Network Theory Test - 2 - PDF Flipbook
Network Theory Test - 2
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GATE
EEE
Network
Theory
Test-02Solutions
NETWORK THEORY
1. For the circuit in FIG. the voltage V0 is
a) 2V
b) 1V
c) –1V
d) None of the above
Answer: (d)
Solution:
The diode is forward biased. Assuming that the diode is ideal,
the Network is redrawn with node A marked as in Fig. 1. Apply
KCL at node A
4 − 0 = 0 + 0+2
2 2 2
3 0 = 1
2
0 = 2
3
1
Common Data for Question 2:
The circuit for Q2 is given in Fig. For both the questions,
assume that the switch S is in position 1 for a long time and
thrown to position 2 at t = 0.
2. At t = 0+, the current i1 is
a) −
2
b) −
c) −
4
d) Zero
Answer: (a)
Solution:
2
The status of the circuit at t = 0- is shown in Fig.1
i1(0-) = 0, i2(0-) = 0
Current through the inductor = i2(0-) – i1(0-) = 0
v1(0-) = V, v2(0-) = V
The status of the circuit at t = 0+ is shown in Fig.2.
v1(0+) = v1(0-) = V
v2(0+) = v2(0-) = V
Current through the inductor at t = 0+
= i2(0+) – i1(0+) = i2(0-) – i1(0-) = 0
Inductor across PQ is equivalent to open circuit.
But individually 1(0+) = 2(0+) = − 2 , as can be seen from
the loop equation
3. Consider the following network:
Which one of the following is the differential equation for v in
the above network?
a) C + = 0
b) G + = 0
c) 1 + = 0
d) C − = 0
Answer: (a)
3
Solution:
=
Applying KVL,
1 + = 0
⇒ + = 0
⇒ + = 0
4. A network N consists of resistors and independent voltage and
current sources. Its value of the determinant based on the node
analysis
1. cannot be negative
2. cannot be zero
3. is independent of the values of voltages and current sources
a) 1, 2 and 3
b) 1 and 2 only
c) 2 and 3 only
d) 1 and 3 only
Answer: (a)
Solution:
Determinant can be negative but it cannot be zero.
5. In a linear circuit, the superposition principle can be applied to
calculate the
a) voltage and power
b) voltage and current
c) current and power
4
d) voltage, current and power
Answer: (b)
Solution:
The superposition principle is not valid for power relationship.
6. In the circuit shown below, current through the inductor is
a) 2 A
1+j
b) −1 A
1+j
c) 1 A
1+j
d) 0 A
Answer: (c)
Solution:
From the given circuit, = 0
=
5
J 1 = 1(1 - )
⟹ = 1 A
1+
7. The half - power bandwidth of the resonant circuit of figure can
be increased by
a) increasing R1
b) decreasing R1
c) increasing R2
d) decreasing and increasing R2
Answer: (a)
Solution:
The given resonant circuit with only L and C elements i.e., R2
→ ∞, R1 → 0 is ideal and has infinite selectivity. As bandwidth
is inversely proportional to selectivity, the half power bandwidth
can be increased by reducing the selectivity and hence by
decreasing the parallel resistance, R2 and increasing the series
resistance R1.
∴ Options (a) and (d) are correct.
6
8. Match List-I with List-II and select the correct answer using the
code given below the lists:
List-I
A. Superposition theorem
B. Thevenin's theorem
C. Kirchhoff's voltage and current laws
D. Maximum power transfer theorem
List-II
1. Impedance matching in audio circuits
2. Linear bilateral networks
3. Large network in which currents in few elements to be
determined
4. Currents & voltages in all branches of a network
Codes:
AB C D
a) 1 4 3 2
b) 2 4 3 1
c) 1 3 4 2
d) 2 3 4 1
Answer: (d)
Solution:
Superposition theorem can be applied only to linear bilateral
networks. To determine current and voltage in few branches, we
7
can replace remaining complex network in its Thevenin’s
equivalent network.
9. The circuit shown in the figure below, will act as an ideal
current source with respect to terminals A and B, when
frequency is
a) Zero
b) 1 rad/s
c) 4 rad/s
d) 16 rad/s
Answer: (c)
Solution:
Internal impedance of the circuit is
= . 1
1
+
The internal impedance of an ideal current source is infinite. In
other words, the internal admittance of an ideal current source is
zero.
=0 ⇒ + 1 = 0
⇒ 1 = 1
16
⇒ 2 = 16
⇒ = 4 /
8
10. Consider the following statements regarding the properties of
an R-L-C series circuit under resonance:
1. Current in the circuit is in phase with applied voltage.
2. Voltage drop across capacitor C and inductance L are equal
in magnitude.
3. Voltage across the capacitor is equal in magnitude to the
applied voltage.
4. Current in the circuit is maximum.
Which of the above statements is/are correct?
a) 1 only
b) 1, 2 and 4
c) 2 and 4
d) 1, 3 and 4
Answer: (b)
Solution:
In series R-L-C circuit under resonance,
| | = | |
So impedance is minimum, i.e., Z = R.
So current is maximum which is also in phase with applied
voltage.
