1. Engineering Circuit Analysis
CH6 Polyphase Circuits
6.1 Polyphase System
6.2 Notations
6.3 Single-phase Three-wire Systems
6.4 Three Phase connection
6.5 The Delta ( ) Connection
6.6 Power measurement

2. Ch6 Polyphsae Circuits 6.1 Polyphase System
Polyphase system : system with polyphase sources
Single source (Vs)
Notice the instantaneous voltage maybe zero
 The instantaneous power will be zero
They all have 120o phase differences
 The instantaneous power will never be zero.
Poly sources ( )

3. Ch6 Polyphsae Circuits 6.1 Polyphase System
The incident with a zero instantaneous power has been exempted.
The source power can be delivered more stably.
The polyphase systems can provide multiple output voltage levels.
Polyphase systems in practice certain sources which maybe approximated
by ideal voltage sources, or ideal voltage sources in series with small
internal impedances.

4. Ch6 Polyphsae Circuits 6.2 Notations a b c d e f g h i j k l
For note c :
For note f :
For note j :

5. The voltage of point a with respect to point b
Ch6 Polyphsae Circuits
6.2 Notations
The voltage of point a with respect to point b
a +; b -;
Similarly, Iab denotes the current from point a to b.
Test with graphical analysis ? (Using the phasor diagram)

6. 6.3 Single-phase Three-wire Systems
Ch6 Polyphsae Circuits
6.3 Single-phase Three-wire Systems
Function: allowing household electronics operating at two levels of voltages to be applied.
Voltage characteristics
1-phase
3-wire
Source
Household electronics may either operate with
or with
Phase characteristics

7. 6.3 Single-phase Three-wire Systems
Ch6 Polyphsae Circuits
6.3 Single-phase Three-wire Systems
Current characteristics
This is no current in the neutral wire.
How if the two
are NOT equal, and all the wires have impedances ?
This is a more practical scenario.

8. 6.3 Single-phase Three-wire Systems
Ch6 Polyphsae Circuits
6.3 Single-phase Three-wire Systems
Example 9.1 (P242)
Determine the power delivered to the
and the
Loads.
rms
② Determine the power lost in the three
lines represented by
respectively.
and
③ Determine the transmission efficiency?
rms
total power absorbed by the loads
η =
total power generated by the sources
Hints: observe a structure with regular meshes and know the impedances, we can
determine the currents I1, I2 and I3 in order to find out the power being lost
and delivered!

9. 6.3 Single-phase Three-wire Systems
Ch6 Polyphsae Circuits
6.3 Single-phase Three-wire Systems
Apply KVL for the three meshes.
Rearranging them in a matrix form as

10. 6.3 Single-phase Three-wire Systems
Ch6 Polyphsae Circuits
6.3 Single-phase Three-wire Systems
It can be calculated:
rms
rms
rms
rms
rms
rms
Hence, the average power delivered to each of the loads are:
Total loaded power

11. 6.3 Single-phase Three-wire Systems
Ch6 Polyphsae Circuits
6.3 Single-phase Three-wire Systems
Power lost in three wires are:
Total lost power
Transmission efficiencyη
Total power generated by the two voltage sources is:
Transmission efficiency

12. 6.4 Three Phase connection
Ch6 Polyphsae Circuits
6.4 Three Phase connection
Voltage characteristics
Balanced three-phase sources
(phasor voltages)

13. 6.4 Three Phase connection
Ch6 Polyphsae Circuits
6.4 Three Phase connection
Positive phase sequence (abc)
(clockwise rotation)
Negative phase sequence (cba)
(Anti-clockwise rotation)

14. 6.4 Three Phase connection
Ch6 Polyphsae Circuits
6.4 Three Phase connection
Line-to-line voltages (take the abc sequence as an example)
Hence
verifies KVL.

15. 6.4 Three Phase connection
Ch6 Polyphsae Circuits
6.4 Three Phase connection
Voltage types
magnitude
Phasor difference
Phase voltages ( )
Line-to-line voltages ( )

16. 6.4 Three Phase connection
Ch6 Polyphsae Circuits
6.4 Three Phase connection
Current characteristics

17. 6.4 Three Phase connection
Ch6 Polyphsae Circuits
6.4 Three Phase connection
Consider three impedances
are connected between each line and
the neutral line.
Hence
When balanced impedances are applied to each of the three lines and
the neutral line carries no current.

