Electrical Circuits III.pdf

Electrical Circuits III: Advanced Circuit Analysis

Electrical Engineering Department | Engr. Gerard Francesco DG. Apolinario 1

Single Phase Power Systems

Depicted above is a very simple AC circuit. If the

load resistor’s power dissipation were substantial,

it is called a “power circuit” or “power system”

instead of regarding it as just a regular circuit. The

distinction between a “power circuit” and a

“regular circuit” may seem arbitrary, but the

practical concerns are definitely not.

One such concern is the size and cost of wiring

necessary to deliver power from the AC source to

the load. Normally, it is not given much thought

for this type of concern if it is merely analyzing a

circuit for the sake of learning about the laws of

electricity. However, in the real world it can be a

major concern. If the source in the above circuit is

given a voltage value and power dissipation

values are also given to the two load resistors, the

wiring needs for this particular circuit can be

attained:

83.33 A for each load resistor in the figure adds

up to 166.66 A total circuit current. This is no

small amount of current, and would necessitate

copper wire conductors of at least 1/0 gage. Such

wire is well over 1/4 inch (6 mm) in diameter,

weighing over 300 pounds per thousand feet.

Bear in mind that copper is not cheap either! It

would be for the best interest to find ways to

minimize such costs for designing a power

system with long conductor lengths. One way to

do this would be to increase the voltage of the

power source and use loads built to dissipate 10

kW each at this higher voltage. The loads, of

course, would have to have greater resistance

values to dissipate the same power as before (10

kW each) at a greater voltage than before. The

advantage would be less current required,

permitting the use of smaller, lighter, and cheaper

wire:

Now the total circuit current is 83.33 A, half of

what it was before. A number 4 gage wire, which

weighs less than half of what 1/0 gage wire does

per unit length, can be used. This is a

considerable reduction in system cost with no

degradation in performance. This is why power

distribution system designers elect to transmit

electric power using very high voltages (many

thousands of volts) to capitalize on the savings

realized by the use of smaller, lighter, cheaper

wire. However, this solution is not without

disadvantages. Another practical concern with

power circuits is the danger of electric shock

from high voltages. Again, this is not usually the

sort of thing pointed out while learning about the

laws of electricity, but it is a very valid concern

in the real world, especially when large amounts

of power are being dealt with. The gain in

efficiency realized by stepping up the circuit

voltage presents us with increased danger of

electric shock. Power distribution companies

tackle this problem by stringing their power lines

along high poles or towers, and insulating the

lines from the supporting structures with large,

porcelain insulators.

At the point of use (the electric power customer),

there is still the issue of what voltage to use for

powering loads. High voltage gives greater

system efficiency by means of reduced

conductor current, but it might not always be

practical to keep power wiring out of reach at the

point of use the way it can be elevated out of

reach in distribution systems. This tradeoff

between efficiency and danger is one that

European power system designers have

decided to risk, all their households and

appliances operating at a nominal voltage of

240 V instead of 120 V as it is in North America.



That is why tourists from America visiting

Europe must carry small step – down

transformers for their portable appliances, to

step the 240 VAC (volts AC) power down to a

more suitable 120 VAC.

Is there any way to realize the advantages of

both increased efficiency and reduced safety

hazard at the same time? One solution would be

to install step – down transformers at the

endpoint of power use, just as the American

tourist must do while in Europe. However, this

would be expensive and inconvenient for

anything but very small loads (where the

transformers can be built cheaply) or very large

loads (where the expense of thick copper wires

would exceed the expense of a transformer).

An alternative solution would be to use a higher

voltage supply to provide power to two lower

voltage loads in series. This approach combines

the efficiency of a high – voltage system with the

safety of a low – voltage system:

The current through each load is the same as it

was in the simple 120 V circuit, but the currents

are not additive because the loads are in series

rather than parallel. The voltage across each load

is only 120 V, not 240, so the safety factor is

better. Take note that a full 240 V across the

power system wires is still operating but each load

is at a reduced voltage. If anyone is going to get

shocked, the odds are that it will be from coming

into contact with the conductors of a particular

load rather than from contact across the main

wires of a power system.

There’s only one disadvantage to this design: the

consequences of one load failing open, or being

turned off (assuming each load has a series on/off

switch to interrupt current) are not good. Being a

series circuit, if either load were to open, current

would stop in the other load as well. For this

reason, the design is modify to:

Instead of a single 240 V power supply, two 120

V supplies (in phase with each other!) in series is

use to produce 240 V, then run a third wire to the

connection point between the loads to handle the

eventuality of one load opening. This is called a

split – phase power system. Three smaller wires

are still cheaper than the two wires needed with

the simple parallel design, so it is still ahead on

efficiency. The astute observer will note that the

neutral wire only has to carry the difference of

current between the two loads back to the source.

In the above case, with perfectly “balanced” loads

consuming equal amounts of power, the neutral

wire carries zero current.

Notice how the neutral wire is connected to

earth ground at the power supply end. This is a

common feature in power systems containing

“neutral” wires, since grounding the neutral wire

ensures the least possible voltage at any given

time between any “hot” wire and earth ground.

An essential component to a split – phase power

system is the dual AC voltage source.

Fortunately, designing and building one is not

difficult. Since most AC systems receive their

power from a step – down transformer anyway

(stepping voltage down from high distribution

levels to a user – level voltage like 120 or 240),

that transformer can be built with a center –

tapped secondary winding:

If the AC power comes directly from a generator



(alternator), the coils can be similarly center –

tapped for the same effect. The extra expense

to include a center – tap connection in a

transformer or alternator winding is minimal.

Here is where the (+) and (-) polarity markings

really become important. This notation is often

used to reference the phasings of multiple AC

voltage sources, so it is clear whether they are

aiding (“boosting”) each other or opposing

(“bucking”) each other. If not for these polarity

markings, phase relations between multiple AC

sources might be very confusing. Note that the

split – phase sources in the schematic (each

one 120˂0° V), with polarity marks (+) to (-) just

like series-aiding batteries can alternatively be

represented as such:

To mathematically calculate voltage between

“hot” wires, we must subtract voltages, because

their polarity marks show them to be opposed to

each other:

𝑃𝑜𝑙𝑎𝑟 𝑅𝑒𝑐𝑡𝑎𝑛𝑔𝑢𝑙𝑎𝑟

120∠0° 120 + 𝑗0 𝑉

−120∠180° − (−120 + 𝑗0)𝑉

240∠0° 240 + 𝑗0 𝑉

If the two sources’ common connection point is

marked (the neutral wire) with the same polarity,

their relative phase shifts will be 180°. Otherwise,

it is denoted with two voltage sources in direct

opposition with each other, which would give 0 V

between the two “hot” conductors. Power

systems in American households and light

industry are most often of the split – phase

variety, providing so called 120/240 VAC power.

The term “split – phase” merely refers to the split

– voltage supply in such a system. In a more

general sense, this kind of AC power supply is

called single phase because both voltage

waveforms are in phase, or in step, with each

other.

The term “single phase” is a counterpoint to

another kind of power system called “polyphase”.

Three Phase Power Systems

Split – phase power systems achieve their high

conductor efficiency and low safety risk by

splitting up the total voltage into lesser parts and

powering multiple loads at those lesser voltages,

while drawing currents at levels typical of a full –

voltage system. This technique, by the way,

works just as well for DC power systems as it

does for single – phase AC systems. Such

systems are usually referred to as three – wire

systems rather than split – phase because

“phase” is a concept restricted to AC.

But from the experience in dealing with vectors

and complex numbers, AC voltages don’t always

add up as they would if they are out of phase with

each other. This principle, applied to power

systems, can be put to use to make power

systems with even greater conductor efficiencies

and lower shock hazard than with split – phase.

Suppose that two sources of AC voltage are

connected in series just like the split – phase

system, except that each voltage source was

120° out of phase with the other:

Since each voltage source is 120 V, and each

load resistor is connected directly in parallel with

its respective source, the voltage across each

load must be 120 V as well. Given load currents

of 83.33 A, each load must still be dissipating 10

kW of power. However, voltage between the two

“hot” wires is not 240 V (120∠0° − 120∠120°) because the phase difference between the two

sources is not 180°.

Instead, the voltage is:

𝐸𝑡𝑜𝑡𝑎𝑙 = (120∠0°) − (120∠120°) = 207.85∠ − 30°

Nominally, the voltage between “hot”

conductors is 208 volts (rounding up), and thus



the power system voltage is designated as

120/208.

The current through the “neutral” conductor will

not be zero, even with balanced load

resistances. Kirchhoff’s Current Law tells us that

the currents entering and exiting the node

between the two loads must be zero:

−𝐼𝑙𝑜𝑎𝑑1 − 𝐼𝑙𝑜𝑎𝑑2 − 𝐼𝑛𝑒𝑢𝑡𝑟𝑎𝑙 = 0

−𝐼𝑛𝑒𝑢𝑡𝑟𝑎𝑙 = 𝐼𝑙𝑜𝑎𝑑1 + 𝐼𝑙𝑜𝑎𝑑2

𝐼𝑛𝑒𝑢𝑡𝑟𝑎𝑙 = −𝐼𝑙𝑜𝑎𝑑1 − 𝐼𝑙𝑜𝑎𝑑2

𝐼𝑛𝑒𝑢𝑡𝑟𝑎𝑙 = −(83.33∠0) − (83.33∠120) = 83.33∠240

= 83.33∠ − 120

It is found out that the “neutral” wire is carrying a

full 83.33 A, just like each “hot” wire.

Note that 20 kW of total power are given to the

two loads, with each load’s “hot” wire carrying

83.33 A as before. With the same amount of

current through each “hot” wire, the same gage

copper conductors is used, so the system cost

over the split – phase 120/240 system is

reduced. However, a gain in safety is realized,

because the overall voltage between the two

“hot” conductors is 32 V lower than it was in the

split – phase system (208 V instead of 240 V).

The fact that the neutral wire is carrying 83.33 A

of current raises an interesting possibility: since

its carrying current anyway, why not use that

third wire as another “hot” conductor, powering

another load resistor with a third 120 V source

having a phase angle of 240°? That way, it could

transmit more power (another 10 kW) without

having to add any more conductors. Let’s see

how this might look:

A full mathematical analysis of all the voltages

and currents in this circuit would necessitate the

use of a network theorem, the easiest being the

Superposition Theorem. Sure enough, there

would be 120 V across each load resistor, with

(approximately) 208 V between any two “hot”

conductors and conductor currents equal to 83.33

A. At that current and voltage, each load will be

dissipating 10 kW of power. Notice that this circuit

has no “neutral” conductor to ensure stable

voltage to all loads if one should open. This

situation is similar to our split – phase power

circuit with no “neutral” conductor: if one load

should happen to fail open, the voltage drops

across the remaining load(s) will change. To

ensure load voltage stability in the event of

another load opening, a neutral wire is added to

connect the source node and load node together.

So long as the loads remain balanced (equal

resistance, equal currents), the neutral wire will

not carry any current at all. It is there just in case

one or more load resistors should fail open (or be

shut off through a disconnecting switch). This

circuit with three voltage sources is called a

polyphase circuit. Since the voltage sources are

all at different phase angles (in this case, three

different phase angles), this is a “polyphase”

circuit. More specifically, it is a three – phase

circuit, the kind is used predominantly in large

power distribution systems.

A single – phase system with three loads

connected directly in parallel would have a very

high total current (83.33 times 3, or 250 A).



This would necessitate 3/0 gage copper wire

(very large!), at about 510 pounds per thousand

feet, and with a considerable price tag attached.

If the distance from source to load was 1000 feet,

it would need over a half – ton of copper wire to

do the job. On the other hand, a split – phase

system with two 15 kW, 120 volt loads could be

design

The current is half of what it was with the simple

parallel circuit, which is a great improvement. It

could get away with using number 2 gage

copper wire at a total mass of about 600 pounds,

figuring about 200 pounds per thousand feet

with three runs of 1000 feet each between

source and loads. However, the increased

safety hazard of having 240 volts present in the

system, even though each load only receives

120 volts must be considered. Overall, there is

greater potential for dangerous electric shock to

occur. Contrasting the two examples against the

three – phase system, the advantages are quite

clear. First, the conductor currents are quite a

bit less (83.33 amps versus 125 or 250 amps),

permitting the use of much thinner and lighter

wire. The use of number 4 gage wire at about

125 pounds per thousand feet, which will total

500 pounds (four runs of 1000 feet each) for the

example circuit. This represents a significant

cost savings over the split – phase system, with

the additional benefit that the maximum voltage

in the system is lower (208 versus 240).

One question remains to be answered: how in

the world do we get three AC voltage sources

whose phase angles are exactly 120° apart?

Obviously you can’t center – tap a transformer

or alternator winding like in the split – phase

system, since that can only give voltage

waveforms that are either in phase or 180° out

of phase. Perhaps, the use of capacitors and

inductors to create phase shifts of 120° could be

considered, but then those phase shifts would

depend on the phase angles of the load

impedances (substituting a capacitive or

inductive load for a resistive load would change

everything!).

The best way to get the phase shifts is to

generate it at the source: construct the AC

generator (alternator) providing the power in

such a way that the rotating magnetic field

passes by three sets of wire windings, each set

spaced 120° apart around the circumference of

the machine:

Together, the six “pole” windings of a three –

phase alternator are connected to comprise three

winding pairs, each pair producing AC voltage

with a phase angle 120° shifted from either of the

other two winding pairs. The interconnections

between pairs of windings (as shown for the

single – phase alternator: the jumper wire

between windings 1a and 1b) have been omitted

from the three – phase alternator drawing for

simplicity.

