1. Balanced poly phase circuits

2. Two and four phase systems  A two phase system is an electrical system in which the voltages of the phases are 90 degree out of time phase.  The emf’s E ac and E bd are 90 degree out of time phase.  A two phase system is the equivalent of two separate single-phase systems that are separated 90 degree in time phase.  If the connection is made between the two windings at n and n`, the system would be called a four-phase system.  The voltages E da, E ab, E bc and E cd are called the line voltages, while voltages E 0a, E 0b, E 0c and E 0d are called the phase voltages or voltages to neutral.  E da =E d0 +E 0a. Thus, the line voltages= times the phase voltages in the four phase star.

3. Two and four phase systems  The line voltages are 45 degree or 135 degree out of time phase from the phase voltages.  For four-phase mesh, the aa` line current is I aa `=I da +I ba.  Thus, the line currents= times the phase currents and 45 degree or 135 degree degree out of time phase in the four phase mesh

4. Three phase four wire system N is the neutral wire. Lighting loads are placed from line to neutral. Motors and other three phase power loads are connected between the three lines.

5. Three phase three wire system

6. Three phase four wire system

7. The current in any line is the same as the current in the corresponding phase. Current in the neutral wire is obtained through the application of KCL. These currents are equal in magnitude and displaced from one another in time phase by 120 degree. Thus the current in the neutral wire is 0 since

8. Three phase three wire system

10. Balanced Wye loads Given the line voltages as 220 volts balanced three phase and R and X of each phase 6 ohms resistance and 8 ohms inductive reactance. Find the line current, power per phase and total power. Draw the vector diagram

11. Balanced Delta loads Given the line voltages as 220 volts balanced three phase and R and X of each phase 6 ohms resistance and 8 ohms inductive reactance. Find the line current, power per phase and total power. Draw the vector diagram self

12. Power calculation in balanced systems Wye connection Dealta connection

13. Volt-Amperes Wye connection Dealta connection Reactive Volt-Amperes Wye connection Dealta connection The sine of the angle between phase voltage and phase current is called the reactive factor of a balanced system.

14. Single phase and balanced three phase power For any phase a Balanced three phase power under steady state conditions is constant from instant to instant. But it is a double frequency variation with respect to time for single phase power..

15. Three wattmeter method  It is not feasible to break into the phases of a delta connected load.  For the Y load it is necessary to connect to the neutral point. This point is not always accessible.

16. Two wattmeter method

18. Watt ratio curve  For each value of, there is a definite ratio of W a /W b.  If the ratio of the smaller to the larger reading is always taken and plotted against the corresponding cos, a curve called the watt-ratio curve results.  At 0.5 power factor, one wattmeter reads 0.  At 0, each wattmeter has the same deflection but the readings are of opposite signs. Fig:46. watt-ratio curve for two wattmeter method of measuring power

19. Sign of wattmeter Method 1  Open line a, all power must be transferred to the load over lines b and c.  If wattmeter W b reads upscale, the power is going to the load.  Reconnect line a and open line b. then connect W a so that it reads upscale.  Reconnect line b.  If at any time after this wither wattmeter needle goes backward against down-scale, power through this wattmeter channel is being transferred to the generator and this power must be of opposite sign to that registered by the other.  Either the potential or current coil will have to be reversed to secure an up-scale reading. Method 2  disconnect the potential coil of W b from line c and connect to line a.  It the needle goes against the down-scale stop, the wattmeter reading was negative.

20. Copper required to transmit power under fixed conditions

22. Unbalanced system  An unbalanced system is due to unbalanced voltage sources or unbalanced load. In a unbalanced system the neutral current is NOT zero. Line currents DO NOT add up to zero. I n = -(I a + I b + I c ) ≠ 0