Exersice 1 Single phase loadampcompensation

Two loads z1 = 100 and z2 = 10+j20 are connected across a 200 V rms 50 Hz source as shown in

figure

a Find real reactive powers and power factor at the source find total current through the circuit

b Find the capacitance of the capacitor across the loads to improve the overall power factor to 08

lagging

Exersice 2 Power and current calculations balanced and unbalanced systems

In the system given below find Ia Ib Ic and apparent power supplied by the source if

a za = zb = zc = j 10

b za = zb = j10 and zc = j 09 (znnrsquo = j01 )

Exersice 3 3-phase loads

A three-phase line has an impedance of 2+j4 The balanced line feeds two balanced load The first

load is Y connected and has an impedance of 30+j40 The second load is connected and has an

impedance of 60-j45 The line is energized by a three-phase supply having a voltage of 20785 V

Taking voltage of phase ldquoardquo as the reference determine

a The current through reactive and active powers from the supply

n

10+j20 100

I

za 10

j 01

j 01

j 01

1120

1-120

j 01

zc

zb

arsquo

brsquo crsquo

nrsquo

a

b c

Ia

Ib

Ic

b The line voltage at which the loads are connected

c The current per phase through each load

d The total active and reactive powers in each load and the line

e Power factor of combined loads

Exersice 4 Ideal single phase two winding transformer

A single-phase two winding transformer is rated at 20 kVA 480120 V 50 Hz A source connected to the

480 V winding supplies an impedance load connected to the 120 V winding The load absorbs 15 kVA at

08 power factor lagging when the load voltage is 118 V Assume that the transformer is ideal and

calculate

a The voltage across the 480 V winding

b The load impedance

c The load impedance transformed to the 480 V winding

d The real and reactive power supplied to the 480 V winding

(Exersices 5 to 9 are taken from ldquoPower System Analysis amp Design SI Versionrdquo J Duncan Glover Mulukutla

S Sarma and Thomas Overbye Publisher Cengage Learning 2012 Cengage Learning Edition 5th Revised

edition ISBN-10 1111425795 ISBN-13 9781111425791)

Exersice 5 Transformer short circuit and open circuit tests

A single-phase two winding transformer is rated at 20 kVA 480120 V 50 Hz

During a short circuit test where rated frequency is applied to the 480 V winding (winding 1) with the

120 V winding (winding 2) shorted the following readings are obtained V1 = 35 V P1 = 300 W During an

open circuit test where rated voltage is applied to the winding 2 with winding 1 is open the following

readings are obtained I2 = 12 A P2 = 200 W

30+j40

30+j40

30+j40

60-j45

2+j4

2+j4

2+j4

60-j45

60-j45 VL= 20785 V

a From the short circuit test determine the equivalent series impedance Zeq1 referred to the

winding 1 Neglect the shunt admittance

b From the open circuit test determine the values of shunt elements Referred to winding 1

Neglect the series impedance

Exersice 6 Per unit impedance single phase transformer

A single-phase two winding transformer is rated at 20 kVA 480120 V 50 Hz The equivalent leakage

impedance of the transformer referred to the 120 V winding (winding 2)is Zeq2 = 005257813 Using

the transformer ratings as base values determine the per unit leakage impedance referred to winding 2

and winding 1

Exersice 7 Per unit circuit three zone single phase network

Three zones of a single-phase circuit are identified in Figure The zones are connected by transformers T1 and T2 whose ratings are also shown Using base values of 30 kVA and 240 volts in zone 1 draw the per-unit circuit and determine the per-unit impedances and the per-unit source voltage Then calculate the load current both in per-unit and in amperes Transformer winding resistances and shunt admittance branches are neglected

Exersice 8 Per unit and actual currents in balanced three phase network

A balanced Y-connected voltage source with V = 480 0 volts is applied to a balanced load with

Zload = 3040 The line impedance between the source and load is Zline = 185 for each phase

Calculate the per-unit and actual current in phase a of the line using Sbase3 = 10 kVA and VbaseLL = 480 volts

