# Network Theory Test - 3 - PDF Flipbook

Network Theory Test - 3

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GATE
EEE

Network
Theory

Test-03Solutions

NETWORK THEORY
1. In the given circuit, the values of V1 and V2 respectively are

a) 5 V, 25 V

b) 10 V, 30 V

c) 15 V, 35 V

d) 0 V, 20 V

Solution:

Nodal ⟹ −5 + 1 + 1 + 2I = 0
4 4

⟹ −5 + 2 1 + 2 × 1 = 0
4 4

⟹ 1 = 5V

By KVL ⟹ 2 – 20 – 5 = 0

⟹ 2 = 25V

2. Consider a circuit which consists of resistors and independent

current sources, and one independent voltage source connected

between the nodes i and j. The equations are obtained for

voltage of n unknown nodes with respect to one reference node
1 ⋮

in the form [∆] = � ⋮ 2�=�⋮⋮�. What are the elements of the ∆?

1

a) All conductance
b) All resistances
c) Mixed conductance
d) Mixed conductance and resistances
3. A network N consists of resistors, dependent and independent
voltage and current sources. If the current in one particular
resistance is I A, it will be doubled if the values of all the
a) independent voltage sources are doubled
b) independent current sources are doubled
c) dependent and independent voltage and current sources are

doubled
d) independent voltage and current sources are doubled
Solution:
If independent voltage sources become double then the current
will also become double.
4. Assertion (A): Maximum power transfer from a source with
complex internal impedance to a complex load will occur if the
source impedance is same as the load impedance.
Reason (R): The efficiency of maximum power transfer cannot
exceed 50%.
a) Both A and R are true and R is the correct explanation of A
b) Both A and R are true but R is NOT the correct explanation

of A

2

c) A is true but R is false
d) A is false but R is true
Solution:
Maximum power transfer occurs when the load impedance is the
complex conjugate of the source impedance.
Since the voltage divides equally across load and source
impedance. Hence efficiency of maximum power transfer is
50%.
5. The current Is in Amps in the voltage source, and voltage Vs
in Volts across the current source respectively, are
a) 13, –20
b) 8, 10
c) –8, 20
d) –13, 20
Solution:

Is = – 13A
Applying KVL in LHS

Vs – 5 × 2 = VA
Vs = VA + 10 = 10 + 10 = 10V

3

6. A 2 mH inductor with some initial current can be represented as
shown below where ‘S’ is the Laplace Transform variable. The
value of initial current is

a) 0.5 A
b) 20 A
c) 1.0 A
d) 0.0 A
Solution:
An inductor with L = 2 mH = 0.002H, through which an initial
current i(0) is flowing at t = 0, is shown in Fig.1. This can be
transformed to s-domain as shown in Fig.2.

Comparing Fig.2 with given circuit,

Li(0) = 1mV = 10−3V

∴ i(0) = 1 mV = 0.5A
2 mH

4

7. In the circuit shown below, the Norton equivalent current in
amperes with respect to the terminals P and Q is

a) 6.4 – j 4.8
b) 6.56 – j 7.87
c) 10 + j0
d) 16 + j0
Solution:
The Norton equivalent current IN = ISC
ISC = current through the short circuit across P and Q as shown
in Fig.1.
Note that (-j 50Ω) is short circuited.

IN = 16×25
25+15+j30

= 1.6(4 − j3) = 6.4 − j4.8

5

8. In the circuit shown, Thevenin's voltage as seen from the
terminals AB is

a) 0 V
b) 1.5 V
c) 6.0 V
d) Indeterminate
Solution:

When no independent source is given in the circuit then the
Thevenin theorem cannot be applied, so VTh = 0.
9. In a series RLC high Q circuit, the current peaks at a frequency
a) equal to the resonant frequency
b) greater than the resonant frequency
c) less than the resonant frequency
d) none of the above is true

6

Solution:

In a series RLC circuit, the impedance Z(jω) has minimum value

equal to R at resonance frequency 0 = √1 .

As I = = , the current always peaks at resonance frequency.

For high �= 0 �, the peak is sharp, selectivity is high,

Bandwidth is small.

