 Hello everyone. Myself, Sanjay Udge, Assistant Professor, Department of Electronics Engineering, Valchand Institute of Technology, Solapur. Today we are going to discuss three-phase delta system. The ring outcome at the end of this session students will be able to analyze concept of three-phase delta system. Guideline, introduction, relation between V-line voltage, phase voltage for delta system, mathematical expression for line current, phase current, numerical based on delta connected system and definitions of active reactive apparent powers. Three-phase delta system. In previous video we had a introduction of three-phase system with a delta connected load or source. Here you will find E, R, Y, then E, Y, B, E, B, R. These are three-line voltages as well as the phase voltages. Whereas this I, R, I, Y and I, B, these are the line currents. I1, 2, I2, 3 and I3, 1, these are the phase currents. Relation between line voltage and the phase voltage for delta system. In delta system, line voltage is equals to the phase voltage. Question. In three-phase system, what is the value of three-phase supply voltage? Is it line voltage or phase voltage? Answer is three-phase system supply voltage is always equal to the light voltage. Mathematical expression for line current and the phase current. Now let us derive the relation for line current and the phase current. This is a delta connected system. I1, I2, I3, I1, I2, I3, these are the phase currents. Whereas I, R, I, Y and I, B, I, R, I, Y and I, B, these are the line currents. Now one example, phase and line voltage. It is the V, R, Y, V, R and V, Y. This is V, R, Y. It is the line voltage as well as the phase voltage. Next, let us draw a vector diagram in which here this is the V, R, V, Y, V, B. I1, I2, I3. Now look at this point, this node, I, R is equal to I1 minus I3. Considering this, let us find out the I, R, that is the line current which is the difference of I1 and I3. This is I1 minus I3 can be shown over here. So this is minus I3. Now the resultant of this can be obtained by using a parallelogram method. So this is line I, R which is equal to I1 minus I3. It is the line voltage. Now let us, this is O, this is Q, this is S and now let us draw a perpendicular online OQ from point S at point P. According to the geometrical concept, we can say that O P is equal to 2 times OQ. O P is equal to 2 times OQ. What is OQ? OQ is equal to 2 times OQ. What is OQ? OQ is the line current I, L or I, R. Next, this OQ is nothing but 2 times O P is equal to 2 times O S cos of 30 degrees. Which is equal to 2 times O S root 3 by 2. What is O S? O S is nothing but the phase voltage Iph root 3 by 2. So we can say that line current is equal to root 3 times phase current. So this is the expression for line current which is equal to root 3 times the phase current. Now coming to the next, let us solve one numerical in which a 3 phase delta connected load draws a current of 20 ampere with a power factor of 0.8 ampere, 0.8, with a supply voltage 3 phase, 400 volts, 400 volts, 50 hertz. Now let us move ahead. It is required to find out the value of R, L and active power P. Since it is a delta system, in delta system line voltage is equal to phase voltage. Here the supply voltage is always a line voltage. So I can say that it is equal to 400 volts. Now since it is required to find out the value of R, L, R, L it is required to find out Zph, Iph, then XLph and so on. Now what is Iph? Iph is equal to what? I L upon root 3. Line current is given as 20 ampere divided by root 3 gives out to be 11.54 ampere. From this we can find out Zph. As you see Zph, Iph and XLph these are the per phase impedance current and reactive inductance respectively. So Zph equals to Vph upon Iph given by 400 volts divided by 11.54 comes out to be 34.64 Zph. Now remember this one is the impedance triangle. In impedance triangle this is Rph, this is XLph and this is Zph. So from this impedance triangle we can find out Rph equals to Zph cos 30. XLph is equal to Zph sin 30. So from this relation we can find out because we are knowing Zph cos 30 sin 30 we can find out from power factor. So Rph comes out to be putting the value of Zph. Zph is equal to 34.64 cos 30 is equal to 0.8 which is power factor. So Rph comes out to be 27.71. XLph we can find out the XLph. XLph also we can find out using this relation Zph sin 30 which comes out to be 21.78. Now from this we can find out XLph is equal to 21.78 is equal to 2 pi FLph. FLph is the frequency given 50 hertz. From this we can find out Lph is equal to 66 milli henry. Acti power is equal to root 3 VL, VLIL cos theta or 3 times VPH Iph cos theta. And it comes out to be 11 kilo Watt. This is 3 phase delta system in this Acti power, Acti power and average power equations are given. Acti power is equal to root 3 VLIL cos 30 or 3 times VPH Iph cos theta. Acti power root 3 times VLIL sin 30 or 3 times VPH Iph sin 30. Apprent power is VL into VL or root 3 times VL into VL or 3 times VPH into Iph. Economics, electrical technology by B. L. Thayaraja. Thank you.