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Mains Exam Solution

SECTION-A

Q–1(a) For the circuit shown in the figure below, derive the expression for output voltage andsketch the nature of the output when V2 = 10V and V1 = 5V.

R1

R1 C1

C1

+ +––

V1 V2

V

Sol: The given circuit in s-domain can be drawn as shown below.

–

+

V (s)1

R1

V (s)x

–

+

R1

V (s)2

1sC1

a V (s)s

1sC2

Applying the concept of virtual ground

Vt = V1(s) and Va = V2(s)

So, 1 1 x

1

1

V s V s V s1 R

sC

= 0

or, 11 1

1

V ssC V s

R =

x

1

V sR

or, Vx(s) = (sC1R1 + 1)V1(s) ...(1)

Similarly,

2 x 2 0

1 1

V s V s V s V sR V / sC

= 0

or, 2 11

1V s sCR

= x

1 01

V ssC V s

R

or, 2 1 1V s 1 sR C = [1 + sC1R1]V1(s) + sC1R1(s)

or, sC1R1V0(s) = (1 + sC1R1)[V2(s) – V1(s)]

Given that V1 = 5V and V2 = 10V

So, V1(s) =5s and V2 =

10s

sC1R1V0(s) = 1 110 51 sC Rs s



Mains Exam Solution

or, V0(s) = 1 1

21 1

5 1 sC Rs R C

or, V0(s) = 21 1

5 1 5·R C ss

Taking inverse laplace transform of above equation

V0(t) =1 1

55 1 tR C

5V

V (t)0

Q–1(b) A continuous LTV system S with frequency response H j is constructed by cascading

two continuous time LTI sytstems with frequency response H1 j and H2 jrespsectively. Figures a and b show the straight line apporximations of bode magnitudeplots of H1 j and H j , respectively, Find H2 j .

6

1 810 40 100 Q(rad/s)

20 dB/decade –40 dB/decade24

20log |H (j )|10 1 20log |H (j )|10 1

8 10 (rad/s)

(a) (b)

Sol. Given Bode plot of H1 j

+20

–40 dB/decade

W (rad/sec)

6

24

20 log 10 |H (jw)|1

Transfer function of H1 (jw) is



Mains Exam Solution

H1 (s) =

sk 11

s s1 18 40

H1(s) =320k(s 1)

(s 8)(s 40)

for k value use line equation

P {i.e. y = mx + c}

Here C = 20 log k

from graph 6 = 0 + 20 log k

k = 2

oo, H1(s) =320 2(s 1)(s 8)(s 40)

=

640(s 1)(s 8)(s 40)

Now Bode plot of H (jw) is given

–40 dB/decade

W (rad/sec)8

–20

20 log |(H (jw)|10

Transfer function of above response is

H(s) = 2 2k 64k

(s 8)s18

for k use line equation

i.e. y = mx + c

–20 = 0 + 20 log k

k = 0.1

so, H(s) = 26.4

(s 8)

In the quesion given that H(jw) cascaded by two (LTI system H, (jw1) & H2 (jw)So, H(s) = H1(s) · H2 (s)

H2(s) =2

1

6.4 / (s 8)H(s)H (s)

640(s 1) / (s 8) (s 40)

Hs(s) =0.01(s 40)(s 1)(s 8)

Ans.



Mains Exam Solution

Q–1(c) Consider a three phase induction motor with the following parameters:Number of poles : 4Supply frequency : 50 HzFull load speed : 1470 rpm

Rotor resistance : 0.12

Standstill reactance : 1.12

Find the(i) Slip for maximum torque(ii) Ratio of maximum torque to full load torque.

Sol: Given, P = 4, f = 50 Hz, Nr = 1470 rpm

r2 = 0.12 , x2 = 1.12

Synchronous speed, Ns =120f 120 50

P 4

or, Ns = 1500 rpm

Full load slip, sf1 = s r

s

N N 1500 1470N 1500

or, sf1 =30 0.02

1500

(i) The slip at which maximum torque occurs is given by

smr =2

2

r 0.12 0.107x 1.12

(ii) We know that

e.f

em

TT

=

mr f

f mr

2 2s s 0.107 0.02

0.02 0.107s s

ore.f

em

TT

=

2 25.357 0.1867 5.544

required ratio,

em

ef

TT

=5.544 2.772

2

Q–1(d) (i) What is smart grid?(ii) Compared to supervisory control and data acquisition (SCADA) system. What are the

advantages of phasor measurement unit (PMU)?(iii) Explain operation of PMU with a neat diagram.



Mains Exam Solution

Sol. (i) A smart grid is an electricity network that uses digital and other advanced technology tomonitor and manage the transport of electricity from all generation sources to meet thevarying electricity demand of end users.

Smart grid system is automated for tracking the electricity consumption at all the location.

Smart grid has following capabilities:-

It can repair itself

It encourages consumer participation in grid operation.

It allow electricity market to grow

It can be operated more efficiently.

(ii) Comparison between SCADA AND PMUs.

Measurment Analogue DigitalResolution 2-4 samples per cycle up to 60 samples per cyclesObservability Steady state Dynamic/TransientMonitoring Local Wide-AreaPhasor angle measurement No Yes

Attribute SCADA PMU

From above table it is concluded that the PMUs are given more accurate & efficient resultsas comparisons to SCADA systems.

(iii) The GPS system consists of 24 satellites in six orbits at an approximate altitude of 10,000miles above the surface of the earth. They are thus approximately at one half the altitudescorresponding to a geo-synchronous orbit. The positioning of the orbital plane and the positioningof the satellites in the orbits is such that at any given instant at least four satellites are in viewfrom any point on the surface of the earth. Often, more than six satellites are visible.

GPS receiver

Analog inputs

Anti-aliasing filter

A/D converter

Phasor microprocessor

Moderms

Phase locked oscillator

PMU

Data concentrator and collator

Advanced applicationSoftware

System control, protection functionsArchival data base

PMU

PMU

PMU

Block diagram of the phasor measurment unit

PMU utilization in a power system



Mains Exam Solution

The civilian-use channel of the GPS system transmits positional coordinates of the satellitesfrom which the location of the receiver station on earth could be determined. The Satellitestransmit a one-pulse per-second signal, along with an identifier for the signal that can beinterpreted by the earth station receivers. The civilian-use transmission of the time signal isprecise to within 1 microsecond, and often in practice is found to be much more accurate.The normal practice is to phaselock a sampling clock to this pulse. The sampling instantwould be identified as the pulse number within a one-second interval identified by the GPStime-tag. The exact format for time-tagging is defined in IEEE standard 1344. It should bementioned that a time standard known as the IRIG-B standard is currently being used by thepower industry for time-tagging digital fault recorders and other substation event monitoringsystems. However, with standard IRIG-B receivers the synchronization accuracy is of theorder of 1 millisecond, which is not enough for precise power measurement (a tolerance of1 millisecond corresponds to an uncertainity of about 20°). The complete block diagram ofPMUs and PMUs utilization in power systems are shown in figure 3.11 and 3.12.

Q–1(e) A PV pane is connected with a single phase fully controlled converter as shown in thecircuit below. The panel is supplying a current of 5A and generated power is 1000W. Theseries inductance in the circuit is large to make the current flat and continuous. Find (i)the triggering angle of the thyristor birdge, (ii) output voltage at rectifier terminal, and (iii)input power factor.

