ITI Basics -2-09

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Instrument Transformers, Inc.

Are you using the right CT's and PT'sfor your application?

John Levine, P.E.

Levine Lectronics and LectricE-mail: [email protected]

GE Multilin/ Instrument Transformers, Inc.

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Current & VoltageTransformer Basics (IEEE

Standards)

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genda

Current TransformersVoltage Transformers

Required Information for Specifying CTs & VTs

Take Home Rules for CTs & VTs

Configure Tool for CTs and VTs

Products and tour of Plant

Applications where used

FT SwitchesConfigure Tool for FT Switches

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Why use Instrument Transformers?

Circuit Isolation

Reduce voltage and currentsto reasonable working levels.

Phasor combinations forsumming and measuring

power

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CURRENT TRANSFORMERS

& VOLTAGE TRANSFORMERS

TYPES OF C.T. AND V.T. CONSTRUCTION

The most common type of C.T. construction is the

`DOUGHNUT' type. It is constructed of an iron toroid, which

forms the core of the transformer, and is wound with secondary

turns.

Secondary Winding Primary Conductor

Iron Core

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Transformer ratio(TR)

Primary Current(100 amps)

SecondaryCurrent(5 amps)

Primary CurrentSecondary

Current

Transformer Ratio =_____________________

1005

___= 100:5 or 20:1

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Polarity

Direction ofPrimary Current

Direction of

SecondaryCurrent

H1

X1

Primary current into polarity =

Secondary current out of

polarity

P1

IEEE

IECPrimaryPolarityMarks

IEEEIECS1

SecondaryPolarity

Marks

emember:

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Polarity

Direction ofPrimary Current

Direction ofSecondary

Current

H1

X1

P1

IEEE

IEC

PrimaryPolarityMarks

IEEEIECS1

SecondaryPolarityMarks

Primary current into non-polarity

Secondary current out of non-

emember:

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G30 typical wiring

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What is your application?

If your application is metering , how high do Ineed to go in current? 2 times ?

If your application is protective relaying, howhigh do I need to go in current? 20 times ?

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CT Rating Factor (RF) -- IEEE

Rated current x (RF) =

Maximum continuous current carryingcapability:

Without exceeding temperaturelimits

Without loss of published accuracyclass

Typical rating factors -- 1.0, 1.33, 1.5, 2.0,

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0.30

0.30

0.60

0.60

Rating

Factor

Accuracy

Class-

%

Rating Factor (RF) -- IEEE

10% 100% 200% 300% 400%

1.0 2.0 3.0 4.0

IEEE C57.13 Accuracy

0.3% @ BX.X RF 4.0

0.3% AccuracyRegion

0.6%Accurac

y

Region

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Short-Time Thermal CurrentRating

One (1) second thermal rating

Expressed as value of RMS primary current

Main influencing factor:CT primary & secondary wire size

Can be converted to thermal rating for anytime period (t) up to five (5) seconds:

I1-sec

t

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T Metering Accuracy

Actualsecondary

current

Ratedsecondary

current

=

Difference in % is known

as the Accuracyof the CT

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EE CT Metering Accuracy

AccuracyClass ( * ) Application

0.15 High Accuracy Metering

0.15S Special High AccuracyMetering

0.3 Revenue Metering

0.6 Indicating Instruments1.2 Indicating Instruments* All accuracy classes defined by IEEE C57.13 or C57.13.6

* Accuracy classes include both ratio & phase angle error

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Burden

Load connected to CT secondary

Includes devices & connecting leads

Expressed in ohms

Standard values = B0.1, B0.2, B0.5, B0.9,B1.8

E0.04, E0.2All burdens defined by IEEE C57.13 or C57.13.6 for 60 Hz only

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Application Burden

Designation

Impedance

(Ohms)

VA @

5 amps

Power

Factor

Metering B0.1 0.1 2.5 0.9

B0.2 0.2 5 0.9

B0.5 0.5 12.5 0.9

B0.9 0.9 22.5 0.9

B1.8 1.8 45 0.9

Standard IEEE CT Burdens (5Amp)(Per IEEE Std. C57.13-1993 &C57.13.6)


E0.2

E0.04

0.2

0.04

5

1

1.0

1.0

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A current transformer for metering purposes may

typically have an accuracy of 0.3%. The C.T. mustmaintain this accuracy for normal load currents,

provided the rated burden on the C.T. is not exceeded.

