ies conventional electrical engineering 2011

8/16/2019 IES Conventional Electrical Engineering 2011

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p ro p e r t ie s

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b)

I< xp

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llow

ing :

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o y n t

in g

v ec t

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n d

i t s

s i

gn i f

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c e

i i ]

L o

ss ta

n g e

n t o

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le c tr

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in w

ave

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i i i) I

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im

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m e

d iu m

1

{c)

W

h a

t i s a

d is

to r t

io n l

e s s

l

i ne?

H o

w to

a c

h ie v

e d i

s to r

t io n

le s s

c o

n d i t

io n

o n

th e

l in e

?

D

eriv

e th

e n

e c e

s sa r

y e q

u a t i

o n s

. 1

2 .

a)

S h o

w th

e e

le c t r

ic a l

c o

n n e

c t io

n

d ia

g ra m

a n d

m

o d

e l

th e

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re

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d

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form

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ssu

m e

th e

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y

v a

r iab

le s

a n d

o b

ta i

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t r an

s fe

r

fu

n c t

io n

fo r

c h

a n g

e

in

p

o s i

t ion

o

f a r

m a

tu re

to

th e c

h a n

g e

in

a rm

a tu

r e v

o lta

ge .

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p re s

s

t h e t r

a n s

fe r

fu n c t io

n

in

s ta n

d a r

d

f

o rm

.

1

2

b) F

o r

m e

s y

s te

m

r e p

re s

e n t e

d b y a

b

lo c k

d

ia g

ra m

s

h o w

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elow

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e v a l

u a t

e t

h e

c l

o sed

- loo

p

t r a

n s f

e r fu n

c t io

n .

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c u la

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p in

g fa

c to r

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f o sc i

l la t i

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b )

i} A

s y s t e m

Ls r e p r e s e n t e d

b y t h e

fo l low ing t r a n s f e r

f u n c t i o n m o d e l

G(s) C (s)

= .

s 5 .

R(s) {s

•

1 (s

2) s 3)

O b t a i n t h e

s t a te s p a c e

m o d e l fo r

t h i s s y s t e m s u c h

th t

t h e s y s te m

m a t r i x

is

i n

d i a g o n a l i z e d

f o r m .

T h e c h o i c e

o f s t a t e

v a r i a b l e s

n e e d s

to b e c l e a r l y

i n d i c a t e d .

5

i i) A s y s t e m i s m o d e l l e d

b y

s t a te s p a c e

m o d e l a s

Y = [ l OJ

E v a l u a t e t h e

s t a t e

t r a n s i t i o n

m a t r i x a n d

t h e

a u t o n o m o u s

r e s p o n s e o f

t h e

s y s t e m

w i t h

i n i t i a l

c o n d i t i o n x O) = {1 2lT- 5

{c)

D r a w

t h e

s c h e m a t ic d i a g ra m

o f

a

2 p h a s e

s e r v o m o t o r

a n d

d r a w

t h e

t o rq u e s p e e d

c h a r a c t e r i s t i c .

W h a t c a r e

i s t a k e n to

o b t a i n l i n e a r

c h a r a c t e r i s t i c ?

D e r i v e t h e l i n e a r i z e d

t r a n s f e r

f u n c t i o n

u n d e r

l o a d c o n d i ti o n .

Part B

4 a)

G i v e

t h e p r o p e r t i e s

o f s u p e r c o n d u c t o rs .

1 0

W h a t a r e

t y p e

I a n d t y p e

II

s u p e r -

c o n d u c t o r s ?

I n d i c a t e th e i r b e h a v i o u r

w i t h

r e s p e c t to a p p l i e d

m a g n e t i c f ie ld .

1 2

D R S R L

RRA/69

4

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5/24

b) Expla in t h e H al l

effect

in s e m i

c o n d u c t o r s a n d d e f in e Hal l c o n s t a n t .

