# Review of 3-Phase Circuits

## Requirements of a Balanced 3-Phase Set

Following are the requirements that must be satisified in order for a set of 3 sinusoidal variables (usually voltages or currents) to be a "balanced 3-phase set"

1. All 3 variables have the same amplitude

2. All 3 variables have the same frequency

3. All 3 variables are 120o in phase

In terms of the time domain, a set of balance 3-phase voltages has the following general form.

va = Vm cos ( t + )

vb = Vm cos ( t + - 120o )

vc = Vm cos ( t + - 240o ) = Vm cos ( t + +120o )

Notice that we have assumed (and will continue to assume) positive (abc) phase sequence, i.e., phase "b" follows 120o behind "a" & phase "c" follows 120o behind phase "b"

Figure 1 below illustrates the balanced 3-phase voltages in time domain.

Figure 1: Balanced 3-Phase Variables in Time Domain

In terms of phasors, we write the same balanced set as follows. Note that the phasors are in rms, as will be assumed throughout this course.

Va = Vm m

Vb = Vm - 120o

Vc = Vm - 240o = Vm +120o

Thus,

Vb = Va (1 -120o) , and Vc = Va (1 +120o)

Figure 2 below illustrates the balanced 3-phase phasors graphically.

Figure 2: Balanced 3-Phase Phasors

## Requirements of a Balanced 3-Phase Circuit

Following are the requirements that must be satisified in order for a 3-phase system or circuit to be balanced

1. All 3 sources are reprensented by a set of balanced 3-phase variables

2. All loads are 3-phase with equal impedances

3. Line impedances are equal in all 3 phases

Having a balanced circuit allows for simplified analysis of the 3-phase 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 "one-line 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 3-phase circuit and some of the naming conventions to be used in this course

Figure 3: A Balanced 3-Phase Circuit

## Terms and Naming Conventions

Phase
describes or pertains to one element or device in a load, line, or source. It is simply a "branch" of the circuit and could look something like this .

Line
refers to the "transmission line" or wires that connect the source (supply) to the load. It may be modeled as a small impedance (actually 3 of them), or even by just a connecting line.

Neutral
the 4th wire in the 3-phase system. It's where the phases of a Y connection come together.

Phase Voltages & Phase Currents
the voltages and currents across and through a single branch (phase) of the circuit. Note this definition depends on whether the connection is Wye or Delta!

Line Currents
the currents flowing in each of the lines (Ia, Ib, and Ic). This definition does not change with connection type.

Line Voltages
the voltages between any two of the lines (Vab, Vbc, and Vca). These may also be referred to as the line-to-line voltages. This definition does not change with connection type.

Line to Neutral Voltages
the voltages between any lines and the neutral point (Va, Vb, and Vc). This definition does not change with connection type, but they may not be physically measureable in a Delta circuit.

Line to Neutral Currents
same as the line currents (Ia, Ib, and Ic).

## Where Does that Come From?

Let us determine the relationships between the line and line to neutral voltages. By applying Kirchoff's Voltage Law (KVL) to the top "loop" of the source section in Figure 3, we get

Vab = Va - Vb = Vm - Vm - 120o

Now, without loss of generality, let = 0o

thus, Va = Vm 0o, and Vb = Vm -120o, so

Vab = Vm 0o - Vm - 120o = Vm (1 - 1 - 120o ) = Vm (1 - (cos 120o - j sin 120o))

= Vm (1 - (-1/2) + j ( / 2 ) ) = Vm (3 / 2) + j ( / 2 ))

Converting to polar form,

Vab = Vm Sqrt[ (3/2)2 + ( / 2)2 ] tan-1 {( / 2) / (3/2) }

= Vm Sqrt[ 9/4 + 3/4 ] tan-1 {1/ }

= Vm tan-1 {(1 / 2) / ( /2) } = Vm tan-1 {(sin 30o) / (cos 30o) }

= Vm tan-1 {tan 30o } = Vm 30o

Thus we have the general equation (for abc sequence anyway)

Vab = Va 30o

The relationships between the currents can be developed similarly. Summing currents at the "A" node in Figure 3 yields the starting equation,

Ia = IAB - ICA

This time choose Ia to be the phasor reference (at 0o). The final result is:

Ia = IAB -30o

These relationships can also be remembered graphically using Figures 4 and 5 below. Figure 4 illustrates the voltage relationship. By looking at the phasor equation as the sum of two vectors (Va and -Vb ) we obtain the resulting Vab shown in the figure.

Since Vab is longer, we know . . . . |Vab| = |Va| ,

and since Vab is ahead of Va, we know that, . . . . (the angle of Vab) = (the angle of Va) + 30o

Figure 4: Graphical Voltage Relationship

Figure 5 illustrates the current relationship. Now the phasor equation is the sum of two vectors (Iab and -Ica ) we obtain the resulting Ia shown in the figure.

Since Ia is longer, we know

|Ia| = |Iab| ,

and since Ia is behind Iab, we know that,

(the angle of Ia) = (the angle of Iab) - 30o

Figure 5: Graphical Current Relationship

## Wyes and Deltas

A summary of the characteristics of the two types of 3-phase circuit connections are given below.

