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Example 1:

A 1200/5, C400 CT with excitation curves shown on above figure, is connected to a

2.0Ω burden. ased on the accurac! c"assification, what is the #axi#u# s!##etrica"

fau"t current that #a! be a$$"ied to this CT without exceeding a 10% ratio error&

Answer:

ased on the criteria that the CT can de"iver 20 ti#es rated secondar! current

without exceeding a 10% ratio error, the #axi#u# fau"t current wi"" be 24000A.

'owever, with a 2.0Ω burden, this wi"" resu"t in a vo"tage be"ow the (nee $oint of the CT

and, as a $ractica" #atter, it wi"" be within 10% accurac! at higher currents. This can on"! be accurate"! deter#ined fro# excitation or ratio correction curves and not fro# the

accurac! c"assification. )or exa#$"e, a CT with characteristics shown in above figure

wi"" $roduce between 1*0+240A without exceeding the 10% ratio error, de$ending on the

$ower factor of the 2.0Ω burden.

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Example 2:

A 1200/5, C400 CT is connected on the 1000/5 ta$. hat is the #axi#u# secondar! burden that can be used and sti"" #aintain rated accurac! at 20 ti#es rated s!##etrica"

secondar! current&

Answer:

-ince the secondar! vo"tage ca$abi"it! is direct"! $ro$ortiona" to the connected ta$, the

CT wi"" su$$ort a vo"tage of 1000/1200 ×400 or . Twent! ti#es the ratedsecondar! current is 100A. Therefore, the #axi#u# burden is /100A or . Ω

Example 3:

Assu#e that secondar! burden in a re"a! circuit is 5Ω. The re"a! setting is 2A and the CT

ratio is 00/5. sing above figure, ca"cu"ate the $ri#ar! current re uired to o$erate the

re"a!&

Answer:

5 Ω ti#es 2A 10

The secondar! exciting current fro# above figure is a$$roxi#ate"! 0.04A. 3ST P I N I =

3S E I I N +=

00/5 0.04 23 A 122A

Example 4:

A re"a! is ex$ected to o$erate for a 6000A $ri#ar! current. The CT ratio is 700/5.

-econdar! burden is .5 Ω. hat is the error for the CT shown in above figure&

Answer:

The tota" secondar! fau"t current is 6000/7003 ×5 5*A. Assu#e that exciting current isneg"igib"e.

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3S BS S R R I V +=

5* .5 0. 13

221

The exciting current wi"" not be neg"igib"e, however, and the ca"cu"ation wi"" not be

iterated.

8ec 9 6

#ax 1 022

dc

N RI N

φ τ =: :

m

line

V Z

122 : :

m

line

V N R N Z

τ 11 22: :

m

line

V N N R

Z N τ = 1 02

2

3dct N RI

t e N

τ −

= 1 02

2

t N I i e

N τ −

=

1 02 2

2 20

13 03 1 3

t t N RI t dt e

N N τ φ φ τ −

− = = −∫ 1 022

3 03 1 3t N RI

t e N

τ φ φ τ −

= + −

8ec+ 61

22

3 sin 3: :

ac m

line

V N t R t

Z N ω φ θ = + − d

dt !ω 2

2ac

V ! N

φ ω

=

122

3 sin 3: : 2

mac

line

RV N t t

Z N π

φ ω φ θ ω

= + − − #ax 122: :

mac

line

RV N Z N

φ ω

=

#ax #ax 1 12 22 2: : : :

m mac dc

line line

RV RV N N Z N N Z

τ φ φ

ω + = +

122

122

: :1 1

: :

m

line

m

line

V N R

Z N V N R

Z N

τ ωτ

ω

= + = + 1 lineline

" Rω = + 1 line

line

# R

= +

1 # R

= +

8ec+7

20 5 100S I A= × = Ω= 71.0S R 1000

4001200S

V = × #ax400

; 152.71

A= = 1200

155

= ×

V V S .10315.052 =+×= 6000

5 5*.700S

I A= × = 7*00 5 7*500S

I A= × =

*1.5*31.05.5* ×=+=S V 7

100 100

7*

E

S

I

I

× = × Ω= 5.2 B R00

2.04

5

×

7* 2.5 0.253S V = + 1 I R = Ω 0.51S R = Ω

< =sin 3 =sin > 3t pi I t I e $ t γ ω δ δ −= + − < =sin cos 3 =cos sin > 3t pi I t e I t $ t