Exersice 9 Voltage calculations balanced Y-Y and -Y connection Three single-phase two-winding transformers each rated 400 MVA 1381992 kV with leakage reactance Xeq = 010 per unit are connected to form a three-phase bank Winding resistances and exciting current are neglected The high-voltage windings are connected in Y A three-phase load operating under balanced positive-sequence conditions on the high-voltage side absorbs 1000 MVA at 090 pf lagging with VAN = 1992 kV Determine the voltage Van at the low-voltage bus if the low-voltage windings are connected

a in Y

b in

V1( 4 0 -=- - -Vz- 120

0 V2 == ~~ V lrcfefeACe)

V1 cmiddot 12 ~ 4-118 = lft2-L Dmiddot

-l 1 2 12112 L-Mmiddot5Sr ( nCgtvc+-ie )

-- -t12s ~ - o9 2~3 ( 36 8~ 0 Q

1 2-1-12 L-3poundamp1- 0 shy

2 l =- ~ ) 1 t=-loacl r_fJ lJ-2 CO 9 z~Ss L3b lrr ) Cbull loOd = c -z-lod Nz

I 2-b(Ad =- 14--~S L3b 810 SL

J $-1 S2 =- 5000 L3G ltis 1- = 12-00 -+ J9000

1 t f ~ uJ vAr

20lD 3

_ 41 661- A (A I =~e~

Ltlt6 0VCA1ed

-

Xe ~ ce11 _ rLe1L =0 ~2-2 f2shy

~4 _Y1_ rro-eJ

Po -==shy 200W

1152 -iZ

(tore loss) 1

3amp0 _ ~ ~ =- ~ ~ Po 200

1 v~ lampoL _ -

r2c 1152

~ Lj~0 - j -1 b ~ 5 b -L

2 91

--Zlif 211middot50

2shy _ J~e 1201

pound_JaQ(se 2 - 2 - = 0 1-2- J2 S bOISE 2010 3

per -lJl+ leCAIpoundCje ~fVpe ~tc e 2 r lA) - 2eq 2 0 0 5 L S L1-8 1~ 0 0 ot 2 C1 L-=t 8 13

0

eqz --t - - -=- J f V

2 bose 2 Omiddott 2

(r-cf~rrecl -0 wilctAI9 2

2 - Vootse1

~ loose

per-lAlr+- l-ealLo3E Nfe~fce

~1 (pV teCj1 _ Of 1-i5j3deg -= 00+29 1+6 1t 11 middot52

- We clAos-e ~ loose b l~es os 0t-1(9 Joltamp-~s of +--e ~fstorVI ershy

b The line voltage at which the loads are connected

c The current per phase through each load

d The total active and reactive powers in each load and the line

e Power factor of combined loads

Exersice 4 Ideal single phase two winding transformer

A single-phase two winding transformer is rated at 20 kVA 480120 V 50 Hz A source connected to the

480 V winding supplies an impedance load connected to the 120 V winding The load absorbs 15 kVA at

08 power factor lagging when the load voltage is 118 V Assume that the transformer is ideal and

calculate

a The voltage across the 480 V winding

b The load impedance

c The load impedance transformed to the 480 V winding

d The real and reactive power supplied to the 480 V winding

(Exersices 5 to 9 are taken from ldquoPower System Analysis amp Design SI Versionrdquo J Duncan Glover Mulukutla

S Sarma and Thomas Overbye Publisher Cengage Learning 2012 Cengage Learning Edition 5th Revised

edition ISBN-10 1111425795 ISBN-13 9781111425791)

Exersice 5 Transformer short circuit and open circuit tests

A single-phase two winding transformer is rated at 20 kVA 480120 V 50 Hz

During a short circuit test where rated frequency is applied to the 480 V winding (winding 1) with the

