Network Theory Test - 5

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GATE
EEE

Network
Theory

Test-05Solutions

NETWORK THEORY
1. Which one of the following gives the V-I characteristic of an

ideal voltage source?

Answer: (b)
Solution:
Ideal voltage source delivers constant voltage irrespective of the
value of current through it.
2. In the circuit shown in the figure, the value of node voltage V2 is

1

a) 22 + j2 V

b) 2 + j22 V

c) 22 – j2 V

d) 2 – j22 V

Answer: (d)

Solution:

By KVL ⇒ 1 − 10∠00 − 2 = 0
⇒ 1 − 2 = 10∠00 ………… (1)

Super nodal equation,

⇒ −4∠00 + 1 + 2 + 2 = 0
− 3 6 6

⇒ −4∠00 + 2 1 + 2 + 2 = 0
6 6 6

⇒ −2 1 + 2(1 + ) = 24∠900 ………… (2)

Equ (1) in (2)

⇒ −2( 2 + 10) + 2(1 + ) = 24∠900
⇒ V2(−1 + j) = 20 + j24
⇒ V2 = �2−0+1+j2j4�

⇒ V2 = (20+ 24)(−1− )
(−1− )(−1− )

= (20+ 24)(−1− )
12+12

= (2 − j22)V

2

3. In the circuit shown, the switch S is open for a long time and is
closed at t = 0. The current i(t) for t > 0+ is

a) i(t) = 0.5 – 0.125e-1000t A
b) i(t) = 1.5 – 0.125 e-1000t A
c) i(t) = 0.5 – 0.5 e-1000t A
d) i(t) = 0.325 e-1000t A
Answer: (a)
Solution:

S is open for −∞ < t ≤ 0−
At t = 0−, the status of the circuit is shown in Fig.1.

iL(0−) = 0.75A, i(0−) = 0
S is closed for 0+ ≤ t < ∞
At t = 0+, iL(0+) = iL(0−) = 0.75A

3

The status of the circuit is shown in Fig.2.

i(0+) = 3 A
8

At t = ∞, the status of the circuit is shown in fig.3.

i(∞) = 0.5A

The given circuit is a first order circuit with time constant, τ =

R = 10 + (10 ∥ 10) = 10 + 5 = 15Ω

L = 15mH = 15 × 10−3H

τ = L = 15×10−3 = 10−3 sec, 1 = 103sec−1
R 15 τ

General formula:

i(t) = IV + (FV − IV) �1 − e−τt� , t ≥ 0+

i(t) = 3 + �12 − 38� (1 − −1000 )
8

= 0.5 − 0.125 −1000 , ≥ 0+

4. Superposition theorem is NOT applicable to networks containing

a) Nonlinear elements

b) Dependent voltage sources

c) Dependent current sources

d) Transformers

Answer: (a)

4

Solution:

Superposition theorem is not applicable to nonlinear

networks, containing nonlinear elements. The theorem is

applicable to linear networks (Time invariant or time

varying) consisting of independent sources, linear

dependent sources (voltage or current), linear passive

elements R, L, C and linear transformer.

5. If the three resistors in a delta network are all equal in values i.e.

RDELTA then the value of the resultant resistors in each branch of

the equivalent star network i.e. RSTAR will be equal to

a) RDELTA
3

b) RDELTA
2

c) 2RDELTA

d) RDELTA
Answer: (a)

Solution:
Delta to star ⇒ Resistance decreases by 3 times

6. Which of the following statements are associated with

Thevenin's theorem?

1. Impedance through which current is required is removed and

open-circuit voltage is found.

2. It is applicable to only d.c. circuits.

3. The network is replaced by a voltage source and a series

impedance remains after removing the load impedance.

5

Select the correct answer using the codes given below:
a) 1and 2 only
b) 1 and 3 only
c) 2 and 3 only
d) 1, 2 and 3
Answer: (b)
Solution:
Thevenin’s theorem is applicable to both a.c and d.c circuits
7. Consider a two-port network with the transmission matrix: T
= � �. If the network is reciprocal, then
a) T-1 = T
b) T2 = T
c) Determinant (T) = 0
d) Determinant (T) = 1
Answer: (d)
Solution:
A two port network is reciprocal in transmission parameters if
AD – BC = 1 i.e., Determinant (T) = 1
8. Maximum power will be delivered from an ac source to a
resistive load in a network when the magnitude of the source
impedance is equal to
a) Half the load resistance
b) Double the load resistance
c) The load resistance
d) Zero

6

Answer: (c)

Solution:

The maximum power will be delivered from ac source to a

resistive load in a network when the magnitude of the source

impedance is equal to the load impedance.