10. The current I in the circuit shown is

a) – j1A

b) j1A

c) 0A

d) 20A

Solution:
For the given values: = 103 /

L = 20 mH, C = 50μF, R = 1Ω,

ωL = 103 × 20 × 10−3 = 20

1 = 1 = 20
103×50×10−6

With reference to the given phasor circuit shown in Fig.1.

7

ZR∥C = −j20
1−j20

Using voltage division rule:

V1 = 20×−j20
j20+1−−jj2200

= −j400 = −Jv
400

I = V1 = −j1A
R

11. In a circuit the voltage across an element is v(t) = 10(t – 0.01)

e –100tV. The circuit is

a) Undamped

b) Underdamped

c) Critically damped

d) Overdamped

Solution:

v(t) = 10(t – 0.01) e –100tV

( ) = 10 − 0.1
( +100)2 +100

On solving, we get the term (s + 100)2 in the denominator, hence

the circuit is critically damped.

8

12. The steady-state response of a network to the excitation
Vcos( + ∅) may be found in three steps. The first two steps
are as follows:
1. Determining the response of the network to the excitation .
2. Multiplying the above response by � = ∅.
The third step is
a) finding the complex conjugate of the expression after step 2.
b) finding the magnitude of the expression after step 2.
c) finding the real part of the expression after step 2.
d) finding the imaginary part of the expression after step 2.
Solution:
Step 1: ejωt ↔ cosωt + jsin ωt
Step 2: Multiply by Vejϕ
Vejϕ. ejωt ↔ V[cosϕ + jsinϕ] [cosωt + jsinωt]
= {[cosϕcosωt − sinϕsinωt] + [cosϕsinωt − sinϕcosωt]}
= V[cos(ωt + ϕ) + jsin(ωt + ϕ)]
To calculate the response of V[cos(ωt + ϕ)], the step 3 should
be finding the real part of the expression after step 2.

13. In a network containing resistances and reactance the roots of
the characteristic equation give for the circuit
a) the force response
b) the total response
c) the natural response
d) the damped response

9

Solution:
i) transient response gives the natural response.
ii) Steady state response gives the forced response.
iii) Natural response + Forced response = Total response
iv) Roots of characteristic equation gives the transient response

i.e., natural response.
14. In the figure shown, the ideal switch has been open for a long

time. If it is closed at t = 0, then the magnitude of the current (in
mA) through the 4 kΩ resistor at t = 0+ is _______ mA.

Solution:

(0− ) = 10 = 1 = (0+ )
10

(0−) = 10 − 5 = 5 = (0+)

At t = 0+: A resistive circuit
So, i(0+) = �54−K0� A = 1.25mA

10

15. A Delta-connected network with its Wye equivalent is shown
in figure. The resistances R1, R2 and & R3 (in ohms) are
respectively

a) 1, 5, 3 and 9
b) 3, 9 and 1.5
c) 9, 3 and 1.5
d) 3, 1.5 and 9
Solution:
Referring to Fig.1, the resistances of the Wye-equivalent are
calculated by using the formula.

e.g., 1 =
+ +

1 = 5×30 = 3
5+30+15

2 = 5×15 = 1.5 ; 3 = 15×30 = 9
50 50

11

16. Voltage and current expressions for the below circuit are given
at t ≥ 0 as v = 125 e–50t V, i = 5 e–50 t A. The value of L will be

a) 0.005 H
b) 0.05 H
c) 0.5 H
d) 5 H
Solution:

Given, v = 125 −50

i = 5 −50

= = 125 −50 = 25
5 −50

As for a source free R-L circuit,

= 0 − /

So, on comparing = 50

∴ = = 25
50 50

= 0.5

12

17. Statement (I): Under steady-state condition, a pure capacitor

behaves as an open circuit for direct voltage.

Statement (II): The current through a capacitor is proportional

to the rate of change of voltage.

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.

Solution:

also = 1

under steady state condition t = ∞ or ω = 0.

