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Title1487180513EE_2017-_Session_2_(Kreatryx_Solutions) (1)
Tags Electricity Physical Quantities Direct Current Power Inverter
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GATE 2017 EE - Session 2

© Kreatryx. All Rights Reserved. 1 www.kreatryx.com











































Kreatryx GATE 2017 EE – Session 2

Answer Key and Solutions





Don’t forget to enter your marks in

the Rank Predictor Form at the end.

Page 2

GATE 2017 EE - Session 2

© Kreatryx. All Rights Reserved. 2 www.kreatryx.com









01. In a load flow problem solved by Newton-Raphson method with polar coordinates, the

size of the Jacobian is 100 × 100. If there are 20 PV buses in addition to PQ buses and a slack

bus, the total number of buses in the system is _________ .



01. Ans: 61

Solution:

Size of Jacobian = 2 × PQ buses + PV buses

100 = 2 × PQ + 20

PQ = 40

Total no. of buses = PQ + PV + Slack bus= 40 + 20 + 1 = 61 bus



02. For a 3-input logic circuit shown below, the output Z can be expressed as















(a) Q R (b) PQ R

(c) Q R (d) P Q R 



02. Ans: (c)





Y QR (Q R)  

Z XYQ (P Q) (Q R) Q   

Z (PQ PR RQ)Q PRQ RQ RQ(P 1) RQ Q R         



03. Let
2

y 2y 1 x   and x y 5  . The value of x y equals _________. (Give the answer

up to three decimal places)



03. Ans: 5.732

x y 5 

x 5 y  �¬�¬�¬�¬�����L��

Since, y2 �² 2y + 1 = x
y2 �² 2y + 1 = (5 �² y)2

2 2
y 2y 1 25 y 10y    

8y = 24

y = 3



X PQ (P Q)  

Page 20

GATE 2017 EE - Session 2

© Kreatryx. All Rights Reserved. 20 www.kreatryx.com









34. A cascade system having the impulse responses 1h (n) 1, 1


  and 2h (n) 1,1


 is shown in

the figure below, where symbol  denotes the time origin.







The input sequence x(n) for which the cascade system produces an output sequence

(n) 1,2,1, 1, 2, 1


   y is

(a) x(n) 1,2,1,1


 (b) (n) 1,1,2,2


x

(c) (n) 11,1,1,1


x (d) (n) 1,2,2,1


x



34. Ans: (d)

 1 11H (z) 1.z 1.z 1 z
 

   

 1 12H (z) 1. z 1. z 1 z
 

   

1 1 2

1 2
H(z) H (z)H (z) (1 z ) (1 z ) (1 z )

  
     

1 2 3 4 5
Y(z) 1.z 2.z 1.z 1z 2z 1z

    
      

3 1 3 2 3
Y(z) (1 z ) 2z (1 z ) z (1 z )

    
     

3 1 2
Y(z) (1 z ) (1 2z z )

  
   

1 2 1 1 2
Y(z) (1 z ) (1 z z ) (1 z )

   
    

1 1 2 1 2

1 1

(1 z ) (1 z ) (1 z z )Y(z)
X(z)

H(z) (1 z )(1 z )

   

 

   
 

 


1 1 2 1 2 3
X(z) (1 z ) (1 z z ) 1 2z 2z z

     
       

 x[n] 1, 2, 2, 1




35. Which of the following systems has maximum peak overshoot due to a unit step input?

(a)
2

100

10 100 s s
(b)

2

100

15 100 s s


(c)
2

100

5 100 s s
(d)

2

100

20 100 s s




35. Ans: (c)

For maximum peak overshoot, 1 

For all 4 options,
n

10 rad/sec 

10
For (a), 0.5

2 10
  




15
For (b), 0.75

2 10
  



Page 21

GATE 2017 EE - Session 2

© Kreatryx. All Rights Reserved. 21 www.kreatryx.com









5
For (c), 0.25

2 10
  




20
For (d), 1

2 10
  




Lowest value of  gives highest overshoot.

Hence, (c) will have maximum peak overshoot.



36. In the circuit shown below, the value of capacitor C required for maximum power to be

transferred to the load is

















(a) 1 nF (b) 1 F

(c) 1 mF (d) 10 mF



36. Ans: (d)

For maximum power transfer ZL = Zs*

Since
s

Z 0.5 

L
Z 0.5 for maximumpower transfer 

L

j
R

c
Z j L

j
R

c

 
 
 

  




[ 100 rad/sec]

L

j
1

100c
Z j100 0.005

j
1

100c

 
 
 

  





L

2

j
j 1

100c
Z j0.5

1
100c 1

10000 c

 
  
 

 
 

 
 



For ZL to be purely real

2

1
0.5 0

1
100c 1

10000 c


 

 
 

 

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