In the example circuit, the three voltage sources

connected together in a “Y” configuration

(sometimes called the “star” or “T” configuration),

with one lead of each source tied to a common

point (the node where the “neutral” conductor is

attached). The common way to depict this

connection scheme is to draw the windings in the

shape of a “Y”.



Phase Rotation

The phase angle shift of 120° is a function of the

actual rotational angle shift of the three pairs of

windings.

If the magnet is rotating clockwise, winding 3 will

generate its peak instantaneous voltage exactly

120° (of alternator shaft rotation) after winding 2,

which will hits its peak 120° after winding 1. The

magnet passes by each pole pair at different

positions in the rotational movement of the shaft.

Where we decide to place the windings will

dictate the amount of phase shift between the

windings’ AC voltage waveforms. If we make

winding 1 our “reference” voltage source for

phase angle (0°), then winding 2 will have a

phase angle of - 120° (120° lagging, or 240°

leading) and winding 3 an angle of - 240° (or 120°

leading). This sequence of phase shifts has a

definite order. For clockwise rotation of the shaft,

the order is 1 – 2 – 3 (winding 1 peaks first, then

winding 2, then winding 3). This order keeps

repeating itself as long as we continue to rotate

the alternator’s shaft.

However, if we reverse the rotation of the

alternator’s shaft (turn it counterclockwise), the

magnet will pass by the pole pairs in the

opposite sequence. Instead of 1 – 2 – 3, we’ll

have 3 – 2 – 1. Now, winding 2’s waveform will

be leading 120° ahead of 1 instead of lagging,

and 3 will be another 120° ahead of 2.

The order of voltage waveform sequences in a

polyphase system is called phase rotation,

phase sequence or phase order. If we’re using

a polyphase voltage source to power resistive

loads, phase rotation will make no difference at

all. Whether 1 – 2 – 3 or 3 – 2 – 1, the voltage

and current magnitudes will all be the same.

Since voltmeters and ammeters would be

useless in telling us what the phase rotation of

an operating power system is, we need to have

some other kind of instrument capable of doing

the job. One ingenious circuit design uses a

capacitor to introduce a phase shift between

voltage and current, which is then used to detect

the sequence by way of comparison between

the brightness of two indicator lamps.

The two lamps are of equal filament resistance

and wattage. The capacitor is sized to have

approximately the same amount of reactance at

system frequency as each lamp’s resistance. If

the capacitor were to be replaced by a resistor of

equal value to the lamps’ resistance, the two

lamps would glow at equal brightness, the circuit

being balanced. However, the capacitor

introduces a phase shift between voltage and

current in the third leg of the circuit equal to 90°.

This phase shift, greater than 0° but less than

120°, skews the voltage and current values

across the two lamps according to their phase

shifts relative to phase 3.

Starting out with a phase rotation of 3-2-1, we can

try all the possibilities for swapping any two of the



wires at a time and see what happens to the

resulting sequence:

No matter which pair of “hot” wires out of the three

we choose to swap, the phase rotation ends up

being reversed (1 – 2 – 3 gets changed to 2 – 1 –

3, 1 – 3 – 2 or 3 – 2 – 1, all equivalent).

The three phases may be numbered 1, 2, 3 or a,

b, c or as is customary, they may be given three

colors. The colors used commercially are red,

yellow (or sometimes white) and blue.

1. Positive Sequence (ABC Sequence)

𝑉𝑎𝑛𝑙𝑒𝑎𝑑𝑠 𝑉𝑏𝑛 𝑙𝑒𝑎𝑑𝑠 𝑉𝑐𝑛 𝑏𝑦 120°

𝑉𝐿 𝑙𝑒𝑎𝑑𝑠 𝑉𝑝 𝑏𝑦 30°

𝐼𝐿 𝑙𝑎𝑔𝑠 𝐼𝑝 𝑏𝑦 30°

2. Negative Sequence (ACB Sequence)

𝑉𝑎𝑛𝑙𝑒𝑎𝑑𝑠 𝑉𝑐𝑛 𝑙𝑒𝑎𝑑𝑠 𝑉𝑏𝑛 𝑏𝑦 120°

𝑉𝐿 𝑙𝑎𝑔𝑠 𝑉𝑝 𝑏𝑦 30°

𝐼𝐿 𝑙𝑒𝑎𝑑𝑠 𝐼𝑝 𝑏𝑦 30°

NOTE: If the phase sequence is not given, you

can use the positive phase sequence.

Alternators are designed to operate with positive

sequence voltage.

Polyphase Motor Design

Perhaps the most important benefit of

polyphase AC power over single – phase is the

design and operation of AC motors.

If the rotating magnet is able to keep up with the

frequency of the alternating current energizing

the electromagnet windings (coils), it will

continue to be pulled around clockwise.

However, clockwise is not the only valid

direction for this motor’s shaft to spin. It could

just as easily be powered in a counter –

clockwise direction by the same AC voltage

waveform.

Notice that with the exact same sequence of

polarity cycles (voltage, current, and magnetic

poles produced by the coils), the magnetic rotor

can spin in either direction. This is a common

trait of all single – phase AC “induction” and

“synchronous” motors: they have no normal or

“correct” direction of rotation. The natural

question should arise at this point: how can the

motor get started in the intended direction if it

can run either way just as well? The answer is

that these motors need a little help getting

started. Once helped to spin in a particular

direction, it will continue to spin that way as long

as AC power is maintained to the windings.

Where that “help” comes from for a single – phase

AC motor to get going in one direction can vary.



Usually, it comes from an additional set of

windings positioned differently from the main set,

and energized with an AC voltage that is out of

phase with the main power.

These supplementary coils are typically

connected in series with a capacitor to introduce

a phase shift in current between the two sets of

windings.

That phase shift creates magnetic fields from

coils 2a and 2b that are equally out of step with

the fields from coils 1a and 1b. The result is a set

of magnetic fields with a definite phase rotation. It

is this phase rotation that pulls the rotating

magnet around in a definite direction.

Polyphase AC motors require no such trickery to

spin in a definite direction. Because their supply

voltage waveforms already have a definite

rotation sequence, so do the respective magnetic

fields generated by the motor’s stationary

windings. In fact, the combination of all three

phase winding sets working together creates

what is often called a rotating magnetic field. It

was this concept of a rotating magnetic field that

inspired Nikola Tesla to design the world’s first

polyphase electrical systems. The line current

and safety advantages of polyphase power over

single phase power were discovered later. What

can be a confusing concept is made much clearer

through analogy. Have you ever seen a row of

blinking light bulbs such as the kind used in

Christmas decorations? Some strings appear to

“move” in a definite direction as the bulbs

alternately glow and darken in sequence. Other

strings just blink on and off with no apparent

motion. What makes the difference between the

two types of bulb strings? Answer: phase shift!

Examine a string of lights where every other bulb

is lit at any given time

When all of the “1” bulbs are lit, the “2” bulbs are

dark, and vice versa. With this blinking

sequence, there is no definite “motion” to the

bulbs’ light. Your eyes could follow a “motion”

from left to right just as easily as from right to

left. Technically, the “1” and “2” bulb blinking

sequences are 180° out of phase (exactly

opposite each other). This is analogous to the

single phase AC motor, which can run just as

easily in either direction, but which cannot start

on its own because its magnetic field alternation

lacks a definite “rotation.”

Now let’s examine a string of lights where there

are three sets of bulbs to be sequenced instead

of just two, and these three sets are equally out

of phase with each other.

If the lighting sequence is 1 – 2 – 3, the bulbs

will appear to “move” from left to right. Now

imagine this blinking string of bulbs arranged into a circle.



Now the lights appear to be “moving” in a

clockwise direction because they are arranged

around a circle instead of a straight line. It should

come as no surprise that the appearance of

motion will reverse if the phase sequence of the

bulbs is reversed.

The blinking pattern will either appear to move

clockwise or counter – clockwise depending on

the phase sequence. This is analogous to a three

– phase AC motor with three sets of windings

energized by voltage sources of three different

phase shifts.

With phase shifts of less than 180° we get true

rotation of the magnetic field. With single phase

motors, the rotating magnetic field necessary for

self – starting must to be created by way of

capacitive phase shift. With polyphase motors,

the necessary phase shifts are there already.

Plus, the direction of shaft rotation for polyphase

motors is very easily reversed: just swap any two

“hot” wires going to the motor, and it will run in the

opposite direction!

Two phase System

Two – phase electrical power was an early 20th

century polyphase alternating current electric

power

distribution

system.

Two circuits

were used,

with voltage

phases differing by one – quarter

of a cycle, 90°. Usually circuits used four wires,

two for each phase. Less frequently, three wires

were used, with a common wire with a larger-

diameter conductor. Some early two – phase

generators had two complete rotor and field

assemblies, with windings physically offset to

provide two – phase power. The generators at

Niagara Falls installed in 1895 were the largest

generators in the world at that time and were two

– phase electric machines. As of 21st century, two

– phase power was superseded with three

phases and is not used in the industry. There

remains, however, a two – phase commercial

distribution system in Philadelphia, Pennsylvania;

many buildings in city center are permanently

wired for two – phase and PECO (the local

electric utility company) has continued the

service. Three phase Y and Delta Configuration

Initially, the idea of three – phase power systems

is by connecting three voltage sources together in

what is commonly known as the “Y” (or “star”)

configuration. This configuration of voltage

sources is characterized by a common

connection point joining one side of each source.

If a circuit is drawn showing each voltage source

to be a coil of wire (alternator or transformer



0

winding) and do some slight rearranging, the “Y”

configuration becomes more obvious.

The three conductors leading away from the

voltage sources (windings) toward a load are

typically called lines, while the windings

themselves are typically called phases. In a Y

connected system, there may or may not be a

neutral wire attached at the junction point in the

middle, although it certainly helps alleviate

potential problems should one element of a

three – phase load fail open.

When we measure voltage and current in three-

phase systems, we need to be specific as to

where we’re measuring. Line voltage refers to the

amount of voltage measured between any two

line conductors in a balanced three – phase

system. With the above circuit, the line voltage is

roughly 208 volts. Phase voltage refers to the

voltage measured across any one component

(source winding or load impedance) in a balanced

three – phase source or load. For the circuit

shown above, the phase voltage is 120 volts. The

terms line current and phase current follow the

same logic: the former referring to current through

any one line conductor, and the latter to current

through any one component.

Y – connected sources and loads always have

line voltages greater than phase voltages, and

line currents equal to phase currents. If the Y –

connected source or load is balanced, the line

voltage will be equal to the phase voltage times

the square root of 3:

However, the “Y” configuration is not the only

valid one for connecting three – phase voltage

source or load elements together. Another

configuration is known as the “Delta”, for its

geometric resemblance to the Greek letter of the

same name (∆).

At first glance it seems as though three voltage

sources like this would create a short circuit,

electrons flowing around the triangle with nothing

but the internal impedance of the windings to hold

them back. Due to the phase angles of these

three voltage sources, however, this is not the

case. One quick check of this is to use Kirchhoff’s

Voltage Law to see if the three voltages around

the loop add up to zero. If they do, then there will

be no voltage available to push current around

and around that loop, and consequently there will

be no circulating current. Starting with the top

winding and progressing counter clockwise, our

KVL expression looks something like this:

120∠0 + 120∠240 + 120∠120 = 0

Another way to verify the fact that these three

voltage sources can be connected together in a

loop without resulting in circulating currents is to



1

open up the loop at one junction point and

calculate voltage across the break:

Starting with the right winding (120 V ˂ 120°) and

progressing counter clockwise, our KVL

equation looks like this:

120∠120 + 120∠0 + 120∠240 + 𝐸𝑏𝑟𝑒𝑎𝑘 = 0

0 + 𝐸𝑏𝑟𝑒𝑎𝑘 = 0

𝐸𝑏𝑟𝑒𝑎𝑘 = 0

Having established that a ∆ - connected three –

phase voltage source will not burn itself to a

crisp due to circulating currents, its practical use

as a source of power in three – phase circuits

can be defined. Because each pair of line

conductors is connected directly across a single

winding in a ∆ circuit, the line voltage will be

equal to the phase voltage. Conversely,

because each line conductor attaches at a node

between two windings, the line current will be

the vector sum of the two joining phase currents.

Not surprisingly, the resulting equations for a ∆

configuration are as follows:

One distinct advantage of a ∆ - connected system

is its lack of a neutral wire. With a Y – connected

system, a neutral wire was needed in case one of

the phase loads were to fail open (or be turned

off), in order to keep the phase voltages at the

load from changing. This is not necessary (or

even possible!) in a ∆ - connected circuit. With

each load phase element directly connected

across a respective source phase winding, the

phase voltage will be constant regardless of open

failures in the load elements.

Perhaps the greatest advantage of the ∆ -

connected source is its fault tolerance. It is

possible for one of the windings in a ∆ - connected

three – phase source to fail open without affecting

load voltage or current!

The only consequence of a source winding failing

open for a ∆ - connected source is increased

phase current in the remaining windings.

Compare this fault tolerance with a Y – connected

system suffering an open source winding.

With a ∆ - connected load, two of the resistances

suffer reduced voltage while one remains at the

original line voltage, 208. A Y – connected load

suffers an even worse fate with the same winding

failure in a Y – connected source.