120 V winding (winding 2) shorted the following readings are obtained V1 = 35 V P1 = 300 W During an

open circuit test where rated voltage is applied to the winding 2 with winding 1 is open the following

readings are obtained I2 = 12 A P2 = 200 W

30+j40

30+j40

30+j40

60-j45

2+j4

2+j4

2+j4

60-j45

60-j45 VL= 20785 V

a From the short circuit test determine the equivalent series impedance Zeq1 referred to the

winding 1 Neglect the shunt admittance

b From the open circuit test determine the values of shunt elements Referred to winding 1

Neglect the series impedance

Exersice 6 Per unit impedance single phase transformer

A single-phase two winding transformer is rated at 20 kVA 480120 V 50 Hz The equivalent leakage

impedance of the transformer referred to the 120 V winding (winding 2)is Zeq2 = 005257813 Using

the transformer ratings as base values determine the per unit leakage impedance referred to winding 2

and winding 1

Exersice 7 Per unit circuit three zone single phase network

Three zones of a single-phase circuit are identified in Figure The zones are connected by transformers T1 and T2 whose ratings are also shown Using base values of 30 kVA and 240 volts in zone 1 draw the per-unit circuit and determine the per-unit impedances and the per-unit source voltage Then calculate the load current both in per-unit and in amperes Transformer winding resistances and shunt admittance branches are neglected

Exersice 8 Per unit and actual currents in balanced three phase network

A balanced Y-connected voltage source with V = 480 0 volts is applied to a balanced load with

Zload = 3040 The line impedance between the source and load is Zline = 185 for each phase

Calculate the per-unit and actual current in phase a of the line using Sbase3 = 10 kVA and VbaseLL = 480 volts

Exersice 9 Voltage calculations balanced Y-Y and -Y connection Three single-phase two-winding transformers each rated 400 MVA 1381992 kV with leakage reactance Xeq = 010 per unit are connected to form a three-phase bank Winding resistances and exciting current are neglected The high-voltage windings are connected in Y A three-phase load operating under balanced positive-sequence conditions on the high-voltage side absorbs 1000 MVA at 090 pf lagging with VAN = 1992 kV Determine the voltage Van at the low-voltage bus if the low-voltage windings are connected

a in Y

b in

V1( 4 0 -=- - -Vz- 120

0 V2 == ~~ V lrcfefeACe)

V1 cmiddot 12 ~ 4-118 = lft2-L Dmiddot

-l 1 2 12112 L-Mmiddot5Sr ( nCgtvc+-ie )

-- -t12s ~ - o9 2~3 ( 36 8~ 0 Q

1 2-1-12 L-3poundamp1- 0 shy

2 l =- ~ ) 1 t=-loacl r_fJ lJ-2 CO 9 z~Ss L3b lrr ) Cbull loOd = c -z-lod Nz

I 2-b(Ad =- 14--~S L3b 810 SL

J $-1 S2 =- 5000 L3G ltis 1- = 12-00 -+ J9000

1 t f ~ uJ vAr

20lD 3

_ 41 661- A (A I =~e~

Ltlt6 0VCA1ed

-

Xe ~ ce11 _ rLe1L =0 ~2-2 f2shy

~4 _Y1_ rro-eJ

Po -==shy 200W

1152 -iZ

(tore loss) 1

3amp0 _ ~ ~ =- ~ ~ Po 200

1 v~ lampoL _ -

r2c 1152

~ Lj~0 - j -1 b ~ 5 b -L

2 91

--Zlif 211middot50

2shy _ J~e 1201

pound_JaQ(se 2 - 2 - = 0 1-2- J2 S bOISE 2010 3

per -lJl+ leCAIpoundCje ~fVpe ~tc e 2 r lA) - 2eq 2 0 0 5 L S L1-8 1~ 0 0 ot 2 C1 L-=t 8 13

0

eqz --t - - -=- J f V

2 bose 2 Omiddott 2

(r-cf~rrecl -0 wilctAI9 2

2 - Vootse1

~ loose

per-lAlr+- l-ealLo3E Nfe~fce

~1 (pV teCj1 _ Of 1-i5j3deg -= 00+29 1+6 1t 11 middot52

- We clAos-e ~ loose b l~es os 0t-1(9 Joltamp-~s of +--e ~fstorVI ershy

a From the short circuit test determine the equivalent series impedance Zeq1 referred to the

winding 1 Neglect the shunt admittance

b From the open circuit test determine the values of shunt elements Referred to winding 1