It is quite acceptable, and in fact desirable, for the C.T.

to saturate when fault current flows. The accuracy for a

typical metering C.T. is specified as:

0.3 M 0.9

O.3% METERING O.9 OHMS BURDEN

This metering C.T. has an accuracy of 0.3% when the

connected burden does not exceed 0.9 OHMS.

CT CLASSIFICATION for MeteringCT CLASSIFICATION for Metering

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Accuracy Class

(0.3, 0.6, 1.2) (*)

Burden

(Ohms)(B0.1, B0.2, B0.5,B0.9, B1.8)

+

Accuracyexpressed as:

0.3B0.2

0.6B0.91.2B1.80.15E0.2

=

TypicalExamples

* Accuracy class is stated at 100% rated current* At 10% rated current, twice the error is allowed (5% for0.15 class)

(0.15*, 0.15S^) E0.2, E0.04)

^ Accuracy class is stated at 100% to 5% rated current

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0.30

0.30

0.60

0.60

Rating

Factor

Accuracy

Class-

%

10% 200% 300%

1.0 2.0 3.0 4.0

IEEE C57.13 Accuracy0.3 @ BX.X; RF 4.0

0.3% AccuracyRegion

0.6%AccuracyRegion


100% 400%

No accuracy guaranteed

at current levels lessthan 10%

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CT Parallelogram IEEE C57.13

Out of 0.3% Class

0.3 classparallelogram


100%10%

Note:Burdenmustbespecified

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Corrected to 0.3% Class

CT Parallelogram IEEE C57.130.3 classparallelogram


100%

10%

Note:Burden mustbe specified

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0.30

0.30

0.60

0.60

Rating

Factor

Accuracy

Class-

%

5% 100% 200% 300% 400%

1.0 2.0 3.0 4.0

IEEE C57.13.6 Accuracy

0.15 @ E0.04, E0.20; 0.15 @ BX.X;RF4.0

0.15% AccuracyRegion

0.3%

AccuracyRegion


0.15

0.1

5

No accuracy guaranteedat current levels less than 5%

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0.30

0.30

0.60

0.60

Rating

Factor

Accuracy

Class-

%

5% 200% 300%

1.0 2.0 3.0 4.0

IEEE C57.13.6 Accuracy

0.15S @ E0.04, E0.20; 0.15S @ BX.X,RF4.0

0.15% AccuracyRegion


0.15

0.1

5

No accuracy guaranteedat current levels less than 5%

400%100%

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EE CT Relay Accuracy

andard Relay Accuracy Classes

C or T100C or T200

C or T400

C or T800

What do these mean?

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EE CT Relay Accuracy

Relay class (C or T___ ) designatesminimum secondary terminal volts

At 20 times rated currentWithout exceeding 10% ratio error

Into a maximum specified burden

Now that everyone is totallyconfused

lets look at some simpleexamples

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T = Determined by test

CT CLASSIFICATION for RELAYINGCT CLASSIFICATION for RELAYING

10 C 40010 C 400

Protection Class CTsProtection Class CTs - Must supply 20 times rated current- Must supply 20 times rated current

C = Calculated

K = CalculatedL = Low internal secondary impedance

H = High internal secondary impedance

Format

LetterAccuracy Voltage at 20 times CT

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rimary current24,000 amps(20 x 1200)

IEEE CT Relay Accuracy

CT1200:5

C orT100

X1X2

Burden ofDevices

( )

Burden ofLeads

()

Secondary current100 amps (20 x 5)

Total ExtBurden1.0

or T100 example

Terminal Volts = (20 times rated) (Totalexternal burden)

=

TerminalVolts=

100

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rimary current24,000 amps(20 x 1200)


CT1200:5

C orT200

X1X2

Burden ofDevices

()

Burden ofLeads

()

Secondary current100 amps (20 x 5)

Total ExtBurden2.0

or T200 example

Terminal volts = (20 times rated) (Total externalburden)

200 Volts = (100 amps) (2.0 )

TerminalVolts=

200

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Standard IEEE CT Burdens (5Amp)(Per IEEE Std. C57.13-1993)


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IEEE CT Relay AccuracyExcitation curve includesvoltage required to overcome

internal resistance (DCR) of CT.Approximately 32 volts.

