: 3 : TEST-10 (Solutions) · 2017. 4. 16. · ACE Engineering Academy...
Transcript of : 3 : TEST-10 (Solutions) · 2017. 4. 16. · ACE Engineering Academy...
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01. (a)
Sol: Given data,
20 kVA, 50Hz, 2000/200 V distribution transformer
Power output = V2I2 cos2
= 200 × 100 × 0.8
= 16000 W
Total loss PL = Pi + k2Pc
= 120 + 1 × 300 = 420 W
=L0
L
PPP
1
=42016000
4201
= 97.44%
For maximum efficiency , K =300120
PP
c
i = 0.632
i.e., at 0.632 pu load
max (cos 2 = 0.8) =i0
i
P2PP2
1
=1202632.016000
12021
= 97.68%.
01. (b)
Sol:
1N
0n
*1N
0n
2 )n(x)n(x)n(x*
1N
0k
nkN
2j1N
0n
e)k(XN
1)n(x
1N
0n
nkN
2j1N
0k
* e)n(x)k(XN
1
1N
0k
* )k(X)k(XN
1
1N
0k
2)k(XN
1
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01. (c)
Sol: Synchronous impedance,currentSC
voltageOCZs
=100
500= 5
Xs =2a
2s RZ = 22 8.05 = 4.936
E0 =2
s2
a )IXsinV()IRcosV(
2 2(2000 0.8 100 0.8) (2000 0.6 100 4.936)
= 1822 V
% Regulation = 1002000
20001822
= – 8.9%
01. (d)
Sol: In above sequential circuit in initial state v, a ‘0’ inputs generates a ‘0’ outputs and the circuit stays
in state v, an input of 1 produces an output 1 and the circuit will move to next state W. In state W
for the input =0, the output also 0 and the next state is Y and this process is continued.
State table:
Present stateNext state
d=0 d =1
Output
d=0 d=1
V V W 0 1
Y V 0 0
Y V 0 0
Y Z V 0 0
Z V X 0 1
In the above table, for both W and X states their next states and outputs also same so, W and X are
equivalent states so, we can reduce one state by replace ‘X’ with ‘W’
W
X
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After reducing X state the minimized state table is
Present stateNext state
d=0 d =1
Output
d=0 d=1
V W 0 1
W Y V 0 0
Y Z V 0 0
V W 0 1
Now again ‘V’ and ‘Z’ are equivalent states So, we can reduce one more state, replace ‘Z’ by ‘V’.
After reducing Z-state , the minimized state table is
Present state Next state
d = 0 d =1
Output
d = 0 d =1
V V W 0 1
W Y V 0 0
Y V V 0 0
Initial given state diagram has 5 states means we require 3 flip-flops, now it reduced to 3 states
means just 2flip-flops required. Here one flip-flop eliminated. So, some complexity will be
reduced.
Z
V W Y
1/0
1/1 0/0
1/0
0/0
0/0
V
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01. (e)
Sol: (i) Air-gap power: The power transferred from stator to rotor across the air gap.
i.e., Pgs
RI3 22
r
Where, Pg = Air gap power
Ir = Rotor current per phase
R2 = Rotor resistance per phase
(ii) Internal mechanical power developed (or) Gross mechanical power out
1
s
1RI3 2
2r
(iii) Shaft power is nothing but net mechanical power output.
i.e., shaft power = Gross mechanical output – mechanical losses.
Rotor input: Rotor cu losses: Gross mechanical power output
1
s
1RI3:RI3:
s
RI3
2
2
r2
2
r
22
r
1
s
1:1:
s
1
1 : s : (1s)
Rotor Cu loss = s Rotor input
Gross mechanical power output
= (1s) Rotor Input
Rotor cu loss = Gross mechanical power output s1
s
01. (f)
Sol: Apply DTFT to given difference equation
)e(Xe2
1)e(X)e(Ye
16
3)e(Ye)e(Y jjjjj2jjj
j2j
j
j
jj
e163
e1
e21
1
)e(X
)e(Y)e(H
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jj
j
j
e41
1e43
1
e21
1)e(H
jj
j
e41
1
21
e43
1
21
)e(H
)n(u4
15.0)n(u
4
35.0)n(h
nn
01. (g)
Sol: A shunt excited d.c generator may fail to self-excite for any of the following reason.
1. Residual magnetism is absent. This difficulty can be overcome by exciting the field winding
from a separated d.c source for a few seconds with the armature at rest.
2. The field connection to the armature is such that the induced emf due to the residual magnetism
tends to destroy the residual magnetism. This condition can be remedied by reversing the field
connection to the armature.
3. For critical resistance Ry line coincides with the linear portion of the magnetization curve.
4. Speed is less than critical speed. At critical speed, the open circuit characteristic (OCC) is
tangential to the Rf line.