In this case, two load resistances suffer reduced

voltage while the third loses supply voltage

completely! For this reason, ∆ - connected

sources are preferred for reliability. However, if

dual voltages are needed (e.g. 120/208) or

preferred for lower line currents, Y – connected

systems are the configuration of choice.

Three – phase transformer circuits

Since three – phase is used so often for power

distribution systems, it makes sense that a three

– phase transformers is needed to be able to



2

step voltages up or down. This is only partially

true, as regular single – phase transformers can

be ganged together to transform power between

two three – phase systems in a variety of

configurations, eliminating the requirement for a

special three – phase transformer. However,

special three – phase transformers are built for

those tasks, and are able to perform with less

material requirement, less size, and less weight

than their modular counterparts.

A three – phase transformer is made of three

sets of primary and secondary windings, each

set wound around one leg of an iron core

assembly. Essentially it looks like three single –

phase transformers sharing a joined core.

Those sets of primary and secondary windings

will be connected in either ∆ or Y configurations

to form a complete unit.

Whether the winding sets share a common core

assembly or each winding pair is a separate

transformer, the winding connection options are

the same:

Primary Secondary

Y Y

Y ∆

∆ Y ∆ ∆

The reasons for choosing a Y or ∆ configuration

for transformer winding connections are the

same as for any other three – phase application:

Y connections provide the opportunity for

multiple voltages, while ∆ connections enjoy a

higher level of reliability (if one winding fails

open, the other two can still maintain full line

voltages to the load).

Probably the most important aspect of connecting

three sets of primary and secondary windings

together to form a three – phase transformer bank

is paying attention to proper winding phasing (the

dots used to denote “polarity” of windings).

Getting this phasing correct when the windings

aren’t shown in regular Y or ∆ configuration can

be tricky. Three individual transformers are to be

connected together to transform power from one

three – phase system to another. First, the wiring

connections for a Y – Y configuration:

Note how all the winding ends marked with dots

are connected to their respective phases A, B,

and C, while the non – dot ends are connected

together to form the centers of each “Y”. Having

both primary and secondary winding sets

connected in “Y” formations allows for the use of

neutral conductors (N1 and N2) in each power

system. Now for a Y - ∆ configuration:

Note how the secondary windings are connected

in a chain, the “dot” side of one winding connected



3

to the “non – dot” side of the next, forming the ∆

loop. At every connection point between pairs of

windings, a connection is made to a line of the

second power system (A, B, and C). Now for a ∆

- Y system:

Such a configuration would allow for the

provision of multiple voltages (line – to – line or

line – to – neutral) in the second power system,

from a source power system having no neutral.

And finally, for a ∆ - ∆ configuration:

When there is no need for a neutral conductor

in the secondary power system, ∆ - ∆

connection schemes are preferred because of

the inherent reliability of the ∆ configuration.

Considering that a ∆ configuration can operate

satisfactorily missing one winding, some power

system designers choose to create a three –

phase transformer bank with only two

transformers, representing a ∆ - ∆ configuration

with a missing winding in both the primary and

secondary sides:

This configuration is called “V” or “Open ∆.” Of

course, each of the two transformers have to be

oversized to handle the same amount of power as

three in a standard ∆ configuration, but the overall

size, weight, and cost advantages are often worth

it. Bear in mind, however, that with one winding

set missing from the ∆ shape, this system no

longer provides the fault tolerance of a normal ∆ -

∆ system. If one of the two transformers were to

fail, the load voltage and current would definitely

be affected. Advantages of Polyphase Systems

1. A polyphase transmission line requires less

conductor material than a single – phase line

for transmitting the same amount of power at

the same voltage.

2. For a given frame size a polyphase machine

gives a higher input than a single – phase

machine. For example, output of a 3 – phase

motor is 1.5 times the output of a single phase

motor of the same size.

3. Polyphase motors have a uniform torque

where most of the single phase motor have a

pulsating torque.

4. Polyphase induction motors are self – starting

and are more efficient. On the other hand,

single – phase induction motors are not self –

starting and are less efficient.

5. Per unit of output, the polyphase machine is

very much cheaper.

6. Power factor of a single – phase motor is

lower than that of polyphase motor of the

same rating.

7. Rotating field can be set up by passing

polyphase current through stationary coils.



4

8. Parallel operation of polyphase alternators is

simple as compared to that of single phase

alternators because of pulsating reaction in

single – phase alternator.

Advantages of Three Phase AC Power Systems

to Single Phase AC Power Systems

1. It is possible to get more power per kilogram

of metal from a three phase machine

2. The power delivered to a three phase load is

constant all the times

3. Three phase systems can use the service of

induction motors easier by allowing them to

start without auxiliary starting windings

4. Two voltages are available

5. Three phase motors are very robust,

relatively cheap, and generally smaller, have

self – starting properties, provide a steadier

output and require little maintenance

compared with single phase motors.

Multiphase Systems

Three – phase system is universally used.

However, attention has been given in recent

years to the use of more than three phases for

power transmission purposes. In particular, six

and twelve phase systems have been studied.

Advantage of six and twelve phase systems

relative to three phase systems are as follows:

1. Thermal loading capacity of lines is more.

2. Corona effects is less because for a given

conductor size and tower configuration the

stress on the conductor surface decreases

with the number of phases.

3. The higher the numner of phases, the

smaller the line – to – line voltage becomes

relative to the phase voltage, resulting in

increased utilization of rights of way

because of less phase – to – phase

insulation requirement.

4. Exisitng double – circuit lines (two three –

phase circuits on each tower) could be

converted to single circuit six – phase lines.

It is always advantageous to describe

multiphase systems in terms of the phase

voltage rather than line – to – line, as in the

case for three – phase systems. The

transmission efficiency is higher.

A six – phase supply can be obtained by

suitable arrangement of the secondary windings

of a three phase transformer. The figure shows

the transformer connections and phasor

diagram.

Balanced Three Phase Circuit

The magnitude of the line currents are the same

and they have 120° phase difference.

Balanced phase voltages are equal in magnitude

and are out of phase with each other by 120°.

A balanced load is one in which the phase

impedances are equal in magnitude and in phase.

A balanced Y – Y system is a three phase

system with a balanced Y – connected source

and a balanced Y – connected load.

𝑉𝑎𝑛 = 𝑉𝑝∠0 𝐼𝑛 = 𝐼𝑎 + 𝐼𝑏 + 𝐼𝑐 = 0 𝑉𝑏𝑛 =

𝑉𝑝∠ − 120

𝑉𝑐𝑛 = 𝑉𝑝∠120 𝑉𝑎𝑏

= 𝑉𝑎𝑛 − 𝑉𝑏𝑛 𝑉𝑎𝑏 + 𝑉𝑏𝑐 + 𝑉𝑐𝑎 = 0 𝑉𝑏𝑐 = 𝑉𝑏𝑛

− 𝑉𝑐𝑛

𝑉𝑐𝑎 = 𝑉𝑐𝑛 − 𝑉𝑎𝑛

𝑉𝑎𝑏 = √3𝑉𝑎𝑛 𝐼𝐿 = 𝐼𝑝

A balanced ∆ - ∆ system is one in which both the

balanced source and balanced load are ∆ -

connected.

𝐼 = 𝐼 ∠0 𝑉 = 𝑉



5

NOTE: To Analyze Delta – Connected system

using one – line equivalent circuits, it is first

transformed into an equivalent wye – connected

load making used off:

Comparison of Star and Delta Connections

1. Loads connected in delta dissipate three

times more power than when connected in

star to the same supply.

2. For the same power, the phase currents

must be the same for both delta and star

connections (since ), hence

the line current in the delta connected

system is greater than the line current in the

corresponding star – connected system. To

achieve the same phase current in a star –

connected system as in a delta – connected

system, the line voltage in the star system is

√3 times the line voltage in the delta system.

Thus for a given power transfer, a delta system

is associated with larger line currents (and thus

larger conductor cross – sectional area) and a

star system is associated with a larger line

voltage (and thus greater insulation). A balanced

∆ - Y system consist of a balanced ∆ - connected

source feeding a balanced Y – connected load.

A balanced Y - ∆ system consist of a balanced Y

– connected source feeding a balanced ∆ -

connected load.



6

BALANCED 3 phase EQUATIONS

where:

θ – angle between VP & IP

Power Factor Improvement

The power factor of different loads are as follows:

Types of load Power factor

(range)

Heating and Lighting 0.95 to unity

Motor loads 0.5 to 0.9

Single phase motors as low as 0.4

Electric welding units 0.2 to 0.3

The higher current due to poor power factor

affects the system and lead to the following

undesirable results:

1. Rating of alternators and transformers are

proportional to their output current hence

inversely proportional to power factor,

therefore large generators and transformers

are required to deliver same load but at a low

power factor.

2. When a load having a low lagging power is

switched on, there is a large voltage drop in

the supply voltage because of the increased

voltage drop in the supply lines and

transformers. This drop in voltage adversely

affects the starting torque of motors and

necessitates expensive voltage stabilizing

equipment for keeping the consumer’s

voltage fluctuations within the statutory limits.

3. The cross – sectional are of the bus bar, and

the contact surface of switch gear is required

to be enlarged for the same power to be

deliverd but at a low power factor.

4. To transmit same power at low power factor,

more current will have to be carried by the

transmission line or the distributor or cable. If

the current density in the line is to be kept

constant the size of the conductor will have to

be increased. Thus more copper is required

to deliver the same load but at a low power

factor.

5. Copper losess are proportional to the square

of the current hence inversely proportional to

the square of the power factor, more copper

losses at low power factor and hence poor

efficiency.

Thus, the capital cost of transformers, alternators,

distributors, transmission lines increase with low

power factor.

Cause of Low Power Factor

1. All AC motors (except overexcoted

synchronous motors and certain types of

commutators motors) and transformers

operate at lagging power factor.

2. Due to typical characteristic of the arc, arc

lamps operate at low power factor.

3. When there is increase in supply voltage,

which usually occurs during low load periods

(such as lunch hours, night hours etc.), the

magnetizing current of inductive reactances

increases and power factor of the electrical

plant as a whole decreases.

4. Arc and induction furnaces operate at a very

low lagging power factor.

5. Due to improper maintenance and repairs of

motor the power factor at which motors

operate fall.

6. Very low lagging power factor of agriculture

motor pump – set.

Methods of Improving Power Factor

1. Use of high power factor motors.

2. Use of induction motors with phase

advancers.

3. Use of static capacitors.

4. Use of capacitance boosters.

5. Use of synchronous condensers.

EARTHING AND GROUNDING



7

The process of connecting the neutral point

of a supply system on the non – concurrent

carrying parts of electrical apparatus to the

general mass of earth in such a manner that

at all times and immediate discharge of

electrical energy takes place without danger

is called earthing. It also means connecting

earth – terminals to electrodes installed

solidly in the mass of earth. It is also a wire

coming from the ground 2.5 to 3 meters

deep from an electrode (plate).

The earth’s potential is always taken as zero

for all practical purposes. The electrical

appliances or machines when connected

with earth attain zero potential and are said

to be earthed.

Good earthing is that earthing which gives

very low resistance to the flow of heavy

current (short circuit current) of a circuit.

Double earth is used to give minimum

resistance to the flow of whole current of the

apparatus in case short circuit to leakage or

any other such fault occurs. Second reason

is, if one earth is out of order, second will do

the work.

The earth resistance for coper wire is 1 Ω

and for GU wire it should not be more than

3 Ω.

The earth resistance should be kept as low

as possible.

Objectives of Earthing

1. To save human life from danger or shock or

death by blowing fuse of any apparatus which

become leaky.

2. To protect all machines fed from overhead

lines from lightning arrestors.

3. To protect large buildings from atmospheric

lightning.

4. To maintain line voltage constant (since

neutral of every alternator, transformer is

earthed).

Methods of Earthing

1. Strip or wire earthing

2. Earthing through water mains

3. Rod earthing

4. Pipe earthing

5. Plate earthing

Earthing of a Power System

Earthing of neutrals of all industrial power

systems is always preferable. Earthing is

necessary as it offers many advantages:

1. Persistent arcing ground is eliminated.

2. Over voltage due to restriking is minimized.

3. The ground faults can be located an isolated

immediately.

4. Steady state voltage stress to earth is

reduced.

5. Sensitive protective apparatus can be used.

6. The maintenance expenditure is reduced.

7. Better safety is ensured.

8. Service reliability is improved.

9. Earthing provides improved lightning

protection. The earthing of systems should be

done at the neutral of the supply transformers

and generators. If the supply transformers

and generators are delta connected, separate

earthing transformers may be used.

Examples:

1. If Vab = 400 V in a balanced Y – connected

three – phase generator, find the phase

voltages, assuming the phase sequence is: a. abc

b. acb

2. Given that the line voltages of a three – phase

circuit are

𝑉𝑎𝑏 = 420∠0°

𝑉𝑏𝑐 = 420∠ − 120°

𝑉𝑐𝑎 = 420∠120°

Find the phase voltages Van, Vbn and Vcn.

3. A balanced Y – connected load with a phase

impedance of 16+j9 Ω is connected to a

balanced three – phase source with a line

voltage of 220 V. Calculate the line current IL.

4. A balanced Y – Y four wire system has phase

voltages

𝑉𝑎𝑛 = 120∠0°

𝑉𝑏𝑛 = 120∠ − 120°

𝑉𝑐𝑛 = 120∠120°



8

The load impedance per phase is 19+j13 Ω

and the line impedance per phase is 1+j2 Ω.