10% ratio error = (20 x 5)(10%)

= (100)(0.10)= 10 ampsHow many terminal

voltswould you estimatethis CT can produce?

d l l i

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T Burden Calculation

PrimaryCurrent

CT

X1X2Burden ofDevices

()

Burden ofLeads()

Secondarycurrent

Total

Burden ZT

How do we calculatethis?

d C l l i

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T Burden Calculation

Assumption: 3 phase CTs are Y connected

ZT = RCT + RL + ZB

ZT =Total burden in ohms (vector summation of

resistance and inductance components)

RCT = CT secondary resistance in ohms @75

deg C (DCR)

RL =Resistance of leads in ohms (Total loopdistance)

ZB = Device impedance in ohms

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GE Multilin Electronic RelayBurden

VA = VI. V= IR, So 0.2 = I*I*R.

.2/25 = .008 ohms

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100:5 C.T. Secondary Winding Resistance = .062 ohm

Resistance of Cable from C.T. to Relay and back = .1 ohms

Resistance of Relay Coil = .02 ohmsTotal Resistance = .182 ohms

If we have a fault of 2,000 amps and the C.T. ratio is 100:5 then the C.T.

secondary current is 100 amps. Therefore we must be able to produce a totalvoltage of 100 amps x .182 ohms = 18.2 Volts. To prevent CT saturation,

select a CT with a knee point above 18.2 Volts.

.

062 .

1

.02

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780-102 is a 1000 to 5 CT, Class C100

1000:5 C.T. Secondary Winding Resistance = .32 ohm

Resistance of Cable from C.T. to Relay and back = .1 ohmsResistance of Relay Coil = .008 ohms

Total Resistance = .428 ohms

If we have a fault of 20,000 amps and the C.T. ratio is 1000:5 then the C.T.

secondary current is 100 amps. Therefore we must be able to produce a total

voltage of 100 amps x .428 ohms = 42.8 Volts. To prevent CT saturation,select a CT with a knee point above 42.8 Volts.

What happens if the fault current is 35,000 amps?

.32

.1

.008

t I fl i CT A

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ctors Influencing CT Accurac

Frequency

Current Ratio

Burden

Low frequency and High accuracy are not friends!!

Low ratio and high accuracy are not friends!!

High burden and High Accuracy are not friends!!

f A idi CT S t ti

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ps for Avoiding CT Saturation

Use higher ratio CTs

Use separate set of high ratio CTs for highfault current tripping

Reduce secondary burdenSelect low burden relays & metersDistribute single phase burdens among

phases

Increase size of secondary leadsReduce length of secondary leads

Use step down auxiliary CTs

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Potential Transformers

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Definitions

Voltage Transformer (VT)

An instrument transformer used to reflect a primaryvoltage into a secondary voltage through a magnetic medium.Always connected in parallel with primary conductor across acircuit load.

Secondary (measuring) voltage is usually 115 or 120volts nominally. The secondary voltage level is selected for easeof measurement and safety.

Control Power Transformer (CPT)

Designed to provide power for contractors, relays anddevices with high inrush currents, Regulation is not as critical.

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POTENTIAL TRANSFORMERSPOTENTIAL TRANSFORMERS

VVPP

VsVs

Relay

4200/120 = 35/1

2400/120 = 20/1

EE VT A Cl

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EE VT Accuracy Class

Metering AccuracyClasses (% error)

0.3

0.6

1.2

0.15

Defined by IEEE C57.13

Applicable from 90% to 110rated voltage

Defined by IEEE C57.13.6

IEEE VT Accuracy Class

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IEEE VT Accuracy Class

These standard burden designations have no significanceat frequencies other than 60 Hz.