γ δ ω δ ω −= + +

= cos sin I t δ ω =cos < sin cos cos sin 3>t I e t α δ ψ ψ ψ ω ψ −− + + =sin cos I t δ ω

2=sin < sin cos cos 3>t I e t α δ ψ ψ ω ψ −− + + =sin t I e γ δ −− −=sin

3t t I

e eα γ δ

α γ α γ

− −− −−

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8ec 9 5

100

pS

S

I I

N I

−×

pS E

I I I

N − = 100 E

S

I I

× 100

pS

S

I I

N

I

−× ?* ° 45 °

8ec + ?

100t% T t%

V V V

− × 3φ 20

13 " ω

ω = ∆ 2

0

1 "&

ω = 20 1 "& ω = < φ ' () ∆ = ±

13

t

e*

di t Ri idt "

& dt −∞= + +∫

2

2

1

e*

d di d i R i "

dt dt & dt = + +

0ω

13 " &

e*

d V V !I "

d & ω ω

ω ω ∆ − ∆ = − ∆ 2

0

12 3

e*

I &

ω ω

= + ∆ 20

1

e* "& ω = 3 I " " ω = + ∆

2 ! "I ω ∆ 3 " & V V ∆ − ∆ 2 2: : 2 2

tan : : " &

R + +

V V "I "I V a R I a R

ω β

− ∆= = = 2

2

+

"a R

ω ∆≈ 72 10m ' "π =

7101* .1

2m " (

π = =

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CT saturation and @C+ offset current

o"e of @C off+set current

T!$ica""! fau"t current consists a s!##etrica" ac co#$onent and a dc offset current. To

understand this conce$t, consider a trans#ission "ine un"oaded exited b! an e uiva"ent vo"tage

source. The fau"t stri(es at ti#e 0t t = . This can be si#u"ated b! c"osing the switch at 1t t =

" ! R ω +

orθ ∠ Z

#ode"s the "ine i#$edance. The fau"t current in the "ine is given b!03 =t i 00 t t ≤≤

−−

+−+= τ θ φ ω 0

0::

3sin3

t t

m e I Z

t V t i 0t t ≥

here B is the ti#e constant of the "ine B 8/ . The fau"t current has two co#$onents in

it. The first co#$onent #ode"s the stead! state sinusoida" ac res$onse whi"e the second current

is the dc offset current due to the $resence of inductive co#$onent in the circuit. eca"" thatcurrent in an inductance can not change instantaneous"!. As t → , the instantaneous dc

current, a conse uence of #aintaining initia" condition 33 00+− = t it i , deca!s ex$onentia""! to

Dero and the current reaches the ac stead! state va"ues. hi"e the dc offset current, wou"d in

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theor! $ersist ti"" infinit!, itEs trace in the actua" wave for# wou"d not be seen be!ond a certain

ti#e constants. Tab"e+; i""ustrates the va"ues of τ t

e−

u$ to 10 ti#e constants.

Ti#e t 0 t B t 2B t 4B t 7B t *B t 10B

τ t

e− 1 0. 76* 0.1 5 0.01* 0.0024 0.000 0.00004

;t is #ore or "ess obvious that, dc offset is not seen in the wavefor# after 5 ti#e constants.

The va"ue of I , can be wor(ed out b! setting the current at -t t = to Dero.

This i#$"ies that

3sin 00 θ ϕ ω −+−= t Z

V I m

Thus

3 0

3sin3sin3 τ θ φ ω θ φ ω t t

mm et Z

V t

Z

V t i

−−−+−−+=

fig.2

C"ear"!, the $ea( va"ue of dc offset current de$ends u$on

• Ti#e at which fau"t stri(es

• Fhase ang"e φ of ac vo"tage

• Z G θ of trans#ission "ine

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)igure 2 shows the wavefor#s of

a3 s!##etrica" ac co#$onent

b3 dc offset current

c3 tota" current for various va"ues of φ θ , G 0t

;t can be seen that severit! of dc offset co#$onent in fau"t current is #axi#u# when

a3 θ φ =

b320π

ω ±=t

)or exa#$"e, if ang"e of trans#ission "ine is *0 0, then with φ *0 0 G50220 ××= π

π t 200

1

sec 5#sec, the severit! of dc offset current wou"d e ua" Z V

I m=0 , which is a"so the $ea(

va"ue of s!##etrica" ac co#$onent of the current. This "eads us to an i#$ortant

conc"usion. iD. $ea( va"ue

13 dc offset current can be as high as the s!##etrica" ac $ea(

23 The dc offset current can be $ositive or negative see fig23

3 @c offset current #a! be tota""! absent

eg. ;f θ ϕ = , 00 =t

43 hi"e, in above ana"!sis, we have considered a sing"e $hase current, a φ fau"t on a

φ trans#ission "ine wou"d a"wa!s induce dc offset current in at"east 2 $hases.

