? The capacitors are connected across the register. These two points are common between resistance and the capacitor. So whatever the potential drop here will be the potential drop across the capacitor. Okay. Okay, thank you sir. Pravdev, you understood? Anyone has any doubts? So we are done then if there is no other doubt. Sir, can you repeat how the 27 micro coulomb would disperse after it reaches X? Dispersement is already there. It is 36 and minus nine. It is there. But I don't care how they are dispersing. All I care is what is the sum of the charges on those two plates initially and finally. Whatever the difference that should come from the wire. So sir, after the 27 micro coulomb reaches X, we can't say what will be the final charge on the plate. Final charges found out before itself. Final charges are arranged in such a manner that potential drops should be equal across three ohm and three micro coulomb. It should be equal. So that is how charges should be arranged. Basically doing the opposite dispersion rate, like calculating the disperse charges then adding it up. No, no, but he is asking a valid question. He is worried about that you know 27 coulomb is given. What happens, you know, because 27 comes from the wire and nine comes from the adjacent plate. So 36 it becomes on the right hand side and nine is given by the adjacent plate. So it become minus nine. Whatever it is, I mean, you can find out how much charge is given by what by just finding out initial charge and final charge. All right then. So that's it from my side. We'll meet again next week and we will take up the fresh chapter. Which chapter is going on in school by the way? The current electricity. Oh, current electricity only. Where they have taught. So meter bridge. Oh, okay, okay. So any idea what they'll do next? So they've been going in order to now. So we are far ahead, I guess. Yes. Okay. So we'll also go in order now. We'll take up the magnetism next. All right guys. So bye for now. Thank you, sir.