Neglect the series impedance

Exersice 6 Per unit impedance single phase transformer

A single-phase two winding transformer is rated at 20 kVA 480120 V 50 Hz The equivalent leakage

impedance of the transformer referred to the 120 V winding (winding 2)is Zeq2 = 005257813 Using

the transformer ratings as base values determine the per unit leakage impedance referred to winding 2

and winding 1

Exersice 7 Per unit circuit three zone single phase network

Three zones of a single-phase circuit are identified in Figure The zones are connected by transformers T1 and T2 whose ratings are also shown Using base values of 30 kVA and 240 volts in zone 1 draw the per-unit circuit and determine the per-unit impedances and the per-unit source voltage Then calculate the load current both in per-unit and in amperes Transformer winding resistances and shunt admittance branches are neglected

Exersice 8 Per unit and actual currents in balanced three phase network

A balanced Y-connected voltage source with V = 480 0 volts is applied to a balanced load with

Zload = 3040 The line impedance between the source and load is Zline = 185 for each phase

Calculate the per-unit and actual current in phase a of the line using Sbase3 = 10 kVA and VbaseLL = 480 volts

Exersice 9 Voltage calculations balanced Y-Y and -Y connection Three single-phase two-winding transformers each rated 400 MVA 1381992 kV with leakage reactance Xeq = 010 per unit are connected to form a three-phase bank Winding resistances and exciting current are neglected The high-voltage windings are connected in Y A three-phase load operating under balanced positive-sequence conditions on the high-voltage side absorbs 1000 MVA at 090 pf lagging with VAN = 1992 kV Determine the voltage Van at the low-voltage bus if the low-voltage windings are connected

a in Y

b in

V1( 4 0 -=- - -Vz- 120

0 V2 == ~~ V lrcfefeACe)

V1 cmiddot 12 ~ 4-118 = lft2-L Dmiddot

-l 1 2 12112 L-Mmiddot5Sr ( nCgtvc+-ie )

-- -t12s ~ - o9 2~3 ( 36 8~ 0 Q

1 2-1-12 L-3poundamp1- 0 shy

2 l =- ~ ) 1 t=-loacl r_fJ lJ-2 CO 9 z~Ss L3b lrr ) Cbull loOd = c -z-lod Nz

I 2-b(Ad =- 14--~S L3b 810 SL

J $-1 S2 =- 5000 L3G ltis 1- = 12-00 -+ J9000

1 t f ~ uJ vAr

20lD 3

_ 41 661- A (A I =~e~

Ltlt6 0VCA1ed

-

Xe ~ ce11 _ rLe1L =0 ~2-2 f2shy

~4 _Y1_ rro-eJ

Po -==shy 200W

1152 -iZ

(tore loss) 1

3amp0 _ ~ ~ =- ~ ~ Po 200

1 v~ lampoL _ -

r2c 1152

~ Lj~0 - j -1 b ~ 5 b -L

2 91

--Zlif 211middot50

2shy _ J~e 1201

pound_JaQ(se 2 - 2 - = 0 1-2- J2 S bOISE 2010 3

per -lJl+ leCAIpoundCje ~fVpe ~tc e 2 r lA) - 2eq 2 0 0 5 L S L1-8 1~ 0 0 ot 2 C1 L-=t 8 13

0

eqz --t - - -=- J f V

2 bose 2 Omiddott 2

(r-cf~rrecl -0 wilctAI9 2

2 - Vootse1

~ loose

per-lAlr+- l-ealLo3E Nfe~fce

~1 (pV teCj1 _ Of 1-i5j3deg -= 00+29 1+6 1t 11 middot52

- We clAos-e ~ loose b l~es os 0t-1(9 Joltamp-~s of +--e ~fstorVI ershy

V1( 4 0 -=- - -Vz- 120

0 V2 == ~~ V lrcfefeACe)

V1 cmiddot 12 ~ 4-118 = lft2-L Dmiddot

-l 1 2 12112 L-Mmiddot5Sr ( nCgtvc+-ie )

