Network Theory Test - 4 - PDF Flipbook

Network Theory Test - 4

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GATE
EEE

Network
Theory

Test-04Solutions


NETWORK THEORY
1. Impedance Z as shown in figure is

a) j29 Ω
b) j9 Ω
c) j19 Ω
d) j39 Ω
Answer: (b)
Solution:
From the given coupled circuit of inductors in series, observe
that the coupling between j5Ω and j2Ω is opposing and the
coupling between j2Ω and j2Ω is aiding. Use the formula:

= L1 + L2 ± 2M
∴ Z = j5 + j2 + j2 – j (10 × 2) + j(10 × 2)

= j9Ω
Note that, because of same mutual reactance, j10Ω, the opposite
nature of couplings is cancelled.

1


2. A connection is made consisting of resistance A is series
with a parallel combination of resistance B and C. Three resistors
of value 100Ω, 50Ω, 20Ω are provided. Consider all possible
permutations of the given resistors into the positions A, B, C and
identify the configuration with maximum possible overall
resistance; and also, the ones with minimum possible overall
resistance. The ratio of maximum to minimum values of the
resistances (up to second decimal place is) ________.
Answer: 2.143
Solution:

Given that, RA or RB or RC = 10Ω
RA or RB or RC = 5Ω
RA or RB or RC = 2Ω

For Req max:

The required combination is

RA = 10Ω and RB = 5Ω or 2Ω

And RC = 2Ω or 5Ω

So, Required = RA + (RB ∥ RC)

= 10 + (2∥5) = 80 = 11.4285Ω
7

= Req max

For Req min:

2


The required combination is

RA = 2Ω and RB = 10Ω or 5Ω

And RC = 10Ω or 5Ω

So, Required = RA + (RB ∥ RC)

= 2 + (10∥5) = 16 Ω = 5.33Ω
3

= Req min.

Hence, = 11.4285 = 2.143
5.33

3. Two coils N1 and N2 turns are wound concentrically on a

straight cylindrical core of radius r and permeability µ. The

windings have length I1 and I2 respectively as shown in figure.

The mutual inductance will be

1. Proportional to N1 N2.
2. Proportional to µ.
3. Inversely proportional to I1
4. Inversely proportional to r2.
Which of these statements are correct?
a) 1, 2, 3 and 4
b) 2, 3 and 4 only
c) 1, 2 and 3 only

3


d) 1, 3 and 4 only

Answer: (c)

Solution:

= 2


4. Consider the following statements:

Any element is redundant if connected in

a) series with an ideal current source

b) parallel with an ideal current source

c) series with an ideal voltage source

d) parallel with an ideal voltage source

Which of the above statements are correct?

a) 1 and 3

b) 1 and 4

c) 2 and 3

d) 2 and 4

Answer: (b)

Solution:

Here, R becomes redundant

Here, R becomes redundant

4


5. For the two – port network shown below, the short – circuit
admittance parameter matrix is

a) �−42 −42�S
b) �−01.5 −01.5�S
c) �01.5 01.5�S
d) �24 24�S
Answer: (a)
Solution:
The given 2-port -network is shown in Fig.1 with admittances.

The Y-parameters are easily identified for a passive reciprocal
π-network.
∴ Y11 = sum of left side admittances = 2 + 2 = 4 mho
∴ Y22 = sum of right side admittances = 2 + 2 = 4 mho
Y12 = Y21 = negative of the admittance of the shunt branch = -2
mho
Y11 = Y22 is the property of symmetric network.

5


6. Which of the following theorems can be applied to any network-

linear or non-linear, active or passive, time-variant or time-

invariant?

a) Thevenin theorem

b) Norton theorem

c) Tellegen theorem

d) Superposition theorem

Answer: (c)

Solution:

i) Tellegen’s theorem is applicable for any lumped network

having elements which are linear or non-linear, active or

passive, time-varying or time-invariant.

ii) The theorem is based on the two Kirchhoff’s laws.

iii) The theorem is completely independent of the nature of

elements and is only concerned with the graph of the

network.

iv) For any network, Tellegen’s theorem states
∑ =1 ( ) ( ) = 0, for all value of t.

This implies that the sum of the power delivered to all

branches of a network is zero.