Solve for the line currents and neutral current.

5. In a wye – delta three – phase circuit, the

source is a balanced, positive phase

sequence with 𝑉𝑎𝑛 = 120∠0° 𝑉. It feeds a

balanced load with 𝑍∆ = 9 +

𝑗12 Ω per phase through a balanced line with 𝑍𝑙

=

1 + 𝑗0.5 𝛺 per phase. Calculate the phase

voltages and currents in the load.

6. A balanced delta – connected load has a

phase current 𝐼𝐴𝐶 = 10∠30° A.

a. Determine the three line currents

assuming that the circuit operates in the

positive phase sequence.

b. Calculate the load impedance if the line

voltage is 𝑉𝐴𝐵 = 110∠0° V.

7. A balanced delta – connected source has

phase voltage 𝑉𝑎𝑏 = 416∠30° V and a positive

phase sequence. If this is connected to a

balanced delta – connected load, find the line

and phase currents. Take the load impedance

per phase as 60∠30° Ω and line impedance

per phase as 1+j1 Ω.

8. Three loads, each of resistance 50 Ω are

connected in star to a 400V, 3 – phase supply.

Determine (a) the phase voltage, (b) the

phase current and (c) the line current.

9. A star – connected load consists of three

identical coils, each of inductance 159.2mH

and resistance 50 Ω. If the supply frequency

is 50 Hz and the line current is 3A determine

(a) the phase voltage and (b) the line voltage.

10. Three identical capacitors are connected in

star to a 400V, 50 Hz 3 – phase supply. If the

line current is 12 A determine the capacitance

of each of the capacitors.

11. Three coils each having resistance 6 Ω and

inductance LH are connected in star to a

415V, 50 Hz, 3 – phase supply. If the line

current is 30A, find the value of L.

12. A 400V, 3 – phase, 4 wire, star – connected

system supplies three resistive loads of

15kW, 20kW and 25kW in the red, yellow and

blue phases respectively. Determine the

current flowing in each of the four conductors.

13. In a balanced three – phase ∆ - Y circuit, the

source is connected in the positive sequence,

with 𝑉𝑎𝑏 =

220∠20° V and 𝑍𝑌 = 20 + 𝑗15 Ω. Find the line

currents.

14. Three 24 μF capacitors are connected in star

across a 400V, 50 Hz, 3 – phase supply. What

value of capacitance must be connected in

delta in order to take the same line current?

15. A 3 – phase, star – connected alternator

delivers a line current of 65 A to a balanced

delta – connected load at a line voltage of

380V. Calculate (a) the phase voltage of the

alternator, (b) the alternator phase current

and (c) the load phase current.

16. Three inductive loads each of resistance 75 Ω

and inductance 318.4mH are connected in

delta to a 415V,

50 Hz, 3 – phase supply. Determine (a) the

phase voltage, (b) the phase current, and (c)

the line current.

17. A delta – connected generator supplies a

balanced wye – connected load with and

impedance 30∠ − 60° Ω. If the line voltages of

the generator have a magnitude of 400 V and

are in the positive phase sequence, find the

line currents IL and phase voltage Vp at the

load.

18. A balanced wye – connected load absorbs a

total power of 5 kW at a leading power factor

of 0.6 when connected to a line voltage of 240

V. find the impedance of each phase and total

complex power of the load.

19. A balanced delta – connected load is supplied

by a 60 Hz three phase source with a line

voltage of 240 V. Each load phase draws 6

kW at a lagging power factor of 0.8. Find:

a. The load impedance per phase

b. The line current

c. The value of capacitor needed to be

connected in parallel with each load to

minimize the current from the source



9

20. A 4200 V, three phase transmission line has

an impedance of 4+j1 Ω per phase. If it is

supplies a load of 1 MVA at 0.75 power factor

lagging. Find: a. The complex power

b. The power loss in the line

c. The voltage at the sending end

21. The following three parallel – connected three

–phase loads are fed by a balanced three –

phase source. Load 1: 250 kVA, 0.8 pf lagging

Load 2: 300 kVA, 0.95 pf leading

Load 3: 450 kVA, unity pf

If the line voltage is 13.8 kV, calculate the line

current and the power factor of the source.

Assume that the line impedance is zero.

22. A three – phase generator supplied 3.6 kVA

at a power factor of 0.85 lagging. If 2500 W

are delivered to the load and line losses are

80 W per phase, what are the losses in the

generator?

23. Determine the total power dissipated by three

20 Ω resistors when connected (a) in star and

(b) in delta to a 440V, 3 – phase supply.

24. A balanced delta – connected load has a line

voltage of 400V, a line current of 8 A and a

lagging power factor of 0.94. Draw a complete

phasor diagram of the load. What is the total

power dissipated by the load?

25. Three inductive loads, each of resistance 4 Ω

and reactance 9 Ω are connected in delta.

When connected to a 3 – phase supply the

loads consume 1.2kW.

Calculate (a) the power factor of the load, (b)

the phase current, (c) the line current and (d)

the supply voltage.

26. The input voltage, current and power to a

motor is measured as 415 V, 16.4 A and 6k

W respectively. Determine the power factor of

the system.

27. A 440V, 3 – phase ac motor has a power

output of 11.25kW and operates at a power

factor of 0.8 lagging and with an efficiency of

84 percent. If the motor is delta connected

determine (a) the power input, (b) the line

current and (c) the phase current.

28. A three – phase 440 V, 51 kW, 60 kVA

inductive load operates at 60 Hz and is wye

connected. It is desired to correct the power

factor to 0.95 lagging. What value of capacitor

should be placed in parallel with each load

impedance?

29. A 3 – phase, star – connected alternator

supplies a delta connected load, each phase

of which has a resistance of 15 Ω and

inductive reactance 20 Ω. If the line voltage is

400V, calculate (a) the current supplied by the

alternator and (b) the output power and kVA

rating of the alternator, neglecting any losses

in the line between the alternator and the

load.

30. Each phase of a delta – connected load

comprises a resistance of 40 Ω and a 40μF

capacitor in series. Determine, when

connected to a 415V, 50 Hz, 3 – phase supply

(a) the phase current, (b) the line current, (c)

the total power dissipated, and (d) the kVA

rating of the load.

31. A professional center is supplied by a

balanced three – phase source. The center

has four balanced three – phase loads as

follows: Load 1: 150 kVA at 0.8 pf leading

Load 2: 100 kW at unity pf

Load 3: 200 kVA at 0.6 pf lagging

Load 4: 80 kW and 95 kVAR (inductive) If

the line impedance is 0.02+j0.05 Ω per

phase and the line voltage at the loads is

480 V, find the magnitude of the line voltage

at the source.

32. A commercially available three – phase

inductive motor operates at full load of 120 hp

(1 hp = 746 W) at 95 percent efficiency at a

lagging power factor of 0.707. The motor is

connected in parallel to a 80 kW balanced

three – phase heater at unity power factor. If

the magnitude of the line voltage is 480 V rms,

calculate the line current.

33. Calculate the total power when three identical

resistances each of 10 Ω are connected wye

when connected across 220 V, 3 phase, 60

Hz source.



0

34. Three identical resistances in star consume

2000 W. If the resistances are connected in

delta across the same supply, what is the

power consumed.

35. Three 50 Ω resistance are connected in wye

across 220 V, 3 phase supply. What is the

equivalent delta connected resistance?

36. Three identical resistance connected in wye

carry a line current of 18 A. If the same

resistances are connected in delta across the

same supply, what is the line current?

37. The apparent power input to a balanced wye

connected load is 30 kVA and the

corresponding true power is 15 kW at 45.45

A. What is the phase voltage?

38. Three star connected impedances, Z1 = 12 +

j16 Ω/phase are connected in parallel with 3

delta connected impedances, Z2 = 30 – j45

Ω/phase. The line voltage is 480 V. What is

the line current?

39. The phase b line voltage and phase a line

current of a balanced three phase system are

V = 220 sin (wt + 210°) V and I = 10 sin (wt –

30°) A respectively. What is the power of the

system?

40. The three phase line has an impedance of 2

+ j4 Ω. The line feeds two balanced three

phase loads that are connected in parallel.

The first load is Y – connected and has an

impedance of 30 + j40 Ω/phase. The second

load is delta connected and has an

impedance of 60 – j45 Ω/phase. The line is

energized at the sending end from a balanced

3 phase supply of line voltage 207.85 V.

Taking the voltage Van as reference

determine:

a. The current, real power and reactive

power drawn

from the supply

b. The line voltage at combined loads

c. The current per phase in each load

d. The real and reactive powers in each load

and the line

UNBALANCED THREE – PHASE SYSTEMS

An unbalanced system is due to unbalanced

voltage sources or an unbalanced load.

NOTE: Unbalanced three – phase systems are

solved by direct application of mesh and nodal

analysis.

Unbalanced Three Phase Systems

Y – Y System

Case 1: If Zneutral is considered zero, each

individual phase current can be independently

determined from the supply voltage in that phase

and the impedance of that phase.

Case 2: If there is a neutral impedance, then

using Millmann’s theorem, we will first have to

determine the voltage of the star point of the

load with respect to the supply neutral.

Another way of solving this case is by Nodal

Analysis. Using the neutral as a node. The nodal

equation would be:



1

Case 3: If the system is a 3 – wire system, rather

than a 4 – wire system, the analysis is the same

as if Zneutral were ∞ (i.e. 1/Zneutral = 0). Thus again

Millmann’s theorem is used to determine VSN

and the load currents are then determined.

𝐸𝐿𝑅 = 𝐼𝐿𝑅𝑍𝐿𝑅, 𝐸𝐿𝑌 = 𝐼𝐿𝑌𝑍𝐿𝑌, 𝐸𝐿𝐵 = 𝐼𝐿𝐵𝑍𝐿𝐵

Another way of solving this problem is by mesh

analysis:

𝐸𝑅 − 𝐸𝑌 − 𝐼1(𝑧𝑠 + 𝑧𝑙𝑖𝑛𝑒 + 𝑍𝐿𝑅 + 𝑧𝑠 + 𝑧𝑙𝑖𝑛𝑒 + 𝑍𝐿𝑌)

+ 𝐼2(𝑧𝑠 + 𝑧𝑙𝑖𝑛𝑒 + 𝑍𝐿𝑌) = 0

𝐸𝑌 − 𝐸𝐵 + 𝐼1(𝑧𝑠 + 𝑧𝑙𝑖𝑛𝑒 + 𝑍𝐿𝑌)

+ 𝐼2(𝑧𝑠 + 𝑧𝑙𝑖𝑛𝑒 + 𝑍𝐿𝑌 + 𝑧𝑠 + 𝑧𝑙𝑖𝑛𝑒

+ 𝑍𝐿𝐵) = 0

𝐼𝐿𝑅 = 𝐼1, 𝐼𝐿𝑌 = 𝐼2 − 𝐼1; 𝐼𝐿𝐵 = −𝐼2

𝐸𝐿𝑅 = 𝐼𝐿𝑅𝑍𝐿𝑅, 𝐸𝐿𝑌 = 𝐼𝐿𝑌𝑍𝐿𝑌, 𝐸𝐿𝐵 = 𝐼𝐿𝐵𝑍𝐿𝐵

𝐼𝑁 = 𝐼𝐿𝐵 + 𝐼𝐿𝑌 + 𝐼𝐿𝑅

NOTE: Converting a Y - ∆ and ∆ - Y Systems to

a Y – Y System is possible. Just remember that it

falls as Case 3 for a Y – Y system. The same

also applies in converting a Y - ∆ and ∆ - Y Systems to a ∆ – ∆ system.

