UNIT 2
Networks
GATE 2011
ONE MARK
MCQ 2.1
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 + j 0
(D) 16 + j 0
MCQ 2.2
In the circuit shown below, the value of RL such that the power
transferred to RL is maximum is
(A) 5 Ω
(B) 10 Ω
(C) 15 Ω
(D) 20 Ω
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Chap 2
Networks
MCQ 2.3
The circuit shown below is driven by a sinusoidal input
vi = Vp cos (t/RC ). The steady state output vo is
(A) (Vp /3) cos (t/RC )
(B) (Vp /3) sin (t/RC )
(C) (Vp /2) cos (t/RC )
(D) (Vp /2) sin (t/RC )
2011
TWO MARKS
MCQ 2.4
In the circuit shown below, the current I is equal to
(A) 1.4+0c A
(B) 2.0+0c A
(C) 2.8+0c A
(D) 3.2+0c A
MCQ 2.5
In the circuit shown below, the network N is described by the following
Y matrix:
0.1 S − 0.01 S
. the voltage gain V2 is
Y=>
0.01 S
0.1 SH
V1
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Chap 2
Networks
(A) 1/90
(B) –1/90
(C) –1/99
(D) –1/11
MCQ 2.6
In the circuit shown below, the initial charge on the capacitor is 2.5
mC, with the voltage polarity as indicated. The switch is closed at
time t = 0 . The current i (t) at a time t after the switch is closed is
(A) i (t) = 15 exp (− 2 # 103 t) A
(B) i (t) = 5 exp (− 2 # 103 t) A
(C) i (t) = 10 exp (− 2 # 103 t) A
(D) i (t) =− 5 exp (− 2 # 103 t) A
2010
ONE MARK
MCQ 2.7
For the two-port network shown below, the short-circuit admittance
parameter matrix is
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Chap 2
Networks
4
(A) >
−2
1
(C) >
0.5
−2
S
4H
0.5
S
1H
1
(B) >
− 0.5
4
(D) >
2
− 0.5
S
1H
2
S
4H
MCQ 2.8
For parallel RLC circuit, which one of the following statements is
NOT correct ?
(A) The bandwidth of the circuit decreases if R is increased
(B) The bandwidth of the circuit remains same if L is increased
(C) At resonance, input impedance is a real quantity
(D) At resonance, the magnitude of input impedance attains its
minimum value.
2010
TWO MARKS
MCQ 2.9
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 − 0.125e−1000t A
(C) i (t) = 0.5 − 0.5e−1000t A
(D) i (t) = 0.375e−1000t A
MCQ 2.10
The current I in the circuit shown is
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Chap 2
Networks
(A) − j1 A
(B) j1 A
(C) 0 A
(D) 20 A
MCQ 2.11
In the circuit shown, the power supplied by the voltage source is
(A) 0 W
(B) 5 W
(C) 10 W
(D) 100 W
GATE 2009
ONE MARK
MCQ 2.12
In the interconnection of ideal sources shown in the figure, it is known
that the 60 V source is absorbing power.
Which of the following can be the value of the current source I ?
(A) 10 A
(B) 13 A
(C) 15 A
(D) 18 A
MCQ 2.13
If the transfer function of the following network is
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Chap 2
Networks
Vo (s)
1
=
2 + sCR
Vi (s)
The value of the load resistance RL is
(B) R
(A) R
4
2
(C) R
(D) 2R
MCQ 2.14
A fully charged mobile phone with a 12 V battery is good for a
10 minute talk-time. Assume that, during the talk-time the battery
delivers a constant current of 2 A and its voltage drops linearly from
12 V to 10 V as shown in the figure. How much energy does the
battery deliver during this talk-time?
(A) 220 J
(B) 12 kJ
(C) 13.2 kJ
(D) 14.4 J
GATE 2009
TWO MARK
MCQ 2.15
An AC source of RMS voltage 20 V with internal impedance
Zs = (1 + 2j) Ω feeds a load of impedance ZL = (7 + 4j) Ω in the figure
below. The reactive power consumed by the load is
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Chap 2
Networks
(A) 8 VAR
(B) 16 VAR
(C) 28 VAR
(D) 32 VAR
MCQ 2.16
The switch in the circuit shown was on position a for a long time,
and is move to position b at time t = 0 . The current i (t) for t > 0 is
given by
(A) 0.2e−125t u (t) mA
(B) 20e−1250t u (t) mA
(C) 0.2e−1250t u (t) mA
(D) 20e−1000t u (t) mA
MCQ 2.17
In the circuit shown, what value of RL maximizes the power delivered
to RL ?
