solutions all mock tests_ee

8/12/2019 Solutions All MOCK Tests_EE

1/64

Chapter 1

EE_Practic Paper A

SOL 1.1 Option (C) is correct.

[ ]y n [ ] [ ]h n x n = *

[ ]y n [ ] [ ]h m x n m m

=

3

3

=

/ k"any integer

[ ]y n kN + [ ] [ ]h m x n kN m m

= + 3

3

=

/

a [ ]x n kN m + [ ] [ ] [ ]h m x n m y n m

= =3

3

=

/So [ ]y nis also periodic with period 'N N=

SOL 1.2 Option ( ) is correct.



SOL 1.5 Option (D) is correct.

1000100110014 2 3 1SSSS


The average value for the set of measurements is given by

XnX

10n

=/

.10

1005 100 5= =

PrecisionX

X X1n

n n=

For the 6th reading

Precision.

..

..

.1100 5

100 100 5 1100 50 5

100 5100 0 995= = = =


Diode will be off if 2 0v >i + . Thus 0vo=

For 2 0v


2/64

Page 2 EE_Practic Paper A Chapter 1

0.816 500 408 V#= =


The pumped storage scheme has the advantage that synchronous machine can be

used as synchronous condenser for VAR compensation. During light load period thegenerator works as synchronous motor.


( )f t 1 2 2cos sina nt b nt n nn 1

= + +3

=

/ 0

2 2T0

= =

k 1 ( )f dt10

1= # tdt 2

1t

0= =#

an 2 ( ) 2cosf t ntdt0

1= # 1n$ (integer)

2 2 0cost ntdt 0

1= =#

bn 2 2sint ntdt n1

0

1

= =

#




SOL 1.14 Option (A) is correct.

Deflecting torque TD NIlBD=

0.2 1.5 10 1 1T mA 00 1 102 2# # ## # #=

3 10 Nm6#=

SOL 1.15 Option (B) is correct.

Closed loop transfer function is

( )T s ( )

( )G s

G s

1=

+

( ) ( )

( )

K s s K

K s

1 3 2

12

2

=+ + + +

+

K

K

K

K

1 0

2 0

1

2

>

>

>

>

&

&

+

+

4 1K>&


Positive sequence component of fault current at Bus-2 is

I1 2f BUS .1.35 .p u

Z j j

1 00 7381 0

221

c c= = =




3/64



T Ia\

Since is constant so

T Ia\

When torque is doubled, armature current drawn by the motor will also be doubled.

Ia1 50 A= and 2 50 100 AIa2 #= =

NowEE

a

a

1

2 NN

1

2= ` is constant

N2 EE

Na

a

1

21#=

.. 500

V I RV I R

N220 50 0 2220 100 0 2

a a

a a

1

21

#

## #=

=

b l

500 476 rpm

210

200#= =


Circuit breakers are the final link in fault removal process of a power system. The

decision made by relays that a fault has been occurred on the line causes tripping

of circuit breakers.


Rs IV

dc

m

=

IV

R2 rms

dcm

#

=

. 500 449.5A100

0 45 100 k#

= =

SOL 1.22Option ( ) is correct.


Per-phase sending and receiving end voltage

VS 19 kVV3

33R= = =

We know that for a loss-less line(R 0= ), real power is

PS sinP X

V VR

S R

= =

33.sin19 1920 10 61

#

# c= = b lSOL 1.24 Option ( ) is correct.




( )y t ( ) ( )x t h t = * ( )x t ( ) ( 1)t t = +

( )h t ( )u t=


4/64


so ( )y t ( ) [ ( ) ( 1)]u t t t = + * ( ) ( ) ( ) ( 1)u t t u t t = + * * ( ) ( 1)u t u t = +

SOL 1.28 Option (C) is correct.SOL 1.29 Option (C) is correct.

Each wattmeter read 60 kW, so total input power

P 60W W kW1 2= + = ...(1)

For leading power factor, we have

tan ( )

( )W W

W W3

1 2

1 2=

+

It is given that power factor has to be changed to 0.866, so

cos .0 866=

.cos 0 866 301 c= =

and, tan tan303

1c= =

3

1 ( )W W120

3 1 2=

120 kWW W1 2a + =

W W1 2 3120 40= =

But W W1 2+ 120= ...(2)

By solving eq (1) and (2)

W1 40 , 80kW kWW2= =

SOL 1.30 Option ( ) is correct.SOL 1.31 Option (B) is correct.

From the combinational logic

Let Dis input, Qnis present state, Qn 1+ is next state, then

R ,D Q S D Q 5 5= =

Characteristic equation of R-S flip-flop is given by

Qn 1+ S RQn= +

So, Qn 1+ ( ) ( )D Q D Q Q n n n5= + +

( ) ( )D Q D Q Q n n n5 5= +

( )(1 ) ( )D Q Q D Q n n n5 5 5= = DQ DQ n n= +

For 0D= , Q Qn n1 =+ 1D= , Q Qn n1 =+So, the circuit function as a T-flip flop.


The thyristor acts as a diode when it is turned-on. By applying KVL in the circuit.

Ldtdi

C idt

1+ # VS=

( )i t sinVLC tS 0=


5/64


Here resonant frequencyLC

10 = =

Let the conduction time of thyristor is t0,

at t t0= ,

( )i t 0 sinVLC

tS 0 0= =

or, t00

=

or, t0 LC=


Let the positive sequence and negative sequence reactances are X1 and X2

respectively.

For a three-phase fault I ,a 3 X

Vt

1= 1V put` =

2000& /

X11000 3

1=

X1 .3 200011000 3 175

#

= =

For a line-to-line fault

I ,a LL X X

V3 t

1 2=

+

2600& ( / )X X

3 11000 31 2

=+

X X1 2+ .260011000 4 231 = =

X2 . . .4 231 3 175 1 056 = =

Base impedance

ZBase ( )( ) ( )

2.42MVA 50

11kV

Base

Base2 2

= = =

X pu2 .. 0.436 pu

ZX

2 421 056

Base

2= = =



First we balance phase angles. We know that to balance the bridge, the sum of

phase angles of opposite arms must be equal. From the figure

Phase angle of arm AB AB 90c= (pure capacitance)

Phase angle of BC & AD BC 0AD c= = (pure resistance)

Phase angle of CD CD 90< c+ (inductive impedance)

The first option is to modify impedance of branch AB(ZAB) so that its phase angle


6/64


is decreased to less than 90c(equal to CD ) by placing a resistor in parallel with the

capacitor as shown in the figure

Condition for balance

Z ZAB CD Z ZBC AD =

YABZ Z Z

Z1AB

CD

2 3= =

where

YAB Rj1

10001= + , jZ 100 500CD= + , Z 500BC= , Z 1000AD=

Substituting above values, we get

R

j110001

+ j

500 1000100 500

#=

+

and R1 5000 =

The second option is to modify the phase angle of arm AD by adding a series

capacitor, as shown in the figure

Balancing condition

ZADZ

Z Z

BC

AB CD =

Substituting the values in above equation, we have

jX1000 C ( )j j500

1000 100 500=

+

or XC 200 =


7/64



When the machine is running as a generator

Field current Ish 1 A250250= =

Load current IL 40 A25010 103#= =

Armature current Ia 40 1 41 AI IL sh= + = + =

Induced Emf Eg 250 4 ( .1) 254.1 V1 0= + =

Now the machine runs as a motor

Load current ILl 40 A25010 103#= =

Field current shIl 1 A250250

= =

Armature currentIal 40 1 39 A= =

Generated emf Em 250 39( . ) 246.1 V0 1= =

Speed as a motorNm .. (800) 775 rpm

EE

N254 4246 1

g

mg .= =




XRA A ; Clear A

MVI B ; 4A"B

SUI 4FH ; A - 4FH"A = B1H

ANA B ; A AND B"

A = 00 HTL

00A= , 4B A=


RMS value of fundamental output voltage.

V0(fund)V

24 dc

= 300V Vdc` =

270.14 V2

4 300#

= =

Power delivered

Pout cosV I01 0 = . cos270 14

2540 45# c= 72.94 kW=


The admittance diagram for the system is shown below:


8/64


YBUS

Y

Y

Y

Y

Y

Y

Y

Y

Y

Y

Y

Y

Y

Y

Y

Y

11

21

31

41

12

22

32

42

13

23

33

43

14

24

34

44

=

R

T

SS

SSSS

V

X

WW

WWWW

.