W h a t

d o

y o u

m e a n b y n e g a t i v e Hal l

c o n s t a n t ?

c)

E x p l a i n t h e

fo l lowing :

i) C o m p l e x

die lec t r ic

c o n s t a n t

ii) C e r a m i c s a s i n s u l a t i n g

m a t e r i a l s

5 .

a)

N a m e t h e

b a s i c

po la r i za t ion

m e c h a n i s m s

w h i c h

o ccu r i n a

dielect r ic .

A

die lec t r i c

mate r i a l c o n t a i n s 2 x 1 0

1 9

p o l a r

molecu le s f

m

3

, e a c h o f d ipo le m o m e n t

1 · 8 x to -

2 7

C - m .

A s s u m i n g t h a t

al l

t h e

d ipo le s

a r e

a l i g n e d in

t h e d i r ec t ion

s-

o f e lec t r ic fiel = 1 0 x V / m f ind

--

P a n d

E r

b) W i t h r e s p ec t t o m a g n e t ic b e h a v i o u r

give t h e c lass i f ica t ion o f m a g n e t i c

ma te r i a l s . D i s c u s s gene ra l e lec t r i c a n d

m a g n e t i c

charac te r i s t i c s

o f

fe r r i t e s a n d

t he i r appl ica t ions .

c) D esc r ibe

in

de ta i l

t h e

t h e r m a l

b r e a k

d o w n

o f

so l id d ie lec t r i c s u n d e r

app l i ca t i on o f

a l t e r n a t i n g a s wel l

a s

1 2

12

1 0

1 6

di rec t ·

vol tages .

1 0

D R S R L R R A /6 9

5

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P .T .O.

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6/24

P

a

rt

e

6

{a )

F

in

d ~ h e

u

r r

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x

w h

ic

h f l

ow

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t

h r

o u

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i l i

e 3

n

re s

is t

o r

in

t h

e

c

i rc

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t o

i

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e

b)

f

igu

re

b

e lo

w

·

12

1 0

.

V

In

i l ie c i rc u i t o f

sw

i t

c h

S

h

a s

be

en

lo

ng

t im

e

;

t h e

f igu re

be low

i n

po

s i t

io

n I

to

r

a

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l

O

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T

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in

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t io

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e

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c

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p

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t

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s i t

io

n

2 .

{ i

i) H

ow

to

ng

d

oe

s i t

t a

k e

in

s e

co

n d

s

or

~ h e

t r

a n

s i e

n t

t

o d i

s a

p p

e a

r

cu

r r

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to

d

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ay

w i

ili

in

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te

rm

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e

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tag

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w

h

ich

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p e

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a

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c

ap

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te

.

12

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R A

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c)

A

s i n g le - p h a s e m o t o r t a k e s 5 0

A a t a

p. .

o f

0 ·5 l ag f r o m

a

2 5 0

V, 5 0 H z

su p p ly .

W h a t

m u s t

b e t h e v a l u e o f t h e s h u n t i n g

c a p a c i t o r

to n u s e t h e overa l l p.f .

to

0 · 9

l a g ?

H o w

d o e s

t h e

c a p a c i t o r

a fec t

t h e l i n e a n d m o t o r c u r r e n t s ? 1 2

7 .

a)

A

f ixed

c a p a c i t a n c e ( Xc

= 2 0 o h m s ) 1s

p l a c e d

in pa r a l l e l

w i t h a se r i e s

c o m b i n a t i o n o f r e s is t a n c e R

=

8 o h m s }

a n d

va r i ab le

i n d u c t a n c e XL

o h m s ) ,

h a v i n g negl ig ib le r e s i s t ance . A n

a l t e r n a t i n g

vol tage

o f 1 2 0 v o l t s i s

app l i ed

a c r o s s t h e p a ra l l e l c o m b in a t io n .

S h o w t h a t t h e v a l u e o f X L w h i c h wil l

p r o d u c e

u n i t y p o w e r f a c t o r r e s o n a n c e

i s g iven

b y

X L = X c

±

X ~ R 2

2 v 4

D e t e r m i n e t h e

m i n i m u m

v a l u e

o f t h e

c u r r e n t

d r a w n

f r o m

t h e

supp ly .