01. (h)
Sol: Given circuit is:
Given circuit has voltage-shunt feedback
To find the input resistance apply some test voltage Vx
Vi–
+Ri gmVi V0
+
–
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Apply KCL at node 1
Ix – gm Vi = I1
Ix – gm Vx =i
x
R
V
xmi
xx Vg
R
VI
im
i
x
xin Rg1
R
I
VR
01. (i)
Sol: 240V/120V, 12kVA has rated current of 50 A/100 A. It’s connected as an autotransformer as
shown in figure.
Auto-transformer rating
= 360 × 100 = 36 kVA
It is 3-times then 2-winding connection.
As 2-winding connection Output, P0
= 12 × 1 = 12 kW
Vi–
+Ri gmVi
Ix 1 I1
–gmVi
Vx
–
+
360V
240V
+
+
120V
150A
240V
100A
L
50A
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=L0
0
PPP
=
0
L
PP
1
1
= 0.962
From which find full-load loss
1 = 0.962 + 0.962
0
L
PP
(or)0
L
P
P=
962.0038.0
; PL=962.0038.0
12 = 0.474 kW
In auto connection full-load loss remains the same. At 0.85 pf
P0 = 36 × 0.85 = 30.6 kW
=
6.30474.0
1
1
= 0.985 or 98.5%
01. (j)
Sol: y(n) = 2y(n–1) + (n)
y(0) = 2y(–1) + (0)
1)1(y24
2
3)1(y
y(–1) = 2y(–2) + (–1)
)2(y223
43
)2(y
y(1) = 2y(0) + (1) = 2(4) + 0 = 8
y(2) = 2y(1) + (2) = 16
y(3) = 2y(2) + (3) = 32
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+
–
0.6
0.2 10
02. (a)
Sol: (i) Resistance of motor = 0.2 + 0.6 = 0.8
Eb1 = 200 – (25 0.8) = 180 V
Let the motor input current be I2 when armature divertor is used.
Series field voltage drop = 0.6 I2
P.D at brushes = 20 – 0.6 I2
Arm. divertor current A10
I6.0200 2
Armature current = I2 –10
I6.0200 2
2a 2
10.6 I 200I
10
As torque in both cases is the same,
1 I1 = 2 I2
22
10.6 I 20025 25 I
10
or
6,250 = 10.6 22I – 200I2
I2 = 35.6 A
P.D at brushes in this case
= 200 – (35.6 0.6)
= 178.6 V
10
2006.356.10I 2a
= 17.74 A
Eb2 = 178.6 – (17.74 0.2) = 175 V
: 11 : TEST-10 (Solutions)
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Now2
1
1b
2b
1
2
E
E
N
N
or2
1
1b
2b
1
2
I
I
E
E
N
N
2
1
N 175 25
N 180 35.6
N2 = 314 r.p.m
(ii) Power equation for salient pole synchronous
Machine: (with zero armature Resistance)
1. IdXd = E Vcos
dd X
cosVEI
2. IqXq = V sin
qq X
sinVI
P = (Id voltage component along d-axis) +
(Iq volt along Q-axis)
cosV
X
sinVsinV
X
cosVE
qd
cossinX
Vcossin
X
Vsin
X
EV
q
2
d
2
d
V
Ia
IqVCos
IqXq
Q-axis
IdXd
m
d-axis Id
E
Vsin
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2sin
X
1
X
1
2
Vsin
X
EVP
dq
2
d
emd
PsinX
EVpowermagneticElectro
2
relq d
V 1 1sin2 Reluctance power P
2 X X
02. (b) (i)
Sol: x(t) = Asin(t) 0 t 1
x(t) = Asin(t) 0 t 1
2T2
1T 0
1
0
t2jnTt
t
tjnn dte)tsin(A
1
1dte)t(x
T
1C
0
0
0
1
0
t)n21(j1
0
t)n21(j dtedtej2
A
1
0
t)n21(j1
0
t)n21(j
n )n21(j
e
)n21(j
e
j2
AC
)n41(
A2C
2n
A2dt)tsin(A
1
1dt)t(x
T
1C
1
0
T
0
0
0nn
tn2j2
e)n41(
A2A2)t(x
02. (b) (ii)
Sol: EFS coefficient T
0
tjnn dte)t(x
T
1C 0
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T
0
00n dt)tnsin(j)tncos()t(xT
1C
T
0
0
T
0
0n dt)tnsin()t(xT
2jdt)tncos()t(x
T
2
2
1C
nnn jba21
C
T
0
tjnn dte)t(x
T
1C 0
T
0
0
T
0
0 dt)tnsin()t(xT2
jdt)tncos()t(xT2
21
nnn jba2
1C
0
T
t
0 adt)t(xT
1C
0
C0 = a0,2
jbaC nn
n
,
2
jbaC nn
n
02. (c) (i)
Sol: Buffer stage using op-amp:
It is also called voltage follower circuit and is shown below
By observing the above circuit we can say that the voltage at the non – inverting input is Ei, the
voltage at the inverting input approaches the voltage at the non – inverting i/p, and the output is at
the same voltage as the inverting input. Hence E0 = Ei and our analysis is complete.
