 Myself Pradheesha, working as an assistant professor in Vyachan Institute of Technology welcome you to this particular video. Learning outcome, at the end of this video lectures student will be able to analyze two port network parameters. Also they will be able to obtain Y parameters for the given two port network. So in the previous video lecture we had discussed that what is actually a port. A port is basically defined as that pair of terminals where you actually supply the energy or actually you withdraw the energy. So over here you can see that a standard configuration of two port network has been shown where this is input port and this indicates an output port. So V1 and I1 and V2 and I2 represent the network variables. So taking two of these network variables at a time and considering rest of the two as independent variables, six different types of network equations can be formed. So all of these network equations have been represented over here, which can be calculated to find out impedance, admittance, hybrid parameters as well as inverse hybrid parameters, transmission parameters and inverse transmission parameters. So in today's lecture we are going to discuss about the short circuit admittance or Y parameters. Again observe the given linear two port network which does not have any independent sources. So the current I1 present at port 1 it can be called as superposition of responses caused by the two voltage sources V1 and V2. So similarly current I2 it can be considered as the superposition of the components caused by the voltage sources V1 and V2. So the equations for both of them you can see over here. The same equations can be again represented in the form of matrices where the centre matrix represents the Y parameters. So the individual Y parameters can be also obtained by setting any of the port voltages equal to 0. Let us look into detail what actually these all four Y parameters are. So observe this particular figure 3A. So in this particular figure we have seen that the port voltage V2 is made equal to 0 which indicates that port 2 has been short circuited. So by making the second port short circuit we can calculate Y11 and Y21 parameter. So Y11 is called as the admittance looking into the port 1 when port 2 is short circuited. Y21 is called as transfer admittance. Now look into the corresponding second figure 3B where the port 1 voltage V1 has been short circuited and then we can calculate V1, Y12 and Y22 values. So similarly Y12 will be called as in transfer admittance while as Y22 is the admittance looking into the port 2 while port 1 has been short circuited. So a tube is said to be symmetrical if the input and output ports can be interchanged without altering the port voltages and currents. What will be the short circuit admittance parameters if the given circuit is symmetrical? Pause the video for a while and think about it. So for a given symmetrical network the condition for Y parameters is Y11 should be equal to Y22. Also if Y12 and Y21 that is nothing but transfer admittances are equal then the corresponding two port network is said to be reciprocal or bilateral. Now let us actually solve one particular problem based on this. So this particular is a two port network and we need to find out the Y parameters for this given network. So the Y parameters can be calculated by using the equations as it has been displayed over here. For calculating Y parameters for the given two port network we need to make the second port as short circuit which indicates that P2 should be made equal to 0. So this is the first condition. Let us redraw the given circuit. Now when the corresponding port voltage V2 has been made equal to 0 which indicates that the second port is short circuited. In that case this resistance A to H resistance it can be neglected as we have a branch in parallel with A to H resistance which indicates that it is a short circuit. That is nothing but it has resistance equal to 0. So the corresponding value of Y11 it can be calculated as ratio of I1 upon V1 when corresponding voltage V2 is made equal to 0. Let us calculate the two currents flowing through the two resistances. Let us name them as this is IB, this is IA, current flowing through 4 ohm is IA, current flowing through 2 ohm is IB. So using current division rule IA can be calculated as I1 multiplied by 2 ohm resistance divided by 6 while as IB is nothing but I1 multiplied by 4 and divided by 6. Now if you see carefully current IB is nothing but negative current minus I2. Let us substitute these two values in our equation Y11. So we have Y11 as ratio of I1 upon V1. Hence I11 is nothing but I1 upon 4 multiplied by IA that is nothing but voltage across 4 ohm is nothing but V1. So we have I1 is equal to 4 into 2 divided by 6 I1 which gives us value as 6 divided by 8 and hence we have value of Y11 as 0.75 or ohm inverse. Now let us calculate the second parameter when we have shorted the port 2. The second parameter that we can calculate when the port 2 has been short circuited is I2 upon V1. So we had discussed that I2 is nothing but current IB flowing through 2 ohm resistances which we have already calculated over here. Let us substitute these values into the equations. We have I2 is nothing but minus 4 I1 upon 6 divided by V1 value we have 4 into 2 divided by 6 into I1. Let us simplify this where I1 gets cancelled the 6 value gets cancelled and we have minus 4 divided by 8 which gives us value as Y21 is minus 0.5 mu or ohm inverse. In this way we have calculated two parameters Y11 is 0.75 mu and Y21 is minus 0.5 mu. Now let us calculate the rest of the two parameters. So we have to calculate Y12 and Y22 values now. So Y12 can be calculated as ratio of current I1 upon V2 and then again let us read out the given circuit by putting short V1 is shorted. This is current I2 and this is current I1. This is 2 ohm resistance and A2 ohm resistance. As in previous case we had calculated two currents IE and IB using current division rule. Let us again recalculate these two values. So this is current flowing through A2 ohms is IA and current flowing through 2 ohms is IB which is nothing but IA we have using current division rule I2 multiplied by 2 ohm divided by 10 which is nothing but 1 upon 5 I2 and IB we have 8 divided by 10 which is nothing but 4 by 5 I2. But since the currents I2 and IB are in exactly opposite directions it has to be taken as negative. So we have now Y12 as I1 which is nothing but current I1 or IB. So let us replace it value by 4 upon 5 I2 and then we have voltage V2. Voltage V2 is nothing but voltage across 8 ohms. So that can be calculated as resistance 8 ohm multiplied by current flowing through it IA which is nothing but 1 upon 5. So it is 1 upon 5 I2. So after simplifying we get the value as minus 0.5 mu or ohm inverse. Now let us calculate the second value Y22 which is ratio of current I2 divided by voltage V2. So we have current I2 divided by voltage V2 is nothing but 8 divided by current flowing through it is 1 upon 5 I2. So this I2 gets cancelled then we have Y22 value adds 5 upon 8 ohm inverse. I have calculated the Y parameters for the given network. Now observe this Y matrix carefully and comment about the given network. Pause the video for a while and think about it. So since the values of Y12 and Y22 are equal we can say that the given network is reciprocal. The video are created by using these two references. Thank you.