1 2

b)

A c i r cu i t h a s

t h e c o n f i g u r a ti o n d e p i c t e d

i n t h e f igure b e lo w

:

e t ) ~

1 4 1

s i n 4 0 0 t

L = O · l H

C = 2 5 J L F

i)

F i n d t h e e q u i v a l e n t

i m p e d a n c e

a p p e a r i n g

to

t h e

r i g h t o f p o i n t s ab

S R l r R R A / 6 9

I P.T .O.

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8/24

i i j

D c t e r r n i n e t h e

v a l u e

o f

n ; a c t a n c e w h i c h make '

s o u r c e

< Urrent i n

p h a s e

w t t h

s o u r c e

vo l t age .

t h e

t h e

t h e

i i i j S h o u l d t h e

r e a c t a n c e

X o f p a r t iiJ

b e i nduc- t ive o r c a p a c i t i v e ?

F i n d t h e

r e q u i r e d v a l u e

o f o r

C.

{ iu j C o m p u t < · t h e effec· t ive

v a l u e o f

t h e

sou1

noe c u r r e n t

for t h e

c o n d i t i o n

d e s c r i b e d in

p a r t

i i). 1 2

c) F i n d ABCD p a r a m e t e r s fo r t h e

two· ·por t

8 . a)

n e t w o r k s h o w n in t h e f igu re b e l o w

12

Part D

E.;xplain, w i t h a diagrru :n , h o w W i e n ·s

b r i d g e

c a n

b e

u s e d

for

e x p e r i m e n t a l

d e t e r m i n a t i o n o f f r e q u e n c y D e r i v e

t h e

e x p r e s s i o n

for f r e q u e n c y in t e r m s

o f

b r i d g e p a r a m e t e r s . 1

D R S R L R R A / 69 8

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9/24

b)

T h e p o w e r fac to r

o f a

c i r cu i t 1s

d e t e r m i n e d b y c o s < > = f VI

w h e r e

1s

t h e p o w e r

in

w a t t , V t h e vo l t age i n

vo l t

a n d

i s

t h e c u r r e n t i n

a m p e r e .

T h e

re la t ive

e r r o r s i n

p o w e r ,

c u r r e n t a n d

vo l t age

a r e re spec t ive ly

± 0 · 5°/o,

1

o o

a n d

+lo/o.

C a l c u l a t e

t h e re la t ive e r r o r

in

p o w e r

fac tor . Also c a l c u l a t e

t h e u n c e r ta i n t y in

p o w e r f ac to r i t h e

e r r o r s w e r e spec i f i ed

a s u n c e r t a i n t i e s .

c)

A n

elec t r ic

ut i l i ty

su p p l i e s

p o w e r

to

M W

l o a d

a t

p o w e r f a c t o r p.f.} 0 · 8 5

a n d

a t

11 kV. T h e u t i l i ty

w a n t s

to m e a s u r e

t h e

vol tage ,

c u r r e n t a n d p o w e r

f a c to r

c o n t i n u o u s l y

u s i n g 2 5 0

V

v o l tm e te r ,

1 0

A

a m m e t e r a n d 2 5 0

V ,

1 0

A

p o w e r

f a c to r m e t e r .

D r a w

c i r cu i t

d i a g r a m

o f

t h e s c h e m e .

D i s c u s s ,

w h y

t h e

p.f .

m e t e r d o e s

n o t

c o m e

b a c k to z e r o r e a d i n g l ike

v o l t m e t e r

a n d a m m e t e r

a f t e r

d i s c o n n e c t i n g t h e

1 0

s u p p l y to i t . 1 0

9 .

a) D e s c r i b e t h e pr inc ip le

o f

f r e q u e n c y

m e a s u re m e n t u s in g

dig i ta l

t e c h n i q u e .

D r a w

i t s

b l o c k d i a g r a m .