PV panel

5AV = 230V

50Hzac

R = 1

L = Large

+

–

Sol. VAC = 230 Volt

f = 50 Hz

Io = 5 A

P0 = 1000 watt

–

+

PVPanel

–

+

T1 T3

T2T4

V0

1

R L = Large

Vp

(a) Single phase fully controlled converter (in inversion mode > 90°)

Panel voltage =1000 200 Volt

5

(i) Vo = Io × R – Vp = (5 × 1 – 200)

oV 195 volt

Vo = m2 V cos

https://iesmaster.org/testseries/ssc-je



Mains Exam Solution

= triggering angle of the thyristor

= 1 195cos 160.342 2 230

(ii) Output voltage at rectifier terminal= – 195 volt

(iii) IPF (input power factor)

Po = Pin

2o1000 I R = VsIsr IPF

IPF =1000 25

230 5

1025IPF 0.891230 5

Q–2(a) The DC-DC converter given below is operating at 30 kHz and drawing an input current of25 A at 48 V DC.

(i) For a load current of 10A, find(i) the duty ratio of the switch,(ii) output voltage

(iii) peak inductor current,(iv) output voltage ripple, and(v) the load current where the inductor just becomes discontinuous.

Sol.

LoadwVin

2000 H+

–

L

(a) Boost converter

C = 1000 F

Given: DC to DC converter

f = 30 kHz

Input current = 25 A

Input voltage = 48 V DC

(i) load current = 10A

(I) For lossless system

Po = Pin

VoIo = VinIin

Vo =48 25

10

= 120 volt



Mains Exam Solution

Vo = inV1

1 =48

120

=48 31 0.6

120 5

Duty ratio of the switch 0.6

(II) Output voltage = 120 volt

(III) Peak inductor current

T T

Io

–Io

Vo

Io

ic

VC

Imm

Iml

t

t

t

l – Ii o

VC

(a) Wave form of current and voltage

Imax = Iin + LI2

= inv252fL

= 3 60.6 4825

2 30 10 200 10

maxI 27.4Amp

(iv) Output voltage ripple CV

CV = oIfc

CV = 3 60.6 10

30 10 1000 10

= 36

30 1000 10

CV 0.2 volt

(v) At the boundary :

TT

Ls

Li

IL

(iL)ay = Is = oI1

= LI2

oI B1

= in3 6

V 0.6 482fL 2 30 10 200 10

IOB = (1 – 0.6) × 2.4 = 0.96 Amp

Q–2(a) (ii) Also find the critical value of L to keep the inductor current just continuous when theinput voltage changes to 60 V with output remaining same. (Assume lossless operation ofconverter components)



Mains Exam Solution

Load8wVin

L = 200 H+

–

C=1000f

Sol.

LoadwVin

L=200 H+

–(a) Boost converter

C

1000 F

D

Given: Vn = 60 volt

Output remain same Po = constant

o

o

V 120voltI 10A

Critical inductance

TT

Iml

IL

Imn = 0

ilag = oL

inII I

2 1

LI = o2I

1

in

Co

1 VL

2I f

...(i)

Duty ratio

Vo = inV

1

1 =in

o

V 60 1V 120 2

12

From equation (i)

LC =3

1 11 602 2 24 F

2 10 30 10

CL 25 H



Mains Exam Solution

Q–2(b) A signal m(t) = 2 cos 20 t – cos 40 t , where the unit of time is millisecond, is ampli-tude modulated using the carrier frequency (fc) of 600 kHz. The AM signal is given by

s(t) = c c5 cos2 f t m t cos2 f t

(i) Sketch the magnitude spectrum of s(t). What is its bandwidth?(ii) What is the modulation index?(iii) The AM signal is passed through a high pass filter with cut-off frequency 595 kHz (i.e

the filter passes all frequencies above 595 kHz, and cuts off all frequencies below595 kHz). Find an explicit time domain expression for the quadrature component ofthe filter output with respect to a 600 kHz frequency refrerence.

Sol: (i) For an amplitude modulated signal the amplitude of modulated signal varies with the mes-sage signal.

Here is message signal we have more than are frequency signal multione AM

SAM(t) = c

m t5 1 cos2 f t

5

SAM(t) = c2 15 1 cos 20 t cos40 t cos2 f t5 5

SAM(t) = c 1 1 2 2 cA 1 cos2 m t cos2 m t cos2 f t

fm1 = [10 Hz] [103]

fm2 = [20 Hz] [103] fi = 600 kHz

Magnitude spectrum

–1/4 –1/4

580 kHz 620 kHz

590kHz 610kHz600kHz

1/2 1/2

5/2

Bw = 2fm1 = 2 fm2 = 2(20) = 40 kHz

(ii) t = 2 21 2 =

2 22 15 5

=1 0.44725

(iii) Before filter the signal was c 1 c 1c c c m1 c t

A AA cos 2 f t cos 2 f f t cos f m t2 2

c 2 c 2c m2 c m2

A Acos2 f f t cos2 f f t

2 2

After filter it will be

c 2 c 2c c c m1 c m2

A AA cos 2 f t cos2 f f t cos2 f f t2 2



Mains Exam Solution

SAM(t) = c c m1 c m215cos 2 f t cos 2 f f t cos 2 f f t2

SAM(t) = 5cos 1200 t cos 1220 t 0.5cos 1240 t

Q–2(c) A 400 V DC shunt motor has armature and fild resistances of 0.2 and 200 respec-tively. It draws current of 6 on no load and 70 A on full load. If its no load and full loadspeeds are the same, determine the field weakening due to load current as percentage ofno load flux.

Sol: Given, Vt = 400 V, ra = 0.2 and rf = 200

ra

Ia

rf

If

I+

–

Vt

DC shunt motor

We know that speed a DC motor

m = a L a a

a a

E V I rk k

no load speed m0 = t a a

a

V I rk

and pull load speed mf1 = 2L a a

a 2

V I r

k

In DC shunt motor I = Ia + If

and If =t

f

V 400 2Ar 200

at no-load, I = 6A

Ia1 = I – If = 6 – 2 = 4A

and at full load, I = 70A

So, Ia2 = I – If = 70 – 2 = 68A

Given that no load speed is same as full load speed

So, 1t a a

a

V I rk

= 2t a a

a 2

V I rk

2

1

= 2

1

t a a

t a a

V I r 400 68 0.2V I r 400 4 0.2



Mains Exam Solution

or,2

1

=

400 13.6 386.4400 0.8 399.2

or,2

1

= 0.968

or,2

11

= 0.968 – 1

or,2 1

1

= –0.032

or,2 1

1100

= –3.2

Field weaken by 3.2% of no-load flux.

Q–3(a) A salient pole star connected altenator is connected to infinite bus operating at 1.0 p.uvoltage. The alternator has Xd = 0.75 p.u and Xq = 0.5 p.u on per phase basis. It is deliver-ing 1.0 p.u power to the infinite bus at 0.8 p.f lag. Calculate (i) the load angle and excita-tion voltage under this condition (ii) the maximum power that can be delivered by thealternator with same excitation and the corresponding load angle (iii) the armature cur-rent and p.f under maximum power condition, and (iv) the theoretical value of maximumpower that the alternator can deliver when its field circuit is suddenly disconnected dueto fault.