Also, | | = | | =
Where Q = Quality factor
9
11. Two series resonant filters are as shown in the figure. Let the 3-
dB bandwidth of Filter 1 be B1 and that of Filter 2 be B2. The
value of B1/B2 is
a) 4
b) 1
c) 1/2
d) 1/4
Answer: (d)
Solution:
For the series resonant filter 1
= 1 = 1
2 � 1 1
Quality factor, Q = Q1 = 1
3 dB Bandwidth 1 =
1
For series resonant filter 2
= 2 = 1
2 � 2 2
10
Where 2 = 1 , 2 = 4 1,
4
2 2 = 1 4 1
4
= 1 1
∴ 2 = 1
2 = 2 = 1 , 2 = 1 1
4 4
2 = 2 = 4 1 = 4 1,
2 1
∴ 1 = 1
2 4
12. An independent voltage source in series with an
impedance ZS= RS + jXS delivers a maximum average
power to a load impedance ZL when
a) = +
b) =
c) =
d) = −
Answer: (d)
Solution:
For maximum power delivery, the load impedance
: = ∗
Where * denotes complex conjugation.
Here = + =Source impedance
∴ = −
11
13. For the circuit shown in the given figure, when the switch is at
position A, the current i(t) = I sin (ωt + 30°) A. When switch is
moved to position B at time t = 0, the power dissipated at the
switching instant in the resistor R remains unchanged
The value of I and the element X would respectively, be
a) 1 A and resistor
b) 2 A and capacitor
c) 3 A and resistor
d) 4 A and capacitor
Answer: (d)
Solution:
When the switch is at position A, the current in R is 1 A
downward.
When the switch is moved to position B, the power dissipated in
R remains same but the current is now 1 A upward.
At t = 0,
( ) = . sin 300 =
2
= 1 − (−1) = 2
2
⇒ = 4
12
14. In the circuit shown below, switch S is closed at t = 0. The
time constant of the circuit and initial value of current i(t) are
a) 30 sec, 0.5 A
b) 60 sec, 1.0 A
c) 90 sec, 1.0 A
d) 20 sec, 0.5 A
Answer: (d)
Solution:
= 3×6 = 2
3+6
= 10
∴ Time constant = = 20
Therefore, option (d) is the correct answer no need to calculate
current i(t).
13
15. If a capacitor is energized by a symmetrical square-wave
current source, then the Steady state voltage across the capacitor
will be
a) a square wave
b) a triangular wave
c) a step function
d) an impulse function
Answer: (b)
Solution:
=
= 1 ∫
Given I = Symmetrical square wave
Then,
∫ = triangular wave
16. The condition AD – BC = 1 for a two-port network implies
that the network is a
a) reciprocal network
b) lumped element network
c) lossless network
d) unilateral element network
Answer: (a)
Solution:
If a two-port network is reciprocal and passive, the A, B, C and
D parameters should satisfy the condition AD – BC = 1.
14
17. The driving point impedance of the following network is given
by Z(s) = 2+00.2.1 +2. The component values are
a) L = 5H, R = 0.5Ω, C = 0.1F
b) L = 0.1 H, R = 0.5Ω, C = 5F
c) L = 5H, R = 2Ω, C = 0.1F
d) L = 0.1H, R = 2Ω, C = 5F
Answer: (d)
Solution:
The s-domain equivalent of the given circuit is shown in below
Fig.1.
( ) = 1 + + 1
= + 2 + = 1 + 2 + 1
= � 2+ 1 + 1 �
( ) = 1
2+ 1 + 1
Compare with the given
15
( ) = 0.2
2+0.1 +2
= 1 = 5
0.2
1 = 2, = 0.1
1 = 0.1, = 2
∴ = 0.1 , = 2 , = 5
18. Which of the following are the conditions for a two-port
passive network to be a reciprocal one?
1. Z12 = Z21
2. Y12 = Y21
3. h12 = – h21
Select the correct condition from the code given below:
a) Only 1 and 2
b) Only 2 and 3
c) Only 1 and 3
d) 1, 2 and 3
Answer: (d)
Solution:
Conditions for reciprocity
12 = 21
1 2 = 2 1
ℎ12 = −ℎ21
12 = − 21
− = 1
′ ′ − ′ ′ = 1
16
19. If the unit step response of a network is (1 – e-αt), then its
impulse response is
a) αe-αt
b) α–1 e-αt
c) (1 – α–1)e-αt
d) (1 – α)e-αt
Answer: (a)
Solution:
Given: Unit step response of the network,
( ) = (1 − − )
∴ Its unit impulse response,
ℎ( ) = ( ) =
20. In the 2-port network shown in the figure, the value of Y12 is
a) –1/3 mho
b) 1/3 mho
c) – 3 mho
d) 3 mho
Answer: (a)
17
Solution:
Short circuiting input port
Applying KCL at node V
+ + − 2 = 0
1 1 1
⇒ 3 = 2 … . ( )
Also = − 1
1
⇒ = − 1
Putting above value in equation (i)
3(− 1) = 2
⇒ 1 � = 1 2
2
1=0
= 1 ℎ
3
18
21. The resistance measured between the two ends of the toroid
shown in the below figure is R. What would be the resistance if
both a and b are doubled?