.

.

.

.

.

.

.

..

..

j

8 5

2 5

5 00

2 5

8 75

5 00

5 0

5 0

22 512 5

0

0

12 512 5

S=

R

T

SS

SSSS

V

X

WW

WWWW

Where Y11 y y y10 12 13= + + ; Y y y y y 22 20 12 23 24= + + +

Y33 y y y y 30 13 23 34= + + + ; Y y y y 44 40 24 34= + +

Y12 Y y21 12= = ; Y13 Y y31 13= =

Y23 Y y32 23= =

and Y34 Y y43 34= = ; Y Y y24 42 24= =

Y14 Y y14 14= =



ID 0.8RV5

D

D=

= mA,.

5R0 86 1

mD=

= k

ID ( )K V Vn GS TN 2=

& 0.8 (0.4)( 1.7) 3.11V VGS GS 2&= = V

VGS , 0, 3.11V V V V G S G S = = = V

ID 0.8= mA. ( )

2.36 kR

R3 11 5

SS& =

=


F2 4Q Q

RR

0

1 2

123

12

=

4

( )( ) ( )u u4 10 5 10

5

3 4x y0

6 6

3# #

#=

+

(4.32 5.76 )u ux y= + mN




9/64




The synchronous speed of the motor is

ns 1000 rPf

120 6120 50 pm# #

= = =

The shaft speed is

nm (1 )s ns= 0.06s` =

(1 0.06)1000 940 rpm= =

The load torque is

L P

m

out

=

508kW Nm940

602

50

#

-

=

SOL 1.49 Option (B) is correct.Power converted from electrical to mechanical form is

Pconv 50 300 600kW W W= + +

50.9 kW=

Induced torque

ind Pconv

m=

6

. 517kW Nm940

02

50 9

#

= =


From Y1to , ,Y P G G G P G G 5 1 1 2 3 2 4 3= =

13 1,YY G G G G G

21

5 1 2 3 4 33

3= = =

+

YY

2

5

YYYY

1

2

1

5

= G H

G G G G G 1 2 2

1 2 3 4 3=+

+


Consider block diagram as SFG

P1 2 2s

2

1#= = ,

P2 2 2 4s s#= =

L1 ( 5) s21 5= = ,

L2 2 ( 1) 2s s

1# #= =

L3 2 2 ( 1) 4s s# #= = ,

L4 2 ( 5) 10#= =

There are no non-touching loop


10/64


3 1s

s5 4 2 10= b l 13 4s s5= + + ,

13 123= = ,

( )T s ss

s

13 4 52 4

= + +

+

( )

s s

s s

4 13 5

2 2 12= + +

+


Average value of output voltage

V av0 ( )D V Vdc T= 250 2V VV, Vdc T` = =

( 2)D Vdc= .D 0 4` =

. ( )0 4 250 2= 0.4(248) 99.2 volts= =

RMS value of the output voltage is

V0 rms ( )D V Vdc T=

. ( ) .0 4 250 2 0 4 248#= = 156.84 volts=


Output power

P0 RV rms0

2

= ( . )

98 . watts25

156 83 44

2

= =

Input power

Pi V Idc 0= 250 V

25av0

#= I RV av

00

` =

250 . 992 watts25

99 2#= =

Chopper efficiency

100input poweroutput power

#= . 100 99. %

990984 06 19#= =


PL 0.001( 70)PG22=

We know that plant factor of plant nis given as

Ln /P P11

L n2 2=

For Plant 1 :

PP

G

L

12

2

0=So

L1 1=

For plant 2 :

PP

G

L

22

2 0.002( 70) 0.002 0.14P PG G2 2= =

L2 ( . . )PP P1

11 14 0 002

1

G

L G

2

2

2

2=

=


11/64


12/64

Chapter 1

EE_Practic Paper B



The resistances R1 6R

I

V

20

120dc2 = = = =

Now, we have the relation for Cfor successful commutation as

C 1.44 1.44 14.4 FRt

660 10off

1

6# #= = =


The sensitivity of 500 A meter movement is given by

S 1/ 1/500 2 /A k VIm = = =

The value of the multiplier resistance can be calculated by

Rs rangeS Rm#= 1R km` =

Rs 2 / 50 1k V V k #= 100 1 99 k= =



SOL 1.6 Option (C) is correct.2 5T1

= & 2T

51

= and 2 7T2

= & 2T

72

=

LCM 2 , 2 25 7 =b l



SOL 1.9 Option ( ) is correct.SOL 1.10 Option (B) is correct.

The gate turn-off thyristor(GTO) has the capability of being turned- off by a

negative gate-current pulse. Also GTO has a faster switching speed than thyristor

and it can withstand larger voltage and current than power transistor or MOSFET.




13/64

Page 2 EE_Practic Paper B Chapter 1



( )C s ( )s s 10

10=

+

s s1

101

= +

& ( )c t 1 e t10=

10a= , Rise time . . 0.22 sTa2 2

102 2

r= = =

Setting time Ts 0.4 sa4

= =


Pattern observed on the screen is an ellipse. So, phase angle

36.9sin531

c= = b l or .143 1cWe can see from the figure that ellipse is in second and fourth quadrants so the

valid value of phase angle is .143 1c.











If a line is terminated in its characteristic impedance, the refilled voltage will be

zero. All the energy sent will be absorbed by the load. It is also called an infinite

line.



Output of the system is ( )y t ( ) ( )h t x t = * ( ) [ ( 1) ( 3)]h t t t = + * ( ) ( 1) ( ) ( 3)h t t h t t = + * * ( 1) ( 3)h t h t = +





14/64


Consider the block diagram as a SFG. Two forward path G G1 2and G3and three

loops , ,G G H G H G H 1 2 2 2 1 3 2 .

There are no non-touching loop.

SOL 1.31 Option (C) is correct.Nominal ratio Knom 2005

1000= =

Secondary burdenRe 1 =

Since the burden of secondary winding is purely resistive therefore, secondary

winding power factor is unity or 0 = . The power factor of exciting current is 0.4,

so we can write

( )cos 90c 0.4=

or 90 0.4 2 .cos 3 571c c= =

Since there is no turn compensation, the turns ratio is equal to nominal ratio or

Kt 200Knom= =When the primary winding carries rated current of 1000 A, the secondary winding

carries a current of 5 A.

Rated secondary winding current,

Is 5 A=

and K It s 200 5 1000 A#= =

Phase angle,

( )cos

K II180

t s

0

= +; E

. 0.0525cos180

1000

23 57cc

= =

: DSOL 1.32 Option ( ) is correct.SOL 1.33 Option ( ) is correct.





By putting ( ) 0R s =

P1 H G2 1= ,

L1 G H H1 2 1= , 13 1= ,

( )T sn G H HH G

1 1 2 12 1=

+

If 1G H H >>1 2 1 , ( )T sn G H HH G

H1

1 2 1

2 1

1=

=


[ ]y n [ 1] 3 [ ] 2 [ 1]x n x n x n = + +


15/64


Linearity :

For two input [ ]x n1 and [ ]x n2[ ] [ ]x b y n 1 1" and [ ] [ ]x n y n 2 2" here

[ ] [ ]Ax n Bx n 1 2+ [ ] [ ]Ay n By n 1 2" +

So, the system is linear.

Casualty :

By taking ztransform on both sides

[ ]Y z ( ) 3 ( ) 2 ( )zX z X z z X z 1= +

System response is

[ ]H z[ ][ ]

3 2X z

Y zz z 1= = +

[ ]h n [ 1] 3 [ ] 2 [ 1]n n n = + +

[ 1]h [ ]y n n0=

So the system is time invariant.SOL 1.39 Option (A) is correct.

For a balanced maxwells bridge, following conditions are true.

Ls C R R3 1 4= 0.1 1.26 500F k # #= 63 mH=

Rs RR R

3

1 4= . 1.34k k470

1 26 500#

= =

QR

Ls

s= .

0.03k

Hz mH1 34

2 100 63# #

= =


VS 132.79 VoltV3

230 kR= = =

Maximum power transferred

Pmax( . )

X

V V

X

V

14132 79S R R

2 2

= = =

1259.5 /MW phase=

For all three phases

Pmax 3 1259.5 MW#=

3778.5 MW=


For a three phase bridge inverter operating in 120c conduction mode, the nth

harmonic component is given by following expression

V ( )linen /cosn V n2 1 3dc

= + ^ h8 BFundamental component

V( )line1 (1.5)V2

dc=

Edc( . )

614.60 V2 1 5415

= =


16/64



Load Current

IL 2 A100400 .0 5

= =b lIn a three-phase fully controlled bridge converter input rms current Isor the current

in each supply phase exists for 120cin every 180c.