Sol. Given xd = 0.75 P.u.

xq = 0.5 P.u.

V = 1 P.u.

Power P = 1 p.u.

P.F. = 0.8 lag

(i) As we know for lagging p.f. alternator

tan = a q

a a

V sin I xVcos I R

Ra = 0

and = +

Id = Ia sin from phasor diagramIq = Ia cos

E = V cos + Id xd for Ra = 0Delivered power = 1

Power = Vpu . Ipu . cosI = 1 × Ia × 0.8

Ia = 1.25 p.u.

So, tan =1 sin(36.86 ) 1.25 0.5

1 cos(36.86 ) 0

= 56.85°

=

=

= 56.85° – 36.86°

https://iesmaster.org/testseries/uppsc-ae



Mains Exam Solution

20 Ans.

then Id = aI sin 1.25 sin 56.85

Id = 1.046 P.u.

So, Ef = V cos + Id Xx

Ef = 1 × cos 20° + 1.046 × 0.75

fE 1.724 p.u. Ans.

(ii) Power equation of salient pole alternator is

P =2

f

d q d

VE V 1 1sin sin2X 2 X X

Here Ef = 1.724 p.u.

V = 1 Pu

then

P =21 1.724 1 1 1sin sin2

0.75 2 0.5 0.75

P = 2.298sin 0.33sin2 …(i)

for maximum power dpd = 0

dpd = 2.298cos 0.66cos2 0

Using numerical method

= 74.98°

Put value in eqn. (i)

Pmax = 2.298 × sin74.98 + 0.3 sin (2 × 74.98)

Pmax = 2.3846 P.u.

(iii) Armature current and p.f. under maximum power condition.

Ef = 1.724

V = 1 P.u.

max = 74.98°

So alternator equation

E = V cos d dVcos I x

1.724 = 1 × cos 74.98° + Id × 0.75

Id = 1.9531 p.u.

Power Pmax = VIa cos {

2.3846 = 1 × Ia cos max

2.3846 = a aI cos cos74.98 I sin sin74.98

2.3846 = Iq × 0.259 + Id × 0.9658



Mains Exam Solution

2.3846 = Iq × 0.259 + 2.253 × 0.9658

Iq = 1.9224 p.u.

so, Ia = 2 22 2d qI I 1.9531 1.9224 2.7405 p.u.

Iq = Ia cos

cos =q

a

I 1.9224I 2.7405

= 45.45°

= = 45.45° – 74.98°

30

p.f. = cos 30° = 0.866 leading

aI 2.7405 p.u.P.f. 0.866 leading

(iv) Max power when Ef = 0

max at =45P =

2

q d

V 1 1 sin22 X X

Pmax =1 1 1 sin 902 0.5 0.75

maxP 0.333 p.u. Ans.

Q–3(b) A closed loop system with unit feedback and having the forwad loop transfer function as

14.4G s

s 1 0.1s

Modify the design using cascaded compensation to satisfy the optimum performancecriterion, so that the transient response to unit step input reaches its final steady statevalue in minimum time without having any overshoot.

Sol: We know that

The transfer function of proportional can derivative controller = kp + kDS

Given:

Forward loop transfer function G(s) = 14.4

S 1 0.15

Open loop transfer function of new system

= p Ds14.4 k k

S 1 0.15

Characteristics equation is S(1 + 0.1 S) + 14.4 kp + 14.4 kDS = 0

2DS p0.1 S S 14.4 k 14.4 k = 0

2D pS 144 k 10 S 144 k = 0

Comparing with standard



Mains Exam Solution

Equation 2 2n nS 2 S 5 = 0

We get,

2n = 144 kp

n = p12 k rad/sec

kp = 1

n = 12 1

n = 12 rad/sec

n2 10144

= kD

Dk 0.0972

Overall system will be:

14.4

S 1 0.1 S(1 + 0.0972 S)R(s)

C(6)

Q–3(c) Two 11 kV, 30 MVA, three phase synchronous generators operate in parallel supplying asub station through a feeder having an impedance of (0.6 + j0.8) ohms to positive andnegative sequence currents and (1.0 + j2.6) ohms to zero sequence currents. Each gen-erator has x1 = 0.8 ohms, x2 = 0.5 ohms and x0 = 0.2 ohms and has its neutral groundedthrough a reactance of 0.2 ohms. Evaluate the fault currents in each line the potentialabove earth occuraence of earth fault on the y and B phases at the sub station.

Sol. Given, two 3-phase synchronous generator operating in parallel.

1 2 0z z z

Feeder (0.6 j0.5) 0.6 j0.8 1 j2.6Each generator j0.8 j0.5 j0.2

Zn = nutral impedance = j0.2 Both the generator are concentred in parallel. (Zg)1 = + ve sequence equivalent impedance of the parallel combination

(za2) = –ve sequence equivalent impedance of the parallel combination(zgo) = Zero sequence equivalent impedance of the parallel combination

(Zg)1 = 9 2j0.8 j0.5j0.4 ,(z ) j0.252 2

(zg)0 =j0.2 3 j0.2 j0.4

2

The net +ve sequence impedence from the generator to the substation(Z1)eq = (zg)1 + (zf)1

= j 0.4 + 0.6 + j 0.8



Mains Exam Solution

= (0.6 + 1.2)

The net –ve and zero sequence impedence from the generator to the substation(Z2)eq = (Z9)2 + (Zf)2 , (Zo)eq = (z9)1 + (Zf)0

= 0.6 + j 0.8 + j 0.25 = (0.6 + j 1.05)

= (j 0.4 + 1 + z 2.6 = 1 + j3)R

Y

B

We have, the +ve sequence currentfor a simultaneous earth fault at Y & B phase of the substitions,

Where 1 ve sequence generated voltage = 11 kV3

IR1 = 1

1 0 2

EZ (y z )

IR2 =0

R10 2

zI

z z

IRo =2

R10 2

zI

z z

111/ 3KVIR

0.6 j1.2 1 j3 0.6 j1.05

IR2 = 1R70 I

70 72

zo z2 = 0 2

0 2

1 j3 0.6 j1.05z z 3.8243 131.82z z (1.6 j4.05) 4.355 68.44

IR1 = 11 1 0.878 63.38

0.6 j1.2 0.3934 j0.7853

6.351 0 6351

0.9935 51.985 2.22 63.414

1RI

= 2.8608 < –63.414 KA

2RI

= 0

R10 2

zI

(z z )

zo + z2 = 4.355 < 68.44 , zo= (1 + f3)= 3.162 < 71.565

2RI

=

3.162 1.565 2.8608 63.414 kA4.355 68.44

= –2.077 < –60.289 kA



Mains Exam Solution

R0I

= 2

R10 2

z (0.6 j1.05)I 2.8608 63.414 kAz z 4.355 68.44

= 1.21 60.255 2.8608 63.414 kA

4.355 68.44

= –0.7948 < –71.6 °KA

R R1 R2 R0I I I I = – 7948 < – 71.6° KA

= 0

Iy = Iy1 + IY2 + IYo = 2R1 R2 R0I I I

where, = < 120°, 2 = < – 120°

Iy = 2.8608 < (–63.414 – 120°) – 2.077 < (120° – 60.289)

– 0.7948 < –71.6°

Iy = (–4.154 – j 0.8689) KA

= 4.244 < – 168.186 kA

IB = IB1 + IB2 + IBo

=2

R1 2 ROI IR I

= (2.8608 < 120° – 63.414 – 2.077 < –120° – 60.289°

– 0.7948 < – 71.6°) kA

= (3.4015 + j 3.32) kA = 4.624 < 42.638 °kA

current through the neutral = 3 Iao

= 3 × (0.7948 < –71.6°) kA

= 2.3844 < 108.4 ° kA

The potential above earth,

= 3 Iao Zn

= 2.3844 < 108.4° × 0.1 < 90°)

= 0.23844 < 198.4° kV

The different branch currents are,

R y B

nutral

I 0; I 4.244 168.186 kA; I 4.624 42.638 kA

I 2.3844 108.4 kA

The potential above earth = 0.23844 < 198.4°kV Ans.

