 Myself, Mrs. Veena Sunilpatki, Assistant Professor, Department of Electronics Engineering, Balchan Institute of Technology, Solaapur. Welcome you all for this session. At the end of this session, students can describe methods of three-phase power measurement. There are number of advantages of three-phase supply. Three-phase power distribution requires lesser amount of copper as compared to single-phase. Size of three-phase motor is smaller than single-phase motor. Three-phase motors are self-starting and single-phase induction motors are not self-starting. For three-phase motors, vibrations are less and the ripple factor of rectified DC produced from three-phase power is less and three-phase motors have better power factor regulation. So due to these number of advantages, generally we use three-phase power supply for industries and for transmission also. Now let us discuss about the generation of three-phase supply. So in this diagram you can see the stator, stator winding is provided, three-phase winding is placed on the stator and the rotor which is nothing but the magnetic field that shaft of the rotor is connected to the turbine. So the magnetic field is rotating, that magnetic field is cut by the stator winding. As the three-phase winding is provided, the three AC cycles are generated. So in second diagram you can see the three-phase voltages that three-phase windings are displaced from each other by 120 degree. So the 120 degree phase difference is there for three-phase voltages. Now we can see the relation between line and phase voltages for star connection. So in star, three line voltages are there vry, vyb, vbr. These three voltages are same for balanced system and vry, vyb are the three-phase voltages that three-phase voltages are also equal for balanced system. And in this system line current equal to phase current that is il equal to iph and line voltage is root three times phase voltage. In phasor diagram you can see ir, iy, ib are lagging to phase voltages vr, vy, vyb by phi angle and the power is given by root three vl, il cos phi. For delta system line voltage and phase voltage both are equal. Again here the three line voltages are there vry, vyb, vbr and three-phase voltages are equal to line voltage but the line current and phase current has the relation il equal to root three iph total active power that is given by root three vl, il cos phi in terms of phase voltages three vph, iph cos phi and the unit for active power is VAT. Total reactive power is given by root three vl, il sin phi in terms of phase voltage and phase current three vph, iph sin phi unit for this reactive power is var and total apparent power is given by s equal to root three vl, il and in terms of phase voltage and phase current three vph, iph. And the unit for apparent power is volt ampere. The power is also given by the formula in root p square plus q square. Now stop the video and think about this what is the relationship of line and phase voltage and current in star and delta. So what is the answer? For star line voltage equal to root three phase voltage and line current and phase current both are same. For delta system line current is given by root three times phase current and line voltage and phase voltage both are same. Now let us see about the three phase power how to measure that. So there are three methods to measure the power one is one wattmeter method, second two wattmeter method and third is three wattmeter method. So in one wattmeter method for balance system only one wattmeter is connected in the system and we can measure the total power by multiplying by three to that one wattmeter reading that will give the total power for this method. Then for three wattmeter method three wattmeters are connected for three phases and the sum of that three wattmeter reading is nothing but the total power for that system. But in this session we are going to discuss about the two wattmeter method the total power is measured by two wattmeter method is nothing but the sum of the two wattmeter readings. Now you can see in this diagram the two wattmeters are connected in star system w one and w two for one wattmeter you can consider the phase voltage, line voltage, phase current, line current. Now this is the phasor diagram for this system so three phase voltages are there and the three line voltages are there that line currents are lagging to three phase voltages by phi angle vrn, vyn, vbn are the three phase voltages and iriyib are the three phase currents. This vrn and the angle between this ir is phi but the line voltage vrb is the sum of vrn and this vbn so angle between this vrn and vrb is nothing but the 30 and the 30 minus phi is the angle between vrb and ir. Just like that the angle between vyb and iy is 30 plus phi. We can calculate the power measured by w one and w two. So current flowing through wattmeter w one is ir, voltage across w one is vrb so w one is given by vrb into ir into cos 30 minus phi and current through wattmeter two is given by iy, voltage across wattmeter two is vyb so wattmeter reading w two is given by vyb into iy into cos 30 plus phi. For balanced load ir equal to iy equal to ib equal to il and vry equal to vyb equal to vr equal to vl and by adding the two wattmeter readings vlil cos 30 minus phi plus vlil cos 30 plus phi. By using mathematical formula for cos 30 minus phi and cos 30 plus phi we can write down the equation w one plus w two equal to vlil cos 30 cos phi plus sin 30 sin phi plus cos 30 cos phi minus sin 30 sin phi. By putting the values here we will get vlil 2 cos 30 cos phi and next we will get vlil cos phi that is nothing but the w one plus w two that is the active power. So the total power is sum of two wattmeter readings now we can find out the reactive power by using this method and power factor by this method. So w one minus w two equal to vlil cos 30 minus phi minus cos 30 plus phi again by using the mathematical formula we can find out w one minus w two equal to 2 vlil sin 30 sin phi so sin 30 is 1 by 2 so w one minus w two equal to vlil sin phi. By dividing equation 2 and 3 we can find out w one minus w two divided by w one plus w two equal to vlil sin phi divided by root 3 vlil cos phi so tan phi equal to root 3 w one minus w two divided by w one plus w two. So power factor we can find out by using this equation that is given by cos phi equal to cos of tan inverse root 3 w one minus w two divided by w one plus w two means we can find out the power factor by using two wattmeter readings w one and w two by using this two wattmeter method. Again we can find out the reactive power by using this method so only by multiplying root 3 to equation 3 is given by root 3 vlil sin phi that is nothing but the reactive power and unit for reactive power is var so by using two wattmeter method we can find out the total active power that is given by root 3 vlil cos phi then reactive power root 3 vlil sin phi and again we can find out the power factor by using these two wattmeter readings you can refer the book electrical technology by B. L. Therese. Thank you.