Therefore rms value of input current

Is 1.15A1802 120 .0 5#= =b l

Input apparent power 400 1.15 796.72 VA3 # #= =

. cos796 72 400=

Power factor

cos 0.5 lagging=

SOL 1.43 Option (B) is correct.Ais singular if 0A =

&

0

1

2

1

0

2

2

3

R

T

SSSS

V

X

WWWW 0=

& ( 1) 2 01

2

2 1

0

2

3

0

2

3

+

+

0=

& ( 4) 2(3) + 0=

& 4 6 + 0= 2& =

SOL 1.44 Option (C) is correct.Let 0.4p1 = , 0.3p2= , 0.2p3= and 0.1p4= P(the gun hits the plane)

P= (the plane is hit at least once)

1 P= (the plane is hit in none of the shots)

1 (1 )(1 )(1 )(1 )p p p p1 2 3 4=

1 (0.6 0.7 0.8 0.9) (1 0.3024) 0.6976# # #= = =


The equations of the given curves are

y2 9x= ...(i)

2x y + 0= (ii)

The curves (i) and (ii) intersect atA(1,3)and B(4,6)

If a figure is drawn the from fig. the required area is

A [ ]dydx y dx xx

x

x

23

1

4

2

3

1

4= = +

+###

[ ( )]x x dx x x x 3 2 221 22

3 2

1

4

1

4

= + = : D# (16 8 8) 2

21 2

21

= =b l


17/64





For CE configuration given in figure, input impedances calculated in previousquestion as

r 2.5R1= = k

R2 [ (1 ) ]r R r RE E = + + = + ( 1)>>a

2.5 100 25 5#= + = k


As calculated in previous question

Av1 g Rm C= (without emitter resistance)

Av2 [ ]g Rg R

1 r m Em C

1= + +

^ h (with emitter resistance)

AA

v

v

2

1 1r

g R1

m E= + +

b l

1 g

g Rm m E= + +b l ( )g rm` =

1 g R1 1m E= + +c m ( 1)>>a

So, 1AA

g Rv

vm E

2

1. + 1

r RE:

= +

1.

22 5 10100 25

3#

#= + =

SOL 1.50 Option (A) is correct. Z ABC ACB AC B = = = +

If AC D=

Then Z D B= +

Therefore one NAND and one NOR gate is required and cost will be 2 unit.


The circuit is as shown below

Thus 4 NAND is required to made a NOR


18/64



The conductors of both transmission lines are arranged in following diagram

Flux linkage of conductor t1due to flux between conductor aand b

t1 2 10 lnI DD7

1

2#=

.

ln2 10 15022 5207

# #=

0.353 10 Wb5#= -T/m

Flux linkage of conductor t2due to flux between conductor aand b

t2 2 10 150 ..ln

23 120 67

# #=

3.43 10 Wb5#= -T/m

Resultant flux linkage

t 0.01 10 Wbt t1 25 #= =

-T/m

Mutual inductance

L (0.01 10 /150) 10 10 /mH km

5 3 3# # #

=

0.00067 /mH km=


Induced voltage in telephone line

3 0.01 10 1077 5 3# # #=

0.03 /V km77=




19/64

Chapter 1

EE_Practic Paper C


Power P vi= 2 2i i ix x x2

#= =

ix 4= A, 32P= W (absorb)

SOL 1.2 Option (A) is correct.Torque speed curve of a 3-phase induction motor is shown below

To achieve a slip greater than 1, the rotor must be coupled to a prime mover and

driven in a direction opposite to that of stator rotating field. Which is shown by

point W in the figure.


Circuit turn off time for main thyristor tc CI

Vs0

=

40 10 76.67 s1202306 #= =


Boolean expression for the given circuit

F [( ) ( ( ))]( ( )A B A A B A A B = + + + + + +

Let A B X+ = and ( ( )A A B Y + + =


20/64

Page 2 EE_Practic Paper C Chapter 1

Then ( )F X Y Y XY Y Y = + = + =

F ( ( ) ( )A A B A A B AB = + + = + =

So, 1F= for 0A = , 1B=

SOL 1.5 Option (B) is correct.Multiply G2and G3and apply feedback formula and then again multiply with G

11.

( )T s ( )G G G H

G G11 2 3 1

2 3=+



The figure is as shown below



The following sketch are not root locus due to the given reason.

(B) On real axis root locus exist to the left of even number of finite poles and zero.

(C) Root locus does not start from poles and does not end on origin.

(D) On real axis root locus exist to the left of even number of finite pole and zero.




Bode plot for the frequency response of op-amp is given following


21/64


Here, f Amax

CL CL# 1f #= fCL "close loop Band-width

f " unity gain bandwidth

Amax

CL "maximum closed loop gain

(20 10 )Amax

CL3

# 1 106

#=

Amax

CL 50=


Resolution Rs 1/10n= 4n` =

1/10 0.0014= =

Resolution on 1 V range 1 0.0001 0.0001V #= =

So, any reading up to the 4th decimal can be displayed. Thus 0.6973 will be

displayed as 0.6973.

Resolution on 10 V range 10 0.0001 0.001V V#= =

So, decimals up to the 3rd decimal place can be displayed. Therefore on a 10 V

range, the reading will be 0.697 instead of 0.6973.


v 3 0.7 2.3= = V

i. ( )

2.652

2 3 3=

= mA



Absolute errorexpected value

expected value measured value=

80 79

. %80 100 1 25#=

=

Relative Accuracy 100expected valuemeasured value

#=

or Relative Accuracy %100 absolute error=

100% 1.25% 98.75%= =




22/64



Mho relay is suitable for long EHV lines as its threshold characteristic is quite

compact enclosing fault area completely.

SOL 1.20 Option (B) is correct.The reverse recovery characteristic of a power diode is shown below. In the figure

reverse recovery time trrcomposed of taand tb . tais the time between zero crossing

and peak reverse current IRR and tb is measured from reverse peak IRRvalue to

. I0 25 RR.

If the characteristics are assumed to be triangular(i.e. abrupt recovery) than from

the figure charge stored is

QR area ABCT= I t21

R R rr= ...(1)

IRR t dtdia=

t ta rrc , than IRR t dtdi

rr= ...(2)

from eq(1) and (2) QR dtdi

t21

rr2= b l


If kis a constant and Ais a square matrix of order n n# then k kA An .

A B A B B B5 5 5 6254&= = = =

& 625=

SOL 1.22 Option ( ) is correct.SOL 1.23 Option (A) is correct.

ib 0.5 0.664 3610=+

+ = A




( )x t1 6 (100 ) (200 )sin cosc t t=


23/64


( )X f1 is convolution of two signals whose spectrum covers 50f != Hz and 100!

Hz. So convolution extends over 150f= Hz and sampling rate 2 300N f1 = = Hz

( )x t2 10 (100 )sin c t2=

Taking Fourier transform

( )x f2 0.1tri f100

= b l B 100= Hz

Sampling rate 2 200N B2 = = Hz

NN

2

1 200300

23= =


Ia 62.98 A3 11 10 1

1200 103

3

# # #

#= =

Vt 6350.8 V3

11 103#= =

Ea ( ) ( )cos sinV I R V I X t a a t a s 2 2 = + + +

For unity pf,cos , sin1 0= =

So Ea ( . . . ) ( . . )6350 8 1 62 98 1 2 6350 8 0 62 98 242 2# # # #= + + +

6601.74 V=

Regulation 100V

E Vt

a t#=

.

6601.74 6350.8 . %6350 8

100 3 95#=

=




The admittance diagram for the system is shown below:


24/64


YBUS

Y

Y

Y

Y

Y

Y

Y

Y

Y

Y

Y

Y

Y

Y

Y

Y

11

21

31

41

12

22

32

42

13

23

33

43

14

24

34

44

=

R

T

SS

SSSS

V

X

WW

WWWW

.

.

.

.

.

.

.

.

..