-- -t12s ~ - o9 2~3 ( 36 8~ 0 Q

1 2-1-12 L-3poundamp1- 0 shy

2 l =- ~ ) 1 t=-loacl r_fJ lJ-2 CO 9 z~Ss L3b lrr ) Cbull loOd = c -z-lod Nz

I 2-b(Ad =- 14--~S L3b 810 SL

J $-1 S2 =- 5000 L3G ltis 1- = 12-00 -+ J9000

1 t f ~ uJ vAr

20lD 3

_ 41 661- A (A I =~e~

Ltlt6 0VCA1ed

-

Xe ~ ce11 _ rLe1L =0 ~2-2 f2shy

~4 _Y1_ rro-eJ

Po -==shy 200W

1152 -iZ

(tore loss) 1

3amp0 _ ~ ~ =- ~ ~ Po 200

1 v~ lampoL _ -

r2c 1152

~ Lj~0 - j -1 b ~ 5 b -L

2 91

--Zlif 211middot50

2shy _ J~e 1201

pound_JaQ(se 2 - 2 - = 0 1-2- J2 S bOISE 2010 3

per -lJl+ leCAIpoundCje ~fVpe ~tc e 2 r lA) - 2eq 2 0 0 5 L S L1-8 1~ 0 0 ot 2 C1 L-=t 8 13

0

eqz --t - - -=- J f V

2 bose 2 Omiddott 2

(r-cf~rrecl -0 wilctAI9 2

2 - Vootse1

~ loose

per-lAlr+- l-ealLo3E Nfe~fce

~1 (pV teCj1 _ Of 1-i5j3deg -= 00+29 1+6 1t 11 middot52

- We clAos-e ~ loose b l~es os 0t-1(9 Joltamp-~s of +--e ~fstorVI ershy

20lD 3

_ 41 661- A (A I =~e~

Ltlt6 0VCA1ed

-

Xe ~ ce11 _ rLe1L =0 ~2-2 f2shy

~4 _Y1_ rro-eJ

Po -==shy 200W

1152 -iZ

(tore loss) 1

3amp0 _ ~ ~ =- ~ ~ Po 200

1 v~ lampoL _ -

r2c 1152

~ Lj~0 - j -1 b ~ 5 b -L

2 91

--Zlif 211middot50

2shy _ J~e 1201

pound_JaQ(se 2 - 2 - = 0 1-2- J2 S bOISE 2010 3

per -lJl+ leCAIpoundCje ~fVpe ~tc e 2 r lA) - 2eq 2 0 0 5 L S L1-8 1~ 0 0 ot 2 C1 L-=t 8 13

0

eqz --t - - -=- J f V

2 bose 2 Omiddott 2

(r-cf~rrecl -0 wilctAI9 2

2 - Vootse1

~ loose

per-lAlr+- l-ealLo3E Nfe~fce

~1 (pV teCj1 _ Of 1-i5j3deg -= 00+29 1+6 1t 11 middot52

- We clAos-e ~ loose b l~es os 0t-1(9 Joltamp-~s of +--e ~fstorVI ershy

1 v~ lampoL _ -

r2c 1152

~ Lj~0 - j -1 b ~ 5 b -L

2 91

--Zlif 211middot50

2shy _ J~e 1201

pound_JaQ(se 2 - 2 - = 0 1-2- J2 S bOISE 2010 3

per -lJl+ leCAIpoundCje ~fVpe ~tc e 2 r lA) - 2eq 2 0 0 5 L S L1-8 1~ 0 0 ot 2 C1 L-=t 8 13

