 myself, Sanjay Udge, assistant professor, department of electronics engineering, Valchand Institute of Technology, Solapur. Today, we are going to discuss DC circuits, delta 2 star conversion. Learning outcome. At the end of this session, students will be able to analyze complicated electrical circuits. Guidelines. Introduction, delta network, star network, delta 2 star conversion are references. Introduction. Review of DC circuits. In the last videos, we discussed basic concepts of DC circuits. It was included Ohm's law, Kirchhoff's current law, Kirchhoff's voltage law, so as to analyze and simplify the given DC circuit. What is the necessity of delta star network? To simplify the DC circuits, comprising of series and parallel resistances, along with the voltage sources, DC voltage sources. We used to apply KCL, KVL in order to find out the currents flowing through a particular branch or voltage across a particular resistance or a total current of the given network or equivalent resistance of a given network. So, there are some networks in which we cannot use Ohm's law and various theorems. In that case, we have to convert the given network successfully between delta 2 star and star 2 delta. The successive conversion between delta 2 star and star 2 delta will simplify the given network. And finally, you can get the equivalent resistance value, then the total current or the voltage drop across various branches current flowing through the various branches. Delta network. This is a delta network consists comprising of three resistances R1, R2, R3 connected between the nodes A, B and C. This is a star network. If you observe one end of R A, R B, R C, they are connected to a common point called as O, whereas other points are going to the different part of the network. So, this is a star network. Conversion from delta 2 star, node 1, this is 2, this is 3, this is a delta network and this one is a star network 1, 2 and 3. Resistance between delta resistance between 1 and 2, it is R1, 2, this is R2, 3 and this is R3, 1. This is R1, R2 and R3. These two electrical circuits are said to be equivalent when the voltage resistance between the two terminals is equivalent. Now, let us find out the condition. First, terminals between 1 and 2 for star network, it is R1 plus R2. It is R1 plus R2. And for delta network, it is R1, 2. R1, 2 is coming in parallel with the series combination of R3, 1 and R2, 3. So, I will write it as R1, 2 parallel R2, 3 plus R3, 1. I will write this equation as R1 plus R2 is equals to R1, 2, R2, 3 plus R1, 2, R3, 1 divided by R1, 2 plus R2, 3 plus R3, 1. That is this equation number 1. Similarly, between terminals 2 and 3, between terminals 2 and 3, it is R2 plus R3, 2 and 3, R2 plus R3. Coming to a delta side, between 2 and 3, this R2, 3 will come in parallel with this. These two are R3, 1 plus R2, 3. I will write it as directly R2, 3 parallel R3, 1 plus R1, 2. I will rewrite this equation as R2 plus R3 is equals to R2, 3, R3, 1 plus R2, 3, R1, 2. Let this be a equation number 2. Now, coming between terminals 1 and 3, for star, it will be R3 plus R1. And for delta, it is R3, 1 parallel R1, 2 plus R2, 3. I will rewrite this equation R3 plus R1 is equals to R1, 2, R3, 1 plus R2, 3, R3, 1. Let this be a equation number 3. Now, I will subtract equation 2 from equation 1. What I will get? I will get R1 minus R3 is equals to R1, 2, R3, 1 minus R2, 3, R3, 1 divided by R1, 2 plus R2, 3 plus R3, 1. Let this be a equation number 4. If I add equation number 3 and 4, we got R3, R3 will get cancelled. Then R2, 3, R3, 1 will get cancelled. It will become 2 times R1 is equals to 2 times R1, 2, R3, 1 upon R1, 2 plus R2, 3 plus R3, 1. 2, 2 will get cancelled. So, R1 will be equals to R1, 2, R3, 1 upon R1, 2, R2, 3, R3, 1. For converting delta, R1, 2, R2, 3, R3, 1 to get this R1, R2, and R3. It is very simple. As we derived one formula for R1, R1 is equal to product of these two resistances divided by summation of this. Similarly, how will I write R2? I will write R2 is equals to R2 is coming over here. This is 2, 1, and 3. R2 is connecting to known junction number 2. To this junction, which delta resistances are connecting, this is R1, 2, and R2, 3. So, to find out R2, it will be product of these two delta resistances, which are connecting to same junction 2. So, it will be R1, 2, R2, 3 plus summation of delta resistances. Similarly, what about R3? R3, you can write, R3 is going to junction number 3, and same junction number 3, which delta resistances are connected, this R2, 3, and R3, 1. So, R3, I can write it as R2, 3, R3, 1 divided by summation of R delta. So, it is very easy, no need to remember the formula. Let us solve one example. Two deltas are shown whose values are 10, 10, 10, 10, 10, 10. This 1, 2, 3. It is required to find out the resistance between 1 and 2. So, what I will do? First of all, I will solve these parallel networks. So, by solving these parallel networks, I will get this as 1, 2, and 3. 10 and 10, it will be a 5, 10 and 10, 5, and this will be a 5. Now, resistance R1, 2 is equals to, it will be 5 ohm, and this 5 ohm, it will come in parallel with the series resistance of these two. So, it will be 5 plus 5. So, it will be 5 into 10 divided by 5 plus 10, it is a 3.66 ohms. So, this is how we can convert, we can get the value of resistances in this. Let us convert this delta network into a star equivalent network. So, this is having value 9, 9 and 9. I want this R1, R2, and R3, 1, 2, and 3. So, R1 will be equals to this R1, which is coming over here. It will be 9 into 9 divided by 9 plus 9 plus 9. 81 divided by 27, it is a 3 ohm. Same is the case for R2, which is coming over here. R2 will be equals to 9 into 9. R2 is coming over. So, it will be a product of 9 into 9 divided by 9 plus 9 plus 9. 27 comes out to be 3 ohm. R3, which is supposed to be coming over here, it will be again 9 into 9 divided by 9 plus 9 plus 9. It is a 27 again, 3 ohm. So, this is the exercise assignment. It is a delta network having values 3, 3, and 3 ohm. It is required to convert into a corresponding star network. To find out here, this is Rx, Ry, and Rz. Rx will be equals to 3 into 3 divided by 3 plus 3 plus 3. So, Rx will be equals to 3 into 3, that is 9 divided by 3 plus 3 plus 9. It is a 9. It is equals to 1. Ry, again it is same. To find out Ry, it will be 3 into 3 divided by 3 plus 3 plus 3. So, it is again 1 ohm. Rz also, 3 into 3 divided by 3 plus 3 plus 3 is equals to 1 ohm. Academic textbook, electrical technology, BL Theraja. Thank you.