Ei Eo
+ +
+
fig (a)
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Fig (b) Equivalent circuit of a buffer amplifier
Eo = Ei
AV = Eo/Ei =1
Since no current flows through Op-Amp, hence the input impedance of the voltage follower is
infinite i.e. Zi = . The output impedance is just that of the ideal operational amplifier itself, i.e
zero Z0 = 0
02. (c) (ii)
Sol:
(A)
Ei
+
-
-++
-Eo
1Ei
+
–
i1
i2
R2
2k
RL=5k
1k
R1 V0
5k
Vs
–
+Vd
R1
1k
2k
R2
Vs
+–
V0
RLAVd
Rs
: 15 : TEST-10 (Solutions)
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21
10sd RR
RVVV
V0= AVd
21
10s0 RR
RVVAV
21
10s
0
RR
RVV
A
V
A [ Ideal op-amp]
21
10s RR
RVV
ss1
20 V
K1
K21V
R
R1V
V0= 3Vs
(B) SRdt
dV0
SRdt
dVA i
CL [ AcL Closed loop voltage gain ]
|ACL Vm m| < SR
322fm < 5105
fm < 13.3kHz
fmax = 13.3kHZ
(C) V0= 3Vs
V0 = 32sint = 6sint
mA2.1K5
6I (max)L
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: 17 : TEST-10 (Solutions)
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03. (a)
Sol: Given p = 1dB, p = 0.2, s = 0.3, s = 15dB
Pre wrapped frequencies: 65.022.0
tan22
tanT2 p
p
02.12
3.0tan2
2tan
T
2 ss
65.0
02.1cosh
110
110cosh
cosh
110
110cosh
N1
1.0
5.11
p
s1
1.0
1.01
p
s
N 3.01
N = 4 (round off to next higher integer)
Calculation of poles: 508.0110 2
11.0 p
17.41 21
2)17.4()17.4(
65.02
a4141N1N1
p
a = 0.237
2
)17.4()17.4(65.0
2b
4141N1N1
p
b = 0.6918
N2)1k2(
2k
, k = 1, 2, 3, 4
1 = 112.5o, 2 = 157.5o, 3 = 202.5o, 4 = 247.5o
S1 = acos1 + jbsin1 = –0.0907 + j0.639
S2 = acos2 + jbsin2 = –0.2189 + j0.2647
S3 = acos3 + jbsin3 = –0.2189 – j0.2647
S4 = acos4 + jbsin4 = –0.0907 – j0.639
The denominator polynomial of H(s) is
H(s) = [(s+0.0907)2 + (0.639)2] [(s+0.2189)2 + (0.2647)2]
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H(s) = (s2 + 0.1814s + 0.4165) (s2 + 0.4378s + 0.118)
As N is even
So, the numerator of H(s) is 04381.01
)118.0)(4165.0(2
Transfer function of)1180.0s4378.0s)(4165.0s1814.0s(
04381.0)s(H
22
The digital filter transfer function is
1
1
1
1
z1
z12
z1
z1
T
2s)s(H)z(H
)z6493.0z5548.11)(z8482.0z499.11(
)z1(001836.0)z(H
2121
41
03. (b) (i)
Sol: SAM(t) = Ac[1+kam(t)] cos 2fct
= Ac[1+cos2fmt] cos2fct
= Accos2fct + Ac cos2fmt cos2fct
= Accos2fct + tfcos2tf2cos22A1.0
cmc
= Accos2fct + 0.05Ac[cos2(fc+fm)t+cos2(fc–fm)t]
If lower side band is removed from above signal
tff2cosA05.0tf2cosA)t(S mcccc*AM
= Accos2fct+0.05Ac cos2fct cos2fmt–0.05Ac sin2fct sin2fmt
= [Ac+Ac0.05 cos2fmt] cos2fct – [0.05Ac sin2fmt] sin2fct
Above signal is given to envelop detector. The output of envelop detector is
2mc2
mcc tsinA05.0tcosA05.0A
tcosA1.0A05.0A m2c
2c
2c
tcos1.00025.01A mc
The maximum value of above signal is
05.11025.1Ac (∵given in question)
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Ac = 1
cmc
m AAAA
= 1(0.1)
Am= 0.1
03. (b) (ii)
Sol: Standard AM signal is given by
S(t) = Ac [1+kam(t)] cos 2fct
S(t) = Ac cos 2 fct + Ackam(t)cos 2fct
When single tone modulation is considered
Then m(t) = Am cos2fmt
S(t) = Ac cos 2fct +AcKaAm cos 2fmt cos 2fct
Taking kaAm =
S (t) = Ac cos 2 fct tff2cos2
Amc
c
tff2cos2
Amc
c
Comparing the above equation with the given signal
S(t) = 20 cos 2000 t + 5 cos 2200t +5 cos 1800 t
(A) c(t) = 20 cos 2000 t
(B) AC = 20, fc= 1000
52
Ac
= 0.5
(C) 5.020AAAA
cmc
m
Am=1
fc+fm=1100
fm=1100–1000
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fm= 1000
m(t) = Am cos2fmt
m(t) = cos200t
(D) power in carrier signal W200
220
2A 22
C
Power in USB W225
2/52
Power in LSB W225
2/52
Power in both side bands W25225
225
8
1
200
25
signalcarrierinpower
bandssideinpower
03. (c)
Sol: (i) Let Is = starting current
Ifl = full load current
2
2
2
2
2
2
s XR
VI
; (s = 1) …….. (i)
2
2
2
f2
2
2
f Xs/R
VI
………….. (ii)
Dividing equation (i) by (ii)
2
f
s
I
I
=
2
2
2
2
2
2
2
f2
XR
Xs/R
= 1ss
ss2
Tmax,2f
2f
2Tmax,
….(iii)
Substituting the values
25 = 1s)04.0(
)04.0(s2
Tmax,2
22Tmax,
(Given that Is = 5Ifl)
or Smax.T = 0.2 or 20%
Tmax = 22
2
s XV5.0
.3
…….. (iv)
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Tfl = 2
2
2
f2
f2
2
sX)s/R(
)s/R(V.