Y - ∆ System

Using Mesh Analysis:

𝐸𝑅 − 𝐸𝑌 − 𝐼1(𝑧𝑠 + 𝑧𝑙𝑖𝑛𝑒 + 𝑍𝐿𝑅 + 𝑧𝑠 + 𝑧𝑙𝑖𝑛𝑒 + 𝑍𝐿𝑌)


𝐸𝑌 − 𝐸𝐵 + 𝐼1(𝑧𝑠 + 𝑧𝑙𝑖𝑛𝑒 + 𝑍𝐿𝑌)


+ 𝑍𝐿𝐵) = 0

Phase Currents: 𝐼𝑅𝑌 = 𝐼1 − 𝐼3

𝐼𝑌𝐵 = 𝐼2 − 𝐼3

𝐼𝐵𝑅 = −𝐼3

Line Currents: 𝐼𝐿𝑅 = 𝐼𝑅𝑌 − 𝐼𝐵𝑅

𝐼𝐿𝑌 = 𝐼𝑌𝐵 − 𝐼𝑅𝑌

𝐼𝐿𝐵 = 𝐼𝐵𝑅 − 𝐼𝑌𝐵

Phase Voltages (LOAD): 𝐸𝑅𝑌(𝐿𝑂𝐴𝐷) = 𝐼𝑅𝑌𝑧𝑅𝑌

𝐸𝑌𝐵(𝐿𝑂𝐴𝐷) = 𝐼𝑌𝐵𝑧𝑌𝐵

𝐸𝐵𝑅(𝐿𝑂𝐴𝐷) = 𝐼𝐵𝑅𝑧𝐵𝑅

∆ - ∆ System

By Using Mesh Analysis:

𝐸𝑅𝑌 = 𝐼1(𝑧𝑠 + 2𝑧𝑙𝑖𝑛𝑒 + 𝑧𝑅𝑌) − 𝐼2(𝑧𝑙𝑖𝑛𝑒) − 𝐼3(𝑧𝑅𝑌)

𝐸𝑌𝐵 = −𝐼1(𝑧𝑙𝑖𝑛𝑒) + 𝐼2(𝑧𝑠 + 2𝑧𝑙𝑖𝑛𝑒 + 𝑧𝑌𝐵) − 𝐼3(𝑧𝑌𝐵)

0 = −𝐼1(𝑧𝑅𝑌) − 𝐼2(𝑧𝑌𝐵) + 𝐼3(𝑧𝑌𝐵 + 𝑧𝑅𝑌 + 𝑧𝐵𝑅)

Phase Currents: 𝐼𝑅𝑌 = 𝐼1 − 𝐼3, 𝐼𝑌𝐵 = 𝐼2 − 𝐼3, 𝐼𝐵𝑅 = −𝐼3

Line Currents: 𝐼𝐿𝑅 = 𝐼𝑅𝑌 − 𝐼𝐵𝑅

𝐼𝐿𝑌 = 𝐼𝑌𝐵 − 𝐼𝑅𝑌

𝐼𝐿𝐵 = 𝐼𝐵𝑅 − 𝐼𝑌𝐵

Phase Voltages (LOAD): 𝐸𝑅𝑌(𝐿𝑂𝐴𝐷) = 𝐼𝑅𝑌𝑧𝑅𝑌

𝐸𝑌𝐵(𝐿𝑂𝐴𝐷) = 𝐼𝑌𝐵𝑧𝑌𝐵

𝐸𝐵𝑅(𝐿𝑂𝐴𝐷) = 𝐼𝐵𝑅𝑧𝐵𝑅

∆ - Y System



2

Using Mesh Analysis:

𝐸𝑅𝑌 − 𝐼1(𝑧𝑠 + 𝑧𝑙𝑖𝑛𝑒 + 𝑍𝐿𝑅 + 𝑧𝑠 + 𝑧𝑙𝑖𝑛𝑒 + 𝑍𝐿𝑌)


𝐸𝑌𝐵 + 𝐼1(𝑧𝑠 + 𝑧𝑙𝑖𝑛𝑒 + 𝑍𝐿𝑌)


+ 𝑍𝐿𝐵) = 0

Phase/Line Currents: 𝐼𝐿𝑅 = 𝐼1, 𝐼𝐿𝑌 = 𝐼2 − 𝐼1; 𝐼𝐿𝐵 = −𝐼2

𝐼𝑁 = 𝐼𝐿𝐵 + 𝐼𝐿𝑌 + 𝐼𝐿𝑅

Load Voltages: 𝐸𝐿𝑅 = 𝐼𝐿𝑅𝑍𝐿𝑅

𝐸𝐿𝑌 = 𝐼𝐿𝑌𝑍𝐿𝑌

𝐸𝐿𝐵 = 𝐼𝐿𝐵𝑍𝐿𝐵

Unbalanced Wye Loads

Unbalanced four – wire Y systems without line

impedance are easily handled using Ohm’s law.

However, for three – wire systems or four – wire

systems with line and neutral impedance, you

generally have to use mesh equations or

computer methods. One of the problems with

unbalanced three – wire Y systems is that you

get different voltages across each phase of the

load and a voltage between neutral points.

Unbalanced Delta Loads

Systems without line impedance are easily

handled since the source voltage is applied

directly to the load. However, for systems with

line impedance, use mesh equations. EXAMPLES:

1. A 3 – phase, 3 – wire, 240 V, CBA system

supplies a delta – connected load in which ZAB

= 25˂90°, ZBC = 15˂30° and ZCA = 20˂0° Ω.

Find the line currents and total power.

2. A 3 – phase, 4 – wire, 380 V supply is

connected to an unbalanced load having

phase impedances of ZR = (8 + j6) Ω, ZY = (8

– j6) Ω and ZB = 5 Ω. Impedance of the neutral

wire is ZN = (0.5 + j1) Ω. Ignoring the

impedances of line wires and internal

impedances of

the emf sources, find the phase currents and

voltages of the load.

3. A 300 V (line) 3 – phase supply feeds a star –

connected load consisting of non – inductive

resistors of 15, 6 and 10 Ω connected to the

R, Y and B lines respectively. The phase

sequence is RYB. Calculate the voltage

across each resistor.

4. A Y – connected load is supplied from a 400

V, 3 – phase, 3 – wire symmetrical system

RYB. The branch circuit impedances are

Determine the current in each branch. Phase

sequence is RYB.

5. Three non – inductive resistances of 5, 10 and

15 Ω are connected in star and supplied from

a 230 V symmetrical 3 – phase system.

Calculate the line currents (magnitudes).

6. A 3 phase, 4 – wire, 400 V symmetrical

system supplies a Y – connected load having

the following branch impedances:

𝑍𝑅 = 100 Ω, 𝑍𝑌 = 𝑗10Ω 𝑎𝑛𝑑 𝑍𝐵 = −𝑗10Ω

Compute the values of load phase voltages

and currents and neutral current. Phase

sequence is RYB. How will these values

change in the event of an open in the neutral

wire?

7. A delta – connected load whose phase

impedances are 𝑍𝐴𝐵 = 50 Ω, 𝑍𝐵𝐶 = −𝑗50 Ω, and 𝑍𝐶𝐴 = 𝑗50 Ω is fed by a balanced wye –

connected three phase source with 𝑉𝑃 = 100 𝑉. Find the phase currents.

8. A balanced three – phase wye – connected

generator with 𝑉𝑃 = 220 𝑉 supplies an

unbalanced wye – connected load with 𝑍𝐴𝑁 = 60 + 𝑗80 Ω, 𝑍𝐵𝑁 =

100 − 𝑗120 Ω, and 𝑍𝐶𝑁 = 30 + 𝑗40 Ω. Find the

total complex power absorbed by the load.

9. In an unbalanced wye system, the phase

voltages are:

𝑉𝑛𝑎 = 132 𝑐𝑖𝑠 0°

𝑉𝑛𝑏 = 130 𝑐𝑖𝑠 235°

𝑉𝑛𝑐 = 133 𝑐𝑖𝑠 117° What

is the voltage at line a, b and c?

10. For the given circuit, Rab = 60 Ω, Zbc = 80 +

j60 Ω. Compute for (a) Phase and line

currents. (b) Power to each phase and total

power.



3

11. For the given circuit, compute the following:

(a) Line currents, magnitudes and angles. (b)

The neutral current.

12. Three impedances, ZR, ZY and ZB are

connected in star across a 440 V, 3 – phase

supply. If the voltage of star – point relative to

the supply neutral is 200˂150° V and Y and B

line currents are 10˂-90° and 20˂90° A

respectively, all with respect to the voltage

between the supply neutral and the R line,

calculate the values of

ZR, ZY and ZB.

Power System Loads

The single – phase

and three phase

loads must be

connected in the

three

– phase system as

accurate as possible

(This is necessary

because residential

and business customers require only single

phase power, while industrial customers

sometimes require both single phase and three

phase power.). Two points should be noted

here:

1. In order to approximately balance the

system, the utility tries to connect one – third

of its single phase loads to each phase.

Three – phase loads generally are

balanced.

2. Real loads are seldom expressed in terms

of resistance, capacitance, and inductance.

Instead, they are described in terms of

power, power factor, and so on. This is

because most loads consist of electric

lights, motors, and the like which are never

described in terms of impedance. (For

example, you purchase light bulbs as 60 W

bulbs, 100 W bulbs, etc., and electric motors

as ½ horsepower, etc. You never ask for a

240 Ω light!) POWER MEASUREMENTS IN 3 PHASE

SYSTEMS For balanced systems: 1. One wattmeter method – used to measure the

total “balanced” three phase power using one

wattmeter only. If delta connected:

𝑾𝒓𝒅𝒈 = 𝑽𝑳𝑰𝑳 𝐜𝐨𝐬 Ɵ

If star connected:

𝑾𝒓𝒅𝒈 = 𝑽𝑳𝑰𝑳 𝐜𝐨𝐬 Ɵ

𝑷𝒕 = 𝟑𝑾𝒓𝒅𝒈

Methods used

a. Potential lead shift method – the current coil

of one wattmeter is inserted in one line, say

line a and one of the other is shifted to line 2

and to line 3 for another reading.

𝑃𝑇 = 𝑊′ + 𝑊′′

b. Artificial neutral method – if three

resistances are connected in star across the

lines of a 3 –phase source, its neutral point will

at every instant of time be exactly the same

electrical potential as the point of the load



4

c. T – method – the current coil is inserted in one

line say line a and one of the potential coil

terminals is connected to the same line. The

other terminal is fastened to the junction

between a pair of equal resistances, the line

ends of which connected to the other two lines.

d. Current transformer (CT) method – this

scheme gives direct measurement of the total

power. It employs a pair of current

transformers and a single wattmeter.

2. Two wattmeter method – used to measure power

drawn by a 3 phase, 3 wire system and 4 wire

system. Delta Connected Load:

where: PT – total real power of the balanced 3

phase load ∅ - pf angle of the balanced 3 phase load

If line ∅ = ∅𝑉𝐿 − ∅𝐼𝐿 If phase

∅ = ∅𝑉𝑝ℎ𝑎𝑠𝑒 − ∅𝐼𝑝ℎ𝑎𝑠𝑒

If 𝑊𝑟𝑑𝑔1 + 𝑊𝑟𝑑𝑔2 is negative, just carry the negative

sign

1. If pf = 1, both watt meters read positive and

equal 2. If pf > 0.5, both

readings are positive,

𝑃𝑇 = 𝑊𝑟𝑑𝑔1 + 𝑊𝑟𝑑𝑔2

3. If pf = 0.5, only one wattmeter gives reading

and the other will not read 𝑃𝑇 = 𝑊𝑟𝑑𝑔1

4. If pf < 0.5, only one wattmeter will read positive

and the other will give reversed reading Star

Connected Load:

For unbalanced systems:

The currents will not be equal nor they will

have 120° phase difference as was the case

of with balanced loads

For unbalanced 3 phase systems we

consider the first watt meters current coil in

line a, and potential coil between lines a and

c. While the second current coil in line b and

potential coil between lines b and c. The

wattmeter readings are:

𝑊𝑟𝑑𝑔1 = 𝑉𝑎𝑐𝐼𝑎 cos(∅) the angle ∅ will be the

difference of 𝑉𝑎𝑐 and 𝐼𝑎.

𝑊𝑟𝑑𝑔2 = 𝑉𝑏𝑐𝐼𝑏 cos(∅) the angle ∅ will be the

difference of 𝑉𝑏𝑐 and 𝐼𝑏.

3. Three wattmeter method – used to measure

power drawn by a 3 phase, 4 wire system.

𝑊𝑟𝑑𝑔1 = 𝑉𝑎𝑛𝐼𝑎 cos 𝜃 𝑊𝑟𝑑𝑔2 = 𝑉𝑏𝑛𝐼𝑏 cos 𝜃

𝑊𝑟𝑑𝑔3 = 𝑉𝑐𝑛𝐼𝑐 cos 𝜃

𝑃𝑇 = 𝑊𝑟𝑑𝑔1 + 𝑊𝑟𝑑𝑔2 + 𝑊𝑟𝑑𝑔3 Watts Ratio Curve The power factor for a balanced load can be obtained

from the wattmeter readings using a simple curve

called the watts ratio curve.



5

where 𝑃ℎ – high reading

𝑃𝑙 – low reading

𝑃𝑇 = 𝑃ℎ − 𝑃𝑙 Blondel's Theorem: Number of Watt meters

Required In general, the Number of Watt meters

required = the Number of Wires – 1

EXAMPLES:

1. A three – phase balanced system with a line

voltage of

202 V rms feeds a delta – connected load with 𝑍𝑃 = 25∠60° 𝛺.

a. Find the line current.

b. Determine the total power supplied to the

load using two watt meters connected to

the A and C lines.

2. A balanced 3 phase load has a line current

of 120 A at 230 V line to line and 0.848 pf

lagging. If the two watt meters are used to

measure the power being drawn by the

system. What are the wattmeter readings?

3. A factory is being supplied by a 3 – phase,

3 wire system with the following

characteristics: 𝑉𝑎𝑏 = 220 𝑐𝑖𝑠 0° 𝐼𝑎 = 110 𝑐𝑖𝑠 − 36.87°

𝑉𝑏𝑐 = 220 𝑐𝑖𝑠 240° 𝐼𝑐 = 125 𝑐𝑖𝑠 53.13° Determine the total power consumed by the

load.

4. A 3 phase, 3 wire feeder has the following

line current and voltages:

𝑉𝑎𝑏 = 220 𝑐𝑖𝑠 0° 𝐼𝑎 = 30 − 𝑗30

𝑉𝑏𝑐 = 220 𝑐𝑖𝑠 240° 𝐼𝑐 = 29 + 𝑗39.8

Determine the reactive power supplied by the

feeder. Let line b as your reference.

5. Given an unbalanced three phase system

with the following loads:

𝑉𝑎𝑏 = 230 𝑐𝑖𝑠 0° 𝑍𝑎𝑏 = 60

𝑉𝑏𝑐 = 230 𝑐𝑖𝑠 120° 𝑍𝑏𝑐 = 30 + 𝑗50

𝑉𝑐𝑎 = 230 𝑐𝑖𝑠 240° 𝑍𝑐𝑎 = −𝑗80

Determine the reading of the two wattmeter’s

properly connected using line c as the

common point.