..

j

8 5

2 5

5 00

2 5

8 75

5 00

5 0

5 0

22 512 5

0

0

12 512 5

S=

R

T

SS

SSSS

V

X

WW

WWWW

Where Y11 y y y10 12 13= + + ; Y y y y y 22 20 12 23 24= + + +

Y33 y y y y 30 13 23 34= + + + ; Y y y y 44 40 24 34= + +

Y12 Y y21 12= = ; Y13 Y y31 13= =

Y23 Y y32 23= =

and Y34 Y y43 34= = ; Y Y y24 42 24= =

Y14 Y y14 14= =




The transient reactance of the generators are

XG1 0.08 0.242060 pu#= =

XG2 0. 0.1 6060 1 pu#= =

XG3 0.0 0.29 2060 7 pu#= =

There values are shown in the equivalent circuit in the figure below.


25/64


As the generator e.m.f.s are assumed to be equal, one source may be used which is

also shown in figure.

The equivalent reactance is

Xeq / . / . / .0.056 pu

1 0 24 1 0 27 1 0 11

=+ +

=

Therefore fault MVA

.1071MVA

0 05660

= =

and fault current

52402 A3 1180

1071 106

#

#= =


The SFG of this system is shown below

L1 G= , L G2 = , L G32= , L L G1 2

2=

3 1 ( ) 1 2G G G G G 2 2 2= + + =

From R1to C1, at 0R2 = ,

P1 G= ,

13 1 ( ) 1G G= =

RC

R1

1

02 =

( )

G

G G

1 2

12=


Bridge is balanced if

LP C R R3 1 4= 0.1 1.26 500F k # #= 63 mH=

Similarly, at balance the unknown resistance

Rp 751.26 500k

RR R

3

1 4

#= = 8.4 k=


26/64


The coil is represented by parallel equivalent, so quality factor is

QL

R

P

p

=

2 100 638.4

Hz mHk

# #= 212=








Let Qnis the present state and Qn 1+ is next state of given X Y flip-flop.

X Y Qn Qn 1+0 0 0 1

0 0 1 1

0 1 0 0

0 1 1 1

1 0 0 1

1 0 1 0

1 1 0 0

1 1 0 0

Solving from K-map we get

Characteristic equation of X Y flip-flop is

Qn 1+ Y Q XQ n n= +

Characteristic equation of a J K flip-flop is given by

Qn 1+ JQ KQ n n= +

By comparing

J Y= , K X=


Peak value of secondary voltage

Vm 4002800 V= =

and 30c=

Average dc voltage is given by

Vdc (1 )cosV2

m

= + (1 30 ) 118.8cos V

2400

c

= + =

RMS voltage

Vrmssin

V4 8

2 /m

1 2

=

+b l


27/64


400 197.1sin V4

30860 /1 2c c

= + =b l



P(none dies) (1 )(1 ) ...p p n= times (1 )p n=

P(at least one dies) 1 (1 )p n=

P(A1dies) {1 (1 ) }n p

1 n=




Only statement (b) is false. For example

S1: [ ] [ ]y n x n b= + , and

S2: [ ] [ ]y n x n b= ,where 0b !

{ [ ]}S x n { { [ ]}} { [ ] } [ ]S S x n S x n b x n 2 1 2= = + =

Hence Sis linear.


For example, consider two LTI system given as

S1: [ ] [ ]y n nx n = and

S2: [ ] [ 1]y n nx n = +

If [ ] [ ]x n n= then { { [ ]}} [0] 0S S n S 2 1 2 = = ,

{ { [ ]}}S S n1 2 { [ 1]} [ 1] 0S n n1 ! = + = +








28/64

Chapter 1

EE_Practic Paper D


( )f cl ( )( ) ( )

1f f

e e1 01 0 c

&=

=

& c ( 1)log e=


x x

dxI

1 =#

Put 2sin sin cosx dx d 2 & = =

Isin sin

sin cossin cossin cos

d d1

2 22

=

=##

I sind c x c 2 2 2 1 = = + = +# I (2 1)sin x c1= +

SOL 1.3 Option (B) is correct.The given equation is ( )D D y e e x x3 = +

P.I. ( ) ( )D D

e e xD

e e1

3 11x x x x

3 2: := + =

+

( ) ( )x e e x e e3 1

12

x x x x :=

+ = +


Let, I2

cosi z

zdz1

11

c

2 =

#

2.2

cos

i z z

zdz1

1

1

1

1

c

=

+b l#

Or I cos cosi z

zz

z dz41

1 10

c

=

+

=a k#SOL 1.5 Option (B) is correct.

% Resolution 1002 1

1n #=

100 6.67%2 1

14 #=

=


The noninverting terminal is at ground level. Thus inverting terminal is also at


29/64

Page 2 EE_Practic Paper D Chapter 1

virtual ground. There will not be any current in 60 k.

Av 1040400= =


f ( ) ( )A BC B A BC CD = + + + AB BC ACD BCD = + + +


D2and D3are ON. If D3is ON, then D1is OFF.

vo 5 0.6 4.4= = V,

iD2 .. . 7.60 5

4 4 0 6k

=

= mA


There is one forward path G G1 2and four loops , ,G G G G G G G G G 1 4 1 2 3 1 2 5 7

and G G G G G 1 2 3 6 7 .

There is no non-touching loop.


In characteristic equation 2s s K3 2+ + , the term sis missing.


Apply the feedback formula to both loop and then multiply.

( )T s G H

G1 1 1

1=+b l G HG1 2 22+b l

G H G H G G H H

G G1 1 1 2 2 1 2 1 2

1 2=+ + +



Number of turns

NAB

V2peak

rms

=

.2 50 20 10 1 8

2 120

4# # # #

#

=

150=


For a dc series motor T Ia2

\

TT1

2

II

1210

2

1

a

a2 2

= =b bl lor T2 . T1 44 1=

the percentage increment in torque is


30/64


TT 100 44%T

T T1

2 1#=

=


The following requirements have to be satisfied prior to connecting an alternator to

an already operating alternator The line voltage of the (incoming) alternator must be equal to the constant

voltage of the already operating alternator.

The frequency of the incoming alternator must be exactly equal to that of the

already operating alternator

The phase sequence of the incoming alternator must be identical to the phase

sequence of the already operating alternator

The prime mover speed of the alternators should be different for operation.


A cage motor shows peculiar behavior at starting because the motor has a certain

relationship between the number of poles and the stator and rotor slots. For some

ratio of rotor-to-stator slots, the machine may run stably at a low speed(1/7 of the

rated speed). This phenomena is called crawling.


Detent torque/Restraining toque:

The residual magnetism in the permanent magnetic material produced.

The detent torque is defined as the maximum load torque that can be applied to

the shaft of an unexcited motor without causing continuous rotation. In case the

motor is unexcited.

SOL 1.18 Option (B) is correct.The input e /2j n must produce the output is the form Ae /2j m . The output in this

case is e 3 /2j n . This violates the Eigen function property of LTI system. Therefore,

Sis definitely not LTI system.


[ ]X k ( ) ,T

A t e dt TA1

/

/jk

T

T

2

2

= =#

10A= , 5T= , [ ] 2X k =


If we go from +side of 1 kthrough 7 V, 6 V and 5 V, we get v1 7 6 5 8= + = V




31/64


RTH 7 9 6.525 6 = + =

For maximum power transfer

RL 6.52RTH = =


Leq L LM

12

2

=

4 22

4= = H


For a star connection

VpV

3L= V Phase voltagep"

V line voltageL"

So Vp 2403

240 3 V= =

Ip 6RV

40240 A

p

p= = =


1. BUS admittance matrix is a sparse matrix.2. GS method is easier but it is less accurate and has a slow convergence rate

compare to NR method .So, GS method is not preferred over NR method.

3. One of the buses is taken as slack bus in power flow studies.


Deflection sensitivity 0.05 /mm V=

Spot deflection 5 mm=

Applied voltage.

100 V0 055

= =



32/64


Output of the inverters

V01 2 400#= and V 2 40002 #=

From the phaser diagram the output voltage is given as

V0 [ ]cosV V V V 2012

012

01 02 21

= + +

( ) (( ) )cos800 800 2 800 800 302 2 2

1

# # c

= + +6 @ 1.54 kV=SOL 1.27 Option (C) is correct.

so

.

( )0 5

500 1500602

#

= ; E .418 67= rad/sec2

and T 40 Nm=

T I=

I.