3
……….. (v)
Dividing equation (iv) by (v)
f
max
TT
=
f22
22
2f
22
sXRXsR
5.0
=
fTmax,
2f
2Tmax,
ss
ss5.0
=04.02.0
)04.0()2.0(5.0
22
= 2.6
or Tmax = 2.6 pu
(ii) As per equation
f
s
f
s
II
TT
sfl = (5)2 × 0.04 = 1
Ts = 1 pu
03. (d)
Sol: Conditions to be satisfied for parallel operation of transformers:
Necessary conditions for possible parallel operation:
1. Voltage ratings mentioned on the name plate of transformers to be connected in parallel must
be same.
2. The transformers must be connected in parallel with correct polarity.
For this, dotted terminals should be connected to same bus bars and undotted terminals should
be connected to another bus bar.
3. The phase sequence of 3- transformers to be connected in parallel must be same.
To get correct phase sequence, the terminals of both secondaries which attain their peak value
simultaneously must be connected to same bus bar, so that potential difference between the
terminals connected to same bus bars equal to zero and there will not be any circulating current.
4. A part from phase sequence matching the phase displacement between the secondaries of both
transformers must be zero. This condition can be fulfil if the two transformers belongs to same
Phasor group.
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Desirable conditions for satisfactory parallel operation:
1. The voltage ratios of transformers to be connected in parallel should be same to avoid
circulating currents.
If voltage ratios are not same, there is a possibility of circulating current given by
BA
B2A2C ZZ
EEI
E2A=Secondary voltage of transformer A
E2B =secondary voltage of transformer B
ZA and ZB are the equivalent leakage impedances in ohms referred to the secondary of
transformer A & B respectively.
2. The two transformers should share the common load proportional to their KVA ratings.
More KVA then more sharing of load.
Less KVA then less sharing of load.
3. TheRx
ratios of transformers to be connected in parallel should be equal to avoid
operation of transformers at different power factors.
B
B
A
A
R
X
R
X
A = tan-1
A
A
R
X, B = tan-1
B
B
R
X
A = B
cosA = cosB = cosL
04. (a)
Sol: Mechanical loss W9003000100
30
Stator core loss = Power input at no-load – Mechanical loss– stator I2R loss at no load
Neglecting stator I2R loss,
Stator core loss = 3000 – 900 = 2100 W
Power input during blocked rotor test
= stator I2R loss + Rotor I2R loss = 4000 W
: 23 : TEST-10 (Solutions)
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Stator I2R loss = Rotor I2R loss
= W20002
4000
(i) At rated load, air-gap power,
Pg = Output power + mechanical loss + rotor I2R loss
Pg =60,000 +900+2000 = 62900W
But rotor I2R loss = 2000 W = s.Pg
Slip at rated load, 0318.062900
2000s
(ii) At rated voltage, power input to motor during blocked rotor test
kW444.4430
1004
2
Air-gap power,
Pg=Power input– stator core loss – stator I2R loss
= 44444 – 2100 – 2000
= 40344 Watt
Synchronous speed,
s/rad504
504
P
f4s
Starting torque .Nm84.25650
40344P
s
g
04. (b)
Sol: The samples of a signal are highly correlated with each other. This is because the signal does not
change fast. This means that its value from present sample to next sample does not differ by large
amount. The adjacent samples of the signal carry the same information with a little difference.
When these samples are encoded by a standard PCM system, the resulting encoded signal contains
some redundant information. To overcome this redundancy, DPCM is used.