;n the re#aining "ecture, we ana"!De the effect of dc offset current on CT $erfor#ance.

@C+ offset current and CT saturation

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e now $"an to show that CT can saturate on dc offset current. A"so, we $"an to show

that the resu"ting distortions in the CT secondar! current can be un+acce$tab"! high. hi"e

doing this ana"!sis, we wi"" neg"ect ac s!##etrica" co#$onent. ;n other words, we rest our

be"ief in su$er$osition theore# at"east ua"itative"! and wi"" fina""! eva"uate effect using it

Hotice that the current that we are dea"ing with is non+"inear, a rigorous a$$"ication of su$er

$osition theore# is si#$"! out of uestion.

)irst consider an idea" CT excited b! the dc offset current source. An idea" CT wi"" faithfu""!

re$"icates $ri#ar! current wavefor# on the secondar! side. 'ence, the secondar! current

wou"d be given b!

τ t

e N

I t i

−

= 02 3

and the vo"tage deve"o$ed across CT secondar! wou"d be given b!

τ t

e N

RI t

−

= 02 3 where1

2

N

N N =

T!$ica" vo"tage wavefor# is shown in fig. 53

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)or si#$"icit!, "et us assu#e that the initia" f"ux in the transfor#er core at t 0 is Dero

030 =φ I Then we can co#$ute the f"ux in the transfor#er core b! using farada!Es "aw

dt d

N V φ 22 = +++++++++ 23

⇒ ∫ =− t

dt t 0

2303 φ φ

−=

−τ τ

t

e N RI

12

0

312

0 τ t

e N "I −−=

⇒ 313032

0 τ φ φ t

e

N

"I t

−

−+=

312

0 τ t

e N "I −−= +++++++ 3

as a conse uence of dc offset current,

Thus, f"ux in the core increases ex$onentia""! to a $ea( va"ue of

2

0#ax

N "I

cd =φ as →∝t

Z V

N " m

Z V m

#axcd φ

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Hote that un"i(e ac vo"tage induced f"ux, which is sinusoida", this f"ux is unidirectiona". The ac

vo"tage induced f"ux has Dero average va"ue. 'owever, dc offset induced does not have this

nice feature. The tota" f"ux in idea" CT core is a su##ation of ac f"ux and dc f"ux.

The ac f"ux in the CT core can be obtained b! substituting o$eratordt d

b! ω ! . 'ence

$hasor re"ationshi$ between $hase 2V G acφ is given b!

2

2

N !V ω

φ =

;f 2 4 3 sin4 3mt V t ω φ = + , then

32sin2

π φ ω ω φ −+= t N

V m

ac

The $ea( va"ue of ac f"ux is given b!

2

#ax

N

V m

ac

ω φ =

'owever #ax02 I RV m =

'ence2

#ax02#ax

N I R

ac ω φ =

and $ea( va"ue of the tota" f"ux is given b!

2

#ax0

2

#ax#ax

N

"I

N

V m

dcac +=+ω

φ φ

;n $ractice, if this f"ux exceeds the (nee+$oint f"ux in the core see fig.3, then the CT core

wi"" saturate.

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As a conse uence of CT core saturation, the secondar! current wou"d not faithfu""! re$"icate

the $ri#ar! current. ;nfact, in $ractice it is observed that CT secondar! current is c"i$$ed. The

c"i$$ing of CT current "eads to Jb"indingK of the re"a! which cannot function further. 'ence,

CT saturation in $resence of dc offset current is a serious $rob"e# which re"a! designers have

to face. Hote that dc f"ux accu#u"ates gradua""!. @e$ends u$on the trans#ission "ine ti#e

constant τ 3. ;t is a$$arent that saturation shou"d not occur i##ediate"! after the ince$tion of the fau"t. Thus, if the re"a! is fast enough in decision #a(ing, it is "i(e"! that a re"a!ing

decision wou"d be underta(en before the CT fu""! saturates. This is another i#$ortant reason

for increasing the s$eed of re"a!ing s!ste#. )or bus+fau"t $rotection, where the dc saturation

due to dc offset current can be a significant contributing factor, uarter c!c"e o$erations LLLLL

s$ecifica""! are i#$osed. -i#i"ar"!, a distance re"a! is ex$ected to o$erate within M+1 c!c"e

ti#e.

