 Hello, everyone. This is Dinkar Patnaik from Valgin Institute of Technology, Sulapur. To begin with, let us recall what we have covered in the last lecture. That is, we have been discussing about the effect of series resonance and the effect of resonance on a series impedance in a series resonance circuit and all. So, it is very important to note that here we have a topic known as parallel resonance. So, in this video, we will be discussing parallel resonance and there are basically two variants of a parallel resonance circuit. So, in this video, I will be focusing on the parallel resonance circuit in a simple parallel resonance circuit as well as tank circuit. So, basically to express, there are two basic variants of a parallel resonance circuit. So, by the end of this video, students will be able to explain the concept of parallel resonance in a single phase AC circuit. So, basically as we have been talking about, even in series resonance circuit, resonance happens whenever the inductive reactance becomes equals to the capacitive reactance. In a similar way, we will be going through the details like below resonant frequency earlier as we have seen in series resonance circuit. The circuit is more dominant in a capacitive way, but in this case, as you can see, case 1 is about where the circuit is becoming more inductive. So, the first case is actually discussing about the condition of impedance in a parallel resonance circuit where the inductive reactivity is more as compared to the capacitive reactance. So, this is exactly the reciprocal of what we have in series resonance circuit. So, let us have a detailed look like how the circuit and the impedance happens and what is going to be the resonant frequency of this. So, as you can see, in the previous video, we have been looking at a series resonance circuit. Now, in this one, let me introduce you something known as a parallel resonance. So, this parallel resonance can be, I mean, in a broader view, can be classified into the first one as simple parallel resonance circuit and the other one, which is more specifically known as a tank circuit. So, let me first introduce you with a simple parallel resonance circuit. That is, we have the same capacitor that we have used in our series resonance circuit along with which we also have another branch, which is with an inductor. But this time, we are connecting this in parallel. So, this is an important aspect to understand. So now, I will be connecting an AC source to this one. Of course, this will be an AC source. Now, I have a capacitor. So, it is going to offer me capacitive reactance. So, I will be calling it xc and similarly, this will be xl. Now, the resistance, which is in series with this xc, will be rc and the resistance, which is in parallel with xl is going to be rl. So now, as soon as you turn on the power supply, the current is going to enter on either of these sides. So, I am considering the clockwise direction. This current is going to enter this node and it will be split into these two. Unlike to what we have in series resonance circuit, where the components like rl and c were in series, where the voltage was being split. Here, we have a special case where we are forming a node. So, whenever you have these rlc elements connected in parallel branches, you need to have these things considered like whether the current is getting split or not. So, considering the same circuit, let us move ahead and identify what is going to be the overall impedance of the circuit. So, if this would have been a series resonance circuit, then we would have called it an impedance. But now, I will be calling, I mean, I will be introducing you to a new thing called as y. So, y is nothing but the reciprocal of impedance. So, y is generally written as g plus jb. This is nothing but, so this is for series. This expression is for a parallel circuit. So, if you compare the series impedance expression with the parallel admittance expression, you can easily identify that the reciprocal of impedance is admittance. Reciprocal of a resistance is conductance and reciprocal of a reactance is nothing but, susceptance. So, similar to the way how we have identified the overall impedance expression for series resonance circuit, later on we equated that to 0, we are going to this time identify the same admittance expression for the given circuit. And then, we will be splitting out the expressions, so that we can split out this imaginary part and then, equate it to 0 to identify the resonant frequency of the circuit. So, let me elaborate this expression for this specific circuit. That is, at resonance, xl becomes equals to xc. So, keeping this thing in mind, what I will be doing is, let me write the expression, y is nothing but, the impedance, I mean the impedance or the conductance offered by this one. So, the conductance is going to be nothing but, the reciprocal of what we exactly have in a normal format in this branch. So, the reciprocal of the resistance that is, I have rl. So, I am writing the expression for this one first, rl plus j omega l. So, I will prefer writing it j omega l directly, instead of writing xl plus, now I am writing the expression for this. It is going to be 1 upon rc minus j by omega c. This can be further simplified as, so this is a mathematical expression, which can be reduced by multiplying them with the complex conjugate of what we have here. So, I will be multiplying this expression with the complex conjugate of the same term, which I have on the denominator. So, rl minus j omega l upon, this will be simply rl square plus omega square l square. So, I am using the mathematical expression of a square plus b square, here to reduce the expression. Similarly, I have the reciprocal of this as rc plus j by omega c, and on the denominator, what I have is rc square plus omega square c square. And of course, this is going to be in the denominator. Now, I will be simply multiplying the complex conjugates, which we have done and simplify the circuit and let us see what we have. So, this is the real part. So, I am simply trying to take the real part, this one and this term here. So, plus what I have is the imaginary susceptance. So, what I have here is 1 by omega c and so the denominator term minus this term, it will be simply omega l by rl square plus omega square l square. Now, here it is important to understand that this part is real, means I am comparing it with this one and this part is imaginary. So, I am talking about this. So, for resonance to occur, this imaginary part must be equated to 0. So, let us take that imaginary term and we will be equating this, so that I will be having either of these terms on LHS and RHS. So, that comes to let me change this. Therefore, at resonance I can call what I have is now instead of simply writing omega l, I will be writing omega rl. On the denominator, I have rl square plus omega square l square. This will be, so I will be simply cross multiplying these things and quickly check what do I have. So, this comes to rl square. So, let me take the omega r terms on one side, so that I have omega r square into rc square plus 1 by omega square c square. So, what we are trying to do is I am simply trying to take all the omega terms on LHS, so that I can bring it to a format like omega r equals to something. So, that later on we can apply the simplification terms like we did last time and identify the resonant frequency. So, this comes to be, I will multiply these terms internally. So, it will be simply equals to 1 by LC plus this will be minus 1 by c square. So, it is not there. So, this comes to omega r square and I am simply trying to take it common. So, that this will be rc square minus l by c equals to 1 by LC and here it will be rl square minus l by c. So, I am taking it common multiplying it and I am trying to take the term 1 by LC common from this one. So, this finally comes to omega r square sorry for this it will be simply omega r equals to 1 by let me directly apply the square root on the RHS and I will be simplifying these terms that is it will be rl square minus l by c upon it will be rc square minus l by c. So, this is the final expression. So, here what I can say in a special condition when rl becomes equals to rc this complete function can be simplified to omega r equals to 1 by LC because these things gets cancelled out and this is nothing but 1 by 2 pi under root LC. So, this is the final expression for resonant frequency of a simple parallel resonant circuit. Now, in this case it is very important to focus that this expression is very much similar to the series resonant resonant frequency. So, because of which you can imagine that the overall effect or the resonance happens at the same frequency whether you are mounting the circuit in a series way or in a parallel way. So, that is all for this video and these are the references used for this particular video and we will be discussing the tank circuit in the upcoming video. Thank you.