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Differential Pulse Code Modulation (DPCM) Transmitter
Working Principle:
The DPCM works on the principle of prediction, the value of the present sample is predicted from
the past samples. The prediction may not be exact but it is very close to the actual sample value.
The sampled signal is denoted by x(nTs) and the predicted signal is denoted by snTx . The
comparator finds out the difference between the actual sample value x(nTs) and predicted sample
value snTx . This is known as Prediction error and it is denoted by e(nTs). It can be defined as,
e(nTs)= x(nTs) – snTx ….. (1)
The error is the difference between unquantized input sample x(nTs) and prediction of it snTx .
The predicted value is produced by using a prediction filter. The quantizer output signal gap
eq(nTs) and previous prediction is added and given as input to the prediction filter. This signal is
called xq(nTs). The prediction is more close to the actual sampled signal. The quantized error signal
eq(nTs) is very small and can be encoded by using small number of bits. Thus number of bits per
sample are reduced in DPCM.
The quantizer output can be written as,
eq(nTs) = e(nTs)+ q(nTs) ….. (2)
x(nTs)
e(nTs) eq(nTs)
snTx
Predictionfilter
Quantizer Encoder DPCMsignal
xq(nTs)
+
+
+
SampledInput
Fig: A Differential PULSE CODE MODULATION Transmitter.
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Here, q(nTs) is the quantization error. The prediction filter input xq(nTs) is obtained by sum snTx
and quantizer output i.e.,
xq(nTs) = snTx + eq(nTs) ….. (3)
substituting the value of eq(nTs) from equation 2 in equation 3, we get,
xq(nTs) = snTx + e(nTs) +q(nTs)
Equation 1 is written as,
e(nTs) = x(nTs) – snTx
e(nTs) + snTx = x(nTs)
xq(nTs) = x(nTs) + q(nTs)
This equation does not depend on the prediction filter characteristics.
Hence, the quantized version of the signal xq(nTs) is the sum of original sample value and
quantization error q(nTs). The quantization error can be positive or negative.
DPCM receiver:
The decoder first reconstructs the quantized error signal from incoming binary signal. The
prediction filter output and quantized error signals are summed up to give the quantized version of
the original signal.
Thus the signal at the receiver differs from actual signal by quantization error q(nTs), which is
introduced permanently in the reconstructed signal.
Predictionfilter
Decoder
DPCMsignal +
+
Output
Fig: DPCM receiver
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Table: Comparison between PCM and Differential Pulse Code Modulation
S.No.
Parameter
of
comparison
Pulse Code
Modulation(PCM)
Differential Pulse Code
Modulation (DPCM)
1.Number of bits. It can use 4, 8 or 16 bits
per sample
Bits can be more than one but less
than PCM
2.Levels and step
size
The number of levels depend on
number of bits. Level size is
kept fixed
Here, fixed number of levels are used.
3.Transmission
Band width
Highest band width is required
since number of bits are high
Bandwidth required is lower than
PCM.
04. (c)
Sol: dt
x(t)d
/2–/2 0
2/
t
–2/
2
2
dt
x(t)d
/2–/2 0
2/
t
–4/
2/
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2t
2)t(
42
t2
dt)t(xd
2
2
Take Fourier transform on both sides
2j
2j
2 e24
e2
)(X)j(
2
2j
2j
ee22
)(X
2
2
2
4sin
82cos1
4)(X
4
Sinc2
)(X 2
04. (d)
Sol: (i) X(pu) = 2)kV(MVA)(X
HV side 0.12 = 2
3
)6.6(1075)(X
X(Ω) = 3
2
1075)6.6(12.0
= 69.696 Ω
LV side 0.12 = 2
3
)4.0(1075)(X
= 0.256 Ω
(ii) Star-Star connection
(x) Line voltage
HV 6.6 3 = 11.43 kV
LV 400 3 = 692.8 V
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Rating = 3 × 75 = 225 kVA
(y) X(pu) = 2)linekV(
)phase3(MVA)(X
=2
3
)36.6(
10225696.69 = 0.12
(z) HV side
X(Ω) = 69.696 Ω/phase
LV side
X(Ω) = 0.256 Ω/phase
(iii) Star-Delta
(x) Line voltages
Star side 6.6 3 = 11.43 kV
Delta side = 400 V
Rating = 3 × 75 = 225 kVA
(y) X(pu), calculated from delta side
X(pu) = 2
3
)14.0(10225)3/256.0(
= 0.12.
X(pu) = 0.12
(z) Star side X = 69.69 Ω/phase
Delta side X = 0.256 Ω/phase
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05. (a)
Sol: The short circuit test has been performed on the delta-connected hv side, with the star connected lv
side is short circuited.
Per phase ratings on the hv side:
Given voltage and current ratings should always be taken as line quantities unless otherwise
specified. Similarly given kVA rating should be taken as 3-phase kVA rating unless otherwise
specified. Per phase ratings can then be obtained based on whether the transformer windings are
connected in star or delta. In this problem; the phase ratings are, rated voltage = 6600 V.
kVA/phase = 500.
rated current = 500 103/6600 = 75.6 A
From the short - circuit test data,
phase voltage (applied in the test) = 300 V
phase current (drawn by the transformer) = 75.75 A
power input per phase = 10,000 W.