CT oversiDing factors

T!$ica""!, an efficient design of transfor#er wou"d corres$ond to choosing the core cross

section such that acmφ shou"d be near the (nee $oint of +' curve. Nne obvious wa! of

avoiding the CT saturation on dc f"ux is to oversiDe the core so that for f"ux 3 #ax#ax dcac φ φ + , the

corres$onding is be"ow the (nee+$oint. 'ence, the factor #ax

#ax#ax 34

ac

dcac

φ φ φ +

is ca""ed core+

oversiDing factor.

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Core+oversiDing factor #ax

#ax

1ac

dc

φ φ +

20

21 N RI

N "I -

ω +=

R "ω +=1

R # +=1

Hote that O/ in above e uation is the trans#ission "ine O/ ratio. )or a 220P "ine O/ ≈ 10.

This wou"d i#$"! that transfor#er core shou"d be oversiDed b! a factor of 11. )or a 400P

"ine, t!$ica" va"ue of O/ ≈ 20. This wou"d i#$"! an oversiDing re uired of about 21 ti#es the

usua" design. C"ear"! this high a#ount of oversiDing is not $ractica". Thus, the i#$ortant

conc"usion is that, $rotection engineers have to "ive with the saturation $rob"e#.

Cautions in CT se"ectionQ hi"e choosing a CT for a $articu"ar a$$"ication, it is necessar! to observe fo""owing

$recautions.

1. The CT rating and continuous "oad current shou"d #atch. )or exa#$"e, if #ax "oad

current is ?0A, a 100Q5 Ct #a! be acce$tab"e but 50Q5 is not acce$tab"e.

2. The #axi#u# fau"t current shou"d be "ess than 20 ti#es the CT rated current. for

exa#$"e 100Q5 CT can be used, so "ong as burden on the CT G #axi#u# $ri#ar! fau"t

current is be"ow 2000A.

. The vo"tage rating of CT shou"d be co#$atib"e. )or exa#$"e, 100Q5 C100 wou"d give

"inear res$onse, u$to 20 ti#es rated current $rovided CT burden is (e$t

be"ow 100/20L5 1 Ω3. ith 2 Ω burden, this CT can be used on"! if #axi#u# current

is "i#ited to 1000A.

4. Fara""e" of CTEs e.g. in differentia" $rotection, or with -8R fau"t can create significant

errors in CT $erfor#ance. Nne shou"d in genera" ascertain that #agnetiDing current is

(e$t #uch be"ow the $ic( u$ va"ue.

)o""owing exa#$"e, i""ustrates this $oint

Exercise pr-+lems:

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;f the current ratio is ade uate for a $rotection, but CT burden is highI then the

$erfor#ance of CT #a! deteriorate due to "arge #agnetiDing current and/or saturation $rob"e#.

The CT $erfor#ance can be i#$roved b! connecting the CTEs in series.

13 -how the dotted ter#ina"s for correct secondar! series connection

23 hat is the A of CT in fig a3 G b3 res$ective"!&

13 S"ectro#echanica" re"a!s tend to saturate at high currents. This reduces the re"a! burden

on CT, and so that the CT $erfor#ance at #oderate"! high currents #a! be considered

better than at re"a!Es rated burden at 5A.