Determine the reading of the two

wattmeter’s properly connected using line b

as the common point. Determine the

reading of the two wattmeter’s properly

connected using line a as the common point.

6. The power input to a 50 hp, 480 V, 3 phase

motor running at full load is measured by 2

watt meters which indicates 25.4 kW and

17.2 kW respectively. What is the operating

power factor and line current?

7. Two watt meters are connected to measure

the input power to a balanced three-phase

load. If the wattmeter readings are 9.3kW

and 5.4kW determine (a) the total output

power, and (b) the load power factor.

8. 8kW is found by the two wattmeter method

to be the power input to a 3 – phase motor.

Determine the reading of each wattmeter if

the power factor of the system is 0.85.

9. When the two wattmeter method is used to

measure the input power of a balanced load,

the readings on the watt meters are 7.5kW

and 2.5kW, the connections to one of the

coils on the meter reading 2.5kW having to

be reversed. Determine (a) the total input

power, and (b) the load power factor.

10. Three similar coils, each having a resistance

of 4 Ω and an inductive reactance of 3.46 Ω

are connected (a) in star and (b) in delta

across a 400V, 3 – phase supply. Calculate

for each

connection the

readings on

each of two

watt meters

connected to

measure the

power by the



6

two – wattmeter method. Symmetrical

Component Method Developed by Dr. C.L.

Fortesque, an

American

scientist showed

that any

unbalanced system

of three phase

currents (or

voltages) may be

regarded as

being composed

of three

separate sets

of balanced vectors namely:

1. A balanced set of 3 phase vectors having

the same phase sequence as the original

unbalanced system of vectors (positive

sequence). These vectors are known as

“Positive Sequence System”.

2. A balanced set of 3 phase vectors having

phase sequence which is opposite of the

original unbalanced system of vectors

(negative sequence). This balanced system

of 3 phase vectors are known as “Negative

Sequence System”.

3. A set of three (3) single phase vectors which

are equal in magnitude and having zero

displacement and which have exactly the

same 3 phase position with respect to any

given reference axis. This system of single

phase vector is known as “Zero Sequence

System”. Hence in mathematical form:

𝐼𝑅 = 𝐼𝑅0 + 𝐼𝑅1 + 𝐼𝑅2 = 𝐼𝑎 = 𝐼𝑎0 + 𝐼𝑎1 + 𝐼𝑎2

𝐼𝑌 = 𝐼𝑌0 + 𝐼𝑌1 + 𝐼𝑌2 = 𝐼𝑏 = 𝐼𝑏0 + 𝐼𝑏1 + 𝐼𝑏2

𝐼𝐵 = 𝐼𝐵0 + 𝐼𝐵1 + 𝐼𝐵2 = 𝐼𝑐 = 𝐼𝑐0 + 𝐼𝑐1 + 𝐼𝑐2

Zero Sequence Component

𝑰𝒂𝟎 = 𝑰𝒃𝟎 = 𝑰𝒄𝟎

𝑰𝑹𝟎 = 𝑰𝒀𝟎 = 𝑰𝑩𝟎

Operator “A”

- is one which when multiplied to a vector,

rotates through a 120° in the counter

clockwise direction.

Let:

𝐴 = 1∠120°

𝐴2 = 1∠240°

𝐴3 = 1∠360° = 1

Positive Sequence Component

𝑰𝒃𝟏 = 𝑨𝟐𝑰𝒂𝟏

𝑰𝒀𝟏 = 𝑨𝟐𝑰𝑹𝟏

𝑰𝒄𝟏 = 𝑨𝑰𝒂𝟏

𝑰𝑩𝟏 = 𝑨𝑰𝑹𝟏

Negative Sequence Component

𝑰𝒃𝟐 = 𝑨𝑰𝒂𝟐

𝑰𝒀𝟐 = 𝑨𝑰𝑹𝟐

𝑰𝒄𝟐 = 𝑨𝟐𝑰𝒂𝟐

𝑰𝑩𝟐 = 𝑨𝟐𝑰𝑹𝟐

Expressing Symmetrical Components of IB and IC

in terms of IA symmetrical components:

𝐼𝑎 = 𝐼𝑎0 + 𝐼𝑎1 + 𝐼𝑎2

𝐼𝑏 = 𝐼𝑎0 + 𝐴2𝐼𝑎1 + 𝐴𝐼𝑎2

𝐼𝑐 = 𝐼𝑎0 + 𝐴𝐼𝑎1 + 𝐴2𝐼𝑎2 Adding

them up:

𝐼𝑎 + 𝐼𝑏 + 𝐼𝑐 = 3𝐼𝑎0 + 𝐼𝑎1(1 + 𝐴 + 𝐴2) + 𝐼𝑎2(1 + 𝐴

+ 𝐴2)

But 1 + 𝐴 + 𝐴2 = 0 also 1 + 𝐴2 + 𝐴4 = 0

Therefore,

𝑰𝒂𝟎 = 𝟏/𝟑(𝑰𝒂 + 𝑰𝒃 + 𝑰𝒄)

Also, Ia0 = Ib0 = Ic0

Multiply Ia by 1, Multiply Ib by A and Multiply Ic by

A2




7

𝐴𝐼𝑏 = 𝐴𝐼𝑎0 + 𝐴3𝐼𝑎1 + 𝐴2𝐼𝑎2

𝐴2𝐼𝑐 = 𝐴2𝐼𝑎0 + 𝐴3𝐼𝑎1 + 𝐴4𝐼𝑎2

Adding them up,

𝐼𝑎 + 𝐴𝐼𝑏 + 𝐴2𝐼𝑐 = 𝐼𝑎0(1 + 𝐴 + 𝐴2) + 3𝐼𝑎1 + 𝐼𝑎2(1

+ 𝐴2 + 𝐴4)

𝑰𝒂𝟏 = 𝟏/𝟑(𝑰𝒂 + 𝑨𝑰𝒃 + 𝑨𝟐𝑰𝒄) Also,

Ib1 = A2Ia1 and Ic1 = AIa1

Multiply Ia by 1, Multiply Ib by A2 and Multiply Ic by

A


𝐴2𝐼𝑏 = 𝐴2𝐼𝑎0 + 𝐴4𝐼𝑎1 + 𝐴3𝐼𝑎2

𝐴𝐼𝑐 = 𝐴𝐼𝑎0 + 𝐴2𝐼𝑎1 + 𝐴3𝐼𝑎2

Adding them up,

𝐼𝑎 + 𝐴𝐼𝑏 + 𝐴2𝐼𝑐 = 𝐼𝑎0(1 + 𝐴 + 𝐴2) + 3𝐼𝑎1 + 𝐼𝑎2(1

+ 𝐴2 + 𝐴4)

𝑰𝒂𝟐 = 𝟏/𝟑(𝑰𝒂 + 𝑨𝟐𝑰𝒃 + 𝑨𝑰𝒄)

Also, Ib2 = AIa2 and Ic2 = A2Ia2 In matrix

notation:

Some Facts about Sequence Currents

1. A balanced 3 phase system consists of

positive sequence components only; the

negative and zero sequence components

being zero.

2. The presence of negative or zero sequence

currents in a 3 phase system introduces

unsymmetry and is indicative of an

abnormal condition of the circuit in which

these components are found.

3. The vector sum of the positive and negative

sequence currents of an unbalanced 3

phase system is zero. The resultant solely

consists of three zero sequence currents.

𝐼𝑎0 + 𝐼𝑏0 + 𝐼𝑐0

4. In a 3 phase, 4 wire unbalanced system, the

magnitude of zero sequence components is

one third of the current in the neutral wire

In the absence of path through the neutral of a 3

phase system, the neutral current is zero and

the line currents contain no zero sequence

components. A delta connected load provides

no path to the neutral and the line currents

flowing to delta connected load can contain no

zero sequence components.

5. In a 3 phase unbalanced system, the

magnitude of negative sequence

components cannot exceed that of the

positive sequence components. If the

negative sequence components were the

greater, the phase sequence of the resultant

system would be reversed.

6. The current of a single phase load drawn

from a 3 phase system comprises equal

positive, negative and zero sequence

components.

Unbalanced Impedances and Admittances

Self – Impedances – Unbalanced impedances

can be resolved into symmetrical components,

although the impedances are vector operators,

and not rotating vectors as are three – phase

voltages and currents. Consider the three star

impedances of (a), which form an unbalanced

load. Their sequence components are:



8

𝑍0 =1

3(𝑍𝑎 + 𝑍𝑏 + 𝑍𝑐)

𝑍1 =1

3(𝑍𝑎 + 𝑎𝑍𝑏 + 𝑎2𝑍𝑐)

𝑍2 =1

3(𝑍𝑎 + 𝑎2𝑍𝑏 + 𝑎𝑍𝑐)

The sequence components of current through the

impedances, and the sequence components of

the line voltages impressed across them are

interrelated by the following equations:

The above equation illustrate the fundamental

principle that there is mutual coupling between

sequences when the circuit constants are not

symmetrical. As the equations reveal, both

positive and negative sequence current (as well

as zero sequence current)

create a zero sequence voltage

drop. If Za = Zb = Zc, the impedances

are symmetrical, Z1 = Z2 = 0, and Z = Za. For this

condition,

0 0 𝟎

𝐸1 = 𝐼1𝒁𝟎

𝐸2 = 𝐼2𝒁𝟎

and, as expected, the sequences are

independent. If the neutral point is not grounded

in (a) I0 = 0 but E0 = I1Z2 + I2Z1 so that there is a

zero – sequence voltage, representing a neutral

voltage shift, created by positive and negative

sequence current flowing through the

unbalanced load. The previous equations also

hold for unsymmetrical series line impedances,

as shown in (b), where E0, E1 and E2 are

components of Ea, Eb and Ec, the voltage drops

across the impedances in the three phases.

Mutual Impedances between phases can also be

resolved into components. Consider Zmbc of (a),

as reference, then

The components of the three – phase line

currents and the components of the three –

phase voltage drops created by the mutual

impedances will be interrelated by the following

equations:

𝒁 𝒎𝟐 = 1

3 𝑍 ( 𝑚𝑏𝑐 𝑎 + 2 𝑍 𝑚𝑐𝑎 + 𝑎𝑍 𝑚𝑎𝑏 )



9

If, as in (b), both self and mutual impedances are

present in a section of a three – phase circuit, the

symmetrical components of the three voltage

drops across the section are:

Again, if both self and mutual impedances are

symmetrical, in all three phases,

𝐸0 = 𝐼0(𝒁𝟎 + 2𝒁𝒎𝟎) = 𝐼0𝑍0

𝐸1 = 𝐼1(𝒁𝟎 − 𝒁𝒎𝟎) = 𝐼1𝑍1

𝐸2 = 𝐼0(𝒁𝟎 − 𝒁𝒎𝟎) = 𝐼2𝑍2

Where Z0, Z1 and Z2 are, respectively, the

impedance to zero, positive, and negative

sequence. For this condition, positive sequence

currents produce only a positive sequence

voltage drop, etc. Z0, Z1 and Z2 are commonly

referred to as the zero sequence, positive

sequence and negative sequence impedances.

Note, however, that this is not strictly correct and

that Z1, the impedance to positive sequence

currents, should not be confused with Z1, the

positive sequence component of self

impedances. Since Z0, Z1 and Z2 are used more

frequently than Z0, Z1, and Z2 the shorter

expression “zero – sequence impedance” is

usually used to refer to Z0 rather than Z. For a

circuit that has only symmetrical impedances,

both self and mutual, the sequences are

independent of each other, and positive

sequence current produces only positive

sequence voltage drops, etc. Fortunately, except

for unsymmetrical loads, unsymmetrical

transformer connections, etc., the three phase

system usually encountered are symmetrical (or

balanced) and the sequences are independent.

Admittances can be resolved into symmetrical

components, and the components used to find

the sequence components of the currents through

a three phase set of line impedances, or star –

connected loads, as functions of the symmetrical

components of the voltage drops across the

impedances.

and

𝐼0 = 𝐸0𝒀𝟎 + 𝐸1𝒀𝟐 + 𝐸2𝒀𝟏 𝐼1

= 𝐸0𝒀𝟏 + 𝐸1𝒀𝟎 + 𝐸2𝒀𝟐

𝐼2 = 𝐸0𝒀𝟐 + 𝐸1𝒀𝟏 + 𝐸2𝒀𝟎 Note,

however, that Y0 is not the reciprocal of Z0, Y1 is

not the reciprocal of Z1, and Y2 is not the

reciprocal of Z2, unless Za = Zb = Zc; in other

words, the components of admittance are not

reciprocals of the corresponding components of

impedance unless the three impedances (and

admittances) under considerations are equal.

Star – Delta Conversion Equations

If a delta arrangement of impedances, is to be

converted to an equivalent star, the following

equations are applicable.



0

When the delta impedances form a three phase

load, no zero sequence current can flow from

the line to the load; hence, the equivalent star

load must be left with neutral ungrounded.

The reverse transformation, from the star

impedances, to the equivalent delta, is given by

An equivalent delta for a star – connected, three

– phase load with neutral grounded cannot be

found, since zero – sequence current can flow

from the line to the star load and return in the

ground, but cannot flow from the line to any delta

arrangement.