0.096T418 67

40 kgm2#= =


Firing angle 25c=

Overlap angle 10c=

so, I0 [ ( )]cos cosLsVm

= +

20 [ ( )]cos cosLs2 50

230 2 25 25 10#

c c c

= +

Ls 0.0045 H=

V0cosV LsI2 m 0

=

. .

. .cos3 14

2 230 2 253 14

2 3 14 50 4 5 10 203# # # # # #c=


33/64


.187 73 9= .178 74c=

Displacement factorV IV I

s s

0 0= .230 20

178 25 20#

#= .0 78=


Impedance Z . .j5 10 11 2 63 4c = + =Line current I

..

440 0 812000 36 86

# c=

34.1 . A36 86c=

Transmission line loss PL (34.1) 5 5814.05 WI R2 2

#= = =

Power at the sending end

PS 12000 5814.05 17814.05 W= + =

Transmission efficiency

PP 100

S

R#=

. . %17814 0512000 100 67 36#= =


Let Base MVA is 40 MVA and Base Voltage is 11 kV, reactances on this base will

be

Generator G1 XG1 0.15 puj=

Generator G2

XG2 0.10 0.20 puj j2040

#= =

Generator G3

XG3 0.10 0.20 puj j2040#= =Transformer T1 XT1 0.15 puj=

Transformer T2 XT2 0.15 puj=

Transformer T3

XT3 0.08 0.64 puj j540

#= =

Line

XLine 40( )

0.367 puj j6640

2#= =

Equivalent circuit diagram is shown below


34/64


Fault current

If .1.37 pu

j j

0 7291 0c

= =

Base current IB 2099.45 Amp3 1140 1000##

= =

Fault current in amperes

If . .1 37 2099 45#=

2.876 kA=


Positive sequence network

Equivalent positive-sequence impedance,

Z eq1 (0.5 0.05) || (0.5 0.05)j= + +6 @ 0.275 puj=

Negative sequence network


35/64


Equivalent negative-sequence impedance

Z eq2 0.275 puZ j1= =

Zero sequence network

Equivalent zero-sequence impedance,

Z eq0 . . .j j j3 0 05 0 5 0 65#= + =

Fault current If Z Z ZV3

eq eq eq

f

1 2 0=

+ +

( . . . )

( )j

j

0 275 0 275 0 653 1 0#

=+ +

+

2.5 puj=SOL 1.32 Option (D) is correct.

The configuration is shown below

Current in voltmeter is given by

IV .E

2000 2000180 09= = = A

I IV+ 2= amp

So I . .2 09 1 91= = V

R.

.IE

1 91180 94 24= = =

Ideally R0 2180 90= =


36/64


% error 100R

R R0

0#=

.90

94 24 90 100#= .4 71= %


The meter current indicated by the instrument in the figure

Im R R REx mb

1=

+ +

To obtain full-scale deflection current put R 0x= in above equation

Ifsd.

kV

0 151 5

=

+ 100 A=

At 50% of full scale deflection

Im100

2A

50 A

= =

R R Rx m1+ + IE

m

b=

Rx ( )IE R Rmb m1= + 501.5 15AV k =

15 k=

At 25% of full-scale deflection

Im100

25A

A4

= =

Rx 251.5 15

AV k

= 45 k=

At 75% of full scale-deflection

Im 0.75 100 75A A #= =

Rx 751.5 15AV k = 5 k1 =


Duration of one cycle of input

T 0.25 sec41 m= =

When the sweep frequency is 8 kHz, duration of one cycle of sweep

Ts 81 0.125 msec= =

no of cycles appeared on the screen.. .0 250 125 0 5= =

When the sweep frequency is 8 kHz, duration of one cycle of sweep

Ts2 21 0.50 msec= =

no of cycles appeared on the screen..

0 250 50 2= =


Secondary circuit phase angle

.. .tan

1 30 8 31 61 c= = b l


37/64


Turn ratio Kt NN

1300 300

p

s= = =

Magnetizing current

Im

sinMagneti g

N

mmf

p= N 1p`

=

90 A190

= =

Loss component

Iwequivalent to iron loss

Nmmf

p= 45 A

145

= =

Actual ratio

Kactsin cos

KI

I It

s

m w = + +

Put, 31.6 0.8517 31.6 0.524cos cos sin sinandc c = = = =

Kact 300. .

590 0 524 45 0 8517# #= + + 317.1=


Assume the transistor is biased in the saturation region

ID I VV1DSS

P

GS2

= b l 8m 18

.1.17m V V1

3 5GS

GS

2

&=

=b l V VD 15 m8= ^ h 0.8 8.6k V=^ h VDS 8.6 1.17 7.43= = V V VGS P 1.17 ( 3.5) 2.33= = V

V V V>DS GS p , Assumption is correct.


For 4.5VCE= V,. 0.5

kI

15 4 5

C= = mA

IB. 0.02

250 5

= = mA

R1 15 k= , 100 kR2 =

IR2. ( )

0.057

k100

0 7 5=

= mA

IR1 I IR B1= + 0.057 0.02 0.077= + = mA

V1 I R VR BE1 1= + (0.077) 15 0.7 1.855= + = V

For 1.0VCE= V

IC 4k15 1

=

= mA,

IB 0.16254= = mA

IR2 0.057= mA


38/64


IR1 0.057 0.16 0.217= + = mA

V1 (0.217) (15) 0.7 3.96= + = V

So 1.86 3.96V1# # V

SOL 1.38 Option (C) is correct.Output of gate 1 is X Xo 1Output of gate 2 is X X X0 1 2+

Output of gate 3 is

( )X X X X 0 1 2 3+ X X X X X 0 1 3 2 3= +

Output of gate 4 would be X X X X X X 0 1 3 2 3 4+ +

Output of gate 5 would be

X X X X X X X X X 0 1 3 5 2 3 5 4 5+ +

So output of gate nwould be... ... ...X X X X X X X X X X X X X X X o n n n n n 1 3 5 2 3 5 4 5 7 1+ + +

SOL 1.39 Option (D) is correct.24m 2+ = and 128m 2 =

On solving we get : 16m= or 8

16 8m 2& = = 8 56m 2& = =

Case I : 16np= and 8npq=

&p21

= and q21

= and 32n=

Case II : 8np= and 56npq=

& 7q= , which is not possible.

The distribution is ( )q p 21 21n

32

+ = +b lSOL 1.40 Option (A) is correct.

:x 0 0.1 0.2 0.3 0.4

Eulers method gives

yn 1+ ( , )y h x y n n n= + ...(i)

n 0= in (1) gives

y1 ( , )y hf x y 0 0 0= +

Here x 00= , y0 1= , .h 0 1=

y1 1 0.1 (0,1) 1 0 1f= + = + =

0n=

in (1) gives ( , )y y hf x y

2 1 1 1= +

1 0.1 (0.1,1) 1 0.1(0.1) 1 0.01f= + = + = +

Thus 1.01y y( . )2 0 2= =

2n= in (1) gives

y3 ( , ) 1.01 0.1 (0.2,1.01)y hf x y f 2 2 2= + = +

y3 1.01 0.0202 1.0302y( . )0 3= = + =

3n= in (1) gives

y3 ( , ) 1.01 0.1 (0.2,1.01)y hf x y f 2 2 2= + = +

y3 1.01 0.0202 1.0302y( . )0 3= = + =


39/64


3n= in (1) gives

y4 ( , ) 1.0302 0.1 (0.3,1.0302)y hf x y f 3 3 3= + = +

1.0302 0.03090= +

y4 1.0611y( . )0 4= =

Hence y( . )0 4 1.0611=


Response of the overall system is given as

( )h t ( ) ( )h t h t 1 2= *

( )h t [ ( ) ( 4)]u t u t t4

2rect= * b l t

42recta b l ( ) ( 4)u t u t = so

( )h t [ ( ) ( 4)] ( ) ( 4)u t u t u t u t = *

[ ( ) ( ) ( ) 4 ( 4) ( 4) ( )u t u t u t t u t u t = * * * ( 4) ( 4)]u t u t + * ( ) ( )u t u t a * ( )tramp=

so, ( )h t [ ( ) ( 4) ( 4) ( 8)]t t t t ramp ramp ramp ramp= +

( )G t ramp(t) 2ramp(t 4) ramp(t 8)= +


Here [ : ]A B

:

:

:

1

3

7

4

8

8

7

2

26

14

13

5

=

R

T

SSSS

V

X

WWWW

:

::

1

07

4

2020

7

2323

14

2993

=

R

T

SSSS

V

X

WWWW R R

R RRR3

2 1

3 1

2

3

&

&

f p

:

:

:

1

0

7

4

20

0

7

23

0

14

29

64

=

R

T

SSSS

V

X

WWWW R R R3 2 3&^ h

( : )A B 3& ( ) 2A= = ,

( )A ( : )A B!