∴ | | =
9. Assertion (A): Power factor is defined as the ratio of apparent

power to the average power in an a.c. circuit.

Reason (R): The magnitude of power factor is always less than

unity.

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

c) A is true but R is false

d) A is false but R is true

Answer: (d)

Solution:

Power factor = average power
apparent power

10. In Fig., the switch was closed for a long time before opening at

t = 0. The voltage Vx at t = 0+ is

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a) 25 V
b) 50 V
c) –5 V
d) 0 V
Answer: (c)
Solution:
At = 0−, 5H behaves as a short circuit as shown in Fig.1. The
inductor current (0−) = 2.5 , (0−) = 0. As the current
through the inductor cannot change instantaneously, (0+) =
(0−) = 2.5 .
At = 0+, the status of the circuit is shown in Fig.2.
From Fig.2, VX = –50 V

8

11. An input voltage v(t) =10√2 cos (t + 100) + 10√5cos (2t + 100)

V is applied to a series combination of resistance R = 1Ω and an

inductance L = 1 H. The resulting steady state current i(t) in

ampere is
a) 10cos(t + 55°) + 10 cos(2t + 10° + tan−1 2)

b) 10cos(t + 55°) + 10 √3 cos(2t + 55°)
2

c) 10cos(t − 35°) + 10 cos(2t + 10° − tan−1 2)

d) 10cos(t − 35°) + 10 √3 cos(2t − 35°)
2

Answer: (c)

Solution:

For RL series circuit, I = H(jω) = 1
V R+jωL

For R = 1Ω, L = 1H, |H(jω)| = 1
R+jωL

For input frequencies 1 r/s and 2 r/s,

H(j1) = 1 , H(j2) = 1
1+j 1+j2

|H(j1)| = 1 , |H(j2)| = 1
√2 √5

∠H(j1) = −450, ∠H(j2) = −tan−1(2)

For input voltage,

( ) = 10√2 cos( + 100) + 10√5 cos(2 + 100)

Steady state current

( ) = 10√2 1 cos( + 100 − 450) + 10√5 1 cos(2 +
√2 √5

100 − tan−1(2))

( ) = 10 cos( − 350) + 10 cos(2 + 100 − tan−1(2))

9

12. In the circuit shown in the figure below, steady state was
reached when the switch S was open. The switch was closed at t
= 0. The initial value of the current through the capacitor 2 C is

a) zero
b) 1 A
c) 2 A
d) 3 A
Answer: (c)
Solution:
The circuit at steady state when the switch was open, is shown
below.

The circuit at t = 0 when the switch is closed, is shown below:

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The equivalent circuit of the above circuit is shown below:

= 10−4 − 4 = 3 − 1 = 2
2 4

13. A step voltage is applied to the circuit shown below. What is the

transient current response of the circuit?

a) Undamped sinusoidal

b) Overdamped

c) Underdamped

d) Critically damped

Answer: (d)

Solution:

( ) = ( )
( )

11

= =

+ + 1 2+ + 1

( ) = 1 = 1
2+2 +1 ( +1)2

So, the transient current response is critically damped.

11

14. In the transformer shown in the figure below, the inductance
measured across the terminal 1 and 2 was 4 H with open
terminals 3 and 4. It was 3 H when the terminal 3 and 4 were
short circuited. The coefficient of coupling would be

a) 1

b) 0.707

c) 0.5

d) indeterminate due to insufficient data

Answer: (c)

Solution:

When terminals 3 & 4 are open, then
V1 = ωLP ∙ I1 ⇒ LP = 4H

When terminals 3 & 4 are short circuited, then

V1 = ωLPI1 + ωMI2

and 0 = ωLSI2 + ωMI1

⇒ I2 = −M I1
LS

So, V1 = 4ωI1 + ωM �−LMS � I1

Given Ls = 2H

⇒ 4 − 2 = 3

12

2 = 1 ⇒ 2 = 2
2

⇒ = √2

Now, = �

⇒ √2 = √4 × 2 = 2√2

⇒ = 0.5

15. For the oriented graph as given below, taking 4, 5, 6 as tree

branches the tie set matrix is

−1 0 0 −1 1 0
a) � 0 −1 0 0 −1 1 �

0 0 −1 1 0 −1
1 0 0 1 −1 0
b) �0 1 0 0 1 −1�
0 0 1 −1 0 1
1 −1 0 −1 1 0
c) �1 0 0 0 −1 1 �
0 0 1 1 0 −1
−1 0 0 −1 0 0
d) � 0 −1 0 0 0 1 �
0 0 0 1 0 −1
Answer: (b)