a) 2R
b) R
c) R/2
d) R/4
Answer: (c)
Solution:
Length of toroid = +
2
Area of cross-section = (a – b)2
∙ ( + )
2
= ( − )2
= ∙ ( + )
2 ( − )2
With ‘a’ and ‘b’ doubled,
′ = ∙�2 +2 2 �
(2 −2 )2
= ∙ ( + ) =
4 ( − )2 2
19
22. For the circuit shown in Figure the total impedance is
a) (7 + j0)
b) (5 + j0)
c) (0 + j8)
d) (7 + j10)
Answer: (a)
Solution:
(3 + 4)‖(3 − 4) = (3+ 4)(3− 4)
3+ 4+3− 4
= 32+42 = 25
6 6
= 17 + 25 = 7
6 6
Total impedance = (7 + 0)
23. A 2 - port network is shown in figure. The parameter h21 for
this network can be given by
a) –1/2
b) 1
2
20
c) – 3
2
d) 3
2
Answer: (a)
Solution:
h-parameter equation to calculate ℎ21:
2 = ℎ21 1 + ℎ22 2
ℎ21 = 21� 2=0
The relevant circuit with 2= 0 is shown in Fig.1 Using current
division rule:
2 = − 1
2
∴ ℎ21 = 2
1
= − 1
2
24. Two identical two-port networks having transmission matrix
� � are cascaded. What will be the resultant transmission
matrix of the cascade?
a) � �
b) �22 22 �
21
c) � 2 + + 2 �
+ +
d) � 22 22�
Answer: (c)
Solution:
When two-two port networks are connected in cascade, the
individual ABCD parameters are multiplied.
Therefore,
� � × � � = � 2 + + 2 �
+ +
25. The number of edges in a complete graph of n vertices is
a)n (n – 1)
b)n(n – 1)/2
c)n
d)n – 1
Answer: (b)
Solution:
Number of edges = nC2 = ( −1)
2
26. A series R-L-C circuit has a Q of 100 and an impedance of
(100 + j0) Ω at its resonant angular frequency of 107rad / sec.
The values of R = 100Ω and L = _____ mH.
Answer: L = 1mH
Solution:
For series RLC circuit under resonance,
= 1 = 107 / sec … . (1)
√
22
= = 1 = 100 … . . (2)
R = Resonant impedance = (100 + j0) Ω
∴ = 100
From (2),
= = 104 = 10−3 = 1
107
= 1
= 1
100×107×100
= 10−11 = 10
27. In a singly connected network if there are b number of
branches and n number of nodes, then the number of
independent meshes M and independent nodes N are
respectively
a) n and b
b) b – n + 1 and n – 1
c) b – n and b
d) b + n – 1 and n + 1
Answer: (b)
Solution:
Given, b = Number of branches
n = Number of nodes
M = Number of mesh
= Total number of tie set
= Number of links
23
= b – (n – 1) = b – n + 1
N = Number of nodes
= Total number of cutset
= Number of trees = (n – 1)
28. What is the open circuit impedance Z'11(S) of the network
shown in the figure given below?
a) 10 + 2S
b) �10 − 4 �
c) �10 + 4 �
d) 10 – 2S
Answer: (c)
Solution:
Open circuit impedance
1′ 1 ( ) = 10 + 1
41
1′1( ) = �10 + 4 �
24
29. Statement (I): Lossless network functions have only
imaginary zeros and poles with only negative real parts.
Statement (II): Lossless network functions obey the separation
property.
a) Both Statement (I) and Statement (II) are individually true
and Statement (II) is the correct explanation of Statement (I).
b) Both Statement (I) and Statement (II) are individually true
but Statement (II) is not the correct explanation of Statement
(I).
c) Statement (I) is true but Statement (II) is false.
d) Statement (I) is false but Statement (II) is true.
Answer: (d)
Solution:
Lossless network is equivalent to LC network.
Properties of LC networks are:
(a) Poles and zeros of the LC function always lie on j -axis.
(b) Poles and zeros are alternate on j -axis.
(c) There is either a pole or zero at origin and infinity.
So, statement (i) is wrong according to above properties.
30. The circuit shown below is driven by a sinusoidal input Vi =
Vp cos(t/RC). The steady state output V0
25
a) (Vp/3) cos(t/RC)
b) (Vp/3) sin(t/RC)
c) (Vp/2) cos(t/RC)
d) (Vp/2) sin(t/RC)
Answer: (a)
Solution:
The transfer function H(j ) s found from the circuit shown:
= cos � 1 � , = 1
o = ( ) = 2
1+ 2
2 = = = (1 − ), at the input frequency, = 1
1+ 1+ 2
Similarly 1 = − = (1 − ) at = 1
1 + 2 = (1 − ) + (1 − ) = 3 (1 − )
2 2
� 1 � = 1 , | | = 1 , ∠ = 0
3 3
For the input = cos( / ), the steady state output
0 = | | cos � + ∠ � = 1 cos � �
3
Note: Recognize that the given circuit is used as the feedback
network in Wein bridge oscillator.
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