Thus system is inconsistent i.e. has no solution.


Since ( )x t is real and odd, [ ]X kis purely imaginary and odd. Therefore [ ] [ ]X k X k = and [0] 0X = . Since { ] 0X k = for 1k > , they only unknown coefficient are [1]X

and [ 1]X

( )T

x t dt 1

T

2# [ ]X kk

2=3

3

=

/

For ( ), ( )x t x t dt 21 2

0

2# [ ]X kk

2

1

1

==

/& [ ] [ ]X X1 12 2+ 1= ,


40/64


2 [ ]X 1 2 1=

[1]X [ 1]X j

2= = or [1] [ 1]X X

j

2= =

Thus there are two solutions

( )x t1 sinj

e j

e t2 2

22 2j t j t2 2 = =

b bl l

( )x t2 sinj

e j

e t2 2

2j t j t22

22

= + =

b bl l


For a separately excited dc generator, terminal voltage is

Vt E I Ra a a=

So the external characteristic is

For a shunt generator

For a series excited generator

For over-compound generator


41/64



For speed of 00 r3 pmfrequency is

f1 15Hzn P120 120

300 6m #= = =

For the speed of 1000 rpmfrequency is

f2 50 Hzn P120 1201000 6m #= = =

The armature current at rated conditions is

Ia .kW A

3 480 1100 120 3# #

= =

The phase voltage is

Vt 480/ 277 V3= =

The internal generated voltage is

Ea ( ) ( )cos sinV I R V I X t a a t a s 2 2 = +

Ea ( ) ( . . )277 1 0 277 0 120 3 1 52 2

# # #= +

. V330 6=Eadecreases linearly with frequency, so at the speed of 00 r3 pm(i.e 15 Hz)

E ,a 300 ( . ) 9.HzHz V

5015 330 6 9 18= =


For s1,

1 2 3 4 + 90tan tan tan23

13 180

131 1 1

c= + b l 70.65 ,c=+ So s1is not on root locus.

For ,s2


42/64


1 2 3 4 + 90tan2 2

11c=

tan1802 2

11 c m 180c= , So s2in on root locus.SOL 1.47 Option (C) is correct.

(0 )iL+ 0= ,

(0 )vL+ 4 12 8= =

( )

dtdv

251 0L

+

(0 ) 0iL= =+

3,26

= =

0 /1 1 25

1 5#

= =

3 3 4j9 25! !

= = ( )v t 12 ( 4 4 )cos sinA t B t e t3= + +

(0)vL 8 12 A= = + ,

& 4A=

( )

dtdv 0L 0 3 4A B= = + ,

& 3B=


( )estep 3 1 ,CA B1= +

A1

2

1 3

1

0=

> H ( )estep 3 1 1 1

2

131

0

0

1+

8 > >B H H 1 31 32= =SOL 1.49 Option (C) is correct.

( )eramp 3 ( ) ( )lim CA B t C A B 1t

1 1 2= + +"3

6 @ 1 CA B1+

32

= ,

( )eramp 3 ( )lim t C A B 32

t

1 23= + =

"3

: DSOL 1.50 Option (A) is correct.

The phase voltage of the alternator

5773 volts3

10000= =

Let %x be the percent winding which remains unprotected. The voltage of the

unprotected portion of the winding

Vunprotected 5773 100x

= a kSince the resistance in the neutral is 10 ohms the fault current will be


43/64


If 5773 100 101 ampx= a bk l .

The current in the pilot wires will be with a CT of 1000/5 amps ratio

Ipilot 5773 100 10

1

1000

5

amps

x

= a b bk l land this current should be equal to 1.8 amps for the operation of the relay. 5773

100 101

10005xa b bk l l .1 8=

or x5773 .3 6 105#=

x.

. %5 773 1036 10 62 363

4

#

#= =


To protect 80% of the winding, the unprotected portion is 20%. The voltage of the

unprotected portion Vunprotected 5773 0.2 1154.6 volts#= =

Let Rbe the minimum value of the earthing resistance; the fault current will be

If. amp

R1154 6

=

The fault current through the pilot wire will be

If.

10005 amp

R1154 6= b l

This should equal the operating current of 1.8 amp

or .1000

5R

1154 6

b l .1 8=

or R1800

5 1154.6 .3 20 #= =


The field current If 120/160 0.75A= =

At no-load condition :

Armature current I ,a nL 2 0.75 1.25 A= =

Induced Emf E ,a nL 120 1.25 0.4 119.5 V#= =

The power developed at no load accounts for the rotational loss in the motor.

From full-load data :

Armature current I ,a fL 14.75 0.75 14 A= =

Induced Emf E ,a fL 120 14 0.4 114.4 V#= =

Speed NfL 2400 rpm=

Hence, the no-load speed:

NnL N EE

,

,fL

a fL

a nL#=

2400.. 2507 rpm

114 4119 5

.#=


With external resistance in the armature circuit and no change in the torque


44/64


developed

Armature current Ia 14 AI ,a fL= =

Induced Emf Ea 120 14 (0.4 3.6) 64 V#= + =

Power developed Pd 64 14 896 W#= =

Speed in this condition

Nm N EE

fLa fL

an#=

.

2400 1343 rpm114 4

64.#=

Developed Power at no-load

P,d nL 119.5 1.25 149.38 W#= =

Rotational loss after adding the external resistance

Pr NN

P,nL

md nL#=

149.38 80 W25071343

#= =

The power output and the efficiency are

Pout 896 80 816 W= =

Input power Pin 14.75 120 1770 W#= =

PP 100

in

out#=

. %1770816 100 46 1#= =



5 0c+ i j j

i j

4 1 14

141 2

= + + b bl l& (8 15) (4 )j i j i1 2+ 20 0+= ...(i)

10 30c+ i j j

i j

1 4 14

142 1

= + + b bl l& (4 ) (8 15)j i j i1 2 + 40 30c+= ...(ii)

[(8 15) (4 ) ]i j j12 2+ (20 0)(8 15) (40 30 ) (4 )j jc+ += +

( 176 248)i j1 + 41.43 414.64j= +

& i1 1.03 0.9 1.37 41.07j += =



45/64


i2( )( . . )

jj j

48 15 1 03 0 9 20 0c+

=

+

0.076 2.04j= +

& i2 2.04 92.13c+=


46/64

Chapter 1

EE_Practic Paper E


Observations in ascending order are 3 , 3, 1 , 0, 2, 2, 2, 5, 5, 5, 5, 6, 6, 6.

Number of observations is 14, which is even.

Median 21= [7 then term 8+ the term] (2 5) 3.521= + =


xv

2

2 2 ( , ), 2 ( , )y h x y yv x g x y

2

2= = = =

by Milnes Method ( )f zl ( ,0) ( ,0) 2 0 2g z ih z z i z = + = + =

On integrating ( )f z z c2= +


Let Ie

dxe

e dx1 1x x

x

=

=

# #

tdt

=# Putting 1 e t e dx dt x x

& = =

(1 )log logt e x= =


( )f xl 3 12c x k2= +

( )f cl 0= 3 12 0c c k2& + =

& c k k6

12 144 1236

144 1231

&!=

=

& 144 12k 12 11k&= = .


It is positive feedback system. So on real axis root locus will exist to the left of even

number of pole and zeroes. Plot will start from pole and end on zeroes. Option (A)

satisfies this condition.


Closed loop transfer function

( )T s ( ) ( )

( )

K s K

K s

1 3 4

42

2

=+ +

The system can be only marginally stable.


47/64

Page 2 EE_Practic Paper E Chapter 1


P1 5 3 2 30# #= = ,

3 1 (3 3) 10#= = , 13 1,=

R

C

310

30

= =SOL 1.8 Option (C) is correct.

During positive cycle when 8v i V,

8vo= V, D1is ON. During negative cycle when 6v i V, D2is on 6vo= V.


2 1x x2 + + 64 5 8 2#= + +

& x 7=


Here Tis, T 1Q Q Tn n&

= + = (always)So, output will toggle at each clock pulse.

The output sequence is 010101010...