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Solution:
The set matrix = [I : BT]

Identity matrix Loop matrix

with respect

to given tree

Only option (b) satisfies above tie set matrix
16. A series LCR circuit consisting of R = 10Ω, | |= 20 Ω

and | | = 20 Ω is connected across an a.c. supply of 200Vrms.
The rms voltage across the capacitor is
a) 200∠-900 V
b) 200∠900 V

c) 400∠900 V
d) 400∠-900 V

Answer: (d)

Solution:

For the series RLC circuit shown in Fig.1., given:

|XL|= |XC| = 20 Ω where XL = jωL and XC = − j i.e., ωL =
ωC

1

ωC

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∴ The circuit is under resonance at ω = ω0 = 1 , �V⃗L =
√LC

−�V⃗C, XL = −XC.

⃗Irms = �V�⃗s(rms) = 200 = 20A
R 10

V�⃗c(rms) = ⃗Irms 1
jω0C

= −j Irms = −j20 × 20
ωc

= −j400 = 400∠−900

Also understand that

= ω0L = 1 = 20 = 2
ω0CR 10

�V⃗L = �V⃗s = 400∠−900 , �V⃗C = − �V⃗s

�V⃗R = V�⃗s = 200

17. A unit step current is applied to a network consisting of only

passive elements. The voltage across the current source

observed is v(t) = (1 + e-t/τ). The simplest possible network will

consist of the elements

a) 1 resistor and 2 capacitors

b) 1 resistor and 2 inductors

c) 2 resistors and 1 capacitor

d) 2 resistors and 1 inductor

Answer: (d)

Solution:

i(t) = 1 ⇒ I(s) = 1
S

v(t) = �1 + e−t/τ�

15

⇒ V(s) = 1 + 1
S S+1τ

⇒ V(s) = τS+1+τS
S(τS+1)

⇒ V(s) = 2τS+1
S(τS+1)

Z(s) = V(s) = 2τS+1 = 2τS+1
I(s) τS+1
S(τS+1)
1
S

Y(s) = τS+1
2τS+1

= τS+1/2 + 1
2τS+1 2(2τS+1)

Y(s) = 1 + 1
2 (4τS+1)

The above expression shows that the network will consist of 2

resistors and 1 inductor.

18. The damping ratio of a series RLC circuit can be expressed as

a) 2
2

b) 2
2

c) �
2

d) 2 �

Answer: (c)

Solution:

In a series RLC-circuit, the damping ratio

= ξ = 1 = R �CL
2Q 2

16

19. The driving point impedance of the network shown in figure is

a) 10 + 2s
b) 10 + 2s + 1/s
c) 10
d) 1/s
Answer: (b)
Solution:
The given circuit can be redrawn as shown here:

Applying KVL,

1 ( ) = �10 + 1 + 2 � ( ) ………. (1)

So driving point impedance, ( ) = 1( )
( )

So from equation (i), ( ) = 10 + 2 + 1

17

20. In the case of above circuit, which one of the following
statements is correct?

a) It can be analyzed by minimum one loop equation.
b) It can be analyzed by minimum two loop equations.
c) It can be analyzed by minimum three loop equations.
d) It cannot be analyzed by loop method
Answer: (a)
Solution:
Apparently there are 3 loops but loops contains current source.
Only one loop equation of super mesh is sufficient.

18

21. In the circuit shown, the switch is opened at t = 0. The circuit
is

a) Critically damped
b) Under-damped
c) Over-damped
d) Undamped
Answer: (c)
Solution:
At t = 0 when switch is opened then circuit can be drawn as
below:

Here R = 4Ω; L = 1H; C = 1F

So damping ratio, ξ = R �CL = 4 �11 = 2
2 2

Since ξ > 1, so the circuit is over damped

19

22. In a series RLC circuit R = 2 kΩ, L = 1H and C = 1/400µF.

The resonant frequency is

a) 2× 104 Hz

b) 1 × 104 Hz

c) 104 Hz

d) 2 × 104 Hz

Answer: (b)

Solution:

In the series R, L, C circuit with R = 2 kΩ, L = 1H, C = 1 μF
400

ωr2es = 1 = 1 = 4 × 108
LC 1×4010×10−6

ωres = 2 × 104r/s

fres = ωres = 1 × 104Hz
2π π

Note that fres does not depend on R. But at resonance f = fres. Zin

is real and minimum equal to R.