T h e u n k n o w n

i n p u t s i g n a l

o f 2 V

s q u a r e w a v e i s

o f

3 ·5 k Hz . D e t e r m i n e t h e d i s p l a y

ind ica t ion i f

t h e

g a t e

e n a b l e

t i m e

1s

i) 0 ·1 s e c o n d ,

i i) s e c o n d a n d

{iii)

1 0

s e c o n d s . 1 0


9

I

P.T.O.

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10/24

f j

S u g g e s t nega t ive

t e m p e r a t u r e

coeffictent: dev ice for

t h e

m e a s u r e m e n t

o f

t e m p e r a t u r e 0

C - 1 5 0

~ C .

D e sc r ib e

t h e

r e , ; i s t ance - t cmpera tu re

c h a r a c

t e r i s t ic .

vol tage--current

charac te r i s t i c

a n d

cu r r en t - t i m e

charac te r i s t i c

o f

t h e r m i s t o r s . D r a w a c i rcu i t for

m e a s u r e m e n t

o f

t e m p e r a t u r e

u s i n g

t he rmi s to r . The o u t p u t r e l a t ion

for

a

ther r r l i s tor

t r a n s d u c e r

i s

g iven b y

l l t L . l . l

R

R o e T 1o

F o r T

· · 3 0 0 K. I} = 3 4 2 0 , R

' ·

1

k.Q

a n d

R

2

k " l , ca l cu l a t e T.

1

c)

D r a w t h e

l ine d i a g r a m

o f

d a t a

acqu i s i t i on s y s t e m f rom p r o ces s p lan t t o

c o m p u t e r s y s t e m . E x p l a i n t h e f unc t ion

o f

e a c h

c o m p o n e n t .

W h a t

d o y o u

u n d e r s t a n d b y s m a r t t r a n s d u c e r ?

D i s c u s s

i t

in

brief .

l .O. a) A paral le l -pLate c a p a c i to r w i th

pla te a r e a

o f 1 0 c m

2

a n d

p la t e

s e p a r a t i o n o f

6

m m

h a s a vo l t age

SOsin( l

0

3

t)

vol t

app l i ed

to

i t s

p la t e s .

Dt. · termine t h e

d i s p l a c e m e n t

c u r r e n t ,

a s s u m i n g

£

=

2t.

0

b)

S t a t e

A m p e r e ' s

c i r cu i t

l aw.

A

ho l low

c o n d u c n n g cy l inde r

h a s

i n n e r r a d i u s a

a n d

o u t e r

r a d i u s b, a n d

ca r r i e s

a

c u r r e n t f a l o n g t h e pos i t ive z -d i rec t ion .

·

1

1

F i n d e v e ry w h e re . 1

D R S R L R R A / 6 9

1 0

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11/24

c) F o r

a l o s s l e s s

two-wi re t r a n s m i s s i o n

l ine

s h o w th t

i)

t h e p h a s e ve loc i ty

u c

k;

L

{ii)

t h e c h a r a c t e r is t ic i m p e d a n c e

1 2 0

h 1 d

0 c o s -

JE; 2a

1 0


BS 14*

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12/24

www.examrace.co


13/24

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14/24

f f1?

f

~ I

(b) I f an

i n d u c t i o n m o t o r

i s

ope ra t . i ng

at 6 0 t ; ~ o l

i t s

s y n c h r o n o u s s p e e d , t h e m a x i m u m e f f ic iency

u n d e r i d e a l conditionos ( theore t i ca l ly

poss ib l e )

is

( i ) 100 >

( ii)

40 '

:.•

(iii}

5 0 £ ~

( i v )

6QC;b

(c) C u r r e n t .