Hence, 75.752 reqhv = 10,000;
where reqhv = resistance/ph (ref hv)
reqhv = 1.74 .
Percentage resistance
= 100voltagerated
)ph/cetanresis)(currentphaserated(
=6600
74.16.75 100 = 2%
zeqhv =75.75
300 = 3.96 /ph
xeqhv =22 74.196.3 = 3.56 /ph.
Percentage reactance =6600
56.36.75 100
= 4.08%
At full load and 0.8 pf lag, percentage regulation= 1006.056.38.074.16600
6.75
= 4.04%
: 31 : TEST-10 (Solutions)
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Total losses = Full load copper loss + Iron loss = 30 kW + 25 kW = 55 kW.
Full load output = 1500 x 0.8 = 1200 kW
100poweroutputlosses
powerputOut%
100551200
1200
= 95.61%
05. (b)
Sol: (i) From (1) & (2) X (s) is of the form
X(s) =)bs)(as(
K
Since x(t) is real & one of the pole of X(s) is –1+j
X(s) =)j1s)(j1s(
K
b = a*
From (5) as X(0) = 8 K = 16
X(s) =2s2s
162
Let R denotes ROC of X(s). From pole location we know there are 2 possible choices of R. It may
be Res< –1 (or) Res> –1
The ROC of Y(s) is R shifted by 2 to the right. Since y(t) is not absolutely integrable, the ROC of
Y(s) should not include j-axis
X(s) =2s2s
162
; Res > –1
(ii)2s
4)s(X
,Y(s)=X(s)H(s)
Y(s) =)2s3s)(2s(
42
=2)2)(1(
4
ss
=)2s(
4
)2s(
4
)1s(
42
y(t) = (4e–t –4te-2t – 4e-2t)u(t)
LTFrom (4) y(t) = e2t x(t) Y(s ) = X(s–2)
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05. (c)
Sol: (i) Asynchronous mod-11 counter
T0
TF/F
Q0
CLR
T1 Q1
CLR
TF/F
T2 Q2
CLR
TF/F
T3 Q3
CLR
TF/F
1 1 1Q0 Q1 Q2 Q31
clk
1 2 3 4 5 6 7 8 9 10 11 12
1 1 1 1 10 0 0 0 0 0
0 0 1 1 0 0 0 01 11
0 0 0 0
1 1 1 1
0 0 0 0
0 0 0 0 0 0 0 0
1 1 1
clk
Q0
Q1
Q2
Q3
0
0
: 33 : TEST-10 (Solutions)
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(ii) Aim: To design a Mod-11 counter
State table:
Present state Next state
Q3 Q2 Q1 Q03Q
2Q 1Q
0Q T3 T2 T1 T0
0 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
0 1 0 1
0 1 1 0
0 1 1 1
1 0 0 0
1 0 0 1
1 0 1 0
0 0 0 0
0 0 0 1
0 0 1 1
0 0 0 1
0 1 1 1
0 0 0 1
0 0 1 1
0 0 0 1
1 1 1 1
0 0 0 1
0 0 1 1
1 0 1 0
States 11,12,13,14,15 are don’t cares in this case
T3 = m(7,10)+ d(11,12,13,14,15)
T2 = m(3,7)+ d(11,12,13,14,15)
T1 = m(1,3,5,7,9,10)+ d(11,12,13,14,15)
T0 = m(0,1,2,3,4,5,6,7,8,9)+ d(11,12,13,14,15)
0
Q3Q2
Q1Q0
00 01 11 10
00
01
11
10
0 1 3 2
4 5 7 6
12 13 15 14
8 9 1011
1
1
T3
T3 = Q2Q1Q0+ Q3 2Q Q1
Q3Q2
Q1Q0
00 01 11 10
00
01
11
10
1
T2
T2 = Q1Q0
1
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To check the self starting counter or not
First take the unused states then check its output
Present sate
Q3 Q2 Q1 Q0
Input
T3 = Q1(Q2Q0+Q3 2Q ) T2 = Q1Q0 T1 = Q0+Q3Q1 T0 = 13 QQ
Output status
3Q
2Q 1Q
0Q
1 0 1 1
1 1 0 0
1 1 0 1
1 1 1 0
1 1 1 1
1 1 1 0
0 0 0 10 0 1 1
0 0 1 1
1 1 1 0
0 1 0 1
1 1 0 1
1 1 1 0
1 1 0 1
0 0 0 1
Q3Q2
Q1Q0
00 01 11 10
00
01
11
10
1
T1
T1 = m(1,3,5,7,9,10)+ d(11,12,13,14,15)
T1 = Q0+Q3Q1
1
1
1
1
1
Q3Q2
Q1Q0
00 01 11 10
00
01
11
10
1
T0 = m(0,1,2,3,4,5,6,7,8,9)+ d(11,12,13,14,15)
130 QQT
1
1
1
1
1
1
11
1
T0 T F/F(LSB)
T1 T F/F
Q0
Q3
Q1
Q0Q1
T2 T F/F
Q0
Q1
T3 T F/F(MSB)
Q2
Q0
Q3
2Q Q1
Q2Q3
3Q
1Q
: 35 : TEST-10 (Solutions)
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From the above state table the unused state 1101 is going to another unused state 1110 and
from there again it goes to 1101 state so, it is locked between these two unused states, Hence it is
not a self starting counter.