Power associated with Sequence Components

The total three phase power of a circuit can be

expressed in terms of the symmetrical

components of the line currents and the

symmetrical components of the line – to – neutral

voltages.

𝑃 = 3(𝐸0𝐼0 cos 𝜃0 + 𝐸1𝐼1 cos 𝜃1 + 𝐸2𝐼2 cos 𝜃2)

where 𝜃0 is the angle between 𝐸0 and 𝐼0, 𝜃1 is the

angle between 𝐸1 and 𝐼1 and 𝜃2 is the angle

between 𝐸2 and 𝐼2. The equation shows that the

total power is the sum of the three components of

power; but the power in one phase of an

unbalanced circuit is not one third of the above

expression, since each phase will contain

components of power resulting from zero

sequence voltage and positive sequence current,

etc. This power “between sequences” is

generated in one phase and absorbed by the

others, and does not appear in the expression for

total three phase power.

Only positive sequence power is developed by

the generators. This power is converted to

negative sequence and zero sequence power by

a circuit dissymmetry such as occurs from a

single line – to – ground or a line – to – line fault.

The unbalanced fault, unbalanced load, or other

dissymmetry in the circuit thus acts as the

“generator” for negative sequence and zero

sequence power.

For reactive power:

𝑄 = 3(𝐸0𝐼0 sin 𝜃0 + 𝐸1𝐼1 sin 𝜃1 + 𝐸2𝐼2 sin𝜃2) and

for Apparent power:

𝑆 = 3(𝐸0𝐼0 + 𝐸1𝐼1 + 𝐸2𝐼2) Examples:

1. In a 3 phase, 4 wire system, the currents in

R, Y and B lines under abnormal conditions

of loading are as under:

𝐼𝑅 = 100∠30° 𝐴

𝐼𝑌 = 50∠300° 𝐴

𝐼𝐵 = 30∠180° 𝐴 Calculate the

positive, negative and zero sequence

currents in the R – line and return current in

the neutral wire.

2. The currents in a 3 – phase unbalanced

system are:

𝐼𝑅 = (12 + 𝑗6) 𝐴 𝐼𝑌 =

(12 − 𝑗12) 𝐴

𝐼𝐵 = (−15 + 𝑗10) 𝐴

The phase sequence in RYB. Calculate the

zero, positive and negative sequence

components of the currents.

3. The sequence voltages in the red phase are:

𝐸𝑅0 = 100 𝑉

𝐸𝑅1 = (200 − 𝑗100) 𝑉

𝐸𝑅2 = −100 𝑉 Find

the phase voltages.

4. The zero and positive sequence

components of red phase are:

𝐸𝑅0 = 0.5 − 𝑗0.866 𝑉

𝐸𝑅1 = 2∠0° 𝑉

If the phase voltage 𝐸𝑅 = 3∠0° 𝑉, find the

negative sequence component of red phase

and the phase voltages EY and EB.

5. The current from neutral to ground

connection is 12 A. Calculate the zero phase

sequence components in phases.

6. A balanced star connected load takes 90 A

from a balanced 3 phase, 4 wire supply. If

the fuses in the Y and B phases are



1

removed, find the symmetrical components

of the line currents:

a. before the fuses are removed

b. after the fuses are removed

7. A 3 φ, 4 wire system supplies loads which

are unequally distributed in the three

phases. An analysis of the current flowing in

R, Y and B lines shows that in R line,

positive phase sequence component is 200

∠0° A and the negative phase sequence

component is 100 ∠60° A. The total

observed current flowing back to the supply

in the neutral conductor is 300 ∠300° A.

Calculate the currents in the three lines.

8. One conductor of a 3 phase line is open. The

current flowing to the Δ – connected load

through the line R is 10 A. With the current

in line R as reference and assuming that line

B is open, find the symmetrical components

of the line currents.

9. Three resistors of 5Ω, 10Ω and 20Ω are

connected in delta across the three phases

of a balanced 100 volts supply. What are the

sequence components in the resistors and

in supply lines?

10. A delta connected load is supplied from a 3

phase supply. The fuse in the B line is

removed and current in the other two lines is

20 A. Find the symmetrical components of

line currents.

11. Three impedances of 5—j10, 6 + j 5 and 3 +

j15 ohms are connected in star to red, yellow

and blue lines of a 3300 V, 3-phase, 3-wire

supply. The phase sequence is RYB. Calculate the line current IR.

12. A star connected load consists of three

equal resistors of 1 Ω resistance. The load

is assumed to be connected to an

unsymmetrical 3 phase supply, the line

voltages are 200 V, 346 V and 400 V. Find

the magnitude of current in any phase by the

method of symmetrical components.

13. The line currents in a 3 phase, 4 wire system

are: 𝐼𝑎 = 300 + 𝑗400 𝐴 𝐼𝑏 = 200 + 𝑗200 𝐴

𝐼𝑐 = 400 − 𝑗200 𝐴 Calculate the

positive, negative and zero components of all

the line currents.

14. The sequence components of phase a

currents are:

𝐼𝑎0 = 0.47 + 𝑗1.49 𝐴

𝐼𝑎1 = 18.4 𝑐𝑖𝑠 − 31.6° 𝐴

𝐼𝑎2 = 3.23 𝑐𝑖𝑠 168.2° 𝐴

Calculate the following: a. The

line current Ia

b. The line current Ib

c. The line current Ic

15. If the currents of a wye – connected

transformer are:

𝐼𝑎 = 10 𝑐𝑖𝑠 − 30° 𝐴

𝐼𝑏 = 12 𝑐𝑖𝑠 215° 𝐴

𝐼𝑐 = 15 𝑐𝑖𝑠 82° 𝐴

What is the phase b positive sequence

component?

16. Three equal impedances (8+j6) ohms are

connected in wye across a 3 phase, 3 wire

supply. The symmetrical components of the

phase A voltage are

𝑉𝑎0 = 40 + 𝑗0 𝑉

𝑉𝑎1 = 220 + 𝑗28.9 𝑉

𝑉𝑎2 = −40 − 𝑗28.9 𝑉

If there is no connection between the load

neutral and the supply neutral. Calculate the

line currents Ia, Ib and

Ic.

Sequence Networks In unbalanced three phase analysis, each piece of

equipment will have three values of impedance – one

corresponding to each sequence current:

i. Positive sequence

impedance (Z1) ii. Negative

sequence impedance (Z2) iii. Zero

sequence impedance (Z0) The

following points may be noted:



2

a. In a 3 – phase balanced system, each piece of

equipment or circuit offers only one impedance –

the one offered to positive or normal sequence

current. This is expected because of the absence

of negative and zero sequence currents in the 3 –

phase balanced system.

b. In a 3 – phase unbalanced system, each piece of

equipment or circuit will have three values of

impedance: positive sequence impedance,

negative sequence impedance and zero

sequence impedance.

c. The positive and negative sequence impedances

of linear, symmetrical and static circuits (e.g.

transmission lines, cables, transformers and static

loads) are equal and are the same as those used

in the analysis of balanced conditions. This is due

to the fact that impedance of such circuits is

independent of the phase order, provided the

applied voltages are balanced. It may be noted

that positive and negative sequence impedances

of rotating machines (e.g. synchronous and induction motors) are normally

different.

d. The zero sequence impedance depends upon the

path taken by the zero sequence current. As this

path is generally different from the path taken by

the positive and negative sequence currents,

therefore, zero sequence impedance is usually

different from positive or negative sequence

impedance. Sequence Impedances of Power System

Elements The concept of impedances of various elements of

power system (e.g. generators, transformers,

transmission lines etc.) to positive, negative and zero

sequence currents is of considerable importance in

determining the fault currents in a 3 – phase

unbalanced system.

a. Transmission Lines – are static devices and

hence the phase sequence has no effect on the

impedance because currents and voltages

encounter the same geometry of the line.

Therefore, positive and negative sequence

impedances of transmission lines are equal.

𝒁𝑻𝑳𝟏 = 𝒁𝑻𝑳𝟐 Zero sequence currents are in phase and flow

through the phases (a, b, c conductors) to return

through the grounded neutral. The ground or any

shielding wire are in the path of zero sequence

and zero – sequence impedance ZTL0, which

includes the effect of the return path through the

ground, is different from Z1 and Z2. Consider a 1 meter length of a three phase line

shown in the figure. The ground surface is

approximated to an equivalent conductor located

at the average distance Dn from each of the three

phases. The phase conductors carry sequence

currents with return paths through a grounded

neutral.

𝐼𝑎0

+

𝐼𝑏0

+

𝐼𝑐0 + 𝐼𝑛 = 0 Since

Since 𝐼𝑎0 = 𝐼𝑏0 = 𝐼𝑐0 and 𝐼𝑛 = −3𝐼𝑎0

Since

Therefore,

𝒁𝑻𝑳𝟎 = 𝒁𝟏 + 𝟑𝒁𝒏

b. Synchronous Machines is designed with

symmetrical windings and it induces emfs of

positive sequence only. The positive sequence

generator impedance is the value found when

positive sequence current flows due to an imposed

positive – sequence set of voltages. Neglecting

armature resistance, the positive sequence

impedance of the machine is

(Sub transient)

(Transient) 𝒁𝟏 = 𝒋𝑿𝒅 (Steady – state value) With

the flow of negative sequence currents in the

stator, the net flux in the air gap rotates at opposite

direction to that of the rotor. Therefore, the net flux

rotates twice the synchronous speed to the rotor.



3

In this case, field winding has no influence

because field voltage is associated with positive –

sequence variables and only the damper winding

produces an effect in the quadrantal axis.

Therefore, the negative sequence impedance is

close to the positive sequence sub transient

impedance.

In a synchronous machine, no zero sequence

voltage is induced. Zero sequence impedance of

the machine is due to the flow of zero – sequence

current. The flow of zero sequence currents

creates three mmfs which are in time phase but

are distributed in space phase by 120°. Therefore,

resultant air gap flux would be zero and there is no

reactance due to armature reaction. Hence,

machine offers a very small reactance due to the

leakage flux. Therefore, the rotor windings present

leakage reactance only to the flow of zero

sequence currents

𝒁𝟎 = 𝒋𝑿𝒍 It may be worthwhile to mention here that any

impedance Ze in the earth connection of a star

connected system has the effect to introduce an

impedance of 3Ze per phase. It is because the

three equal zero – sequence currents, being in

phase, do not sum to zero at the star point, but

they flow back along the neutral earth connection.

c. Two Winding

Transformers In power transformers, core losses and

magnetizing current are on the order of 1 % of

rated value and hence magnetizing branch is

neglected. The transformer is modelled with

equivalent series leakage impedance.

Transformer is a static device and if the phase

sequence is changed, leakage impedance will not

change. Therefore, positive, negative and zero

sequence impedances are the same and equal to leakage impedance of the

transformer

𝑍1 = 𝑍2 = 𝑍0 = 𝑍𝑙 The equivalent

circuit for the zero – sequence impedance

depends on the winding connection and also upon

whether or not the neutrals are grounded.

a. Wye (grounded) – Wye (grounded)

connection: Both neutrals are grounded and

there is a path for the zero sequence current to flow in the primary

and secondary.

b. Wye (grounded) – Wye connection:

Primary neutral is grounded and secondary

neutral is isolated. Zero – sequence current

in the secondary is zero. Consequently, zero

– sequence current in the primary is zero.

c. Wye – Delta connection: In this case

neutral is isolated. As neutral is isolated, zero

sequence current cannot flow.

d. Wye (grounded) – Delta connection: In this

case neutral is grounded. Primary current can

flow because there is zero sequence

circulating current in the delta connected

secondary and a ground return path for the

wye connected primary. Also note that no

zero sequence current can leave the delta

terminals, therefore, there is an isolation

between primary and secondary side.

e. Delta – Delta connection: In this case zero

sequence current circulate the delta

connected windings, but no current can leave

the delta terminals.



4

d. Three Winding

Transformers

Three phase,

three winding

have an

additional tertiary

winding, and may

be represented by

a single line

diagram

corresponding to

the ampere – turn balance, or power balance.

𝑁𝑃𝐼𝑃 + 𝑁𝑆𝐼𝑆 + 𝑁𝑇𝐼𝑇 = 0 or 𝑉𝑃𝐼𝑃 + 𝑉𝑆𝐼𝑆 + 𝑉𝑇𝐼𝑇 = 0

which in per unit quantities would yield the

common equation

𝐼𝑃,𝑝𝑢 + 𝐼𝑆,𝑝𝑢 + 𝐼𝑇,𝑝𝑢 = 0 This

may be represented

by three

reactances

connected in T,

giving the general single line

diagram for fault studies for the 3

winding transformer, as shown. The positive sequence and negative sequence

diagrams would have a direct connection to the T

connection of reactances from P, S and T. The zero sequence network would again be built

up from the single winding arrangements

described and would yield the single line diagrams

given:

A particular point to keep in mind is that what is

generally available from measurements for a 3

winding transformer would be the impedances

across a pairs of windings. (ie. ZPS, ZPT, and ZST), with the third winding on open

circuit. Thus we could relate the values to the

effective primary, secondary and tertiary

impedances (ZP, ZS and ZT) as follows, with

reference to the equivalent circuit.