Q–4(a) A signal g(t) band limited to B Hz is sampled by periodic pulse train pT(t) made up of a

rectangular pulse of width 1

8B sec (centered at origin) repeating at the nyquist rate (2 Bpulse per sec). Show that the sampled signal gs(t) is given by

sn 1

1 2g t g t g t cos 4n Bt4 n

How will you recover g(t) from the signal gs(t)?



Mains Exam Solution

Sol: Given,

Rectangular pulse of width =1 sec

8B

Nquist rate = 2B pulsed per sec

s1T

2B

n(t)

12B

1

16B

116B

12B

t

Fouriers series of T t can be written as

n t = o n o n ox 1

a a cosn t b sin t

ao = nT

1 t dtT

D.ax = n oT

2 t cosnw tdtT

and bx = n oT

2 t sinnw tdtT

Given signal is even signal

bn 0

ao = Tno

1 t dtT

=1 16

116B

1 1 dtT

o1a4

an = T 2n oT 2

2 t cosnw t dtT

an = 116B

1 16B

2 1 cos 4B nt dt1 2B

= sin n 4 sin n 4

4B4B n

an =2 nsinn 4

https://iesmaster.org/testseries/gate-2021



Mains Exam Solution

PT(t) = n 1

1 2 sin n 4 cos 4B nt4 n

gs(t) = ng t t

=

n 1

g t 2 nsin g t cos 4B nt4 n 4

the g(t) is,

g(t) =

g(t)

–B B t

The fourier tranform of g(t)

–B B t

2 sinB

= –B B

From multiplication property of fourier transform.

1 2 1 2m t m t M f M f

FT2 sin g t cos4 Bt4

–3B B–B 3B–2B 2B

When

LPF g(f)g(t)

the low pass filter output

= B–B t

Ho(f)

i.e fH f 4rect2B

Q–4(b) A 3-phase half controlled rectifier with free wheeling diode is supplying a separately ex-cited DC motor for speed control purpose. The AC input to the converter is 415 V, 3-phase,50 Hz. The motor parameters are: V = 220 V DC, P = 10.5 kW

Rated speed = 1100 rpm, armature resistance ra = 0.4

The field current is kept constant at rated value. The motor is operated at rated speed



Mains Exam Solution

delivering half rated torque.(i) Find motor terminal voltage and triggering angle of thyristor bridge.(ii) Find the speed of the motor if one of the input phases to the converter is out due to fault

and triggering angle is kept as before with same load torque.(iii) Also find the new triggering angle if the motor speed is to be maintained at rated value

with same load torque. (Neglect losses in the machine)

Sol: (i) Given:

3 half controlled rectifier:

Eb

+

–

V0

I0

T1

D2 D6 D4

T3 T5RYB415 V50 Hz

La

ra

Vo = 220 V

ra = 0.4

N = 1100 rpm

P0 = 1.5 kW

Field current is kept constant at rated value.

i.e constant

P0 = V0 I010.5 × 103 = 220 × I0

Rated current I0 =310.5 10 47.73A

220

V0 = Eb1 + I0ra

Eb1 = 220 – I0ra

= 220 – 47.473 × 0.4

Eb1 = 220.91 V

For seperately excited DC motor

i.e constant

I

at rated torque rated 0I ...(i)

at half rated torque rated1I2

...(ii)

From equation (i) & (ii)

1

0

II =

12



Mains Exam Solution

I1 = 0I2

I1 =47.73

2I1 = 23.865 Amp.

So, V0 = Eb1 + I1ra

V0 = 200.91 + 23.865 × 0.4

0V 210.456 V

As we know for half controlled rectifier

V0 = mL3V 1 cos2

210.456 = 23 f 415 1 cos2

cos = –0.2493

104.44

(ii) If one of the input phase to the converter is out due to fault then its works like 1 halfcontrolled rectifier.

EbD2 D6

T3T1R

B

415 V

50 HzLa

ra

For this b/w V0 = mV 1 cos

V0 = 2 415 1 cos104.44

V0 = 140.3 V

Eb2 = V0 – I1 ra

Eb2 = 140.3 – 23.865 × 0.4

= 130.755 V

As we know, bE N N speed rpm

b1

b2

EE =

1

2

NN

200.91130.755

=2

1100N

N2 = 715.895 rpm

(iii) If the motor speed is to be maintained at rated value with same load torque

i.e Eb1 = 200.91 V

I1 = 23.865 (Half rated torque)



Mains Exam Solution

V0 = Eb1 + I1 ra

V0 = 200.91 + 23.865 × 0.4

V0 = 210.456 V

So, V0 = mV 1 cos

210.456 = 2 415 1 cos

82.76

Q–4(c) The figure below shows single line diagram of a power system with generators at bus-I

and bus-3. The voltage at bus-1 is 1.05 0 p.u and at bus-3, V = 1.04 p.u. Line imped-ances are in p.u and line charging susceptances are neglected. Obtain state vector usingfast decoupled load flow (FDLF) for one iteration.

z = (0.02 + j0.04)p.u12

3

1 2

P = 400 MWQ = 250 MW

2

2z = (0.0125 + j0.025)p.u

23

z = (0.01 + k0.03)p.u13

Slack bus

P = 200 MW|V | = 1.04 p.u

3

3

Sol. Given: The current diagram as:1 2Z = (0.02 + j 0.04)pu12

P = 400 mW2

Q = 250 mVAR2

V = 1.05 < 0°1

Z = (0.01 + j 0.03)pu13

3 Z = (0.0125 + j 0.025)p.u23

P = 200 mW3

|V | = 1.04 pu3

y12 = 12

1 1 1 22.36 63.435 (10 j20)z (0.02 j0.04) 0.04472 63.435

y23 =23

1 1 1 35.778 63.435 (16.45 j32)z (0.0125 j0.025) 0.02795 63.435

y13 =13

1 1 1 31.6456 71.565 10 j30z (0.01 j0.03) 0.0316 71.565

Y11 = y12 + y13 = 20 – j50 Y22 = y12 + y23 = 26 – j52

Y12 = –y12 = –10 + j20 Y23 = –y23 = –16 + j 32

Y13 = –y13 = –10 + j30 Y21 = –y12 = –10 + j20



Mains Exam Solution

Y33 = y13 + y23 = 26 – j 62

Y31 = y13 = –10 + j30

Y32 = –y23 = –16 + j32 p.u

[y]Bus=

20 j50 10 j20 10 j3010 j20 26 j52 16 j3210 j30 16 j32 26 j62

The expression of power P at Bus i is

Pi = n

i j ij ij i jj i

v v y cos

…(i)

Qi = n

i j ij ij i jj i

v v y sin

…(ii)

In fact decompled load flow soln.