06. (a)
Sol: (i)
(x) X(z) = 11
1
212
11
21
ZZ
Z
1Z
2
11
1)z(X
If x(n) is absolutely summalbe Z >2
1 (ROC includes unit circle)
x(n) = )n(u2
1n
(y)
2
21
1
1
2111
Z2
1
Z2
1Z
Z
Z2
11
Z2
1Z1(Z
2
11)Z
2
11
X(Z) = 1–Z-1 + 32 Z4
1Z
2
1 + -----
x(n) = (n) (n1) + 34
12
2
1nn
(z)21
1
z8
1z
4
11
z3)z(X
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X(Z) =
11
1
Z4
11Z
2
11
Z3
=11 Z
41
1
4
Z21
1
4
If x(n) is absolutely summable Z >2
1 (ROC includes unit circle)
x(n) = 4
)(2
1nu
n
)(4
14 nu
n
(ii) Y(Z)+ 1.5[Z-1Y(Z) + y(–1)] + 0.5[Z–2Y(Z) + Z-1y(–1) + y(–2)] =1Z5.01
1
Y(Z)[1+1.5Z-1+0.5Z-2]+3+Z-1–2=1Z5.01
1
)Z5.0Z5.11)(Z5.01(
1
)Z5.0Z5.11(
)Z1()Z(Y
21121
1
=)Z1()Z5.01(
1
Z5.01
11211
Solve for Partial fraction expansion Inverse Z-transform is
y(n) = [4(–1)n – 4(–0.5)n +0.5n(–0.5)n–1]u(n)
: 37 : TEST-10 (Solutions)
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06. (b) (i)
Sol : No.of chips required =capacityAvailabe
designedbetoMemory
Available ROM chips = 20488 = 210248 = 22108 = 2k8
Memory to be designed = 8kB
No.of chips required = 4kB2
kB8
So for 4 chips address lines assigned A0 to A10
So decoder required is 2:4
Total no.of address lines of Micro processor = n =16
No.of address lines required for memory to be interfaced = x [8kB = 2x = 213 x =13]
Lines for decoding logic = y
n–x =y 16–13 = 3 (decoding logic)
2×4
A11
A12 y0y1y2
y3
2kB
1CS
RD WR
2kB
2CS
RD WR
2kB
3CS
RD WR
2kBA0
A10
C000H
C7FFH
C800H
CFFFH
D000H
D7FFH
D800H
DFFFH
A14
13A
E
Decoder
A150CS
RD WR
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06. (b) (ii)
Sol:
T-state
MVI B, XX 7 without loop
LOOP: DCR B 4 with loop
JNZ LOOPTrue (10)/ False (7) with loop
HLT 5 Without loop
Let no.of count value = n
Given Texe= 500s, fop = 5106 Hz
Texe = TwoL +TwL = 500s
valuecountstatesTof.Nof
1T
clkexe
valuecountT105
1T woLstate6woL
s4.2112105
1175
105
1T
66woL
TwL = Texe–TwoL = 500 –2.4 = 497.6 s
497.610–65106 = 14n–14+11
497.65 = 14n–3
H10 2B17892.17714
356.497n
XXB
TwL = T(condition satisfied)+ T(condition not satisfied)
(True) (False)
)False()True(
111)1n(14105
1106.497
66
Tn–1
F1
n
T=4+10=14F= 4+7 = 11
loop
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06. (c)
Sol: Total induction motor load
= 4 25 = 100 kW
Total load on transformers is kVA
=)866.0()95.0(
100
= 121.55 kVA
The open delta must be able to supply 121.55 kVA to the load.
(i) 866.0CapacityinstalledtotalrsTransforme
kVAinCapacityAvialble
kVA rating of two transformers
= kVA34.140866.0
55.121
Rating of each transformer =2
34.140
= 70.17 kVA
(ii) Secondary line current
= 1000400
17.70 = 174.43 A
Primary line current (H.V side)
=11000
40043.174
= 6.34 A
Load power-factor angle
= cos–1 (0.866) = 30
One of the transformer power factor
= cos (30 – 30)
= 1
Second transformer power factor
= cos (30 + 30)
= 0.5
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(iii) Real power supplied by one
= 70.17 1 = 70.17 kW
and another transformer
= 70.17 0.5 = 35.085 kW
Total power supplied open delta
= 70.17 + 35.085
= 105.26 kW
(iv) When third transformer is installed the available capacity is 70.17 3
= 210.51 kVA
06. (d)
Sol: 5 kW, 200V dc shunt motor
Back emf Eb = VIa Ra= 200 4 1 = 196V
Power developed (P) = EbIa= 1964 = 784W
This developed power is used in over coming the rotational Loss of the machine.
