EXAMPLE 1: Balanced Y-Y Connection

Calculate the line currents in the three-wire Y-Y system as

shown below

0

pca

0

pbc

0

pab

210V3V

90V3V

30V3V

Line to line voltages or line voltages:

Magnitude of line voltages:

pL V3V

cabcabL VVVV cnbnanp VVVV

SOLUTION

• Due to the three-phase circuit is balanced; we may replace it

with its single-phase equivalent circuit

• Phase “a” equivalent circuit:

21.86.8121.816.155

0110I

8.21155.16)810()25(Z

;Z

VI

Aa

T

T

ANAa

jj

• Since the source voltage are in positive (ABC) phase

sequence, the line currents are also in positive sequence:

A2.986.811.8266.81

024II

A141.86.81

120II

AaCc

AaBb

EXAMPLE 2: Balanced Y-Delta Connection

A balanced Y-connected source with

is connected to a Δ-connected balanced load (8+j4)Ω per phase. Calculate the phase and line currents.

V10100V 0

an

CA

0

pca

BC

0

pbc

AB

0

pab

V210V3V

V90V3V

V30V3V

Line voltages:

903

1503

303

ABBCCAc

ABABBCb

ABCAABa

IIII

IIII

IIII

Line currents:

Z

VI

Z

VI

Z

VI

CACA

BC

BC

ABAB

Phase currents:

**NOTES:

SOLUTION

• Line Voltage:

• Phase Current

• Line Current

EXAMPLE 3: Balanced Delta-Delta

Connection

A balanced Δ connected load having an impedance

20-j15 Ω is connected to a Δ connected, positive sequence

generator having

Calculate the phase currents of the load and the line

currents.

V0330V 0

ab

Z

VI

Z

VI

Z

VI

CACA

BC

BC

ABAB

CAca

BCbc

ABab

VV

VV

VV

Line voltages: Line currents:

3II pL

Magnitude line currents:

3

ZZY

Total impedance:

Phase currents:

903

1503

303

ABBCCAc

ABABBCb

ABCAABa

IIII

IIII

IIII

SOLUTION

• Line Voltage= phase voltage

• Phase current

• Line Current

EXAMPLE 4: Balanced Delta-Y Connection

A balanced Y connected load with a phase resistance

of 40 Ω and a reactance of 25 Ω is supplies by a

balanced, positive sequence Δ connected source with

a line voltage of 210 V. Calculate the phase currents.

Use Vab as reference.

A single phase equivalent circuit

Y

p

Y

an

aZ

V

Z

VI

0303

SOLUTION

• Phase voltage = Line voltage (generator side)

• Phase Voltage (load side)

• Phase Current

EXERCISE 1

A Y-connected balanced three-phase generator with an impedance

of 0.4+j0.3 Ω per phase is connected to a Y-connected balanced load

with an impedance of 24 + j19 Ω per phase. The line joining the

generator and the load has an impedance of 0.6 + j0.7 Ω per phase.

Assuming a positive sequence for the source voltages and that

Find: (a) the line voltages

(b) the line currents

VVan

030120

EXERCISE 2

One line voltage of a balanced Y-connected

source is If the source is connected

to a Δ -connected load of

Find: (a) the phase currents

(b) the line currents

EXERCISE 3

A positive-sequence, balanced -connected source

supplies a balanced Δ-connected load. If the

impedance per phase of the load is 18+j12 Ω

and , find IAB and VAB

In a balanced -Y circuit,

and ZY = (12 + j15) Ω.

Calculate the line currents.

EXERCISE 4

VVab

015240

EXAMPLE 5: Unbalanced Y Connected

load

The unbalanced Y-load has balanced voltages of 100 V.

Calculate the line currents. Take

SOLUTION:

EXAMPLE 6: Unbalanced delta

Connected load

The unbalanced -load is supplied by balanced line-to-line voltages

of 240 V in the positive sequence. Find the line currents. Take VAB as

reference.