Frequency of output isf2

i


The conduction loss v/s MOSFET current characteristics of a power MOSFET is

best approximated by a parabola.


Output of this

Here the inductor makes T1and T3in ON because current passing through T1and

T3is more than the holding current.

SOL 1.13 Option (D) is correct.Wagner earth connection eliminates the capacitance existing between the detector

terminal and ground.


a f 50 Hz=

So total time 20f1

501 msec= = =

Conduction time for each feedback diode in a cycle is being given by


48/64


tconduction 2.5820 msec= =


Frequency of vertical input

fy tantan

Vertical genciesHorizontal genciesfx #=

Where fxis the frequency of horizontal input

fy 1,000 2,500 Hz25

#= =


Systematic errors may be instrumental error, environmental error or observational

errors.


After killing all source equivalent resistance is R

Open circuit voltage v1=


( )v tC ( ) [ (0) ( )]v v v e C C C RCt

3 3= +

0 (10 0)e t= + V 10e t= V


Z Z1 4 Z Z3 2=

300Z4 (300 600)(200 100)j j= +

& Z4 400 300j=

SOL 1.20 Option (A) is correct. ( )X s ( ) ( )x t e dt u t e dt 2st st

00= = +

33 ##

e dts1st

0= =

3 #SOL 1.21 Option (B) is correct.

( )y t ( ) ( 2 )x t Y j e X j 2

2 j2& = = b l

& ( )Y j sine22 2j2

=

sine 2j2

=


The function 1 does not describe the given pulse. It can be shown as follows :


49/64



Emf

E1 4.44 voltsB A f N m c 1# # #=

or 400 . B4 44 50 10 50 350m4

# # # # #=

so Bm 1.0296 T=


ASCR implied Aluminium Conductor Steel Reinforced conductor which are mostly

used in overhead transmission lines.


The positive sequence impedance of a transformer is equal to its leakage reactance.

Transformer is a static device so the leakage impedance does not change with changes

in phase sequence of appliead voltages. Thus the negative sequence impedance is

also equal to its leakage reactance.



IL 100500 5= = A,

VL 100 5= V

Is 101001000

= = A,

V1 100 4 10 60#= = V

P25 1000 500 10 4 1002

#= = W

I2 225100

= = A,

I2 bI1=

I2 2b 5= = ,

b 0.8952= =

In center mesh, 25aV Ib

VL1 2= +

60a 2 25/

5d

a5

100 560300

= + = =


[ ]x n1 is right-sided signal

2,z z21

> >1 1 gives 2z >1


50/64


[ ]x n2 is left-sided signal

2,z z21<


51/64


pf (41.84 ) 0.74cos laggingc= =


As shown on the figure

Phase voltage on primary side (Y)

VPY3

6600= V

On the secondary side (T) line voltage and phase voltage will be same that is

VP V3 126600

L

#= = V

Phase current on secondary side

IP 10 12 120 A#= =

Line current on secondary side

IL 120 3= A

Output kVA 1203 3 12

6600

3## #

=

.66 3 114 3= = kVA


The current Imainand Iauxare shown in the following figure

The impedance angle of the main winding is

main .. .tan

4 53 7 39 61 c= = b l

To produce 90cphase difference between the currents in the main and auxiliary


52/64


winding, the impedance angle of the auxiliary winding circuit (including the starting

capacitor) must be

39.6 90.0 .50 4c c c= =

The combined impedance of the auxiliary winding and starting capacitor is equal

to

Ztotal 9.5 (3.5 )Z jX j X aux C C = + = + +

Where XC1

C = is the reactance of the capacitor to be added.

.

.tan X9 5

3 5 C1 + b l .50 4c=

.. X

9 53 5 C+ ( . ) .tan 50 4 1 21c= =

or XC 1.21 9.5 3.5 15.0 #= =

The capacitance Cis then

C( . )

177 FX1

377 15 01

C #

= =

=


The power flow diagram is

Air-gap power Pag P Pin SCL=

40 1.5 38.5kW kW kW= =

Rotor copper loss PRCL sPag=

0.04 38.5 1.54 kW#= =

Developed power Pd P Pag RCL=

38.5 1.54 36.96 kW= =

Pout P Pd r=

36.96 0.8 36.16 kW= =

Efficiency . 0.904PP

4036 16

in

out= = =


Given circuit


53/64


Convert delta-to-star(i.e in star each impedance is 15/3 5 /phase= )

Impedances in parallel per phase

( )j

j5 8 65 8 6

+ ++

jj

jj j

13 640 30

13 613 6

205700 50

#= ++

= +

. . . .j3 41 0 732 3 49 12 1c = + =

Total impedance per phase

Z 2 5 3.41 0.73j j= + + +

5.41 5.73 7.88 .j 46 65c = + =

Current drawn at supply

I ./

8.06A7 88

110 3= =

Line-to-line voltage

V2 . . 48.72 V3 8 06 3 49##= =


The admittance diagram is shown below:


54/64


Y11 . .j j j j0 5 5 5 10 5= = ;

Y22 0.5 2.5 5 8.0j j j j= =

Y33 . . .j j j j j0 5 5 10 2 5 18 0= = ;

Y44 .j j j j5 10 5 20 0= =

Y12 Y 021= = ; 5.0Y Y j13 31= = ; .Y Y j5 014 41= =

Y23 2.5Y j32= = ; 5Y Y j24 42= = ; 10.0Y Y j34 43= =

Hence the bus admittance matrix is given by

YBUS

.

.

.

.

.

.

.

.

.

.

.

.

.

.

j

j

j

j

j

j

j

j

j

j

j

j

j

j

10 5

0

5 0

5 0

0

8 0

2 5

5 0

5 0

2 5

18 0

10 0

5 0

5 0

10 0

20 0

=

R

T

SSSS

SS

V

X

WWWW

WWSOL 1.35 Option (B) is correct.

Let the base MVA is 6 MVA.

Per-unit impedance of transformer

XT . .j0 01 0 08= +

Per-unit impedance of each feeder

XF( )

( )(0.04958 0.09916)pu

jj

33

9 18 62

#=

+= +

Per-unit impedance of load

XL.

..

jj

5 8

0 2 60 207#= =

Thus, equivalent impedance diagram of the system is shown below


55/64


Equivalent impedance upto fault point

Z ( )eq pu ( . . ) ( . . ) (0.01 0.08)j j j0 04958 0 09916 0 04958 0 09916= + + + +8 B ( . . ) ( . . )j j0 02479 0 04958 0 01 0 08= + + +

Z ( )eq pu (0.03479 0.12958) 0.13417 . puj 74 97c= + =

Short-circuit rating of circuit breaker

.

44.72Base MVA MVAZ 0 13417

6( )eq pu

= = =


Given Synchronous generator of 500 MW, 21 kV, 50 Hz, 3-, 2-pole

0.9pf= , Moment of inertia 27.5 10M 3#= kg-m2

Inertia constant H ?=

Generator rating in MVA

G.

500 555.56cos

P0 9MW MVA

= = =

N 30002120 50pole120 f # #= = = rpm

Stored K.E 221

21

60M M N2

2

= = b l 27.5 10

21

602 3000 MJ3# # # #

= b l

1357.07 MJ=

Inertia constant ( )H Rating of Generator (MVA)

Stored K.E=

H..

555 561357 07

=

2.44 sec=


( )T s ( ) ( )

( )

s s s K s K

K s

3 3 3 2 4

24 3 2= + + + +

+

Routh table is shown below

s4 1 3 2 4K

s3 3 3K+

s2

( )K 3 12 + 2 4K

s1 ( )K

K K1233

++

s0 2 4K

( )

0K

312

> +

12K


56/64


2 4 0K > 2K>& and 33K> ,

These condition can not be met simultaneously. System is unstable for any value

of K.

SOL 1.38 Option (A) is correct.1. Over damped response (a, b)

Poles : Two real and different on negative real axis.

2. Under damped response (c)

Poles : Two complex in left half plane

Poles : Two complex in left half plane

3. Undamped response (d)

Poles : Two imaginary

4. Critically damped (e)

Poles : Two real and same on negative real axis.