0

eqz --t - - -=- J f V

2 bose 2 Omiddott 2

(r-cf~rrecl -0 wilctAI9 2

2 - Vootse1

~ loose

per-lAlr+- l-ealLo3E Nfe~fce

~1 (pV teCj1 _ Of 1-i5j3deg -= 00+29 1+6 1t 11 middot52

- We clAos-e ~ loose b l~es os 0t-1(9 Joltamp-~s of +--e ~fstorVI ershy

2shy _ J~e 1201

pound_JaQ(se 2 - 2 - = 0 1-2- J2 S bOISE 2010 3

per -lJl+ leCAIpoundCje ~fVpe ~tc e 2 r lA) - 2eq 2 0 0 5 L S L1-8 1~ 0 0 ot 2 C1 L-=t 8 13

0

eqz --t - - -=- J f V

2 bose 2 Omiddott 2

(r-cf~rrecl -0 wilctAI9 2

2 - Vootse1

~ loose

per-lAlr+- l-ealLo3E Nfe~fce

~1 (pV teCj1 _ Of 1-i5j3deg -= 00+29 1+6 1t 11 middot52

- We clAos-e ~ loose b l~es os 0t-1(9 Joltamp-~s of +--e ~fstorVI ershy

--

--

-- --12DVrDt1edZ

Vl Vz

N1 x V6otse 2 VbOtSe 2 N2

SbOse bose 2

middot vJ~ O~ ore CDMfOle+ ~~ cOMlclered we use fIlNltefb+-e

fOt1(3S OS bCSE vOuts

W~ severOI compone(1s orp i(lVD Jed i-Vle S8sreM base

vOtues ~j be clAfferw+ floM the VepLote rcx-Hr9S

~ (Egt0VEiCT ptA VoJJJes fom Its MYYgtEf tote roHIl9S i-c-l-he

Sjs+eM Ioose I

middot vJ~ O~ ore CDMfOle+ ~~ cOMlclered we use fIlNltefb+-e

fOt1(3S OS bCSE vOuts

W~ severOI compone(1s orp i(lVD Jed i-Vle S8sreM base

vOtues ~j be clAfferw+ floM the VepLote rcx-Hr9S

~ (Egt0VEiCT ptA VoJJJes fom Its MYYgtEf tote roHIl9S i-c-l-he

Sjs+eM Ioose I

So~+iol

-

L I 2shy middot3 o

Vs =-2202

rB---3 I I~ xIle=Zfl -~-----lt

clood O9kJO2 Q3 (

(~~ - T2 -shy-- Ttl

2o~vA30~vA 4bO115 24D49O V )lt9 ~O1 ptA)lt~ = 0 pA

l eec IS (EBlec+ed)

D Iv cle lAe

reg 1le-i-erV e ~e voLve ( eOtC~ =2L)(e

S = 30 iL vA (fer ~re fie+vJu k ) b~e

3ofe1

VbGSe 1 == 24D V

i + ~e2

2lt0 (~80 ) -=- q80 V Vbosez 2-qo

3D IDsS60Seurobase3 -250 A V6CSe3 120

shy

I

- -

CegtcA O+e fE -Vftt r()pedovlce~

for T t1 ~ bse vOLues =rotrltjS bf ~e T 1

~ feq = KT1 -pA

t=o euro ~

1-- (e 2shy~Ife pLA - - - = 0 2 b 04 pe r v(Ii

z6(XSe2 1 middot b~

~ ~lDOd o ~+J 0 24-0 tOCXd ~ ~hXA~ pIA - - =18tS+j O 416-=J- pL

-tbOlSe 5 0 middot48

t-O( sOlK (euro ~ iPV shy

--~------------------------~eC pft j lD-1+0 2604 +O 13=i8~O 4 1pound t ) +1 gi-5

_ o 91pound1- LJL -= o - 3 g 1-26010 208~ f 2bO to

TI _I II _J TI_~ _ --(C)1395-26 010) 2bO - OCia = OQCiPlA - -~e 6 - -

~ cleO -ttofStOMel U)i()ciA(9S ~~ loeeA ciiMffOted b~

p roper ~elec+fDn of bOse JClues

v=- VLA CDnnect-ed V==-VLL~~~480LZ V

LoOd ~ ()flfeurocteci z =- 2-6 ~ 30 (400 J2

Z Irfe =1 25 0 SL I 1 __

- PVIVISe

Vbotsf2L

Vbose =- Yis 0 (llne +0 III(e )