2

3

2

3

PPQQ

=

2 2

2 3

23 3

2 3 3

2 2 2

2 3 3

3 2

2 3

P P0 0

P P0 0

Q Q V0 0V V V

0 0V V

2

2

P =

3 22 j 2j 2j 2 j 2 22 22

j 1

V V V sin V y sin

or,2

2

P = – Q2 – |V2|

2 B22

B22 = |y22| sin 22

ii i2

i i

assuming, B Q and

v v

We obtain,

2

2

P = – 2 22V B

2

3

P = 2 3 23 23 2 3V V Y sin

* 2 3 3assuming, is too small & V 1.04 1P.u

2

3

P = 2 23 23V Y sin

= 2 23V B



Mains Exam Solution

similarly,3

3

P =

33 33 3 32

2

PV B and V B

P

2

2

3

3

PV

PV

= 22 23 2

32 33 3

B BB B

…(I)

B22 = 12 22y sin 52

B23 = 23 23y sin 32

B32 = 32 32y sin 32

B33 = 33 33y sin 62

[B] = 152 32 0.028182 0.014545B

32 62 0.014545 0.023636

2

2

QV

= 2 22 22 2v y sin Q

where, Q2 = 3

2 i 2j 2j 2 jj 1

V V Y sin

22 22 22 2assuming, y sin

then, 2

2

QV

= i 22 22y y sin

and 2

3

QV

= 0

2Q =32

22

V 0QV

V as Bus 3 is PV Bus

or,2

2

QV

= 22 2B V …(ii)

assuming a base of 100 MVA

(400 + j250) =400 j250 P.u

100

= (4 + j2.5) p.u

2

2 2(Scheduled) 22

P (sch) 0 4 4 p.uP P P (cal)

Q (sch) j(0 2.5) j2.5

P2 (Cal) = n

2 j 2j 2 j ijj 1

V V cos ( ) y

assuming, similarly 2 3 20 & 0 & V 1.0 p.u

P2 (cal) = 21 2 1 12 23 2 3 23 22 221.05cos y 1.04cos y cos y

21 23 22116.656, 116.565, 63.435

https://iesmaster.org/testseries/ese-prelims-2021



Mains Exam Solution

= – 1.14 p.u

2P 4 (1.14) 2.86 p.u.

similarly,

Q2(cal) = n

2 j 2j 2 j 2j i

V V sin Y

Q2 (cal) = –2.28 p.u.

2Q = 2(sch.) 2(cal.)Q Q

= – 2.5 – (–2.8) = –0.22

(1)2

(0)3

= 1 2 2

3 3

P / VB

P / V

as obtained earlier, from eqn. (1)

[B]–1 = 0.028182 0.0145450.014545 0.023636

(1)2 = – 0.060483

(1)3 = – 0.008909

Similarly, from equation (ii)

2V = 2

2 22 2

Q1V V

= 1 2.2 0.0042308 p.u

52 1.0

(1)2

(1)3

(1)2

0 ( 0.060483 0.060483

0 0.008909 0.008909

V 1 ( 0.0042308) 0.995769

There are the various after 1st interal

SECTION-BQ–5(a) A thryristor is having the I-V characteristic as given in the figure below. It is used in a half

wave rectifier circuit with resistive load operating at 30 and carrying a peak loadcurrent of 100 A. Determine the average conduction loss in the thyristor.

id

200 A

1.0 V 2.0 V Vd



Mains Exam Solution

Sol.

ACsupply

(a)

RLoad

id

200 A

1.0 V 2.0 volt

RD

From figure (b)

Voltage drop across thyristor = 1 V

Resistance of thyristor = 1/200

3DR 5 10

Given Ip = 100 A

= 30°

output average current

Io = m1 I sin t d t

2

= mI cos t2

mo

II 1 cos2

I0ay = 100 1 cos30 39.69 Amp2

Ip

Lo

Vo

Vs

Output rms current

I0ms = 1/2

2m

1 I sin t d t2

= 1/22

m 2Isin t d t

2

I0rms = 1/2

mI 1sin222

= 1/2100 1/ 6 sin60

22

0mI 49.27 Amp

Conduction loss of thyristor



Mains Exam Solution

Ploss = 20rm D 0ay TI R I V

Ploss = (49.27)2 × 5 × 10–3 + 29.69 × 1

lossP 41.82 watt

Q–5(b) A three-phase equilateral transmission line has a total corona loss of 55 kW at 110 kV and100 kW at 114 kV. What is the disruptive critical voltage between lines ? What is the coronaloss at 120 kV ?

Sol. Given,

Sl. Voltage Corresponding CronaNo. loss1. 110 kV 55 kW2. 114 kV 100 kW

Corona loss Pc = 2 5

df 25 r243.5 V V 10

D

kN/km/hour

where, V = Phase voltage in kv (rms value)

Vd = Disruptive critical voltage in kv (rms value)

r = Radius of conductor in metres

D = Spacing between conductors in metres

f = System frequency

So, 2c dP V V

1

2

c

c

PP =

21 d

22 d

V V

V V

or55

100=

2d

2d

110 V

114 V

or (110 – Vd) = d0.55 114 V

or Vd – 0.7416 Vd = 110 – 84.5447

or 0.2584 V2 = 25.4553

or Vd(L) – L = 98.51 V, VD = (Ph) = 98.51

3 = 56.876

Corona loss at 120 kV

cP100

=

2 2

2 2120 98.51 21.49

114 98.51 15.49

or Pc =461.8201100239.9401

= 192.473 kW

Let power loss at 120 kV be ‘P’



Mains Exam Solution

P55 =

2

2120 98.52 35

(110 98.52)

PL = 3.5 × 55 kW = 192.55 kW

Pcorona at 120 kV = 192.55 kW Ans.

Q–5(c) A Gaussian pulse is specified by

g(t) = 2 2tAe ,

where is an arbitrary attenuation coefficient and A is constant. Show that the Fouriertransform of g(t) is also Gaussian.

Sol. Gaussian pulse

g(t) = 2 2tAe

As we know, fourier transform of g(t):

G(j ) =j tg(t) e dt

G(j ) =2 2t j tA e e dt

=2 2( t j t)A e dt

... (i)

Let 2 2t j t = 2P 2pq2 2t j t = 2 2(p q) q

2 2 2p t 2pq = j t

p = t q =j t j t2P 2 t

q =j2

So, 2 2(p q) q =2 2j jt

2 2

Let dt = du

dt =du

So, from equation (1)

G(j ) =

22

2u4 duA e

=

2

2 2u 4 duA e .e

=

2

22 u4A e e du

=

2

22 u du4A 1e . e



Mains Exam Solution

2

24AG(j ) e

2u1 e du 1

G( ) =

2 2

2f

1 A e

This is also a gaussian function.So, we can say fourier transform of g(t) is also gussian.