7. For t > 0, the voltage across the resistor is

a) 1 � −√23 − −12 �
√3

b) −21 �cos �√23 � − 1 sin �√23 ��
√3

6


c) 2 −21 sin �√23 �
√3

d) 2 −12 cos �√23 �
√3

Answer: (b)

Solution:

Voltage across 1Ω resistor = current through it = current through

1F capacitor

= C dvc(t) = dvc(t)
dt dt

Using vc(t) obtained

( ) = 2 � − √3 cos �√23 � − 1 − sin �√23 ��
√3 2 2 2 2

= − �cos �√23 � − 1 sin �√23 ��, t > 0
2 √3

(or)

Let the voltage across the resistor R be VR, Then

VR(s) = 1 = s
1s+s+1 s2+s+1

VR(s) = �s+21� − 1 × 2 √3
�s+12�2+�√23�2 2 √3 2

�s+21�2+�√23�2

Applying Inverse Laplace Transform on both sides

VR(t) = −t cos �√23 t� − 1 −t sin �√23 t�
√3
e2 e2

∴ VR(t) = −t �cos √3 t − 1 sin √3 t�
2 √3 2
e2

7


8. What is the current I in the circuit given below?

a) 0 A
b) 1 A
c) 2 A
d) 3 A
Answer: (b)
Solution:

8


(1 + 3) ∥ (1 + 3) = 2

IT = 6 = 2A
3

I = IT = 2 = 1A
2 2

9. A ramp voltage v(t) = 100t volts, is applied to an RC

differentiating circuit with R = 5kΩ and C = 4µF. The

maximum output voltage is

a) 0.2 volts

b) 2.0 volts

c) 10.0 volts

d) 50.0 volts

Answer: (b)

9


Solution:

RC differentiating circuit is shown in Fig.1

( ) = ( ) = 100 ( )
Time constant, τ = RC = 5 × 103 × 4 × 10−6 = 0.02 sec

Formula: 0( )| = 0(∞)= Final value
= =
= 100 × 0.02 = 2

Detailed analysis is given below:

Transfer function, 0( ) =
( ) + 1

= =
+ 1 +50

( ) = 100
2

0( ) = 100 = 2 − 2
( +50) +50

0( ) = 2(1 − −50 ) ( )

Maximum v0(t) occurs as t → ∞

Maximum v0(t) = Final value = 2V

Final value also can be obtained from final value theorem:

→lt∞ 0( ) = →lt0[ 0( )] = 2

10


v0(t) is sketched in fig.2.

10. The maximum power that a 12 V DC source with an internal

resistance of 2 Ω can supply to a resistive load is

a) 72 W

b) 48 W

c) 24 W

d) 18 W

Answer: (a)

Solution:

From maximum power transfer theorem

Pmax = VT2 h = 12×12 = 18W
4RL 4×2

11. Which one of the following statements is not correct for the

circuit shown below at resonant frequency?

a) The current is maximum
b) The equivalent impedance is real

11


c) The inductive and capacitive reactance are equal in
magnitude.

d) The quality factor equals 1 � .


Answer: (d)
Solution:
i) At resonance frequency, impedance Z(jω) = R which is

real.
ii) Since minimum impedance at resonance, hence maximum

current.
iii) At resonance, imaginary term is zero which means

inductive and capacitive reactances are equal in magnitude.

iv) Quality factor, = 1 �


12. For the series R-L-C circuit of fig. 1, the partial phasor
diagram at a certain frequency is shown in fig.2. The operating
frequency of the circuit is

a) equal to the resonance frequency
b) less than the resonance frequency
c) greater than the resonance frequency
d) not zero

12


Answer: (b)

Solution:
For the given series R, L, C circuit, with current ⃗

�V⃗R = R⃗I, �V⃗L = jωL⃗I,

�V⃗C = − ⃗I � ⃗ = �V⃗R + V�⃗L + V�⃗C


� ⃗ is lagging V�⃗R in the given partial phasor diagram. This is

possible if �V�⃗C� > ��V⃗L�, as shown in the complete phasor

diagram, Fig.1.

. . , 1 >


2 < 1 (= 02) or < 0


∴ The operating frequency is less than the response frequency.

13. For the R-L circuit shown, the current i(t) for unit step input

voltage will rise to 0.63 in

a) 1 s
b) 2 s

13


c) 0.5 s

d) 1.5 s

Answer: (c)

Solution:

The time at which the current rises to 63% of its final value is

nothing but the time constant.

= = 1 = 0.5
2

14. An iron-cored choke of large inductance is connected to a d.c.

supply as shown in the circuit below. A capacitor C is also

connected across the switch. The role of C is to also connected

across the switch. The role of C is to

a) Improve the power factor of the circuit
b) Minimize the current drawn from supply
c) Prevent the arcing across switching under switching

conditions
d) Increase the magnetic flux in the core
Answer: (c)

14


15. The switch has been in position 1 for a long time and abruptly
changes to position 2 at t = 0

If time t is in seconds, the capacitor voltage VC (in volts) for t >
0 given by
a) 4 (1 – exp (– t/0.5))
b) 10 – 6exp (– t/0.5)
c) 4 (1 – exp (t/0.6))
d) 10 – 6exp (– t/0.6))
Answer: (d)
Solution:

(0−) = �150� . 2 = 4 = (0+)
(∞) = 5 × 2 = 10v

= . = 6 × 0.1 = 0.6
( ) = 10 + (4 − 10) − /
= 10 − 6 −0 . 6 for 0 ≤ ≤ ∞

16. Time constants of R-L and R-C circuits are respectively: R = 1
Ω; L = 1 H and C = 1 F
a) 1 sec and 1 sec
b) 1 sec and 2 secs
c) 2 sec and 3 secs
d) 2 sec and 4 secs

15


Answer: (a)

Solution:

Time constant of RL circuit,

=


⇒ = 1

Time constant of RC circuit,

τ = RC sec = 1sec

17. Consider the building block called ‘Network N’ shown in the

figure. Let C = 100 µF and R=10 kΩ.

Two such blocks are connected in cascade, as shown in the
figure.

The transfer function V3(s)/V1(s) of the cascaded network is

a)
1+

b) 2
1+3 + 2

16


c) �1+ �2

d)
2+

Answer: (b)

Solution:

Nodal, ( )− 1( ) + ( ) + ( )− 3( ) = 0

1 1


( ) �2 + 1 � = [ 1 ( ) + 3( )] ………… (1)

Nodal, 3( ) + 3( )− ( ) = 0

1


( ) = 3( ) �1 + 1 � ………… (2)

Equ (2) in equ (1)

3( ) ��1 + 1 � �2 + 1 � − � = . 1 ( )
3( ) ��1 + 1 � �2 + 1 � − 1� = 1 ( )

Here τ = RC = 1sec

3( ) = 1 = 2
1( ) �1+ 1 ��2+ 1 �−1 ( +1)(2 +1)− 2

= 2
2+3 +1

17


18. Two two-port networks are connected in parallel. The

combination is to be represented as a single two-port network.

The parameters of this network are obtained by addition of the

individual:

a) Z-parameters

b) h-parameters

c) Y-parameters

d) ABCD parameters

Answer: (c)

Solution:

Parallel connection of two port network

� 12 11 12 22� = � 12 11 + 1 1 1 2 + 1 2 �
+ 2 1 2 2 + 2 2

19. Consider the following statements:

If the differential equation for the circuit shown in the given

figure is v(t) = -4v(t) + 4, then the set of values of E, R and C

are respectively

1. 1V, 1Ω and 0.25F
2. 2V, 0.5Ω and 0.5F
3. 2V, 1Ω and 0.5F
4. 1V, 0.25Ω and 1F

18


Which of these statements are correct?

a) 1 and 2

b) 2 and 3

c) 3 and 4

d) 1 and 4

Answer: (d)

Solution:

= −4 ( ) + 4


Laplace transform,

( ) = −4 ( ) + 4


( ) = 4 = 1 − 1
( +4) +4

( ) = 1 − −4

For a RC circuit,
( ) = (1 − −4 )

So, E = 1

RC = 1 = 0.25
4

= 0.25 , = 1 or R = 1Ω, C = 0.25F

20. A reciprocal two-port network is symmetrical if

a) ΔA = 1

b) A = C

c) Z11 =Z22

d) ΔY = 1

Answer: (c)

19


Solution:
Conditions of symmetry for various parameters are given below:

Z – para ⇒ Z11 = Z22
Y – para ⇒ Y11 = Y22

h – para ⇒ ∆h = 1
AB – para ⇒ A = D
21. The circuit shown in figure' with R: 1/3, L = 1/4H, C = 3F has

input voltage v(t) = sin2t. The resulting current i(t) is

a) 5sin (2t + 53.10)

b) 5 sin (2t – 53.10)

c) 25sin (2t + 53.10)

d) 25 sin (2t – 53.10)

Answer: (a)

Solution:

For the sinusoidal input, v(t) = 1 sin (2t), with Vm = 1, ω = 2,

and R = 1 , L = 1 , C = 3F, the phasor circuit is shown in
3 4

Fig.1.

20


� ⃗ = 1. 0

Yequ = 3 + 2 + j6 = 3 − j2 + j6 = 3 + j4
j

⃗I = �V⃗Y = (3 + j4) = 5ejθ

Where θ = tan−1(4/3) = 53.10

( ) = 5 sin(2 + 53.10)

22. The transmission parameter matrix [T] for an ideal transformer

with a turns ratio of n1: n2 is given by

1 1
2�
a) � 2
0 1

1 0
2�
b) � 2
0 1

1 1
2�
c) � 2 −
0 1

1 0
2�
d) � 2 −
0 1

Answer: (b)

Solution:

The transmission parameters are represented as

1 = 2 − 2
1 = 2 − 2

21


Conditions of an ideal transformer.