23 se of instantaneous over current re"a!s has the $otentia" to overco#e this $rob"e# of

saturation of CTEs

3 @ifferentia" $rotection can o$erate on externa" fau"ts due to the un e ua" saturation of

CTEs

8ecture+7Sxa#$"es

7. ;f a 00Q5 c"ass C CT is connected to a #eter with resistance Ω= 1 I R andsecondar! current in the CT is 4.5A find out the $ri#ar! current vo"tagedeve"o$ed across the #eter and % rate error. 8ead wire resistance Ω= 02.0 " Rsecondar! resistance S R of a 00/5 CT Ω= 15.0

@iagra#Ω= 1 I R , Ω= 02.0 " R Ω= 15.0S R A I S 5.4=

Tota" secondar! resistance S " I T R R R R ++= Ω= 16.1-econdar! vo"tage T S R I ×=

16.15.4 ×= V 275.5=

)ro# )ig 5.6,Sxciting current ; S for 5.275

0.0 A

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Turns ratio H 00/5 70 3 E S p I I N I += 70 4.5 0.0 3 261.*A

o"tage across #eter I S R I ×=

14.5 ×= 4.5

atio error 1005.4

0/.0100 ×=×=

S

E

I

I

0.76%+ R

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8ecture 9 *

Sxa#$"es1. @esign a CC T for a 1 2( trans#ission "ine using the fo""owing data.

esistive urden VA1503/ =−φ

() ' /=∆

, $hase ang"e error β 40 #inConsider 4 choices of 2 as ( , 11( , 7.7( and . (@iagra# 1 @iagra# 2Trans#ission "ine vo"tage 1 2( . -u$$ose 2 F H3 be the vo"tage to be

$roduced b! the ca$acitive $otentia" divider with ca$acitance va"ues C 1 and C 2and 8 the va"ue of tuning inductor. The standardiDed T secondar! vo"tage is 110vo"ts 8+83.'ere s$ecification for $hase ang"e error β is 40 #inutes variation in fre uenc!can be u$to () ' /=∆ . Fhase ang"e error for change in ω b! ω ∆ in the abovee uation circuit, is given b!

ω ω

∆+= 312&

"

At tuning fre uenc! "&

12 =ω

-ubstituting 2 1 "& ω =

Fhase ang"e error ω ∆+= 3 " "ω ∆= "2

% $hase ang"e error β + Ra

"2

2 ω ∆= +++ 13

sing this e uation the va"ue 8 for different va"ues of 2 is found out.13 8et 2 be ( 8 + H3

22= 150+

V R

=

Ω×== 52= 10*.216++ Ra R/22 ×Π=∆×Π=∆ ' ω

rad 01174.0701*0

40#in40 =

××Π==β

)ro# e n 13

2210*.21601174.0

2

5=

×Π×××=∆×

×=ω

β + R "

( 2.7622= . .

"& & µ

ω ?

221 1051.11051.11 −− ×=×==+

23 3112 N "/V V −=2

= 411 10 3 242 10150+

R × ×= = × Ω

= 40.01174 242 102 2 2

+ R " β

ω π × × ×= =

× ∆ × ×

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( 2.646=

2.64634/11

2221 ×==+ "& &

ω . µ 210/7.1 −×=

3 /V V 7.72 =2

= 7.7 10 3150+

R × ×=

Ω×= 41012.*6= 40.01174 *6.12 10

2 2 2+ R "

β ω π

× × ×= =× ∆ × ×

,27? ( =

. & & µ 221 1066. −×=+

43 /V V /./2 =2

= . 10 3

150+ R

× ×=

Ω×= 4106*.21= 40.01174 21.6* 10

2 2 2+ R "

β ω π

× × ×= =× ∆ × ×

( 25.76=

1 2 0.151& & . µ + =The va"ues of 8, 21 & & + for different va"ues of 2 are tabu"ated be"ow.

2 8 21 & & +( 7622.2' 0.00151 . µ

11( 646.2' 0.01 7 . µ

7.7( 27?' 0.0 66 . µ . ( 76.25' 0.151 . µ

)ro# the above tab"e it is c"ear that s#a""er the va"ue of 2, the s#a""er is theva"ue of 8 and higher the va"ue of C 1 and C 2 for tuning condition. ;f we se"ect too "owva"ue of 2 and 8 then ca$acitance va"ues wi"" be be!ond avai"ab"e "i#its, and if wese"ect higher va"ue of 2 and 8, then CC T and inductor wi"" beco#e bu"(!. -o aco#$ro#ise is necessar! and "et us se"ect 2 7.7(

)or 2 7.7( 8 27?'

. & & µ 0 66.021 =+

How,1

21

2 & & &

V V +=

1

7

/

/ 100/66.0

107.7/

101/2&

−×=

××

×

7

1 101 210107.70 66.0

×××××=

− . & µ

. µ 00//.0=

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. & µ 0 44.02 =;n this design, we ex$"ained the basic conce$t for CC T design and we assu#edthe transfor#er to be idea". ut in actua" design $ractice the va"ue of #agnetiDingi#$edance of transfor#er, resistance of reactor etc have to be ta(en into account,as ratio error α and $hase ang"e error β wi"" a"so get affected b! these va"ues.