18. 6.4 Three Phase connection
Ch6 Polyphsae Circuits
6.4 Three Phase connection
Example 9.2 (P247)
Phase voltages:
line-to-line voltage:
Line currents:
Power absorbed by the three loads

19. 6.4 Three Phase connection
Ch6 Polyphsae Circuits
6.4 Three Phase connection
Example 9.2 (P247)
How about the instantaneous power?
Note: Van = 200V rms
Similarly , the instantaneous total power absorbed by the loads are :
The total instantaneous power is NEVER ZERO.

20. 6.4 Three Phase connection
Ch6 Polyphsae Circuits
6.4 Three Phase connection
Example 9.3 (P249)
A balanced three-phase system with a line voltage of 300Vrms is supplying a balanced Y-connected load with 1200W at a leading power factor (PF) of 0.8. Determine line cuurent IL and per-phase load impedance Zp.
The phase voltage is: Vp = 300/ Vrms.
The per-phase power is: 1200W/3 = 400W.
Therefore , and IL = 2.89Arms
The phase impedance is:
A leading PF of 0.8 implies the current leads the voltage, and the impedance angle is: -argcos(0.8) = -36.9o
and Zp = o Ω
Note: the apparent power of a Y-Y connected load is P = Van × IAN
(phase voltage × line current)

21. 6.5 The Delta ( ) Connection
Ch6 Polyphsae Circuits
6.5 The Delta ( ) Connection
The neural line dose not exist. Balanced impedances are connected
between each pair of lines.

22. ﹠ Ch6 Polyphsae Circuits 6.5 The Delta ( ) Connection
Voltage characteristics
Phase voltages
Line voltages ﹠
Current characteristics
Phase currents
Line currents

23. 6.5 The Delta ( ) Connection
Ch6 Polyphsae Circuits
6.5 The Delta ( ) Connection
connections
connections
Phase voltages √
Line voltages √
Phase currents √
Line currents √

24. 6.5 The Delta ( ) Connection
Ch6 Polyphsae Circuits
6.5 The Delta ( ) Connection
Example 9.5 (p251)
Determine the amplitude of line current in a three-phase system with a line voltage of 300Vrms that supplies 1200W to a Δ-connected load at a lagging PF of 0.8.
The per-phase average power is: 1200W/3 = 400W
Therefore, 400W = VL ∙ IP ∙ 0.8 = 300V ∙ IP ∙ 0.8, and IP = 1.667Arms
The line current is: IL = IP = A = 2.89Arms
Moreover, a lagging PF implies the voltage leads the current by argcos(0.8) = 36.9o
The impedance is:
Note: the apparent power of a Δ connected load is P = Vab × IAB
(line voltage × phase current)

25. Ch6 Polyphsae Circuits 6.6 Power measurement Wattmeter measured by
current coil
measured by
potential coil
current coil
Passive
Network
potential coil
E.g.

26. Ch6 Polyphsae Circuits 6.6 Power measurement 1 2
Validate the power meter reads the actual power absorbed/delivered by the three impedances.

27. Ch6 Polyphsae Circuits 6.6 Power measurement reactive (PF=0)
capacitive / inductive (0<PF<1)
resistive (PF=1)

28. . Ch6 Polyphsae Circuits 6.6 Power measurement 12 N
6.6 Power measurement
Example 9.7 (p256)
with positive phase sequence.
(1) Find the reading of each wattmeter.
(2) The total power absorbed by the loads.
With positive phase sequence , we know :
Wattmeter 1 reads and :

29. . Ch6 Polyphsae Circuits 6.6 Power measurement 1 Wattmeter 1 reads : 2
Example 9.7 (p256)
Wattmeter 1 reads :
Wattmeter 2 reads and :
Hence ,
Q: Please try to prove the two wattmeters read the power associated with the three impedances.