 So, we have like did any numerical on it, fine so write down this LCR circuits see at times impedance find what is impedance by the way, what it is? Impedance is the equivalent resistance sort of time, just the equivalent resistance is in the case of direct current when it you find the equivalent resistance in case of alternating current we call it impedance, fine. So, impedance for series is what for series LCR, impedance is divided by z and that is given as under root of X L minus X C whole square plus R square, fine. So, you know what is X L and X C right, X L is what? X L is omega L and X C 1 by omega C. So, in case of series LCR circuit we have derived this, can you find out the impedance of the parallel circuit? For example, circuit is like this, there is a capacitor, inductor and the resistance RLC connected across an AC voltage V naught sin omega t. So, basically I want to find out what is this current I here, you will use phasors to solve this, just like we have used phasors in series circuit this is for series, can you use phasors to find the solution of the parallel circuit? It is not in our school servers, but something which you should know, do the phasor diagram, what will be common among all these three? Voltage voltage, because they are in parallel voltage will be common, in series the current was common. So, what you did in series connection some of the voltages is equal to supply voltage, you drawn the phasor like that, here what you will do? Some of the currents is equal to this current, fine and voltage across resistance and current across resistance they will be in the same phase, now do it, draw the phasor diagram first, if you do not come with revision of what you have done earlier, then come very difficult. So, voltage across resistance and current across resistance, they will be in the same phase, this is the voltage V naught and this is also the current across the resistance, it is correct, both the phases will have same angle, current across resistance and voltage across resistance, does not change any phase. And I know that if this is I R that is I L and this is I C, then phasor I is sum of all the phasors I R plus I L plus I C, now how will you draw the current through the inductor downwards or upwards, will it be same phase 90 degree, it will be 90 degree ahead or behind of this, behind because voltage of inductor is 90 degree ahead, so current automatically is behind, voltage is same across all, so this V naught is same for inductor also, but current across inductor whose amplitude is V naught by X L, will be 90 degree behind the voltage, so this is I L, V naught by X L, now how will you find I C, 90 degree ahead, so it will be like this, so this is V naught by X C. Now, sum of all these three will be the current, total current, can you quickly find out, others, you just need to add these three vectors, you will get a total current, if you add these two vectors, what you will get, you will get this vector, which is equal to what, this is equal to V naught by X C minus V naught by X L, and then I will add this vector with that, this is the vector, we will obtain this one, what is the magnitude of this vector, this is 90 degree root over this length square plus that length square, what is I R value, V naught by R, so this the amplitude of the current will be root over V naught by X C minus V naught by X L whole square plus V naught by R whole square, so this is the value of I naught, hence if I take V naught outside, this is I naught, now can you tell me what is impedance, is this the impedance, this one, no, 1 by 1, because whatever is the equivalent resistance, V naught by equivalent resistance should be equal to current, so 1 by impedance should be equal to 1 by X C minus 1 by X L whole square plus 1 by R square, so using phasor you can solve this particular circuit also, but what if like you know the situation is inductor and resistance are parallel and then capacitor in series and then again capacitor and one more inductor they are in parallel, so situation becomes very complicated then, so you cannot use phasor all the time, so these are the simplistic scenarios, so to tackle those scenarios we will be using something called complex numbers, so first time you will be probably seeing how you can use complex number to analyze the physical scenario, write down usage of complex number in analysis of AC circuits, see what is X, C, X, X, 1 by omega C, X L is omega L and resistance is just R, now I mean see complex number is used to analyze the circuit, does not mean that if I take X C as imaginary, it is imaginary does not mean that I am just using it as a tool, so what I will do wherever I see omega, I just multiply it with I, so what I will do is that I see is 1 by I times omega C and X L is I times omega L, so X C is imaginary, X L is imaginary, but R is real, see this will help us to analyze the circuit, in Arvind plane real and imaginary number they are 90 degree to each other, so 90 degree to each other they should be treated separately, because this will not affect that, that will not affect this, now if it is 1 by omega C it automatically becomes minus I by omega C, so when you add these two they will automatically subtract and that is how they should be, getting it, now if you just take a series circuit 1 R then C and 1 L, like this if you connect, see I can understand there will be slight trouble in assuming X C and X L to be imaginary numbers, but we are not getting into too much of detail on to it, like why and how it is done, what I am telling you if you use complex number you know whatever you understand here if you use that you will be able to analyze it completely, so I am not getting into why complex number is getting utilized here, just like we understood how phasors can be used without getting into why phasors are used, similarly here also we will be using complex number without getting into why complex number, fine, now if you use