c h o p p i n g p h e n o m e n o n 1s

a s s o c i a t e d w i t h

( i ) o v e r c u r r e n t r e l a y s

--ii)

s y n c h r o n o u s

g e n e r a t o r

( i i i ) i n d u c t i o n m o t o r

(iv) p o w e r tr a n s fO r m e r

(d) O v e r reach ing · a n d

u u n d e r

r e a c h in g ' '

a r e

e x p e r i e n c e d i n

( i ) f r e q u e n c y r e l a y s

( i i )

o v e r c u r r e n t r e l a y s

( i i i )

d i s t a n c e r e l ay s

( iv)

u n d e r vol tage

r e l a y s

(e} In

a

fu l l w a v e

r ec t i f i e r

c i r cu i t w i t h

c e n t r e - t a p

t ran>


15/24

A m o n g D T L , R T L , E C L a n d

C M O S

logic fami l ies ,

E C L h a s t h e propaga t ion de lay a n d p o we r

d is s ipa t ion /ga te respec t ive ly a s

i)

l ow

h igh

ii)

low low

iii)

high , low

iv)

h igh ,

h igh

g) W h i c h o n e o f

t h e fo l lowing

h a s t h e

h i g h e s t

pr io r i ty

?

( i)

RST

7·5

( i i)

RST

5·5

iii)

TRAP

( iv )

H O L D

(h) In P A M , t h e c a r r i e r wave cons i s t s o f a per iodic

tra in o f

rectangular

p u l s e s

a n d

t h e carrier

f requency

i s

equa l to

i ) t h e b a n d w i d t h o f t h e m o d u la t i n g s ig n a l

i i ) t h e

s a m p l i n g

r a t e

o f t h e m o d u la t i n g s ig n a l

( i i i ) a t l eas t

t e n t imes

g r e a t e r t h a n

t h e

b a n d w i d t h

o f t h e mo d u la t i n g s igna l

iv) twice

t he b a n d w i d t h o f t h e m o d u l a t i n g

s igna l

( i ) T h e

m a i n

d r aw b ack o f a p h a se contro l o f 1-phase

controlled rectif ier

circuit i s

i ) i t req u i res m o re gate cu rren t

( i i ) radio f requency in terference

i i i )

more

power loss

iv) eff ic iency i s

less

[Con td . ]

www.examrace.co


16/24

cj) T h e i m p o r t a n t

f e a t u r e s o f r o t o r O N · - O F F cont.rnl

I . B

a :

b

are

(i)

( .

lL

iii_}

iy)

fa:1t r e s p o n s e ~ 1SffiOoth v a r i a t i o n s i n ospecd

b e t t e r p o w e r fac tor a t

low s p e e d s

a n d w i d e

r a n g e


17/24

S E C T I O N

(a)

E xp la in

N o

vol tage re lease NVR) a n d O v e r load

re lease OLR) coils

provided in

a D C m o t o r

starter

(b) 400

V s h u n t m o t o r d r a w s 3

whi l e

su p p l y i n g

t h e

r a t e d load

a t

a

speed

o f 120 rad / s . T h e

a r m a t u r e

r e s i s t ance

is

1·0

o h m a n d t h e f ie ld

w i n d i n g r e s i s t a n c e

i s

250

ohms . De te rm ine t h e

external res i s tance that m u s t b e inser ted in

ser i e s wit h t h e armat u re circuit so t h a t armatu re

c u r r e n t

does no t exceed

150

o f i t s r a t ed

va lue

w h e n

t he

m o t o r is

plugged .

F i n d

t h e

b r a k i n g

to rque a t t he i n s t an t o f plugging.

c) 6600 V/400 V/110 V Sta r /S ta r /Me s h connec t ed

t h r e e -pha s e t r a ns fo rm e r h a s a m a gne t i z ing

c u r r e n t

of · 5·5

a n d b a l a n c e d

t h r e e -pha s e

loads o f 1000 kVA a t 0·8

l ag

on secondary

a n d

200 kVA, 0·5 l ead ing power f ac to r l oad on t he

tert iary

F i n d t h e

primary current

a n d p o wer

factor . Neglec t losses .

l-fow will t he breakdown s l i p

a n d

b r e a k d o w n

torque

e

affected w h e n t h e

rotor res i s tance

i s

i n c r e a s e d ?

b)

4-pole, 5 0

Hz ,

3 -phase

induct ion

m o t o r

del ivers

a s ha f t

t o rque o f 110 N - m

a t full load

a n d

r u n n i n g t 9 5 0

rpm. Ca lcu la te

i )

rot-or

c o p p e r

loss ii) power i npu t to t he rotor . T h e m e c h a n i c a l

l o s ses accoun t for 100 W.

