 Hello everyone, myself Sanjay Utke, Assistant Professor, Department of Electronics Engineering, Valchandra Institute of Technology, Solapur. Today, we are going to discuss three-phase star system. Learning outcome, at the end of this session, students will be able to analyze concepts of three-phase star circuits. Guideline, introduction, relation between line current and phase current for star system, mathematical expression for line voltage and the phase voltage, numerical based on star connected system, acknowledgement. Introduction, basic concepts of three-phase balance of the system. In the last video, we discussed delta types three-phase system. A system is said to be a balanced or rather than a system is said to be a balanced supply when the three voltages, three phases are having same magnitude, but differs in a phase difference of 120 degree. Similarly, a load is said to be balanced when these three loads are having the same nature that is R L, R C or R L C carries the same phase currents. So, in three-phase systems, you will find delta type connections or star type connections. Definition, line current, it is the current flowing between the two lines, phase current. It is the current flowing to a particular phase, line voltage. It is the voltage between two lines, phase voltage. It is the voltage across individual phase. The phase sequence, the order in which all these three voltages attains their maximum values. Generally, the phase sequences are y and then b. In between line and the phase currents for star system, for star system, line current is equals to the phase current exercise. In three-phase balanced load, what is the value of three-phase impedances? In the three-phase balanced load, all these three loads are having same value and the same nature also. In three-phase balanced load system, each impedances will have equal values. They will carry same currents, line or phase currents because in star system, we will find both line current and the phase currents are same. Mathematical expression for line voltage and the phase voltage. Here, I will draw a three-phase system. This is r, y and this is b. This is i r, i b and i y. This is a neutral fourth line. This is v r n. This is v y n and this is v b n. Here, I will write which are the phase voltages v r n, v y n, v b n. These are the phase voltages. What about line? Line voltages are v r y, v y b, v b r. These are the line voltages. Now, it is required to find out the relation between this line voltage and the phase voltage. For that, I will draw a phasor diagram or vector diagram. This will be v r n, 120 degree delayed by v y n and this is v b n. All these are 120 degree. For the sake of simplicity, I will not show the angle. So, all these three are 120 degree apart. It is required to find out the line voltage. For example, v r y, v r y is equals to v r minus v r minus v r minus v y. For minus v y n, I will draw this being minus v y n. I will find out resultant of these two vectors using a parallelogram method. So, this is line voltage v r y. Now, let us denote o, this point q, this point s. Let us draw a perpendicular line voltage. From point s, a perpendicular s p on line o q. This is say p. From the geometry, o q is equals to 2 times o p, where o q is equals to v l or v r y. o q is equals to 2 times o p is equal to o s cos of 30 degree. V line that is v r y is equals to 2 times o s is what it is v p h cos 30 is root 3 by 2. So, finally, I can write this v l is equal to root 3 times v p h. It means that line voltage is equals to root 3 times the phase of the line voltage for a star connected system. A three phase star connected balanced load having each impedances z p h is equal to 8 plus j 10 supplied with a voltage 400 volts 50 hertz supply. It is required to find out the I p h phase current, line current, phase voltage, power factor, then active power, reactive power and apparent power. Since the system is balanced, it means that all the impedances, three impedances are having same value and nature. Let us concentrate our numerical with respect to a single phase. So, to find out I p h, it is required to know v p h and z p h. So, what is z p h? It is 8 plus j 10 plus j 10 plus j 10 plus j 10 plus 10 j z p h which is in a rectangular form. Since we want I p h which is equal to v p h upon z p h, it is a division. Convert this rectangular form into polar form comes out to be 12.81 angle 51.34 degrees. So, next what is v p h? For v p h, it is v l upon root 3 and so 400 divided by root 3 comes out to be 231 volts. Now, we can find out I p h. I p h is equal to v p h 231 upon z p h 12.81 angle 51.34 degree. So, in this way, using calculator we got the value of I p h which is equal to 18.03 angle minus 51.34 degree. So, this is I p h which is also equals to line current because in star connected system both I p h and I l are same. Now, power factor. Power factor it is equals to cos of 51.34 degree comes out to be 0.62 lagging. Here the power factor is lagging because the current is lagging. Why the current is lagging? Because it is an inductive circuit. So, voltage leads currents by this angle 51 degrees. Now, coming towards active power root 3 v l I l cos of theta comes out to be putting value of v l 400 volts I l 18.03 cos of 51 degree comes out to be 7.8 kilo watt. Q reactive power root 3 v l I l sin theta comes out to be again putting value of v l and I l comes out to be 9 kilo volt ampere reactive. Similarly, apparent power is equal to root 3 v l into I l put value of v l I l comes out to be 12 kilo volt ampere. So, in this way we got first of all I p h then power factor then p q s. Acknowledgment, text book, electrical technology by B L Theraja. Thank you.