In terms of the time domain, a set of balance 3phase voltages has the following general form.
Figure 1 below illustrates the balanced 3phase voltages in time domain.
Figure 1: Balanced 3Phase Variables in Time Domain
Thus,
Figure 2: Balanced 3Phase Phasors
Having a balanced circuit allows for simplified analysis of the 3phase circuit. In fact, if the circuit is balanced, we can solve for the voltages, currents, and powers, etc. in one phase using circuit analysis. The values of the corresponding variables in the other two phases can be found using some basic equations. This type of solution is accomplished using a "oneline diagram", which will be discussed later. If the circuit is not balanced, all three phases should be analyzed in detail.
Figure 3 illustrates a balanced 3phase circuit and some of the naming conventions to be used in this course
Figure 3: A Balanced 3Phase Circuit
V_{ab} = V_{a}  V_{b} = V_{m}  V_{m}  120^{o}
Now, without loss of generality, let = 0^{o}
thus, V_{a} = V_{m} 0^{o}, and V_{b} = V_{m} 120^{o}, so
V_{ab} = V_{m} 0^{o}  V_{m}  120^{o} = V_{m} (1  1  120^{o} ) = V_{m} (1  (cos 120^{o}  j sin 120^{o}))
= V_{m} (1  (1/2) + j ( / 2 ) ) = V_{m} (3 / 2) + j ( / 2 ))
Converting to polar form,
V_{ab} = V_{m} Sqrt[ (3/2)^{2} + ( / 2)^{2} ] tan^{1} {( / 2) / (3/2) }
= V_{m} Sqrt[ 9/4 + 3/4 ] tan^{1} {1/ }
= V_{m} tan^{1} {(1 / 2) / ( /2) } = V_{m} tan^{1} {(sin 30^{o}) / (cos 30^{o}) }
= V_{m} tan^{1} {tan 30^{o} } = V_{m} 30^{o}
Thus we have the general equation (for abc sequence anyway)
I_{a} = I_{AB}  I_{CA}
This time choose I_{a} to be the phasor reference (at 0^{o}). The final result is:
Since V_{ab} is longer, we know . . . . V_{ab} = V_{a} ,
and since V_{ab} is ahead of V_{a}, we know that, . . . . (the angle of V_{ab}) = (the angle of V_{a}) + 30^{o}
Figure 4: Graphical Voltage Relationship
Since I_{a} is longer, we know
I_{a} = I_{ab} ,
and since I_{a} is behind I_{ab}, we know that,
(the angle of I_{a}) = (the angle of I_{ab})  30^{o}
Figure 5: Graphical Current Relationship
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each phase is connected between a line and the neutral  each phase is connected between two lines  
Phase voltages = Line to neutral voltages (V_{a}, etc.) Phase currents = Line currents (I_{a}, etc.)
 Phase voltages = Line voltages (V_{ab}, etc.) Phase currents = currents from line to line (I_{ab}, etc.)

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Similarly, the two loads given below are the same in terms of the resulting power, line currents and line voltages and can usually be substituted as desired. Note that the Y connection is the one needed for the oneline diagram!
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I_{N} = I_{a} + I_{b} + I_{c}
Why must this be so? Because the sum of a balanced set of 3phase variables is equal to zero. This can be verified mathematically using the definition, or visually by considering using vector addition to add the balanced set in Figure 5.
Because the neutral current is zero, this means that if the neutral in the load is connected to the neutral in the source, no current will flow. Thus, the voltage at each of the neutrals must be the same. This means they can be considered to be the same point.
Now consider the circuit of Figure 12. In general, any circuit with a source, load, and line configuration can be converted to a circuit of this type by replacing any Delta connected sources or loads with the equivalent Wye connected sources or loads.
Figure 12: Completely YConnected Circuit Including Neutral
Figure 13: ReDrawn AllY Circuit
I_{a} = V_{a} / ( Z_{line} + Z_{Y}) , I_{b} = V_{b} / ( Z_{line} + Z_{Y}) , and I_{c} = V_{c} / ( Z_{line} + Z_{Y})
Notice that these equations are VERY similar.
Recall that in balanced set of variables, once we know one variable, the other two can be found by simply adding and subtracting 120^{o}. Thus, we only need to consider and solve one loop of Figure 13  this is the oneline diagram!
Figure 14 shows the oneline diagram for the circuit of Figure 13. Usually the one line that is considered is the "a" phase. The "b" phase quantities are then found by subtracting 120^{o}, and the "c" phase quantities are found by adding 120^{o}.
Figure 14: The OneLine Diagram
S_{3} = S_{a} + S_{b} + S_{c}
and for a Delta circuit, the equation is
S_{3} = S_{ab} + S_{bc} + S_{ca}
Another adavantage of having a balanced circuit is that each phase has the same power. That is,
S_{} = S_{ab} = S_{bc} = S_{ca} = S_{a} = S_{b} = S_{c}
so that,
S_{3} = 3 S_{} = 3 S_{ab} = 3 S_{a}
Just in case you didn't know, right now you should be thinking "This is very cool!"
The single phase power can be found using either
S_{} = V_{a} I_{a}* or S_{} = V_{ab} I_{ab}*
We can do some interesting rearrangements to get the power in terms of the line voltage (V_{ab}) and line current (I_{a}) only.
S_{} = V_{a} I_{a}* = V_{a}  I_{a} = {V_{ab} / } I_{a} = S_{}
Thus, S_{3} = 3 S_{} = 3 {V_{ab} / } I_{a} = V_{ab}  I_{a}
In general, a unbalanced threephase circuit requires that you draw the complete circuit including all 3phase and singlephase loads and perform a circuit analysis of the whole thing. Normal methods such as "meshes" or "node voltages" may be used. If you have the simple case in which a balanced 3phase load is connected directly to a source and a single phase load is connected in parallel to the same source, you may calculate the currents in the balanced load using a oneline method. The single phase current is calculated separately and then individual line currents can be found by summing the currents at certain nodes in the system.
Remember any circuit that does not have all loads with the same impedance in all three branches is an unbalanced circuit.
Figure 15: The Basic Wattmeter
Under balanced conditions and conditions in which there are only three wires in the system, we can measure the power in all three phases of a load (or source) by using only two meters. This is called the "Two Wattmeter Method."
This method is quite convenient when all you have access to are the three wires going into a threephase motor (for example). You want to measure P_{3}, where do you connect your meter?
connect the positive terminals of the voltage coils to the same two phases (where you're measuring the current)
connect both of the negative voltage terminals to the third phase.
Figures 16 and 17 below show two possible connections with phases "b" and "c" respectively, used as the voltage reference. Note that the "plusminus" symbol marks the positive voltage terminal & the negative terminal is generally unmarked.
Figure 16: Two Wattmeter Connection with "b" as Reference
Figure 17: Two Wattmeter Connection with "c" as Reference
W_{2} = I_{c} V_{cb} cos (_{Vcb}  _{Ic} )
W_{2} = I_{b} V_{bc} cos (_{Vbc}  _{Ib} )
similarly, for the balanced condition, the magnitude three phase reactive power can be found using . . . Q_{3} = W_{1}  W_{2}
the sign of Q may vary depending on how the wattmeters are connected. So, it is generally safer to determine the sign using other means.
September 18, 1997