23. Relative to a given fixed tree of a network,

a) Link currents form an independent set

b) Branch currents form an independent set

c) Link voltages form an independent set

d) Branch voltages form an independent set

Answer: (a)

Solution:

Link currents form an independent set, as each link with other

tree branches forms a unique loop (fundamental loop). This

20

gives rise to loop method of Network analysis using KVL to
each loop.
If number of links = l simultaneous independent equations are
obtained in ‘l’ currents (i1, i2, i3,... ). Once link currents are
obtained by solving these equations, the other branch currents
(i.e., tree branch currents) can be obtained by superposition of
link currents.
24. A two-port network has the ABCD parameters �37 84�. Two
such identical networks are cascaded. The ABCD parameters of
the overall cascaded network will be
a) �164 186�
b) �7333 8408�
c) �11 11�
d) �499 6146�
Answer: (b)
Solution:
The ABCD parameter of overall cascaded network will be given
by

= �37 84� �37 84� = �3733 4808�

21

25. In the circuit shown in the figure below, the switch S is closed
at t = 0. Which one of the following gives expression for the
voltage across the inductance as a function of time?

a) e−t
2

b) (1−e−t)
2

c) (1 − e−t)

d) (e−t)

Answer: (d)

Solution:
( ) = (∞) + � (0+) − (∞)� −

(0−) = 0
(0−) = (0+)

As inductor does not allow sudden change in current

(∞) = 1
1

As t → ∞, inductor acts as short circuit,

( ) = 1 + (0 − 1) −

( ) = 1 − −

( ) = = 1( − ) = −

22

26. Match List-I (Excitation) with List-II (Two-port parameters)

and select the correct answer using the code given below the

lists:

List-I List-II

A. I1, I2 1. Z

B. V1, V2 2. Y

C. V1, I2 3. G

D. I1, V2 4. H

Codes:

AB C D

a) 1 2 3 4

b) 4 2 3 1

c) 1 3 2 1

d) 4 3 2 1

Answer: (a)

Solution:

Z-parameter: Independent variables = I1, I2

Y-parameter: Independent variables = V1, V2

h-parameter: Independent variables = V1, I2

23

27. In the RLC circuit shown in the figure, the input voltage is
given by Vi(t) = 2cos(200t) + 4sin(500t). The output voltage
v0(t) is

a) cos(200t) + 2sin(500t)
b) 2cos(200t) + 4sin(500t)
c) sin(200t) + 4cos(500t)
d) 2sin(200t) + 4cos(500t)
Answer: (b)
Solution:
Given Vi(t) = 2 cos(200t) + 4 sin(500t)
Let us apply SPT [super position theorem only consider
2cos200t, then circuit becomes

So, 0′( ) = 2 cos 200
Now only consider 4sin500t, then circuit becomes

24

So again V0"(t) = 4 sin 500t
Finally according to SPT,

V0(t) = V0′(t) + V0"(t)
V0(t) = 2cos(200t) + 4 sin(500t)
28. The conditions under which a passive two-port network
represented by ABCD is reciprocal and symmetrical are
a) AD – BC = 1; A = C
b) AD – BC = 0; A = D
c) AD – BC = 1; D = A
d) AD – BC = 0; C = B
Answer: (c)
29. The impedance parameters Z11 and Z12 of the two - port
network in Fig. are

a) Z11 = 2.75Ω and Z12 = 0.25Ω
b) Z11 = 3Ω and Z12 = 0.5Ω

25

c) Z11 = 3Ω and Z12 = 0.25Ω
d) Z11 = 2.75Ω and Z12 = 0.5Ω
Answer: (a)
Solution:
Convert middle π-network to T-network as shown in Fig.1.

Simplify to get overall T-NW as shown in Fig.2.

Z11 is calculated by keeping I2 = 0 (Output open)
The relevant circuit is shown in Fig.3.

26

11 = 1 1� 2=0 = 2.5 + 0.25 = 2.75
Z12 is calculated by keeping I1 = 0 (Input open)
The relevant circuit is shown in Fig.4.

12 = 2 1� 1=0 = 0.25
Note: The z-parameters of T-network can be easily identified:
Z11 = Impedance of the input series arm + Impedance of the
shunt arm.
Z22 = Impedance of the output series arm + Impedance of the
shunt arm.
Z12 = Z21 = Impedance of the shunt arm.
Another method:
For I2 = 0, the circuit is shown in Fig.5 and Fig.6.

27

11 = 1 1� 2=0

Z11 = 2 + [1∥3] = 2 + 3 = 2.75
4

For I1 = 0, the circuit is shown in Fig.7

12 = 2 1� 1=0
From Fig.7, V1 = 0.25 I2, Z12 = 0.25Ω
30. The equivalent capacitance across 'ab' will be

a) 0.2 μF
b) 0.1 μF

28

c) 0.5 μF
d) 0
Answer: (b)

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