CWT–04
A
Booklet No.:
Booklet Series:
03092014
Network System
Student Name:
Roll Number:
Duration: 90 Minutes
PAPER
MAXIMUM
MARKS: 60
INSTRUCTIONS
1.
IMMEDIATELY AFTER THE COMMENCEMENT OF THE EXAMINATION, YOU SHOULD CHECK
THAT THIS TEST BOOKLET DOES NOT HAVE ANY UNPRINTED OR TORN OR MISSING PAGES
OR ITEMS ETC. IF SO, GET IT REPLACED BY A COMPLETE TEST BOOKLET.
2.
This Test Booklet contains 30 questions. Each question comprises four responses (answers).
You will select the response which you want to mark on the Answer Sheet. In case you feel that
there is more than one correct response, mark the response which you consider the best. In any
case, choose ONLY ONE response for each item.
3.
You have to mark all your response ONLY on the separate Answer Sheet provided.
4.
All items carry equal marks.
5.
Before you proceed to mark in the Answer Sheet the response to various items in the Test
Booklet, you have to fill in some particulars in the Answer Sheet as per instructions.
6.
Each questions 2 marks and 2/3 negative mark is assigned for the wrong answer.
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Ph(011) -26194869, Cell: 9873000903, 9873664427 , 8860182273:
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1
QH Engineers Zone, 65/C, Near Prateek Market, Canara Bank, Munirka, New Delhi-110067,
P h ( 0 1 1 ) - 2 6 1 9 4 8 6 9 , C e l l : 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7 : E - ma i l : q h e n g i n e e r z o n e @ g ma i l . c o m
Network
(1.)
A Network N feeds a resistance R as shown
combination
in circuit below let the power consumed by
inductance of 1H the steady-state current
R the 4 watts if an identical network is
i t  in ampere
added as shown in figure
of
resistance
of
1
and
(a.) 10 2 sin t  25  6.324sin  2t  43 43 
(b.) 5 2 sin t  25   4.472sin  2t  43 43 
(a.) Equal to 4 watts
(c.) 10sin t  25  6.324sin 2t  43 43
(b.) Less than 4 watts
(d.) None
(c.) Between 4 and 16 watts
Ans: (c)
(d.) More than 16 watts
(2.)
(5.)
The current value form in figure is following
Ans: (c)
through a 5  resistor find the average
Find I
power absorbedly the resistor
(a.) 2.13 43.50 A
(a.) 50.4 W
(b.) 2 43.50 A
(b.) 59.3 W
(c.) 2.13 43.50 A
(c.) 53.3 W
(d.) 2 43.50 A
Ans: (a)
(3.)
(d.) 54.3 W
Find Vs  ?
Ans: (c)
(6.)
Find the average power absorbedly the 1  
resistor in the network should in fig.
(a.) 301.37 7.89V
(b.) 200 7.89V
(a.) 4.5 W
(c.) 130 7.89V
(b.) 3.0 W
(d.) 400 7.89V
(c.) 4 W
Ans: (a)
(4.)
An
input
(d.) 6 W
voltage V t   10 2 sin t  20 
10 2 sin  2t  20  is applied to a series
2
Ans: (a)
(7.)
Find V0  ? if I x  6 45 A
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Network
Common Data For 10, 11, 12
The switched in fig has been closed for a
long time it is open at t = 0
(a.) 13.8911V
(b.) 12.2110.09V
(c.) 13.3111.31V
(d.) None
(10.)
Ans: (c)
Find i 0  , v 0 
(a.) 2 A, 4 V
(b.) 2 V, 4 A
(8.)
Find out V0 t  at t  0
(c.) 6 A, 8 V
(d.) 2 A, 3 V
Ans: (a)
(11.)
Find
di  0  dv   0 
,
dt
dt
(a.) 20 A/S, 20 V/S
(b.) 0 A/S, 20 V/S
(c.) 20 A/S, 0 V/S
(a.) 1.85 e
(b.) 1.9 e
5.54t
(d.) None
4.5 t
Ans: (b)
(12.)
(c.) 2 e 5.44 t
(a.) 0 A, 12 V
(d.) 1e 5.53 t
(b.) 12 A, 0 V
Ans: (d)
(9.)
i , v 
(c.) 0.001 A, 12 V
Find V0 t  for t  0 when switch 2 is closed
(d.) None
t=0
Ans: (a)
Common Data for 13, 14, 15
In the circuit of figure.
(a.) V0 t  
8
1  e 0.96t V , t  0
3
(b.) V0 t  
16
V
3
(c.)
3
t 0
8
1  e 0.96t V , t  0
3
(13.)
Find iL  0  , vc  0  ,vR  0 
(a.) 0 A, 0 V, –20 V
(b.) 0 A, –20 V, 0 V
(d.) None
(c.) 0 A, 0 V, 0 V
Ans: (c)
(d.) None
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Network
Ans: (b)
(14.)
Find
(c.)
diL    dvC    dv R   
0 ,
0
0
dt
dt
dt
8 e 4t A
(d.) 16 e 4t A
Ans: (a)
(a.) 0 A/S, 2 V/S 2/3 V/S
(18.)
(b.) 2 A/S, 0 V/S 2/3 V/S
(c.) 0 A/S 2 V/S 2/15 V/S
(d.) None
Ans: (a)
(15.)
i L    , v R    , vC   
(a.) 1 A, 4 v, –20 V
(b.) 0 A, 16 V, –20 V
The Norton’s equivalent across ab is
(c.) 0 A, 4 V, –20 V
(a.)