Q–5(d) What are the advantages and limitations of lead and lag networks in a practical controlsystem ?

Sol: Lag NetworkAdvantages:

A phase lag Network offers high gain at low frequency. This it performs the function of a lowpass filter.

Increases the steady state performance of the system.

The lag network offers a reduction in bandwidth & this provides longer rise time & settingtime.

Disadvantages

Its shows more sensitivity towards variations in the parameter.

Due to the addition of external network, some attenuation is introduced. So the overall gainmust be increased to handle the attenuation. But this will increase the requirement of moreelements & so the cost & space requirement.

Adversely affect the stability of the system.

Decrease bandwidth, because response is quite slow.

rBW t 0.35

Lead Network:Advantages

Improves damping of the overall system.

Transient respones gets improved.

Improve the phase margin.

Quick response therefore increase bandwidth.

It maximizes the velocity constant of the system.

Disadvantages:

Its also adds some attenuation to it.

Thus to compensate the attenuation there must be an additional gain enhancement. But withan increase in gain the requirement of more element increases. This leads to cause costenhancement as well as more weight & greater space.

The lead network increases B.W but with increased B.W the system becomes more suscep-tible to noise.

This sometimes makes the system conditionally stable.

Q–5(e) For a Scott connected transformer, prove that the number of turns on primary of the

teaser transformer is 3

2 times the number of turns in primary of main transformer..



Mains Exam Solution

Sol. Scott-connection and its schematic connection diagram are shown in

Teaser Transfer

MainTransfer

A3

A1

B3

B2

B1C

B

N

a1

b2

a3

A2

b1

Scott-connection

b

a

IaA

IaB

O

A ac

LoadN

C B

c b

Load

d

IcC D

Its schematic connection diagram

Let the positive direction of primary voltage and current be taken arbitrarily along the arrows fromA to D and from B to C as shown in figure (i).

Neglecting the transformer internal impedance drops the voltage Vab or VAB must be equal to thevoltags induced in AD and DB.

abV

= AB AD DBV V V

ADV

= AB DB AB BDV V V V

If VAB = VBC = VCA = VL (line voltage) then with VBC as reference phasor as shown in figure (ii)

Vab

Vao

Voc

Vbc

Vbo

Voa

Vca

Vco

Vob 90°

120°

120°

Fig. Supply voltage phasor diagram. Scott connection phasor diagram for the primary voltages and the secondary voltages.

Vad

Vbc

90°

VAB VAD

VCA

VBD

VBC120°60°

BC L BD BC L AB L1 1V V 0 ,V V V 0 ,V V 1202 2

and CA LV V 120

Now, ADV

= L L1V 120 V 02

or ADV

= L L L1 3 1 3j V V j V2 2 2 2

or ADV

= L3 V 90

2

https://iesmaster.org/results/ese



Mains Exam Solution

Hence, VAD the voltage across the teasear, is 32

of that across the main and leads it by 90°

Since, VAD = 32

VL and VBC = VL the primaries of both the transformers would have equal voltage

per turn only if the teaser primary turns are 32

times the primary turns of the main transformer..

Q–6(a) A 15 kW, 400 V, 3-phase, star connected synchronous motor has synchronous impedanceof 0.4 j4 . Find the motor excitation voltage for full load output at 0.866 leading powerfactor. Take the armature efficiency of 95%.

Sol. Given: Zs = 0.4 + j4 = 4.02 84.28

Efficiency ( ) = 95%

output power = 15 kW

VL = 400 V

=Output powerInput power

Input power =o / p power

=15000 15789.47 watt0.95

So Armature current

at full load Ia =15789.47

3 400 0.866

= 26.316 30As we know for motor

Eph = Vph – Iazs

Eph =400 26.316 30 4.02 84.28

3

= 290.88 19.36

So Eline = ph3 E = 3 290.88

= 503.818 19.36 Ans.

Q–6(b) A synchronous machine is connected to an infinite bus through a transformer and a doublecircuit line shown in the figure. The infinite bus voltage is V 1.0 0 p.u. The direct axistransient reactance of the machine is 0.20 p.u., the transformer reactance is 0.10 p.u. andthe reactance of each of the transmission lines is 0.4 p.u., all to a base of the rating of thesynchronous machine. Initially the machine is delivering 0.8 p.u. power with a terminalvoltage of 1.05 p.u. The inertia constant H = 5 MJ/MVA. All resistances are neglected.Determine the equation of motion of the machine rotor.

Line-2

Line-1V = 1.0 0° p.u.

Infinite Bus-BarE = |E|



Mains Exam Solution

Sol. Given,

xd = j.2 p.u, xime j0.4 p.u

xtr = j 0.1 p.u., Vbar 1 < 0°p.u

V = 1 < 0° p.ub

j 0.2 p.u. j 0.1 p.u.j 0.4 p.u.

j 0.4 p.u.

Pe =1t b

t t mV V 0.8 0.3sin or,

(0.1 0.2) 1.05

L = 13.2 B°

i =1.05 13.213 0

0.3 90

=0.0222 j0.24 0.241 84.715

0.3 90 0.3 90

= 0.8033 5.258

Ef = tv i jxd

= 1.05 < 13.213 + 0.8033 × 0.2 < 90° – 5.285°

= 1.05 < 13.213 + 0.161 < 84.715°

= 1.04 + j0.4 = 1.1143 < 21.04 °

Xeq = j0.20 + j0.10 + j 0.20

j 0.50 p.u.

Pe =f t

eq

E v 1.114 1sin sin 2.228 sinx 0.5

Solving equation of the roler is given as

2

2gH d

f dt

= pm – pe

= 0.8 – 2.228 sin

or,2

21 5 d

50 dt

= 0.8 – 2.228 sin

or,2

2ddt

= 2rad31.4 0.8 0.228 sin

sec Ans.

Q–6(c) The open loop transfer function of unity feedback control system is given by

G(s) = K 0 a b

s s a s b

(i) Find the range of the gain constant K (> 0) for stability using Routh-Hurwitz criterion.

(ii) What type of control do you use if the system is required to have zero steady-state errorfor ramp input ? Let ‘A’ be the parameter that can be varied in the introduced control. Findthe range of ‘K’ for stability in terms of parameters a, b and A using Routh-Hurwitz criterion.



Mains Exam Solution

Sol. (i) Characteristic equation figure given open loop function(S2 + Sa) (S + b) k = 0

3

2

1

0

S 1 ab

S a b kkS ab 0

a bS k

k > 0

ab >k 0 k ab(a b)

a b

(ii) Transfer of PI controller =A1s

Characteristic equation is:

2

A s k1

s s a s b

= 0

4 3 2s a b s abs ks ak = 0

4

3

2

1

0

s 1 ab ak

s a b kab a b k

s Aka b

ab a b ka b Ak

a bs

a b ab ka b

s Ak

For stability: ak > 0k > 0

and a b ab k

a b A k 0a b

(a + b) ab – k – (a + b)2 A > 0

(a + b) ab – (a + b)2 A > 0

k < (a + b) – (a + b)2 AA

and a b ab k

0a b

k (a + b) ab

k < (a + b) ab – (a + b)2 AA

Q–7(a) A system consists of two plants connected by a transmission line and a load is at powerplant-2 as shown in the figure. Data for the loss equation consists of the information that200 MW transmitted from plant-1 to the load results in transmission loss of 20 MW. Find



Mains Exam Solution

the optimum generation schedule considering transmission losses to supply a load of204.41 MW. Also evaluate the amount of financial loss that may be incurred if at the time ofscheduling transmission losses are not coordinated. The incremental fuel cost character-istics of plant-1 and plant-2 are given by

1

1

dfdP = 0.025 P1 + 14 Rs/MW-hr

2

2

dfdP = 0.05 P2 + 16 Rs/MW-hr

1 2

LoadTransmission line

Plant-1 Plant-2

Sol.