Active Power in Single Phase

• Active Power (P) delivered to or absorb by

resistive component to each phase, Pᶲ:

where

is the phase angle between and

or

(W)IVPV

I cos

(W)RIP 2

(W)R

VP R

2

V

I V I

Reactive Power in Single Phase

• Reactive Power (Q) delivered to or absorb by

reactive component to each phase, Qᶲ:

where

is the phase angle between and

or

(VAR)IVQV

I sin

(VAR)XIQ 2

(VAR)X

XQ

x

2

V

I V I

• Apparent Power (S) - is the product of voltage and current that deliver to each phase.

or

Apparent Power in Single Phase

(VA)IVS

(VA)ZIS 2

Power in Y-Connection

3

sin 3

co 3

cos3

3

cos3

cos

3

LLT

LLT

LL

LLT

T

VIS

VIQ

sVI

VIP

VIPP

VIP

The angle θ is angle between the voltage and current in any phase of

the load ( it is the same in all phases), and the power factor of the load is the cosine of the impedance

angle θ

Power in Δ-Connection

3

sin 3

co 3

cos3

3

cos3

cos

3

LLT

LLT

LL

LL

T

T

VIS

VIQ

sVI

VI

P

VIPP

VIP

The angle θ is angle between the voltage and current in any phase of

the load ( it is the same in all phases), and the power factor of the load is the cosine of the impedance

angle θ

V

I

T

T

S

Pcospf,FactorPower

Displacement angle between V and I

Power Factor, pf

Defined as the ratio of the real power flowing to

the load over the apparent power in the circuit.

Active Power, P (kW) Rea

ctiv

e Po

wer

, Q

(kV

AR

)

• Defined as 'the cosine of the angle between the voltage and

current'.

• In AC circuit, the voltage and current are ideally in phase.

• But practically, there exists a phase difference between them.

• The cosine of this phase difference is termed as power factor.

• It can be defined and mathematically represented as follows:

• Power factors are usually stated as "leading" or "lagging" to show the sign of the phase angle. Capacitive loads are leading (current leads voltage), and inductive loads are lagging (current lags voltage).

Phase Voltage Phase Current

Phase Current Phase Voltage

•Lagging power factor ( inductive loads)

•leading power factor • (capacitive loads)

EXAMPLE 7

A 208-V three-phase power system is shown in above figure. It consists of an

ideal 208-V Y-connected three-phase generator connected through a three-

phase transmission line to a Y-connected load. The transmission line has an impedance of 0.06+ j 0.12Ω per phase, and the load has an impedance of 12 + j9Ω per phase. Find

(a) the active, reactive and apparent powers consumed by the load

(b) the power factor of the load

SOLUTION

(a) The active power consumed by the load is

The reactive power consumed by the load is

The apparent power consumed by the load is

(b) The load power factor is

3 cos

3(120 )(7.94 )cos37.1

2280

load L LP V I

V A

W

3 sin

3(120 )(7.94 )sin 37.1

1724 var

load L LQ V I

V A

3

3(120 )(7.94 )

2858

load L LS V I

V A

VA

cos

cos37.1

0.8

loadPF

lagging

TWO-WATTMETER METHOD

To measure the power delivered by a three-phase, 4-wire system,

three single-phase wattmeter could be connected to measure

power in each of the phases and the readings added to obtain the total power.

However is not necessary because two single-phase wattmeter

connected as shown in Figure 1 will gave the same result.

The total power is the algebraic sum of the two wattmeter readings

and this method of power measurement is known as the two-

wattmeter method.

Figure 1: The Two-Wattmeter Method of Measuring Three-Phase Power

Figure 2 shows a wattmeter connected to measure the power delivered to a load and the equivalent circuit connections of the DAI

to obtain the same result with the Metering system.

Figure 2: Measuring Power with a Wattmeter

Two Wattmeter Method Alternative hookup for the Two

Wattmeter Method

EXAMPLE 8

An unbalanced Δ load is connected to a three-phase, Y-connected generator having a line voltage of EAB, EBC and ECA. Calculate the readings of the wattmeters W1 and W2. Find PTotal

SOLUTION