2. @iagra#The e uiva"ent circuit of a CC T is shown in fig *. . The va"ues of C 1 and C 2 are0.001* . µ and 0.01* . µ res$ective"!. Tuning inductor has an inductance of 4?6' and resistance of 4720 Ω .O# of the 7.7( T is 1U Ω , core "oss 20 watts $er $hase, A burden 150 A $er $hase. a"ue of C# for co#$ensating the current drawn b! # is e ua"to . ?101*. −× .

a3 erif! the a$$ro$riateness of choice of 8 and C#.AnsQ ;f . & µ 001*.01 = and . & µ 01*7.02 = then the va"ue 8 of tuning inductor is given b!

31

212 & &

"+

=ω

where 2 ' ω π = and ' tuning fre uenc!

2 7

12 503 0.001* 0.01*73 10

"π −

=× + ×

4?7.6' which is e ua" to the given va"ue of 8. How 71 10m # = × Ω1

mm

# & ω

=

7

1 1

2 503 1 10m

m

& # ω π

= =× × ×

?.1* 10 . −= ×The va"ue is a"so sa#e as the se"ected va"ue of C# 'ence the se"ection of both 8and C# is a$$ro$riate.

b3 )ind out the no#ina" va"ue of / 2

AnsQ 1 22 1

0.001* 0.01*70.001*

& & V V &

+ += =

11.

11. x 7.7

1 2/V =

c3 ;f the fre uenc! dro$s fro# 50'D to 46'D, what wou"d be the va"ues of ratio

error and $hase ang"e error&AnsQ Core "oss 20w

22 20m

V 0

R=

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2 22 77003

20 20mV

R = =

72.1* 10= × ΩA burden 150 A resistive3

22

150+

V R =

2 22 77003

150 150+V

R = =

52.?04 10= × ΩThe e uiva"ent circuit can be re$resented as shown be"ow.

@iagra# *.12710m # = Ωat f 50'D

72 10m ' "π =710

1* .12 50m " ( = =Π ×The fre uenc! of interest is 46'D. 'ence va"ues of O # and other i#$edance can

be ca"cu"ated at 46'D. The above circuit can be si#$"ified as@iagra# *.1

here1 1 1

mm m +

! ! &

Z R # Rω = − + +

?7 5

1 12 46 .1* 10

2.1* 10 2 46 1* .1 2.?04 10 !

! π π

−= − + × × × +× × × ×

7 7 7 70.45? 10 1.074 10 0.?4 10 .44 10 ! !− − − −= × − × + × + ×7 7.?02 0.1243 10 .?04 10 1.*2 ! − −= − × = × − °

7

1.?04 10 1.*2

Z −= × − 257146.5 1.*2= °Ω

25701*. 2 *1 5.15 != + Ω

t%t%

V I

! R ! " Z

& ω

ω

=+ − +

7

7700 0

4720 2 46 4?6 25701*. 2 *1 5.152 46 0.0204 10

! ! !π

π −

= + × × − + +

× × ×

[ ]7700 0

4720 14767*.? 175??4 25701*. 2 *1 5.15 ! ! !=

+ − + +7700 0

2707 *. 2 110*?.*4 !=

−

1*

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7700 0270*64.14 2.44

A= − °

T t%V I Z = ×7700 0

257146.5 1.*2

270*64.14 2.44

= × °

− °74*0.42 4.27= °

'ence % ratio error7700 74*0.423

1007700

−= ×

1.*1%Fhase ang"e error 4.27 °

2 7.7V /V =r

α = 2 2ω α

+

0t t ≥2

2

10

e*

d i R dii

dt " dt "& + = + = 1 2 50nw

"& π = = × 2 n

R Iw

"=

22

2 2 0n nd i

w w idt

α + + = 1α <

8ec 9 *1

2 50ne* "&

ω π = = × 2 n R "

ζω =2

22 2 0n n

d ii

dt ζω ω + + = ζ nω 1ζ <

t e τ − 0 0sin 3: :

m

line

V I t Z

ω φ θ − + − φ θ =

1?

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20/20

20

ta duty

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