Land

plant-2plant-2

The expression of the transmission line cross is gain as

P1 = 2 211 1 22 2 12 1 2P P 2 P P

The load is concluded to plant-2, 22 n2 0

PL = 211 1P

Given, when P1 = 200 mW, PL = 20 mW

PL = 20 = 211 11

20(200) or (MW)40000

4 2L 1P 5 10 P = 5 × 10–4 (MW)

Given1

1

dfdp = 0.025 P1 + 14,

22

2

df0.05P 16

dp ,

When the losses are coordinated, then

1

1

L

1

dfdp

P1P

=

2

2

1

2

dfdp

P1P

2

1

PP

= 10–3 P1

2

2

P 0P

13

1

0.025P 141 10 P

= 20.05P 161

L

2

P 0P



Mains Exam Solution

or, 0.025 P1 + 14 = (0.05 P2 + 16) (1 – 10–3 P1) …(1)

again, P1 + P2 – PL = Pload = 204.11 …(ii)

or, P2 = 204.41 – P1 + 211 1P

= 204.41 – P1 + 5 × 10–4 21P

from (1) & (ii), we obtain

0.025 P1 + 14 = 4 21 10.05 204.41 P 5 10 P 16 {1 – 10–3 P1}

= 6 21 110.2205 25 10 P 0.05P 16 (1 – 10–3 P1)

= 4 2 31 1 126.2205 0.25 10 P 0.05P 1 10 P

0.025 P1 + 14 = 26.2205 – 0.0262205 P1 + 0.25 × 10–4 P12

– 0.25 × 10–7 P12 – 0.05 P1 + 0.05 × 10–3 P1

2

0.025 P1 + 14 = 26.2205 + 7.5 × 10–5 P12 – 0.25 × 10–7 P1

3

– 0.0762 P1

or, 0.25 × 10–7 P13 – 7.5 × 16–5 P1

2 + 0.1212 P1 – 12.2205 = 0

or, P1 = 133.35 mW

PL = Bu P12 = 5 × 10–4 × (133.35)2 mW

= 8.891 mW

P2 = Pload + PL – P1

= 204.41 + 8.891 – 133.35

= 79.951 mW

When the losses are coordianted then for optimum performance,

1 2

L

P 133.35 MW; P 79.951MW &P 8.891MW

When the losses, are not coordianted but included,

1

1

dfdp =

2

2

dfdp

0.025 P1 + 14 = 0.05 P2 + 16 …(i)

P1 + P2 – PL = Pload

P2 = 204.41 + 5 × 10–4 P12 – P1 …(ii)

showing the equations (i) and (ii), we get

0.025 P1 + 14 = 0.05 (204.41 + 5 × 10–4 P12 – P1) + 16

0.25 × 10–4 P12 – 0.075 P1 + 126.2205 = 0

or P1 = 2827.1 MW , P2 = 172.91 MW

PL = Bu P12 = 5 × 10–4 × (172.91)2 MW

PL = 14.96 MW

P2 = Pload + PL – P1

= (204.41 + 14.95 – 172.91) MW = 46.451 MW



Mains Exam Solution

1 2

L

P 172.91MW; P 46.451MWP 14.95 MW

When the losses are coordinates)

Total savings indluced when the losses are coordinated is given as:

= 133.35 79.9531 1 2 2172.91 46.451

0.025P 14 dP 0.05P 116 dP `

=

2 2 2 20.025 0.05133.35 172.91 14 133.35 172.91 79.953 46.4512 2

16 79.953 46.451

= Rs.63.4 63Rs

Total savings incurred when the losses are coordianted.

= Rs 63. Ans.

Q–7(b) A continuous-time integrator has a system function Ha(s) = 1s .

(i) Design a discrete-time integrator using bilinear transformation and find the differenceequation relating the input x[n] to the output y[n] of the discrete-time system.

(ii) Find the frequency response of the discrete-time integrator found in part (i) and deter-mine whether or not this system is a good approximation of the continuous time system.

2For 1, sin and cos 1

2

Sol: Given,

Ha(s) = 1/sTo convert s-domain to z-domain by bilinear transformation

Put S =1

1s

2 1 zT 1 z

Ts = Sampling interval

H(z) = 1

1s

12 1 zT 1 z

C zR z =

1s

1

T 1 z

1 z

1

s s

2 2z z C zT T

= 1R z z R z

Applying inverse z-tranform on both sides.

s s

2 2C n C n 1 r n r n 1T T

(ii) Now, put z = je

jH e = j

j

s

1 e2 1 eT

https://iesmaster.org/results/gate



Mains Exam Solution

je = cos jsin

jH e =

s

1 cos jsin2 1 cos jsinT

jH e =

s

1 cos jsin 1 cos jsin2 1 cos jsin1 cos jsinT

jH e =2 2

sT 1 sin 2jsin cos2 2 2cos

jH e = sT 2sin

2 2 1 cos

=

T sinj2 1 cos

jH e = sT cos 2j2 sin 2

Since by is given that if 1 then sin

cos =2

12

So, by using the above approximation,

cos2

= 21 22

=

21

8

sin2

=2

jH e = 2

s1 8T

j2 2

=21 8Tj

jH e =21 8

j0, H e

j1, H e 7 8

j2, H e 1 4

By seeing the magnitude response of discrete time integrator is seems to be the response ofcontinuous time intervals.

Q–7(c) For a 3-phase, 50 Hz, 415 V, 4-pole induction motor, the standstill resistance and reac-tance are 3.0 and 5.0 at 50 Hz respectively. The machine has magnetising induc-tance of 350 mH and stator resistance of 1.2 . The machine is supplied from a 3-phasevoltage source inverter with quassi square wave output voltage waveform per phase asshown in the figure below. The DC bus voltage is 500 V.



Mains Exam Solution

If the machine is operating at 4% slip, find (i) the fundamental input current, (ii) harmoniccopper losses in the machine up to 13 harmonics, and (iii) input power factor.

Assume negligible core losses, equal distribution of stator and rotor leakage reactancesand linear magnetic circuit.

