THREE PHASE CIRCUITS
 A three-phase system is equivalent to three single-phase circuits
 Balanced phase voltages are equal in magnitude and are out of phase with each other by 120◦
Three – Phase Voltage Source
*** double subscript notation.
Phase voltages
Line voltages
Balanced voltages
Phase sequences - is the time order in which the voltages pass through their respective maximum
values
 abc or positive sequence - produced when the rotor rotates counter-clockwise
 acb or negative sequence - produced when the rotor in the clockwise direction
 A balanced load is one in which the phase impedances are equal in magnitude and in phase
Possible Connections of Source and Loads
Y-Y connection
Y-∆ connection
∆-∆ connection
∆-Y connection
BALANCED WYE-WYE CONNECTION
 is a three-phase system with a balanced Y-connected source and a balanced Y-connected load
Phase Voltages: (assuming the positive sequence)
Source
Load (assuming no line impedance):
Line-to-line voltages
Line Currents:
Note: for a balanced system, the neutral current is always equal to zero.
Single – phase Equivalent Circuit
Example 1:
Example 2:
BALANCED WYE-DELTA CONNECTION
 consists of a balanced Y-connected source feeding a balanced -connected load.
Line voltages: (Assuming the positive sequence: phase voltages)
Source
Load
Phase Currents:
 Phase currents have the same magnitude but are out of phase with each other by 120◦
Line Currents:
Note: alternative way of analyzing the Y-∆ circuit is to transform the ∆-connected load to an equivalent Yconnected load
Example 1:
BALANCED DELTA-DELTA CONNECTION
 both the balanced source and balanced load are ∆ - connected.
Phase Voltages : (assuming a positive sequence)
Source:
Load: (assuming no line impedance)
Phase Currents:
Line Currents:
Note: alternative way of analyzing the ∆-∆ circuit is to transform both the load and the source to
their Y - equivalents
Example
1:
BALANCED DELTA-WYE CONNECTION
 consists of a balanced ∆ - connected source feeding a balanced Y-connected load.
Phase Voltages : (assuming a positive sequence)
Source:
Load: (assuming no line impedance)
Line Currents:
Note: Another way to obtain the line currents is to replace the delta - connected source with its
equivalent wye-connected source
Example 1:
Example 2:
POWER IN A BALANCED SYSTEM
Power per Phase
Real
Reactive
Apparent/ Complex
Three - Phase Power
Real
Reactive
Apparent/ Complex
Example 1:
Example 2:
UNBALANCED THREE-PHASE SYSTEMS
 is due to unbalanced voltage sources or an unbalanced load
 Unbalanced three-phase systems are solved by direct application of mesh and nodal analysis
 Total power is not simply three times the power in one phase but the sum of the powers in the
three phases.
Example 1:
Three-Phase Power Measurement
1. Three-Wattmeter Method - well suited for power measurement in a three-phase system
where the power factor is constantly changing.
 In a three-wattmeter method, the algebraic sum of three-wattmeter readings simply
gives the total electric power consumed in the circuit, whether it is balanced or not
Where: P1, P2, & P3 correspond to the readings of wattmeters W1, W2, and W3,
respectively
 The one end of the three wattmeters’ potential coils is connected to the common point
o. If a neutral point is available in the three-phase system, then the common point o
should be connected to that neutral point.
2. Two-Wattmeter Method - most commonly used method for three-phase power measurement.
 Notice that the current coil of each wattmeter measures the line current, while the
respective voltage coil is connected between the line and the third line and measures
the line voltage
 Although the individual wattmeters no longer read the power taken by any particular
phase, the algebraic sum of the two wattmeter readings equals the total average
power absorbed by the load, regardless of whether it is wye- or delta-connected,
balanced or unbalanced
 The total real power is equal to the algebraic sum of the two wattmeter readings,
 Consider the figure




Assume the source is in the abc sequence and the load impedance ZY
θ.
Each phase voltage leads its respective phase current by θ
Each line voltage leads the corresponding phase voltage by 30°.
Thus, the total phase difference between the phase current Ia and line voltage Vab is θ
+ 30°
 The average power read by wattmeter W1 is
 Similarly, we can show that the average power read by wattmeter 2 is
 We now use the following trigonometric identities to find the sum and the difference
of the two wattmeter reading P1 and P2
 Thus, the sum of the wattmeter readings gives the total average power
 Similarly,
 Thus, the difference of the wattmeter readings is proportional to the total reactive
power,
 the apparent power can calculated as
 the tangent of the power factor angle is
 Thus the power factor is,
 Thus, the two-wattmeter method not only provides the total real and reactive powers,
it can also be used to compute the power factor.
 Although these results are derived from a balanced Y-connected load, they are equally
valid for a balanced Δ-connected load
 The two-wattmeter method cannot be used for power measurement in a 3-phase 4wire system unless the current through the neutral line is zero.
 We can use the three-wattmeter method to measure the real power in a 3-phase 4wire system
Example 1:
Calculate the readings of the two wattmeters ( W1 & W2 ) connected to measure the total
power for a balanced star-connected load shown in the figure below, fed from a three-phase,
400 V balanced supply with phase sequence as abc. The load impedance per phase is (20 + j15)
Ω. Also find the line and phase currents, power factor, total power, total reactive VA and total
VA
Calculate the readings of the two wattmeters (W1 & W2) connected to measure the total power for
a balanced delta-connected load shown in the figure, fed from a three=phase, 200 V balanced supply with
phase sequence as abc. The load impedance per phase is (14 – j14) Ω. Also find the line and phase currents,
power factor, total power, total reactive VA and total VA