complex number like this then you do not need to treat it in any special manner, you can add it exactly like the way you have added in DC scenarios, what was the DC scenario if three resistance are in parallel we just have to add them up, fine, so equivalent resistance will be what, equivalent resistance which is impedance will be some of this, this that, so it will be I times omega L minus I by omega C plus R, so if I take R here and take I common it will be I times omega L minus omega C, so this is impedance, fine, now this is a number whose magnitude will be what, if I just take magnitude of Z how much is this, what is the magnitude of complex number A plus I B, under root A square plus B square, right, so if you take mod of Z it will be equal to R square plus X C minus X L whole square, same thing you are getting or not, okay and the phase in R-built plane will be what tan of phi will be equal to imaginary part divided by real part, you remember all this, so this is X C minus X L divided by R, okay, now can you use the complex number in case of the parallel circuit and see whether you are getting the same thing which you have derived just now, so are you guys getting the same thing, sure, you are not getting, so in case of Greenwood sponsor, yeah, what happened to you back, we said it to you, we said it to us, I forward, I forward, I forward, when is this, 29, 30, 31st October, October, okay, and what is this about, sorry, sir, we have a sports and cultural fest going on and more than 25 schools coming, so we are looking for Greenwood high, 3 days, yeah, 3 days, okay, and then we have packages, yeah, we said to you all the details, keep, so if it is connected in parallel through an AC supply, this is R, this is L and that is C, okay, now what should be the equivalent resistance now, if I consider the situation to be very similar to how it was in DC, what should be it, then in parallel, right, in DC how you deal with parallel resistances, 1 by Z is equal to 1 by X C plus 1 by X L plus 1 by R, okay, what is X C, now it is an imaginary number, okay, X L is I omega L and R is just R, are you getting it, what I am doing here, so this will become what, I times omega L minus I, but I times omega C minus I by omega L plus 1 by R, okay, this is 1 by Z, fine, so mod of 1 by Z, magnitude of 1 by Z is under root of 1 by R square plus omega C minus 1 by omega L the whole square, fine, what is omega C, 1 by X C and 1 by omega L is what, 1 by X L, okay, so you get the same thing, fine and what is the phase difference between current and voltage will be tan inverse of this divided by that, I mean are you able to appreciate it is very straight forward if you use complex number, so the phase difference between current and voltage will be such that tan of phi will be equal to imaginary part divided by real part, imaginary part is what, omega C minus 1 by omega L divided by real part that is 1 by R, so like this you can analyze any circuit, for example, so how do you get the tan phi value, so tan phi is equal to imaginary part divided by real part, let us take one example here, inductor capacitor, then you have a resistance and another capacitor like this, find out the impedance of this circuit and the phase difference between current and voltage, this is an AC voltage, this capacitor and this capacitor they are of let us say different values C 1 and next one is C 2, so you also use complex number tan, yeah I expect you to remember that, what is the value of the X or the for the gamma omega L, this is X C 2 will be with R, there is this, this is what you get, this is how you are done or not, I have done it directly with this complex, yeah basically yeah, see X C 1 is what, X C 1 is you have complex number over here, X C 1 is 1 by I times omega C 1, X L is what, I times omega L divided by I times omega L plus 1 by I times omega C 1, getting it, this will become slightly you know bigger expression but do not, so this will be what 1 by I times omega C 2 into R divided by R plus 1 by I omega C 2, so this you just simplify omega, omega get cancelled, I also got cancelled, so you will get this expression, separate real part and imaginary part of this combined thing and then take a square root of real part square plus imaginary part square, so I got this, okay you have taken first the mod of this, yeah and then take a mod of that, no you can't do that, so I just called it, I just called each particular one and then added it, what you have done, tell me now, I took this, I added them both, yeah that's all, how you have taken impedance of this like this, sir, you can take it, take this, take out the value of the, same thing I have done, same thing I have done, this is what, 1 by X L plus 1 by X C 1 inverse of that, it is same thing, okay, it's like you have written 2 plus 2 as 1 plus 2 plus 2 and I have directly did 2 plus 2 as 4, you have done the same thing, okay, this is 1 by of 1 by X C 1 plus 1 by X L, this is also same thing, fine, so like this you can analyze any circuit and the phase difference between voltage and current will be what, tan inverse of imaginary part divided by real part, okay, now it should be very straight forward, okay, but then you can't use this in any school exams, getting it, you can't use complex number of course, but this you can use as a shortcut to solve some objective questions which comes in exams like jail. So how important is AC chapter in means? It's very important, AC is a huge chapter as good as moving child magnetism and also see the thing is in other chapters other than AC, questions have been coming year after year and they have exhausted almost every variety, so if some very nice or innovative questions come, there is more chance that it will be from AC chapter because not many questions were asked from this chapter in earlier days, now they are asked in increasingly more, okay, this chapter in semiconductors, fine, so let us proceed further, write down LC oscillation.