Burden VA PF

W 12.5 0.10

X 25 0.70

M 35 0.20

Y 75 0.85

Z 200 0.85ZZ 400 0.85

MeteringAccuracyClass Burdens

EE VT A Cl

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EE VT Accuracy Class

Expressed as:Accuracy Class + Burden Code

0.3 W,X,Y0.6 Z

1.2 ZZ

These standard designations have no significanceat frequencies other than 60 Hz.

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Installation Guidelines

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Installation Guidelinesution:Rated voltage: Do not operate above 110%

Line to ground rated:

Do not connect line to line

Do not use on ungrounded systemsw/o consulting factory

Rated Frequency: Do not operate below ratfrequency w/o consulting

factory

EEE VT Groups

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EEE VT GroupsVT

GroupNo. of

BushingConnection

Method

NeutralGrounding

Notes

1 2 open

Y-Y possible

Any Withstand 25% over rated voltageon an emergency basis

2 2 open

Y-Y possible

Any Withstand 10% over rated voltagecontinuously. Primary rated forline to line voltage.

3 1 Y-Y-Y Any Outdoor, two secondary windings.Withstand 10% over rated voltagecontinuously.

4A 1 Y-Y Effectively Withstand 10% over rated voltagecontinuously & 25% on anemergency basis. For operation at100% rated voltage.

4B 1 Y-Y

Y-BrokenCorner

Non-

effectively

Withstand 10% overvoltage

continuously. For operation at58% rated voltage.

5 1 Y-Y Effectively Outdoor. Withstand 40% overrated voltage for 1 minute and10% over rated voltagecontinuously

T Typical Connections

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T Typical Connections

Open Delta Connection(2) Double Bushing VTs

Y Y Connection(3) Single Bushing VTs

erroresonance

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erroresonancePossible with Y connected grounded VTs

on ungrounded power systems

A VT is an inductive component

Capacitance to ground exists in thesystem

When they match ferroresonance may

occurMay cause higher VT voltages &

saturation

Possible results -- High VT currents

erroresonance

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erroresonance

Recommended reading:

Ferroresonance of Grounded Potential

Transformers on Ungrounded PowerSystems

AIEE Power Apparatus & Systems,

Aug 1959, pg 607-618, by Karlicek andTaylor

erroresonance Damping

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Y Broken Corner VTconnection

Ferroresonancedamping resistor R

FR

Voltage relay

todetectgroundfault

Non effectively

grounded system

0 volts @ normal condition

3 times normal L-N phasevoltage @ ground fault

referred method


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Y/Broken Corner Connection

Double secondary VT

(1) Relaying / Metering

(1) Ground fault detectionand ferroresonance dampin

Preferred method

erroresonance

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erroresonanceamping

Ferroresonancedamping resistorRFR value

Based on 2 variables:Air core inductance of primary

winding (La)

VT ratio (N)RFR = 100 La / N

2 Power rating (watts) of the resistor is a system relatedproblem.

General recommendation is 50% of VA rating of a singleVT.

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Required Information

for Specifying CTs & VTs

Current Transformer (CT) RFQ Specification

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p

Environment: ___Indoor ___Outdoor

System Voltage (kV)0.60.723.65.07.28.7121524

2534.534.5

Power Frequency (kV)4310192026283450

407070

BIL (kV)10

4060607575110125

125150200

Standard (Check one)IEEE __IEC __IEC __ IEEE __IEC __IEEE __IEC __ IEEE __IEC __

IEEE __IEEE __IEEE __

CT Application: ___Metering ___Protection

Dimensions: ___ Inches ___ mm

Max. Outside: L ______ x W ______ x D ______

CT Window: Round: ______ Diameter; Rectangular: L ______ x W ______ ; PrimaryBar: _____

Current Ratio: _______ : 5 _______ : 1

Continued next slide

Current Transformer (CT) RFQ Specification (Continued)

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pAccuracy:

Indication Only: _____ %, _____VA (Skip metering & protection selections)

Metering Class:

IEEE: ___0.3 ___0.6 ___1.2 ___2.4 ________Other

IEC: ___0.2 ___0.5 ___1.0 ________Other

Metering Burden:IEEE (Ohms): ___B0.1 ___B0.2 ___B0.5 ___B0.9 ___B1.8 ______OtherIEC (VA): ___2.5 ___5.0 ___10 ___15 ___30 ______Other

Protection Class: C______(IEEE) ___VA, ___P___(IEC)

Operating Frequency: ___60HZ ___50HZ

Rating Factor: ___1.0 ___1.33 ___1.5 ___2.0 ______Other

Secondary Connections: ___Terminals ___24 inch leads

Outer Insulation: ___Standard ___Cotton tape & varnish ___Polyester tapeInsulation Class: ___105 0C (Standard) _____Other

Other Special Requirements (dimensional constraints, mounting requirements, etc):____________________________________________________________________________________________________________________________________________________________

Voltage Transformer (VT) RFQ Specification

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Environment: ___Indoor ___Outdoor

System Voltage (kV)0.60.723.65.07.28.7121524

2534.534.5

Power Frequency (kV)4310192026283450

407070

BIL (kV)10

4060607575110125

125150200

Standard (Check one)IEEE __IEC __IEC __ IEEE __IEC __IEEE __IEC __ IEEE __IEC __

IEEE __IEEE __IEEE __

Continued next slide

perating Frequency: ___60HZ ___50HZ

Accuracy:

IEEE: ___W ___X ___M ___Y ___Z ___ZZ ________Other(Enter 0.3, 0.6, 1.2, or leave blank)

IEC: ___10VA ___25VA ___50VA ___100VA ___200VA ___500VA______Other

(Enter 0.2, 0.5, 1.0, or leave blank)

Voltage Transformer (VT) RFQ Specification (Continued)

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hermal Rating: _______VA (Optional)

rimary Voltage: ___1 bushing _________VAC - Phase to neutral

___2 bushing _________VAC - Phase to phase

econdary Voltage: ___120V ___115V ___110V ___100V

___120/3 ___115/3 ___110/3 ___100/3___Other _________________

ated Voltage Factor (RVF) (1 bushing only): ___1.9 for 30s ___1.9 for 8 hours ___Other________

uses:

___Primary ___Secondary ___None (600 720 V)

___Primary ___Live parts only ___Switchgear Style ___Unfused (2.5kV 15kV)

Note integral fusing not available above 15kV

Take Home Rule

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Take Home Rule# 1

CTs are intended to be proportionalcurrent devices. Very high voltagescan result from open circuiting thesecondary circuit of an energizedCT. Even very small primarycurrents can cause damageConsult the factory if you have

questions.

Never open circuit acurrent

transformer secondarywhile the primary isenergized

ake Home Rule # 2

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Never short circuit thesecondary of an energized

VT

ake Home Rule # 2

VTs are intended to be used as proportional

voltage

devices. Damaging current will result fromshort

circuiting the secondary circuit of an energized

ake Home Rule # 3

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ake Home Rule # 3

Metering applicationsdo not require a C classCT

C class ratings are specified for

protection purposes only. With someexceptions metering class CTs are typicallysmaller and less expensive.

ke Home Rule # 4

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ke Home Rule # 4

CT secondary leads mustbe added to the CTburdenElectronic relays usually represent

very little burden to the CTsecondary circuit. In many casesthe major burden is caused by theCT secondary leads.

ake Home Rule # 5

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ake Home Rule # 5

Never use a 60 Hz ratedVTon a 50 Hz System

60 Hz VTs may saturate at lower

frequencies and exceed temperaturelimitations. VT failure is likelysevereequipment damage is possible.

ake Home Rule # 6

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ake Home Rule # 6

Exercise caution whenconnecting grounded VTsto ungrounded systems

Line to ground voltage on any VT may exceed

the primary voltage rating during a faultconditionVT must be designed for application.

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QUESTIONS?

Documents

ITI Basics -2-09