-SbOlse

LoOd

6 CD(()ec+eJ -----~ convert +0 ) 2- 2-d a

3- 10 40_ nA =

6 ItIeuro _ 1 I 85 i) _ ltgt

=0 043 40 186 re lAnt~6se 23 04

10 L 400 =0 434 40 C)

fee- v(1shy2 3 Dtt

tn or clec- to do +lAe colClAOIt-Ivf D(le-pkO(se equIvotleuroltl

WE So~D LA Qj AS e LIle +e (leLA+-rod va H-0I3e

- -

CegtcA O+e fE -Vftt r()pedovlce~

for T t1 ~ bse vOLues =rotrltjS bf ~e T 1

~ feq = KT1 -pA

t=o euro ~

1-- (e 2shy~Ife pLA - - - = 0 2 b 04 pe r v(Ii

z6(XSe2 1 middot b~

~ ~lDOd o ~+J 0 24-0 tOCXd ~ ~hXA~ pIA - - =18tS+j O 416-=J- pL

-tbOlSe 5 0 middot48

t-O( sOlK (euro ~ iPV shy

--~------------------------~eC pft j lD-1+0 2604 +O 13=i8~O 4 1pound t ) +1 gi-5

_ o 91pound1- LJL -= o - 3 g 1-26010 208~ f 2bO to

TI _I II _J TI_~ _ --(C)1395-26 010) 2bO - OCia = OQCiPlA - -~e 6 - -

~ cleO -ttofStOMel U)i()ciA(9S ~~ loeeA ciiMffOted b~

p roper ~elec+fDn of bOse JClues

v=- VLA CDnnect-ed V==-VLL~~~480LZ V

LoOd ~ ()flfeurocteci z =- 2-6 ~ 30 (400 J2

Z Irfe =1 25 0 SL I 1 __

- PVIVISe

Vbotsf2L

Vbose =- Yis 0 (llne +0 III(e )

-SbOlse

LoOd

6 CD(()ec+eJ -----~ convert +0 ) 2- 2-d a

3- 10 40_ nA =

6 ItIeuro _ 1 I 85 i) _ ltgt

=0 043 40 186 re lAnt~6se 23 04

10 L 400 =0 434 40 C)

fee- v(1shy2 3 Dtt

tn or clec- to do +lAe colClAOIt-Ivf D(le-pkO(se equIvotleuroltl

WE So~D LA Qj AS e LIle +e (leLA+-rod va H-0I3e

TI _I II _J TI_~ _ --(C)1395-26 010) 2bO - OCia = OQCiPlA - -~e 6 - -

~ cleO -ttofStOMel U)i()ciA(9S ~~ loeeA ciiMffOted b~

p roper ~elec+fDn of bOse JClues

v=- VLA CDnnect-ed V==-VLL~~~480LZ V

LoOd ~ ()flfeurocteci z =- 2-6 ~ 30 (400 J2

Z Irfe =1 25 0 SL I 1 __

- PVIVISe

Vbotsf2L

Vbose =- Yis 0 (llne +0 III(e )

-SbOlse

LoOd

6 CD(()ec+eJ -----~ convert +0 ) 2- 2-d a

3- 10 40_ nA =

6 ItIeuro _ 1 I 85 i) _ ltgt

=0 043 40 186 re lAnt~6se 23 04

10 L 400 =0 434 40 C)

fee- v(1shy2 3 Dtt

tn or clec- to do +lAe colClAOIt-Ivf D(le-pkO(se equIvotleuroltl

WE So~D LA Qj AS e LIle +e (leLA+-rod va H-0I3e

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