∴ = 0 (open circuited)

18. A 2 μF capacitor is charged by connecting it across a 100 volt

DC supply. The supply is then disconnected and another 1μF

uncharged capacitor is connected across this capacitor. If there

is no leakage, which of the following gives the correct voltage

across the plates of each of the capacitors?

a) 100 volts
3

b) 100 volts
2

13

c) 200 volts
3

d) 100 volts

Solution: (c)

Solution:

Initially,

= = (100)(2 × 10−6)
= 200

As per conservation of charge,

q = q1 + q2

= 200 × 10−6

= 1 = 2
1 2

⇒ 1 = 2 2

⇒ 2 = 200
3

= 2 = 2030×10−6 = 200
2 1×10−6 3

14

19. Consider the following statements associated with reciprocal

two-port networks:

1. Z12 = Z21 2. Y12 = Y21

3. h12 = h21 4. AD – BC = 1

Which of the statements given above are correct?

a) 1, 2 and 3

b) 2, 3 and 4

c) 1, 3 and 4

d) 1, 2 and 4

Solution:

Conditions for reciprocity

Z12 = Z21

Y12 = Y21

h12 = – h21

g12 = –g21

A′D′ – B′C′ = 1

20. The parallel RLC circuit shown in figure is in resonance in this

circuit

a) | | < 1mA
b) | + | > 1mA

15

c) | + | < 1mA
d) | + | > 1mA
Solution:
Understand the relations in parallel RLC network at resonance:
Refer to the phasor circuit in Fig.1.

At resonance YL = –YC

−j = −jωC,
ωL

Resonance frequency, ω0 = 1

= = 0
0

The circuit simplifies to fig.2

⃗IR = ⃗IS, V�⃗ = ⃗ISr

⃗IL = −j �V⃗ = −j R ⃗IS = −jQ0⃗IS
ω0L ω0L

⃗IC = ω0C V�⃗ = ω0C ⃗IS = Q0⃗IS

16

∴ �⃗IL� = �⃗IC� = Q0�⃗IS�,
�⃗IR + ⃗IL� = �⃗IS − jQ0⃗IS�

= |1 − jQ0|�⃗IS�
= �1 + Q20�⃗IS� > �⃗IS�
Similarly, �⃗IR + ⃗IC� = �⃗IR + ⃗IL� > �⃗IS�
⃗IL + ⃗IC = 0, �⃗IL + ⃗IC� = 0
∴ �⃗IR + ⃗IL� > 1mA
21. The h11 and h22 of a standard T-network with series
impedances 2 Ω and 7 Ω, and shunt branch impedance of 3 Ω
are
a) 5 Ω and 10 mho respectively
b) 10 Ω and 5 mho respectively
c) 5 Ω and 0.1 mho respectively
d) 10 Ω and 0.2 mho respectively
Solution:

Applying KVL at port 1 and port 2
1 = 5 1 + 3 2 ………. (i)
2 = 3 1 + 10 2 ………. (ii)

17

ℎ11 = 1 1� 2=0 and ℎ22 = 2 2� 1=0

So put V2 = 0 in equ (ii)

2 = − 3 1
10

Put this value in equ (i)

1 = 5 1 + 3 �− 130� 1
ℎ11 = 1 1� 2=0 = 4.1 ≈ 5

ℎ22 = 2 2� 1=0

Put I1 = 0 in equation (ii)

So, 2 = 1 = 0.1mho
2 10

h22 = 0.1mho

22. Z-parameters for the network shown in the figure are

a) Z11 = Z, Z22 = Z, Z12 = Z21 = Z
b) Z11 = 1/Z, Z22 = 1/Z, Z12 = Z21 = 1/Z
c) Cannot be determined
d) Z11 = Z, Z22 = Z, Z12 = Z21 = 1/Z

18

Solution:

Z-parameter for the above network cannot be determined as I1
and I2 are dependent.
23. Which one of the following statements is not correct?
a) A tree contains all the vertices of its graph
b) A circuit contains all the vertices of its graph
c) The number of f-circuits is the same as the number of chords
d) There are at least two edges in a circuit
24. If two identical first order loss-pass filters are cascaded non-
interactively, then the unit step response of the composite filter
will be
a) critically damped
b) underdamped
c) overdamped
d) oscillatory
Solution:
Let the first order response of LPF be 1/(s + τ). Since two
identical LPF are cascaded non-interactively, so the unit step
response of composite filter will be 1/(s + τ)2, which is critically
damped response.