1

1

2

10

1

[Contd.J

www.examrace.co


18/24

1c:

A n indwn. r ia l load o f 4 0 0 0

k W i s s u p p l i e d a t

11 kV, t h e po" ' ' ' r fac tor b e i n g 0 · 8 agl{ing. A

s y n c h r o n o u ~ XlOt.o.r r e q u i r e d

to

o::.P-t:t

a n

add i t i ona l load

of· 1-500

H P

<

I

103 ·25

k

W ; u n d


19/24

4.

5.

S E C T I O N

(a ) S t a r t i n g

f rom

fundaTnentals

derive

t h e

eq u at ion s for s en d in g en d vo l t age a n d s e n d i n g

end

current

for

a

long transmiss ion

line_

U s e

the

d i s t r i b u t ed

p a ra fu e t e r fo rm o f

r e p r e s e n t a t i o n

o f

t h e l ong l ine . Ob t a i n t h e A , B , C , D p a r a m e t e r s

o f

t h e

l ine .

(b)

A

sy n c h r o n o u s g en e r a to r

1s

r a t e d 2 0 M V

A ,

13 ·8 kV. t h a s X

1

= 0·25 p.u . , X

2

=

0·35

p .u .

a n d X

0·10

p .u .

T h e n e u t r a l 1s sol id ly

g r o u n d e d . T h e m a c h i n e i s on no l o a d a n d is

o p e r a t i n g a t r a t e d vol tage

w h e n

a D - L - G f a u l t

occurs a t i t s t e rmina l s . F i n d t h e s u b t r a n s i e n t

c u r r e n t

in

al l

t h e

p h as e s

a n d t h e

f au l t (a l l

in

a m p s )

a n d

t h e l ine- to - l ine vo l tages in

al l

t h e

18

p h a s e s

( in kV).

2

{c )

a )

S k e t c h typica l

sw in g c u r v e s for a

s y n c h ro n o u s

m a c h i n e ( i ) sh o w in g t h a t t h e m a c h i n e is s t a b l e

a f t e r

a

d i s tu rbance

(ii)

s h o w i n g

t h a t

t h e

m a c h i n e is u n s t a b l e

af te r

a d i s tu rbance .

A

r i n g

f eede r wi th five sec t ions

a n d

fed a t one

po in t is to b e pro tec ted u s i n g Direc t iona ) ove r

c u r r e n t (DOC) R e l a y s a n d

O v e r

c u r r e n t (OC)

R e la y s w i t h su i t ab le t i m e ; , Tading. E x p l a i n t h e

w o r k i n g o f t h e sc h e m e . S h o w t h e

loca t ion

o f

D.O.C s

a n d

O.C s

a n d

t h e i r

t i m e

o f

ope ra t ion .

A s s u m e a t im e g ra d i n g o f 5 rns b e t w e e n t h e

r e lays . T h e fa s t e s t

re l ay needs

5 m s for t to

operate

15

0 RSR L RRB

7

[Con td

.]

www.examrace.co


20/24

b ) i )

A s u r g e

o r

100

k V is i n c i d e n t o n

a

l i ne

h a v i n g

a s u r g e

i m p e d a n c e

o r 4 0 0 ohrns . f t

m e e t s

cable h a v i n g

~ u r g e

i m p e d a n c e nr

4 0 o h m s D e r i v e

exp res s ion for t h e

t . rano;mit ted

vo l t age

a n d

reflect ed

vo l t age

a n d

con1pute

the ir va lu es

i i ) E x p l a i n t h e prac t i ca l i m p o r t a n c e o f t h i s

s i tuat ion

c ) A

3 - p h a s e

d o u b l e c i rcu i t

l ine

I S a r r a n g e d as

s h o w n b e lo w

I 4 m - - j

a

G-----

0

c ·

m

I

y ·

b ~

'

