 on analytical solutions. This is first of all second order differential equation, is that it? Have you seen second order differential equation before where SHM simple harmonic motion is what acceleration is equal to minus omega square x, so a is what d square x by dt square x plus omega square x equals to 0, instead of x, q is there. This is second order differential equation, you know that solution of this differential equation is what x equal to a sin omega t plus theta. So, any second order differential equation will have a generic solution as a sin or cosine function, this is the assumption actually. So, we will assume that the solution of this equation, this differential equation is given as q is equal to q m sin of omega t plus phi, this omega and this omega they are same, because this frequency is the property of a system. So, the system is oscillating with this frequency omega, so omega will not change, frequency does not change just like that. So, q is equal to q m sin omega t plus phi is the solution of this, but we do not know what is q m and we do not know what is phi, how will you find out then, what is the value of q m and phi, just got to find out if this is a solution of this differential equation, then you can substitute this value here, so no, this if this solution of that it should satisfy this equation. So, put it here and try to rearrange the term and get it properly, you arrange sin of term, you get sin and cos both, you take sin separately and cos separately are done, so d q by d t will be q m omega cos omega t plus phi, so these values just substitute here, you will get minus of q m, so your equation will become very dirty, I mean it becomes now horrible, but then deal with it, I mean there is no as such, what I say you just have to be careful, so that you do not make any silly error, there is no conceptual error probabilities you will be making here, just silly error you will be making, because this equation becomes very big, sin of omega t plus phi plus r times q m omega cos of omega t plus phi plus q by c, so 1 by c times q 1 here, q m sin omega t plus phi, this is equal to q m sin omega t plus v m sin omega t, now let us take sin omega t common, if you take sin omega t common q m omega, I will take q m omega sin omega t plus phi common, I will get 1 by omega c minus omega, you know why I have taken q m omega common, because this is familiar to me, plus you can take q omega common throughout, you will get plus r times cos of omega t plus phi is equal to v m sin omega t, so this is what x c, so x c minus x l into sin of omega t plus phi plus r times cos of omega t plus phi is equal to v m divided by q m omega sin omega t, any doubt, you know what is z impedance is what, whole square plus r square under root, so now what you do, you divide throughout by z divide and multiply by z, so if you do that, there is a reason why we do that, this is a common trick, if you get c 1 sin theta plus c 2 cos theta, what you do you multiply and divide by root over c 1 square plus c 2 square, this will let you write it as a single trinometric function, this is c 1 square plus c 2 square, this will be equal to c 1 square plus c 2 square sin of theta plus phi, where tan of phi is equal to c 2 by c 1, this is what you assume as cos phi, this automatically becomes sin phi, it is a common trick, so this will let you write in a simpler manner, similarly here z is what root over this square plus that square, so I can write down this as z into, is it very overwhelming, are we able to handle it, any doubt you let me know, this is like, here you repeat the x, what you do, here you can you repeat something and then it is fine, tell me if you have any doubts, no, impedance is just incidental, it happens to be impedance, the trick is this only, root over c 1 square plus c 2 square happens to be impedance, and it is good that it is impedance because it is familiar to us, so I can say that this entire thing as cos of theta, this is sin of omega t plus phi, if this is cos theta that will become sin theta plus sin theta cos of omega t plus phi is equal to something, so this what I can write that as z times sin of omega t plus theta plus phi, this equals to V m times sin of omega t plus phi, this V m divided by Q m omega sin omega t, so basically impedance should be equal to V m divided by Q m times omega, so Q m should become V m divided by omega z, we got the value of Q m, comparing these two, and what is theta, theta is such that tan theta is equal to you assume that to be sin theta, it is to be cos theta, tan theta will be equal to r divided by x c minus x n, it does x c minus x n, so theta plus phi should become what, theta plus phi should become 0 or 2 pi or 4 pi whatever, so best thing is to take 0, so theta will be equal to minus, what solution will be 5, 5 will be equal to minus of theta, so phi will be equal to minus of tan inverse r divided by x c minus x n, in the textbook what they have done, they have assumed Q is equal to Q m cos of omega t plus phi, so when you read from the textbook, you just consider the fact that you have taken sin, not the cos, so if you take as cos, everything will rearrange and you will get tan of theta to be x c minus x n by r, fine. So, here rather than sin, you will get cos, here you will get sin, so this you will assume at sin theta, this you will assume at cos theta, fine. So, this is analytical solution, any doubt will not, anything, we will do some numerical on whatever you will learn till now, this is considered to be the difficult derivation from this, you do it one or twice, one is the same thing can be derived using phasor or using analytical method, you got the value of charge Q, you can differentiate it, you will get the value of current, using phasor we got the value of current, do this question, I will register, what is the unit of impedance, find out current V RMS across resistance and V RMS across capacitance, three things you have to find out, current voltage across resistance, voltage across capacitance, see for impedance it does not matter whether inductor is there or not, if inductor is not there, it will have only x c, x l will be 0, did you get the answer, anything, get the value of impedance first, how much it is, in this case impedance will be root over R square plus x c square, x c is 2 and 2, this is under root 200 square, so 4 0 2 1 2 root over 4 0 2, how much is that roughly, that is approximately 291.50, so I am telling you the value, Z is 291.5, now get the value of current V RMS across R and across C, that is RMS voltage, so V is equivalent, no, no, no, no, no, total V is that, it is not V across R, for current you need to find the answer, no, how can I see, you are getting impedance less 291 is less than 220, you got the maximum current, when I say just find out the current, I get RMS current, what do you get higher RMS value, I want higher RMS only, higher RMS is 220 divided by 291.5, this is 0.7 or something, now get the value of voltage across resistance and across capacitance, okay, did you get V R and V C, you called me right, V R and V C, how much is that, V R is simply I into R, V C is what, I into X C, how much is that, 151, 151 volts, 151 volts, V C is, X C is 212 and current is this, what is V C, 212.1, 160, okay, but isn't it weird that some of these two is more than 220, ideally, there is this and a resistance connected to a voltage source, some of this voltage plus that voltage should give you this, right, source is 220, voltage across resistance is 151, voltage across capacitance is 160, some of these two voltage is like 311, which is more than 220, how come, and nothing is wrong, so what is, how will you visualize what is happening, some of this plus this should be equal to that or not, yes or no, no, why, why no, no, that is not the reason, no, you can't add them like that, they are, they are 90 degree away, right, V C is 90 degree away from V C, we are or not, so basically what should happen, root over 151 square plus 160 square should give you 220 and which is correct, fine, because V R is this and V C is this, some of these two voltages will be taken as a Pythagoras current because they are 90 degree, any doubt till now, we are halfway this chapter, we will finish it off next class and we will solve few questions, but you guys have to come better prepared, I am still feeding that you are not putting the hard work which is required, you think that you might be working hard or you don't have time and all, but then that's not true, there are many students who are working very, very hard and there are lakhs in number, not hundreds, you are competing with them, pull your socks, it's very important and the worst is you will not get second chance to correct your mistake, the mistake you do right now stays with you throughout your life, very percussions, you will not get low, I made a mistake, let me correct it, that should not happen, but I am telling you, take it seriously, don't make an error, okay, so solve as many questions as possible and you start revising class 11th also, every week take one chapter of physics, chemistry and math and try to revise that, do questions from H C 1 or any book, whatever you have.