 The Wye = Y = "Star" connection ____ The Delta = connection each phase is connected between a line and the neutral each phase is connected between two lines Figure 6: A Y Circuit Figure 7: A Circuit Phase voltages = Line to neutral voltages (Va, etc.) Phase currents = Line currents (Ia, etc.) Neutral connects the 3 phases Phase voltages = Line voltages (Vab, etc.) Phase currents = currents from line to line (Iab, etc.) Neutral is not present

## Y to Conversions

In terms of power, currents & line voltages, the following sources are the same and may be used interchangably in most cases. Note, the Y connection should be used in a one-line diagram.

 Wye connected source ____ Delta connected source Figure 8: A Y Source Figure 9: A Source VA = Vab / ( 30o) Vab = VA ( 30o)

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 one-line diagram!

## The One-Line Diagram

If the circuit is balanced, all corresponding sets of 3-phase voltages and currents are balanced, and the neutral current will be zero.

IN = Ia + Ib + Ic

Why must this be so? Because the sum of a balanced set of 3-phase 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 Y-Connected Circuit Including Neutral

If the Neutral points in Figure 12 are actually the same point, Figure 12 can be redrawn as shown in Figure 13.

Figure 13: ReDrawn All-Y Circuit

From this figure we see that each of the phase currents depends only on the source in the corresponding loop. In other words,

Ia = Va / ( Zline + ZY) , Ib = Vb / ( Zline + ZY) , and Ic = Vc / ( Zline + ZY)

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 120o. Thus, we only need to consider and solve one loop of Figure 13 --- this is the one-line diagram!

Figure 14 shows the one-line 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 120o, and the "c" phase quantities are found by adding 120o.

Figure 14: The One-Line Diagram

## 3-Phase Power

The 3-phase (3) power of a circuit is simply the sum of the power in the three individual phases. Thus for a Wye circuit, the equation is

S3 = Sa + Sb + Sc

and for a Delta circuit, the equation is

S3 = Sab + Sbc + Sca

Another adavantage of having a balanced circuit is that each phase has the same power. That is,

S = Sab = Sbc = Sca = Sa = Sb = Sc

so that,

S3 = 3 S = 3 Sab = 3 Sa

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 = Va Ia* or S = Vab Iab*

We can do some interesting rearrangements to get the power in terms of the line voltage (Vab) and line current (Ia) only.

S = Va Ia* = |Va| | Ia| = {|Vab| / }| Ia| = S

Thus, S3 = 3 S = 3 {|Vab| / }| Ia| = |Vab| | Ia|

Note:
In balanced systems, all the S's and S3 have the same power factor (pf) and thus the same power factor angle = impedance angle = .

## Collection of Important 3-Phase Equations

If Xa, Xb, and Xc are a balanced set,

Xb = Xa (1 -120o) , and

Xc = Xa (1 +120o)

In general,

S3 = Sa + Sb + Sc

S3 = Sab + Sbc + Sca

For a balanced system,

S3 = 3 S

S = Va Ia*

S = Vab Iab*

For the balanced and positive sequence case,

Vab = Va 30o

Ia = IAB -30o

## What to Assume

If you are given a voltage, current, or power value and not told specifically which variable it is, you should assume that you have been given a "line" value. That is, assume the following:

Voltage => Line voltage = |Vab|

Current => Line current = |Ia|

Power => Three Phase Power = S3, P3, or Q3

## Unbalanced Circuits

When we have an unbalanced circuit, we CANNOT use the one-line diagram to solve for "a" phase values and then get the answers for the other phases by adding or subtracting 120o.

In general, a unbalanced three-phase circuit requires that you draw the complete circuit including all 3-phase and single-phase 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 3-phase 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 one-line 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.

## Wattmeters

The schematic for a wattmeter is given in Figure 15 below. Note that in order to measure power, we need to measure a current and a voltage. The wattmeter doesn't care which current or which voltage you use. It will give you a reading regardless of whether or not it means anything. It is up to the user (you) to make sure the meter is sensing the correct voltage and current to give a meaningful measurement!

Figure 15: The Basic Wattmeter

W = I V cos (V - I )

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 three-phase motor (for example). You want to measure P3, where do you connect your meter?

To measure the 3-phase power correctly using two meters:
connect the current coils in two of the phases

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 "plus-minus" 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

The meter readings for Figure 16 are:

W1 = |Ia| |Vab| cos (Vab - Ia )

W2 = |Ic| |Vcb| cos (Vcb - Ic )

The meter readings for Figure 17 are:

W1 = |Ia| |Vac| cos (Vac - Ia )

W2 = |Ib| |Vbc| cos (Vbc - Ib )

The three phase real power is found using . . . P3 = W1 + W2

similarly, for the balanced condition, the magnitude three phase reactive power can be found using . . . |Q3| = |W1 - W2|

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.

Shelli Starrett

September 18, 1997