(d.) None
Ans: (a)
(16.)
An inductance load of resistance 40 
inductance 0.1 H, is switched on at t = 0 an
ac
voltage
V  10 sin t   
find
(b.)
the
switched angle  such that there is no
transient   3.14 rad/sec
(a.) 0.4497°
(b.) 1.4497°
(c.)
(c.) 2.4497°
(d.) None
Ans: (a)
(17.)
In the circuit of fig below the switch is
closed at t = 0 after a long time. The initial
(d.)
condition on the capacitors are vC1  0    4V
and vC2  0    2V the current iR t  for t  0
is
Ans: (a)
(19.)
R and C are connected in parallel across a
sinusoidal voltage of 220 V if the currents
through the source and the capacitor one
10 A and 6 A respectively what is the value
of R ?
(a.) 4 e 4t A
(b.) 14 e 4t A
4
(a.) 27.5 
(b.) 10 
(c.) 22
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Network
(d.) None
(22.)
Ans: (a)
(20.)
Peak of the supply voltage is 100V and
peak of supply current is 50 2 and lags
Find the r.m.s value of currents
from supply voltage by 30°
Determine peak value of I L and also find
power dissipated by 1
(a.) 99.55 105.98 A , 25W
2A
(a.)
(b.) 50 30 A, 50W
(b.) 2 A
(c.)
(c.) 100 45 A, 100W
1
A
2
(d.) 99.55 2 105.98 A, 100W
(d.) None
Ans: (d)
Ans: (a)
(21.)
The Y- parameters of a z-port network are
1 2
S
3 4 
y   
A resistor of 2 is connected as shown in
fig they new y-parameter would be
(23.)
The RL circuit the figure is feed from a
constant
magnitude,
variable
frequency
sinusoidal voltage source v in at 100 Hz the
R and L element each have a voltage drop
Vrms if the frequency of the source is
changed to 50 Hz the new voltage drop
across R
3
2
(a.) 
3
 2
5
2
S
9
2 
3
2
(b.) 
5
 2
3
2
S
9
2 
1 2 
(c.) 
S
5 6 
(d.) None
(a.)
2
Vrms
5
(b.)
8
Vrms
5
(c.)
5
Vrms
8
(d.) None
Ans: (b)
Ans: (b)
5
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Network
(24.)
Series RLC is overdamped its quality factor
In the linear network containing resistors
Q at resonance is
only, if Vs  2V , and I s  4 A, I is found to
(a.) Q > 1
be 8A, when Vs  2V and I s  4 A, I is 0A
(b.) 1 < Q
the value of I when Vs  2V and I s  2 A
(c.) Q > 0.5
(a.) 6A
(d.) Q < 0.5
(b.) 3A
Ans: (d)
(25.)
(c.) 4A
In the circuit of figure below the equivalent
(d.) –3A
resistance seen by the inductor is
Ans: (a)
(28.)
Z  s  for the nf network shown in the above
figure is
(a.) 400
(b.) 100
(c.) 200
(d.) iL  0  is required
(26.)
3  s 2  6s  8 
Ans: (c)
Z s  
In the circuit below shown what value of R L
Determine the value of C and R
maximizes the power delivered by R L
(a.)
2
9
F and 
9
2
(b.)
2
F and 1
3
(c.)
1
F and 4 
6
(d.)
1
F and 1 
2
Ans: (c)
(a.) 4
(b.)
8

3
(c.) 2.4
(d.) 6
s 2  4s  3
(29.)
For
input
v t   u t 
unit
i  0   1 A, vc  0   1
Determine I  s 
Ans: (a)
(27.)
6
(a.)
s2
s  s 1
(b.)
s4
s  s 1
2
2
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
step
Network
(c.)
s 1
s  s 1
2
(d.) None of thse
Ans: (d)
(30.)
In circuit of figure
(a.)
3
R
4
(b.)
2
R
3
(c.) R
(d.)
4
R
3
Ans: (a)
Determine R AB
7
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m
Network
Answer Key
Civil Engineering
(1.)
(d)
(9.)
(a)
(17.)
(b)
(25.)
(b)
(2.)
(d)
(10.)
(b)
(18.)
(c)
(26.)
(b)
(3.)
(c)
(11.)
(d)
(19.)
(c)
(27.)
(c)
(4.)
(a)
(12.)
(a)
(20.)
(b)
(28.)
(c)
(5.)
(a)
(13.)
(d)
(21.)
(a)
(29.)
(b)
(6.)
(b)
(14.)
(a)
(22.)
(b)
(30.)
(b)
(7.)
(b)
(15.)
(a)
(23.)
(d)
(8.)
(a)
(16.)
(b)
(24.)
(d)
8
EN G IN EER S ZO N E, 6 5 / C , P r at ee k Ma r k et , Ne ar C an a ra B an k, Mu n i r ka, N ew Del h i -1 1 0 0 6 7 ,
P h (0 1 1 ) -2 6 1 9 4 8 6 9 , C e l l: 9 8 7 3 0 0 0 9 0 3 , 9 8 7 3 6 6 4 4 2 7: E - ma i l : q h en gi n e er zo n e @ g m ai l . co m