5·5

m

. ~ r

3 Ill

c Q

l


21/24

S E T I O N

a) T h e c i rcu i t

s h o wn

uses a si l icon

transistor

w i th J =

50,

V

BE

= 0-6

V,

V CC = 22-5 V a n d

R c 5-6 K. F in d

t he

va lues o f

t he re s i s to rs RE,

R

1

a n d R

2

so t h a t Q

poin t i s

se t

a t

V

CE

12

V

a n d Ic = 1·5 rnA. Th e s tab i l i ty factor ,

S

m u s t

be

< 3.

I

:?

R c

c

_

2

D R SR L R R B

9

[C o n td . ]

www.examrace.co


22/24

7.

b ) A n i n p u t

ot·

E · t 5 0 V

i s a p p l i e d

to

t h e c l ipp ing

circuit

T h e o u t p u t current f rom t h e circuit

is

t o

b e

IL 2 0 xnA a n d

t ;he

n e g a t i v e o u t p u t

v o l t a g e

i>

n o t t o e x c e e d

0 · 5

V

n

[ > - - - - , - - - - - = = ~

C a l c u l a t e

t h e v a l u e of R

1

. Spec i fy t h e d i o d e

in

t erms

o f

forward current

power diss ipat ion

and

p e a k

r e v e r t ~ - e

voltage.

A s s u m e

t h e reverse

s a t u r a t io n c u r r e n t

I ~ i s j i A a n d

f o r w a r d vo l t age

o f

diode ,

V f'

0·7 V.

c )

a )

:b

c )

Co n Bt r u c t l og ic c i rcu i t

to

g i v e a n

X

=

A B + A C l A D + C } w i t h o u t a n y

o u t p u t

r e d u c t i o n

in n u m b e r

o f

gates

G i v e

t.he logic circuitti

in

s t e p

b y s t e p .

E x p l a i n

M e m o r y m a p p e d

110.

E x p l a i n

I n t e r r u p t , ;

a n d

S e r i a l

110

o f 8085 .

L i s t

t h e

m a c h i n e cy c le s

o f

8 0 8 5 .

0 RSR l RRB 1 0

[ C o n t

www.examrace.co


23/24

a )

W i t h

a

explain

s ignaL

S E T I O N

neat c ircuit diagram and w v e f o r m s ~

h o w

enve lope de tec tor de tec t s A M

b)

A s inuso ida l m odu la t ing wave

o f a mp l i t u d e

5 V

a n d f r equency

1

k H z

i s appl ied

to

a

f r equency

modula to r . T h e f r equency

sens i t iv i ty o f

t he

mo d u la to r i s 40

Hz/V. T h e

c a r r i e r

f requency i s

100

kHz .

Calcu la te

i)

f requency devia t ion

ii)

modula t ion

index

iii) i n s t a n t a n e o u s f requency

o f

t h e F M wave .

6

Also wri te t he

express ion

of t he

F M

waveform.

6

c) Cons ider a n F D M s y s t e m us ing

A M S S B

m o d u l a t i o n to t r a n s m i t

24

i n d e p e n d e n t vmce

i npu t s . A s s u m e

a

b a n d w i d t h

o f

4

k H z for

each

votce

input

Determine

the

transmiss ion

bandwid th . C o m p a r e t h e

b a n d w i d t h

w i th t h a t o f

a

s t a n d a r d 8 b i t

P C M w i t h T D M . A s s u m e b i t

dura t ion

is

0·64

7

J. S.

11

LContd. ]

www.examrace.co


24/24

9 .

v

,.

f n l ' ' ig 9•

.a,,

th,·

~ o a d

f ' t : ~ - : > l ~ ~ r : - a n c < : ·

~ o u r c e

vo] tage

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