𝑍𝑝𝑠 = 𝑍𝑝 + 𝑍𝑠 𝑍𝑝𝑡

= 𝑍𝑝 + 𝑍𝑡

𝑍𝑠𝑡 = 𝑍𝑠 + 𝑍𝑡 The values of ZP, ZS and ZT can then be

determined as

𝑍𝑝 = 0.5(𝑍𝑝𝑠 + 𝑍𝑝𝑡 − 𝑍𝑠𝑡) 𝑍𝑠 = 0.5(𝑍𝑝𝑠 + 𝑍𝑠𝑡 − 𝑍𝑝𝑡)

𝑍𝑡 = 0.5(𝑍𝑝𝑡 + 𝑍𝑠𝑡 − 𝑍𝑝𝑠)

As in the case of the 2 winding transformer, 3Zn

is included wherever earthing of a neutral point

is done through an impedance Zn. Sequence Network of Loaded Synchronous

Machines



5

A – Positive, B – Negative and C – Zero Sequence

Network Points to remember:

1. The three sequence networks are independent.

2. The neutral of the system is the reference for

positive and negative sequence networks but

ground is the reference for the zero sequence

network.

3. There is no voltage source in the negative or

zero – sequence networks. Only the positive

sequence network has a voltage source.

4. The grounding impedance is reflected in the zero

sequence network as 3Zn. Examples: 1. Draw the three sequence networks for the

transmission network shown.

2. Draw the positive sequence network of the power

system.

3. Draw the zero sequence network of the power

system.

4. Draw the zero – sequence network of the sample

power system shown. Data are given as: G: xg0 =

0.05 pu M: xm0 = 0.03 pu T1: xt1 = 0.12 pu T2: xt2 = 0.10 pu Line 1: xl10 = 0.7 pu Line 2: xl20 = 0.7 pu

5. Draw the Zero sequence network. Data is given

below: G1: 100 MVA, 11 kV, Xg10 = 0.05 pu G2: 100 MVA, 11 kV, Xg20 = 0.05 pu T1: 100 MVA, 11/220 kV, XT1 = 0.06 pu T2: 100 MVA, 220/11 kV, XT2 = 0.07 pu Line 1: XL10 = 0.3 pu Line 2:XL20 = 0.3 pu

6. Draw the positive, negative and zero sequence

component of the power system.

7.

Draw the zero – sequence network of the

sample power system.

8. Draw the positive, negative and zero – sequence

networks of the sample power system shown in

the diagram.



6

9. A 50 MVA, 11 kV, synchronous generator has a

sub transient reactance of 20%. The generator

supplies two motors over a transmission line with

transformers at both ends as shown below. The

motors have rated inputs of 30 and 15 MVA, both

10 kV, with 25% sub transient reactance. The

three – phase transformers are both rated 60

MVA, 10.8/121 kV, with leakage reactance of 10%

each. Assume zero – sequence reactances for

the generator and motors of 6% each. Current

limiting reactors of 2.5 ohms each are connected

in the neutral of the generator and motor No. 2.

The zero sequence reactance of the

transmission line is 300 ohms. The series

reactance of the line is 100 Ω. Draw the positive,

negative and zero sequence networks.

10.

Draw the zero – sequence network of the sample

power network.

Introduction to Power System

Generation,

Transmission and

Distribution

systems are the

main components

of an electric

power system.

Generating

stations and

distribution

systems are

connected through

transmission lines.

Normally, transmission lines implies

the bulk transfer of power by high – voltage

links between main load centers. On the other

hand, distribution system is mainly responsible

for the conveyance of thos power to the

consumers by means of lower voltage

networks.

Electric power is generated in the range

of 11 kV to 25 kV, which is increase by

stepped up transformers ot the main

transmission voltage. At sub -

stations, the connection between various

components are made, for example, lines and

transformers and switching of these

components is carried out. Transmission level

voltages are in the range of 66 kV to 400 kV (or

higher). Large amounts of power are transmitted

from the generating stations to the load centers

at 220 kV or higher.

The power supply network can be divided int two

parts, transmission and distribution systems. The

transmission system may be dividided into

primary and secondary distribution system. Most

of the distribution networks operate radially for

less short circuit current and better protective

coordination.



7

Distribution networks are different than

transmission networks in many ways, quite apart

from voltage magnitude. The general structure or

topology of the distribution is different and the

number of branches and sources is much higher.

A typical distribution system consists of a step –

down transformer at a bulk supply point feeding a

number of lines with varying length from a few

hundred of meters to several kilometers. Several

three – phase step – down transformes are

spaced along these feeders and from these, three

– phase four – wire networks of consumers are

supplied which gives 230 V, single phase supply

to houses and simular loads.

A complete circuit diagram of a power system for

all the three – phases is very complicated. It is

very much practical to represent power system

using simple symbols – for each component

resulting in what is called a single line diagram.

Power system engineers have devised the per

– unit system such that different physical

quatities such as current, voltage, power and

impedance are expressed as a decimal fraction

or multiple of base quantities. In this system, the

different voltage levels disapper and a power

network consisting synchronous generators,

transformers and lines reduces to a system of

simple impedances.

Power System Representation

Equivalent Circuit and Reactance Diagram

The power system components are: generators,

transformers, transmission lines, and loads.

Equivalent circuits of these components may

then be interconnected to obtain a circuit

representation for the entire system. In other

words, the one line diagram may be replaced by

an impedance diagram or a reactance diagram

(if resistances are neglected).



8

In the equivalent circuit of the components are

based on the following assumptions:

1. A generator can be represented by a voltage

source in series with an inductive reactance.

The internal resistance of the generator is

negligible compared to the reactance.

2. The motor load is inductive.

3. The static load has a lagging power factor.

4. A transformer is represented by a series

impedance on a per phase basis.

5. The transmission line is of medium length and

can be represented by a T section.

The reactance diagram is drawn by neglecting all

the resistances, static loads, and capacitances of

the transmission line. Reactance diagrams are

generally used for short circuit calculation,

whereas the impedance diagram is used for

powerflow studies.

Single Phase Representation of Balanced Three

Phase System

1. Three – phase two winding Y – Y

transformer bank

2. Three – phase two winding Y – ∆

transformer bank



9

The Per Unit System

An interconnected power system typically

consists of many different voltage levels given a

system containing several transformers and/or

rotating machines. The per – unit system

simplifies the analysis of complex power

systems by choosing a common set of base

parameters in terms of which, all systems

quantities are defined. The different voltage

levels disappear and the overall system reduces

to a set of impedances. The primary advantages

of the per – unit system are:

1. The per – unit values for transformer

impedance, voltage and current are

identical when referred to the primary and

secondary (no need to reflect impedances

from one side of the transformer to the other,

the transformer is a single impedance).

2. The per – unit values for various

components lie within a narrow range

regardless of the equipment rating.

3. The per – unit values clearly represent the

relative values of the circuit quantities. Many

of the ubiquitous scaling constants are

eliminated.

4. Ideal for computer simulations

The definition of any quantity (voltage, current,

power, impedance) in the per – unit system is

The complete characterization of a per – unit

system requires that all four base values be

defiend. Given the four base values, the per unit

quantities are defined as

Note that the numerator terms in the previous

equations are in general complex while the base

values are real – valued. Once any two of the four

base values are defined, the remaining two can

be determined according to their fundamental

circuit relationships. Usually the base values of

power and voltage are selected and the base

values of the current and impedance are

determined according to

It can also be assume that:

1. The value of Sbase is constant for all points in

the power system, and

2. The ratio of voltage bases on either side of a

transformer is chosen to equal the ratio of the

transformer voltage ratings,

then the transformer per – unit impedance

remains unchanged when referred from one side

of a transformer to the other. This will allow us to

eliminate the ideal transformer from the

transformer model.

In three phase per unit anlysis, the per phase

base voltage, current , apparent power of a circuit

is related by the following equations in single

phase



0

Thus, the final equations for three phase per unit

analysis are:

Change of Base Formula

Examples:

1. A single phase transformer is rated 20 kVA,

480/120 V, 60 Hz. The equivalent leakage

impedance of the transformer referred to the

120 V winding, denoted by winding 2, is 𝑍𝑒𝑞2

= 0.0525∠78.13 Ω. Using the transformer

ratings as the base values, determine the

per unit leakage impedance referred to

winding 2 and referred to winding 1.

2. A single phase two – winding transformer is

rated 25 kVA, 1100/440 volts, 50 Hz. The

equivalent leakage impedance of the

transformer referred to the low voltage side

is 0.06 < 78° Ω. Using transformer rating as

base values, determine the per – unit

leakage impedance referred to low voltage

winding and referred to high voltage

winding.

3. The figure shows single line diagram of a

single – phase circuit. Using the base values

of 3 kVA and 230 V, draw the per – unit

circuit diagram and determine the per – unit

impedances and per – unit source voltage.

Also calculate the load current both in per

unit and in Amperes.

4. The figure shows single – line diagram of a

power system. The ratings of the generators

and transformers are given below: G1: 25

MVA, 6.6 kV, xg1 = 0.2 pu

G2: 15 MVA, 6.6 kV, xg2 = 0.15 pu

G3: 30 MVA, 13.2 kV, xg3 = 0.15 pu

T1: 30 MVA, 6.6 ∆ - 115 Y kV, xT1 = 0.1 pu

T2: 15 MVA, 6.6 ∆ - 115 Y kV, xT2 = 0.1 pu

T3: Single – phase unit each rated 10

MVA, 6.9/69 kV, xT3 = 0.1 pu

Draw the per – unit circuit diagram using base

values of 30 MVA and 6.6 kV in the circuit of

generator 1.

5. A 100 MVA, 33 kV, three phase generator

has a reactance of 15%. The generator is

connected ot the motors through a

transmission line and transformers as

shown. Motors have rated inputs of 40 MVA,

30 MVA and 20 MVA at 30 kV with 20%

reactance each. Draw the per – unit circuit

diagram.

6. Draw the per – unit impedance diagram of

the system shown. Assumed base values

are 100 MVA and 100 kV.

G1: 50 MVA, 12.2 kV, xg1 = 0.10 pu



1

G2: 20 MVA, 13.8 kV, xg2 = 0.10 pu

T1: 80 MVA, 12.2/132 kV, xT1 = 0.10 pu

T2: 40 MVA, 13.8/132 kV, xT2 = 0.10 pu

Load: 50 MVA, 0.8 pf lagging operating at 124

kV.

7. The figure shows a sample power system

network. Find the current supplied by the

generator, the transmission line current, the

load current, the load voltage and the power

consumed by the load.

8. The single line diagram of a three – phase

ower system is shown. Select a common

base of 100 MVA and 13.8 kV on the

generator side. Draw per – unit impedance

diagram.

G: 90 MVA, 13.8 kV, xg = 18%

T1: 50 MVA, 13.8/220 kV, xT1 =

10% T2: 50 MVA, 220/11 kV, xT2

= 10%

T3: 50 MVA, 13.8/132 kV, xT3 = 10%

T4: 50 MVA, 132/11 kV, xT4 = 10%

M: 80 MVA, 10.45 kV, xm = 20%

Load: 57 MVA, 0.8 pf (lagging) at

10.45 kV.

xline1 = 50 Ω

xline2 = 70 Ω

9. The simple power system contains a 480 V

generator connected to an ideal 1:10 step –

up transformer, a transmission line, an ideal

20:1 step – down transformer, and a load.

The impedance of the transmission line is

20 + 𝑗60 𝛺, and the impedance of the load

is 10∠30°. The base values for this system

are to be 480 V and 10 kVA at the generator.

Find the base voltage, current, impedance

and apparent power at every point in the

power system. Convert the power system to

its per unit equivalent circuit and find the

power supplied to the load and power lost in

the transmission line in this system.

10. A 13.8 kV, 100 MVA, 60 Hz, 3 – phase

synchronous generator has a nameplate

resistance R of 10% or 0.10 pu and a

reactance X of 80% or 0.8 pu. These values

are specified on the base of the machine’s

rating. The base quantities of the power

system it is connected to are VLLbase = 14.4

kV and Sbase = 500 MVA. Find the pu

impedance of the generator on the base of

the power system.

REFERENCES:

[1] Alexander, Charles K. and Sadiku,

Michael N.O. (2013). Fundamentals of Electric

Circuits pp. 503 – 554. McGraw Hill Higher

Education.

[2] Bird, John (2007). Electrical and

Electronic Principles and Technology pp. 311 –

326. Elsevier. [3] Boylestad, Robert L. (2007).

Introductory Circuit Analysis pp. 1029 – 1066.

Pearson Prentice Hall.

[4] Glover, J. Duncan, Sarma, Mulukutla S. and

Overbye, Thomas (2012). Power System:

Analysis & Design pp. 1 – 158. CENGAGE

Learning.

[5] Hobson, J.E. and Whitehead, D.L.

Symmetrical Components [PDF Document].

Retrieved from

https://www.scribd.com/doc/63947448/Sym

metrical-

Components

[6] Lucas, J.R. (2001). Three Phase Theory &

Symmetrical

Components [PDF Document]. Retrieved from

http://www.elect.mrt.ac.lk/EE201_3phase_sym

_comp.pdf [7] Nahvi, Mahmood and Edminister,

Joseph (2003). Schaum’s Outlines: Electric

Circuits pp. 248 – 272. McGraw Hill Higher

Education.



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[8] Rajput, R.K. (2006). A Textbook of Electrical

Technology pp. 129 – 183. Laxmi Publications

LTD. [9] Theraja, B.L. and Theraja, A.K. (2003).

A Textbook of Electrical Technology pp. 665 –

854. S. Chand & Company LTD.

Documents

Electrical Circuits III.pdf