–1/3 Vd

–2/3 Vd

1/3 Vd

2/3 Vd

V

0 /3 2/3 4/3 5/3

2

t

Sol:

The input voltage can be written as

VR = K 0

s

n 6K 1

2V sin n tn

Per phase voltage

Solving in per phase

Harmonics present 5, 7, 11, 13

Zfundamental

Vfundamental = 2 500 Volt

2

= 225.08 V

3.696 –37.3

3.696 –37.3

35 j5 j

225.08

2.9474 –3.75

76.2

2.047 90

For 5th harmonic

Slip =s r

s

5N N5N

[ as rotation of 5th harmonic is opposite of rotor rotation]

S5 = 1.192



Mains Exam Solution

Nr = 1440 rpm, Ns = 1500 rpm

1.862 –81.91°

0.0818 –90°

175i25i

3 1.2 3.71671.192

45.016

1.781 81.54°

V5 = 2 500

5 2 = 45.016 Volt

For 7th harmonic

35j

31.2 4 676

0.8628

0.0417 –90°

245 i32.154V

0.9519 –82.72° 0.9105 –82.39°

Slip7 =S r

s

7N N 0.86287N

V7 = 2 500 32.154 V7

for 11th haromonic

Slip =s r

s

11N N 1.08711N

V11 = 500 2 20.461 Volt11

55i

31.2 3.959

1.0870.0172 –90°

385 i20.461V

0.3878 –86.05° 0.371 –85.88°



Mains Exam Solution

For 13th harmonic

Slip = s r

s

13N N 0.926113N

65i

31.2 4.4390.9261

0.0122 –90°

455 i17.313V

0.2772 –86.26° 0.265 –86.09°

V13 = 500 2 17.313 Volt13

(i) Ifundamental = 3.696 Amps

(ii) Harmonic Copper Loss = 3I2R

R = 1.2 + 3 = 4.2

Harmonic Loss = 2 2 2 24.2 3 1.781 0.9105 0.371 0.265

Harmonic Loss = 53.031 Watt

(iii) Vsrms Isrm Pf = Pout = I2R 21

RR R5

Vsrm = 500 2 235.7Volt3

Isrm = 2 2 2 23.696 1.862 0.9519 0.3878

Isrm = 4.2642 Amp

Pout = 216.501 2.9474 76.2 678.463 watt

Pf = 678.463

4.2642 235.7 = 0.675

Q–8(a) A 50 Hz, 3-phase induction motor has a slip of 0.2 for maximum torque, when operated onrated frequency and rated voltage. If the motor is run on 60 Hz supply with application ofrated voltage, find the ratio of

(i) Starting currents

(ii) Starting torques

(iii) Maximum torques

With respective values at 50 Hz

Neglect the stator impedance.



Mains Exam Solution

Sol. Let rated volstage is v volt rotor leakage reactance is x2 at 50 Hz and 1x = 260 x50

= 1.2 x2

It is stated that neglect stator impedance. So, the equivelent circuit is

Now, starting curvent

V

r2 x2

Ist = 2 22 22 2 22 1

V V as r xxr x

Starting torque,

Test =2 2

2 22 2 2s s2 2 2

3 V 3 vr rw wr x x

Maximum torque,

Tem =2

r 2

3v 1w 2x

(i)Starting current at 60 HzStarting current at 50 Hz = 2 2

2 2

x xVVx x

=2

2

x 1 0.83331.2x 1.2

(ii)Starting current at 60 HzStarting current at 50 Hz =

2 22 22 2

2 2

3V r 50x60x 3V r

=

22

22

x5060 1.2x

= 3

1 1 0.57871.7281.2

(iii)Maximum torque at 60 HzMaximum torque at 50 Hz =

22

22

50 2x3V60 2x 3V

=2

2

x5060 1.2x

= 21 1

1.441.2 = 0.6944

Q–8(b) The current of an induction motor is sensed through a suitable arrangement and con-verted to equivalent voltage. The current contains fundamental and higher order 5th and7th harmonics. In order is separate the fundamental, the equivalent voltage waveform ispassed through the following circuit as given in figure. Find the (i) cut-off frequencies ofeach section, (ii) overall gain attenuation in dB for fundamental, 5th and 7th harmonics,and (iii) overall phase shift of the measured fundamental current.

7.5 k

Vi 0.33 F 12 k

0.33 F

10 k10 k

Vo



Mains Exam Solution

8 (b) (i)

C

R2

R3

V0

Vi

R2

R1V01 C

–+

–+

01V =i c i

2 2 21 C 1

V X V

R X 1 (R C)

Section–1 is LPF

10 w 0(V ) = Vi

01(V ) =i

21

V

1 (R 2 (f )

10 f cutoff(V ) =1 i

21

V V2 1 (R (2 (f ) )

2B, (f 1) fcutofff =1

12 R C

For sation-2

V0 =1 10 3 0 s

2 2 23 C 3

V R V R C

R X 1 (R C)

It is a high pass filter

0 w(V ) = 10V

( 0 cuttoff(V ) =10 301

2 23 L

V RV2 R x

2 23 LR x = 2

32R2Lx = 2

bR L 3x R

3 cutoffC 3

1 1R ff 2CR

f1cuttoff = 3 61 64.3 Hz

(2 ) (7.5) (10 ) (0.33) (10 )

2 cuttofff = 6 31 40.19 Hz

(2 ) (0.33) (10 ) (12) (10 )

(ii) for fundamental with frequency = 50 Hz

01V = (0.78944) 1fV



Mains Exam Solution

V0 = 1 10 f(0.77941)V 0.6153 V

Gain in dB = 20log (0.6153) = –4.21 dBfor 3rd harmonic f = 150 Hz

10V = 3f(0.39401) V

V0 = 1 50 f(0.965928) V 0.2459 V

for 5th harmonic f = 250 Hz

10V = 5f(0.24911)V

V0 = 1 50 f(0.9873)V 0.2459 v

Gain in dB = 20log (0.2459) = –12.184 dB

for 7th harmonic f = 350 Hz

10V = 0.18 7fV

V0 = 100.993 V = 7f0.178 V

14.99 dB

(iii) 10V = 10 3i c0

1 c c 3

V R(V )jX VR jX jX R

V0 =3 c

i3 c 1 c

R jX VR jX R jX

XC = 9645

V0 = 1f(0.7794 38.79) (0.7894 37.866) V

Phase shift = 0.924°Q–8(c) Given the following facts about a real signal x(t) with Laplace transform X(s):

A : X(s) has exactly two poles

B : X(s) has no zeros in the finite s-plane

C : X(s) has a pole at s = – 1 + j

D : e2t x(t) is not absolutely integrable

E : X(0) = 8

Determine X(s) and specify its region of convergence.

Sol: Point A: X(s) has exactly two poles

x(t) is real signal therefore poles & zeros will occurs in conjugate pairs (if they are complex)

Point C: X(s) has a pole at s = –1 + j

Poles are present in conjugate pairs

So, P1 = –1 + j

P2 = –1 – j

Now,

Transfer function

X(s) = 1 2

Ks P s P =

Ks 1 j s 1 j



Mains Exam Solution

= 2K

s 1 1

X(s) = 2K

s 2s 2 ...(i)

Point E: X(0) = 8

From equation (i)

X(0) = 8 = K

0 0 2

K = 16

So, 216X s

s 2s 2

From point D: g(t) = e2t x(t)

G(s) = X(s – 2)

–i –i

+j +jx x

x x–j –j

Pole zero plot of X(s) Pole zero plot of G(s)

ROC of g(t) : > 1Right sided

g(t) = e2t x(t)

mains exam solution · 2020. 10. 27. · f-126, katwaria sarai, new delhi - 110 016 ph:...

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