• For ideal transformer internal resistance of each coil is equal

to zero.

• Primary and secondary coil are electrically isolated and hence

it is not possible to find transfer parameter values.

The equations for an ideal transformer are

1 = 1 . 2
2

2 = − 2 . 1
1

� 11 � = 1 0 �− 2 2 �
2�
� 2
0 1

Therefore transmission parameters are

1 0
2�
� 2
0 1

23. Consider the graph and tree (dotted) of the given figure

22


The fundamental loops include the set of lines
a) (1, 5, 3), (5, 4, 2) and (3, 4, 6)
b) (1, 2, 4, 3), (1, 2, 6), (3, 4, 6) and (1, 5, 4, 6)
c) (1, 5, 3), (5, 4, 2), (3, 4, 6) and (2, 4, 3, 1)
d) (1, 2, 4, 3) and (3, 4, 6)
Answer: (a)
Solution:
The number of fundamental loops = b – n + 1
Here, b = 6 and n = 4
So number of fundamental loops = 6 – 4 + 1 = 3
24. In the circuit shown, the average value of the voltage Vab (in
volts) in steady state condition is _____.

Answer: 5
Solution:
For 5V dc source, in steady state, ′ = 5
For 5πsin(5000 )V a.c source, in steady state

= sin(5000 + )
By SPT, = ′ + "

= 5 + sin(5000 + )
= 1 ∫0 ( ) = 5 + 0 = 5

23


25. Consider the following statements with regard to a complete
incidence matrix:
1. The sum of the entries in any column is zero.
2. The rank of the matrix is n – 1 where n is the number of
nodes.
3. The determinant of the matrix of a closed loop is zero.
Which of the statements given above are correct?
a) 1 and 2 only
b) 2 and 3 only
c) 1 and 3 only
d) 1, 2 and 3
Answer: (d)

26. The Z parameters Z11 and Z21 for the 2-port network in Fig. are

a) 11 = − 6 Ω; 21 = 16 Ω
11 11

b) 11 = 6 Ω; 21 = 4 Ω
11 11

c) 11 = 6 Ω; 21 = − 16 Ω
11 11

d) 11 = 4 Ω; 21 = 4 Ω
11 11

Answer: (c)

24


Solution:
Z-parameters equations:

E1 = Z11I1 + Z12I2
E2 = Z21I1 + Z22I2
Z11 & Z21 are calculated by making I2 = 0 (Output port open)
The relevant circuit is shown in Fig.1.

11 = 1 1� 2=0
Write KVL for the left loop

E1 = 6I1 – 10 E1

Or 1 = 6 1 ………. (1)
11

∴ 11 = 6
11

21 = 1 2� 2=0

Write KVL for the right loop.

E2 = 4I1 – 10 E1

Substitute for E1 from (1)

Or 1 = 6 1 ………. (1)
11

25


2 = 4 1 − 10 �161 1�

= �4 − 6101� 1 = − 16 1
11

21 = − 16
11

27. An R-L-C circuit for the driving-point admittance function

�1 1 ⁄ + 1 + � is

Answer: (b)

Solution:

Driving point admittance function

( ) = 1/ +
1 + 1

= 1/ + = 1 +
+
+



26


28. Which one of the following driving point functions does not

represent an LC network?

a) Z(S) = ( +3)
( 2+1)( 2+9)

b) Z(S) = ( 2+25)
( 2+36)

c) Z(S) = � 2+10�� 2+36�
( 2+5)( 2+25)

d) Z(S) = � 2+16�
( 2+25)

Answer: (a)

Solution:

Properties of LC functions:

i) LC functions are the ratio of either even to odd or odd to

even polynomials.

ii) Poles and zeros of the LC functions always lie on jω-axis.

iii) Poles and zeros are alternate in jω-axis.

iv) There is either a pole or a zero at origin and infinity.

v) The highest and lowest powers of sin numerator and

denominator can differ at the most by 1.

Option (a) violates property.

29. Two coupled coils connected in series have an equivalent

inductance of 16 mH or 8 mH depending on the interconnection

then the mutual inductance m between the coils is

a) 12 mH

b) 8√2 mH

c) 4 mH

27


d) 2 mH
Answer: (d)
30. For the two-port network shown in the figure, the impedances
(Z) matrix (in Ω) is

a) �462 294�
b) �89 284�
c) �96 264�
d) �462 660�
Answer: (c)

Solution:

( ) = � − + − � ℧
+

= �110−+310310 610−+310310� ℧

4 −3310� ℧

= �−30310 60

28


( ) = 4 × 3 − 1 × 1 = 1
30 60 30 30 180

3 1

( ) = ( )−1 = �610 340� 1

30 1

30 180

= �69 264�

29


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