For full scale deflection current 100 mA

Ifsd1 100 mA= Vm 100 1 100A k mVI Rm m #= = =

I1 I Ish m1= +

Ish1 I Im1= 100I I mAfsd1 1` = =

100 100 99.9mA A mA= =

Rs .1.001

mAmV

IV

99 9100

sh

m

1

1 = = =

Now for full scale deflection current of 1 A

Ifsd2 1 A=

Vm 100 mVI Rm m= =

Ish2 I Im2= 1I I Afsd2 2` = =


57/64


1 100 999.9A A mA= =

Rs .0.10001

mAmV

IV

999 9100

sh

m

2

= = =

Ratio R

R

sh

sh

2

1

.

.

100 10001

1 001

= =SOL 1.40 Option (B) is correct.

Secondary circuit phase angle

.

.tan1 51 33 691 c= = b l

cos 3 .6 0.8cos 3 9 32c= = , and

or, sin 3 .6 .sin 3 9 0 555c= =

Turn ratio Kt NN

1300 300

p

s= = =

Magnetizing current

Im 90sinMagneti g AN mmf 1100p= = =

Secondary circuit burden impedance ( . ) ( . ) .1 5 1 0 1 82 2 = + =

Secondary induced voltage

Es 5 1.8 9 V#= =

Primary induced voltage

Ep9E

300 300s= =

Loss component Iwiron loss

Ep=

( / ). 40

9 3001 2 A= =

Phase angle cos sinK II I180 t sm w

= b l . .180

300 5100 0 832 40 0 555

#

# #

= b l

2.34c=


Let 3 kR1 = , 6 kR2 = , 50 nFC=

RsC

vR

v v

1

i i o

1

2+

0=

&

sR CRv

Rv

1

i i

1

1 2

+

+b l vo=

( )vRR

sR C1 1i1

21+ +: D vo=

Rv

R R sR R C i1

2 1 1 2+ +6 @ vo=

vv

i

oR

R RR RsR R C1

1

2 1

1 2

1 2= +

++: D


58/64


&vv

i

o (1 ( ) )RR

s R R C 11

21 2= + +b l

f dB3 ( )R R C2

1

1 2=

( ) (

1.59 kHz2 50

12 2

13k 6k n k)50n

= = =


Since the DC gate current is zero, v Vs GSQ =

IDQ ( )I K V V Q n GSQ TN 2= =

& 0.5 1( 0.8)VGSQ2=

VGSQ 1.51V vs= =

VDSQ 5 (0.5 )(7 ) ( 1.51) 3.01m k= = V

The transistor is therefore biased in the saturation region.

The small-signal equivalent circuit is shown below

vo (7 )g v km gs=

vgs , (7 )v vv

A g kii

ov m= = =

gm 2 ( )K V Vn GS TN = 2(1 )m= (1.51 0.8) 1.42 = mS

Av (1.42 )(7 )m k= 9.9=


Let Qnis the present state and Qn 1+ is next state of given X Y flip-flop.

X Y Qn Qn 1+

0 0 0 1

0 0 1 1

0 1 0 00 1 1 1

1 0 0 1

1 0 1 0

1 1 0 0

1 1 0 0

Solving from K-map we get


59/64


Characteristic equation of X Y flip-flop is

Qn 1+ Y Q XQ n n= +

Characteristic equation of a J K flip-flop is given by

Qn 1+ JQ KQ n n= +

By comparing

J Y= , K X=


Required error in MSB /1 2# LSB

Let %x! is the tolerance in resistance

RV

R xV

1100

+a k R

V21

2n 1#

x

x

1 100

1100

1

+

+

aa

kk

2

1n

#

2x100

n# 1 , (2 1) 1

x x100 100

n# #+

x 1002 1

1n# #


MVI A, DATA1 ; DATA1"A

ORA A ; Set flag

JM DSPLY ; If negative jump to DSPLY

OUT PORT1 ; A"PORT1

DSPLY : CMA ; Complament A

ADI 01H ; A + 1"A

OUT PORT1 ; A"PORT1

HLT

This program displays the absolute value of DATA1. If DATA is negative, it

determine the 2s complements and display at PORT1.


Equation xA B= is consistent only if ( ) ( )A A B: =

Otherwise system is said to be inconsistent i.e. possesses no solution. Now,

[ : ]A B ::

:

11

1

12

2

13

610

12

=

R

T

SSSS

V

X

WWWW

1

1

1

1

1

2

1

2

1

6

4

2

=

R

T

SSSS

V

X

WWWW

R R

R R

R

R

2 1

3 1

2

3

&

&

f p

1

0

0

1

1

0

1

2

3

6

4

2

=

R

T

SSSS

V

X

WWWW ( )R R R3 2 3&


60/64


& ( : )A B 3=

As one of the minor 0

1

0

0

1

1

0

6

4

2

!

Now, system is inconsistent if

( )A ( : )A B! i.e. ( ) 3A !

It is possible only when 3 0 = i.e. 3=


:x 0 0.2 0.4 0.6

On calculation we get

( )f x x y2=

( )f x1 .0 1996=

( )f x2 .0 3937=

( )f x3 .0 5689=Using predictor formula

y( )p4 (2 2 )y h f f f 34

0 1 2 3= + +

Here 0.2h=

y( )p4 . [ (. ) (. ) (. )]03

0 8 2 1996 3937 2 5689= + +

y( )c4 ( 4 ),y h f f f

3*

2 2 3 3= + +

f *4 ( , ) (. , . ) .f x y f 8 0 3049 07070( )p

4 4= = =

y( )c4 . [. (. ) . ] .0795 302 3937 4 5689 7070 3046= + + + =


RMS value of fundamental component is

Vrms1 10.8 VV

22 dc

= = 24V Vdc=

RMS harmonic voltage V, ,

/

nn

2

3 5 7

1 2

rms=

3

=; E/

[12 (10.8) ]V V /rms02

12 2 2 1 2= =

5.23V=

Total harmonic distortion (THD) THD

V V

1,

/

rmsn

n1

2

2 3

1 2

rms=

3

=; E/

V

V V

1

0

2

1

2

rms

rms

= -

THD.

. . . %10 85 23 0 484 48 4= = =


We have, RMS value of fundamental component


61/64


Vrms1 10.8VV

22 dc

= =

The lowest harmonic is third harmonic. Third harmonic voltage is

Vrms3 3.6 V

V

3 2

2 dc= = 24V Vdc=

HF for the third harmonic

HF3 .. 33.33%

VV

10 83 6

rms

rms

1

3= = =

DF of the third harmonic

DF3( / )

.. /

0.037 3.7%V

V 310 83 6 9

rms

rms

1

32

= = = =


The incremental fuel cost of G1

dP

dC

1

1 2 0.01P1= +


dPdC

1

2 2 0.02P2= +

For optimum load division the two incremental costs should be equal, that is

dPdC

1

1 dPdC

1

2=

. P2 0 01 1+ . P2 0 02 2= +

P1 P2 2= ...(i)

Total load by the two generators

P P1 2+ 450= ...(ii)From of equation (i) and (ii), we get

P1 300 , 150MW P MW2= =



dPdC

1

1 2 0.01P1= +

2 0.01 300 5 /Rs MWh#= + =


dPdC

1

2 2 0.02P2= +

2 0.02 150 5 /Rs MWh#= + =Hence the incremental fuel cost of the plant for most economic operation is 5 Rs/

MWh.


x2o 5s x u421

2 2= + ,

x1o x2= ,

y 5 4x x1 2= +


62/64


x

x

1

2

o

o> H x

x u

0

5

1

421

0

11

2= +> > >H H H , y xx5 4 12= 8 >B H


OMC

CA520

41

= => >H H

det 0OM= . Thus system is not observable

CM B AB0

1

1

421= = 8 >B H

det 1CM= . Thus system is controllable.


GeneratorG1:

Voltage drop per ampere 1 A40

280 240= =

Let output current is IG1, then output voltage will be

VG1 280 1 IG1#=

GeneratorG2:

Voltage drop per ampere 1.2 A50

300 240=

=

Let output current is IG2, then output voltage will be

VG2 1. I300 2 G2#=

For parallel operation, output of each generator would be same i.e.

VG1 VG2=

280 1 IG1# 1. I300 2 G2#= or .I I1 2G G1 2 + 20= ...(i)

Total current delivered

I IG G1 2+ 60= ...(ii)

Solving equation (i) and (ii) we get

IG1 23.6A= and 36. AI 4G2=

So the output voltage

VG1 256.4V VG2= =


PG1 256.4 23.6 6.05V I kWG G1 1 #

= = = PG2 256.4 . .V I 36 4 9 3 kWG G2 2 #= = =


63/64

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