Rotational Loss = 784W
P = 78460
NT2
No load Torque(T) =10002
60784
=7.48N-m
8A
200V
+
2A
Rsh=100
6A
Ra= 1
: 41 : TEST-10 (Solutions)
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07. (a)
Sol: Three-phase problems can be solved in per unit by treating them as single-phase problems.
Here Xs = 0.8pu, ra = 0, Ef = 1.2 pu,
Vt = 1.00 pu. For an input kVA of 100% at Vt = 1.00, VtIa = 1.00 and therefore
Ia = 1.00 p.u.
As Ef = 1.2pu is more than Vt = 1.00, synchronous motor is working at a leading pf.
2sat2
aat2f XIsinVrIcosVE
222 )8.0(sincos)2.1(
sin6.164.0sincos44.1 22
Or 125.06.1
2.0sin
Power factor, cos = 0.9922 leading.
Mechanical power developed by the motor
.pu9922.09922.011cosIV at
When excitation emf is reduced to 1.00 pu for the same load, then
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9922.0sinX
VE
s
tf
o54.52or9922.0sin8.0
11
with ra = 0,
cosVE2VEXI tf2t
2f
2sa
o222a 54.52cos112)1()1(I8.0
or Ia = 1.1065 p.u.
Input kVA 1065.11IV at
.kVA%65.110or.u.p1065.1
07. (b)
Sol: (i)
1120 R
K211
R
K112
R
K10V
1R
K10
2
2R
K10
2R
K1010 R2 = 1k, R1 can be any value.
+
–R2
R2
10k
10k
11I
12I
20 R
K21V
R
K11V
R
k10V
–
–
+
+
VI1 = 1V
VI2 = 2V
R1
R1
2k
1k
11I R
k21V
12I R
k11V
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(ii)
159K2
K10V0
= 5[–6]
= –30V
Op-amp saturates
V0 = –15V
07. (c)
Sol: x(n) = 0.5, 0.5, 0.5, 0.5, 0, 0, 0, 0
+
–2k
2k10k
10k
V0
–
–
+
+
5V
4.5V
R1
R1
2k
1k
151
215
9)11(5.4
1W08
1W08
1W08
1W08
1W08
1W08
1W08
jW 28
jW 28
707.0j707.0
W18
707.0j707.0
W 38
jW 28
X(0)
X(1)
X(2)
X(3)
X(4)
X(5)
X(6)
X(7)
x(0)
x(4)
x(2)
x(6)
x(1)
x(5)
x(3)
x(7)
–1
–1
–1
–1–1–1
–1
–1
–1
–1
–1
–1
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Input Stage-1 outputs Stage-2 outputs Stage-3 outputs
0.5 0.5 1 2 = X(0)
0 0.5 0.5 – j0.5 0.5 – j1.207 = X(1)
0.5 0.5 0 0 = X(2)
0 0.5 0.5 + j0.5 0.5 – j0.207 = X(3)
0.5 0.5 1 0 = X(4)
0 0.5 0.5 – j0.5 0.5 + j0.207 = X(5)
0.5 0.5 0 0 = X(6)
0 0.5 0.5 + j0.5 0.5 + j1.207 = X(7)
07. (d)
Sol: Number of poles, 4P
Total lamp load, W500010050p
Terminal voltage Volts200V
Field resistance, 50R sh
Armature resistance 2.0R a
Voltage drop/brush = 1V
(i) Armature current, Ia:
Load current,Voltageterminal
ConsumedPowerI
25A200
5000
V
P
100W Lamps(50 No.s)
Ia
V=
200V
Ish
Rsh=50
+
Ra=0.2
I
: 45 : TEST-10 (Solutions)
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Shunt field current,sh
sh R
VI
A450
200
Armature current,sha
III
.A29425
(ii) Current per path:
Current per path4
29
a
Ia = 7.25 A.
[ ,4pa generator being lap wound]
(iii) General e.m.f., Eg:
dropbrushRIVE aag
V8.207122.029200
Hence, generated e.m.f. = 207.8 V.
(iv) Power output of D.C. armature:
Power output of D.C. armature
kW026.6kW1000
298.207
1000
IE ag
: 46 : ESE OFFLINE TEST-2017
ACE Engineering Academy Hyderabad|Delhi|Bhopal|Pune|Bhubaneswar| Lucknow|Patna|Bengaluru|Chennai|Vijayawada|Vizag|Tirupati|Kukatpally |Kolkata