19

25. The transfer function 2( ) of the circuit shown below is
1( )

a) 0.5 +1
+1

b) 3 +6
+2

c) +2
+1

d) +1
+2

Solution:

Taking Laplace transformation of the circuit

0 = + 1 = +1
+ 1 + 1 +2

0 = 10×103×1000×10−6 +1
10×103×100×10−6 +2

= 106×10−6 +1 = +1
106×10−6 +2 +2

20

26. Which one of the following functions is an RC driving point

impedance?

a) ( +3)( +4)
( +1)( +2)

b) ( +3)( +4)
( +1)( +2)

c) ( +3)( +4)
( +1)( +2)

d) ( +2)( +4)
( +1)( +3)

Solution:

Properties of RC driving point impedance function:

i) Poles and zeros are alternate on negative real axis.

ii) The entity nearest to origin is a pole and nearest to infinity

is zero.

A two-port network shown below is excited by external dc

sources. The voltages and the currents are measured with

voltmeters V1, V2 and ammeters A1, (all assumed to be ideal), as

indicated. Under following switch conditions, the readings

obtained are

i. S1 - Open, S2 - Closed

A1 = 0 A, V1 = 4.5 V,

V2= 1.5V, A2 = 1 A
ii. S1 - Closed, S2 - Open

A1 = 4 A, V1 = 6 V,

21

V2 = 6V, A2 = 0A

27. The h-parameter matrix for this network is

a) �−−13 0.367�
b) �−33 0−.617�

c) �13 0.367�

d) �−33 −01.67�

Solution:

Inserting the values of z-parameters in (1) & (2),

V1 = 1.5I1 + 4.5I2 ………. (3)

V2 = 1.5I1 + 1.5I2 ………. (4)
The h parameters are described by

V1 = h11I1 + h12V2 ………. (5)

I2 = h21I1 + h22V2 ………. (6)

h11 & h21 are calculated by keeping V2 = 0

∴ From (5) & (6),

1 = ℎ11 1 � ………. (7)
I2 = h21I1

From (4), I2 = -I1 ………. (8)

From (3) & (8)

22

1 = 1.5 1 − 4.5 1
= −3 1 ………. (9)

Comparing (7) with (8) & (9),

ℎ11 = −3 , ℎ21 = −1

h11 & h22 are calculated by making I1 = 0

∴ From (5) & (6)

1 = ℎ12 2 � ………. (10)
I2 = h22V2

From (3) & (4)

1 = 4.5 2 � ………. (11)
V2 = 1.5I2

Comparing (10) & (11)

h22 = 1
1.5

= 2 mho
3

= 0.67mho

ℎ12 = 4.5 = 3
1.5

∴ h-parameter matrix is given by

�ℎℎ1211 ℎℎ1222� = �−−31 0.367�

23

28. A part of an electrical network has the configuration shown in
the figure below. The voltage drops across the resistances are 20
V, 30 V and 65 V with respective polarities shown in the figure.
Which one of the following is the correct value of the resistance
R3?

a) 13Ω
b) 5Ω
c) 65Ω
d) 65/17 Ω
Solution:

I1 = 20 = 2A; I2 = 30 = 5A
10 6

Applying KCL at node A,

10 + I2 = I1 + I3 ⇒ I3 = 13A

24

65 = R3I3
(13)(R3) = 65 ⇒ R3 = 5Ω
29. The ABCD parameters of an ideal n : 1 transformer shown in
fig. are � 0 0 �. The value of X will be

a) n
b) 1/n
c) n2
d) 1/n2
Solution:
For the given ideal transformer shown in fig.1,

V1 = n = −I2
V2 1 I1

V1 = V2 ………. (1)

I1 = −1 I2 ………. (2)
n

25

The A, B, C, D parameters are defined by

V1 = AV2 + B(−I2) ………. (3)

I1 = CV2 + D(−I2) ………. (4)

Compare the equations (1) & (3) and (2) & (4):

∴ A = n, B = 0, C = 0, D = 1

A, B, C, D parameter matrix

= � � = � 0 01� ………. (5)

Comparing (5) with the given matrix in question,

� 0 0 � , = 1

30. H(s) = ( ) = 4 +6 when i(t) is a unit step function, as t tends
( ) +2

to infinity is given by the value of v(t) is

a) 0

b) 3

c) 4

d) ∞