(A) 2.4 Ω
(B) 8 Ω
3
(C) 4 Ω
(D) 6 Ω
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Chap 2
Networks
MCQ 2.18
The time domain behavior of an RL circuit is represented by
L di + Ri = V0 (1 + Be−Rt/L sin t) u (t).
dt
For an initial current of i (0) = V0 , the steady state value of the
R
current is given by
(A) i (t) " V0
(B) i (t) " 2V0
R
R
(C) i (t) " V0 (1 + B)
R
(D) i (t) " 2V0 (1 + B)
R
GATE 2008
ONE MARK
MCQ 2.19
In the following graph, the number of trees (P) and the number of
cut-set (Q) are
(A) P = 2, Q = 2
(B) P = 2, Q = 6
(C) P = 4, Q = 6
(D) P = 4, Q = 10
MCQ 2.20
In the following circuit, the switch S is closed at t = 0 . The rate of
change of current di (0+) is given by
dt
Page 46
(A) 0
(R + Rs) Is
(C)
L
(B) Rs Is
L
(D) 3
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Chap 2
Networks
GATE 2008
TWO MARKS
MCQ 2.21
The Thevenin equivalent impedance Zth between the nodes P and Q
in the following circuit is
(A) 1
(B) 1 + s + 1
s
(C) 2 + s + 1
s
2
(D) s2 + s + 1
s + 2s + 1
MCQ 2.22
The driving point impedance of the following network is given by
Z (s) = 2 0.2s
s + 0.1s + 2
The component values are
(A) L = 5 H, R = 0.5 Ω, C = 0.1 F
(B) L = 0.1 H, R = 0.5 Ω, C = 5 F
(C) L = 5 H, R = 2 Ω, C = 0.1 F
(D) L = 0.1 H, R = 2 Ω, C = 5 F
MCQ 2.23
The circuit shown in the figure is used to charge the capacitor C
alternately from two current sources as indicated. The switches S1
and S2 are mechanically coupled and connected as follows:
For 2nT # t # (2n + 1) T , (n = 0, 1, 2,..) S1 to P1 and S2 to P2
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Chap 2
Networks
For (2n + 1) T # t # (2n + 2) T, (n = 0, 1, 2,...) S1 to Q1 and S2 to Q2
Assume that the capacitor has zero initial charge. Given that u (t) is
a unit step function , the voltage vc (t) across the capacitor is given by
(A)
3
/ (− 1) n tu (t − nT)
n=1
3
(B) u (t) + 2 / (− 1) n u (t − nT)
n=1
3
(C) tu (t) + 2 / (− 1) n u (t − nT) (t − nT)
n=1
(D) / 60.5 − e− (t − 2nT) + 0.5e− (t − 2nT) − T @
3
n=1
Common data question 2.23 & 2.24 :
The following series RLC circuit with zero conditions is excited by a
unit impulse functions δ (t).
MCQ 2.24
For t > 0 , the output voltage vC ^ t h is
(A) 2 ^e
3
Page 48
−1
2
t
− e th
3
2
−1
(C) 2 e 2 t cos c 3 t m
2
3
−1
(B) 2 te 2 t
3
−1
(D) 2 e 2 t sin c 3 t m
2
3
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Chap 2
Networks
MCQ 2.25
For t > 0 , the voltage across the resistor is
3
1
(A) 1 _e 2 t − e− 2 t i
3
(B) e
−1 t
2
3
1 sin 3 t
c 2 mG
=cos c 2 t m −
3
−1
(C) 2 e 2 t sin c 3 t m
2
3
−1
(D) 2 e 2 t cos c 3 t m
2
3
Statement for linked Answers Questions 2.25 & 2.26:
A two-port network shown below is excited by external DC source.
The voltage and the current are measured with voltmeters V1, V2 and
ammeters. A1, A2 (all assumed to be ideal), as indicated
Under following conditions, the readings obtained are:
(1) S1 -open, S2 - closed A1 = 0,V1 = 4.5 V,V2 = 1.5 V, A2 = 1 A
(2) S1 -open, S2 - closed A1 = 4 A,V1 = 6 V,V2 = 6 V, A2 = 0
MCQ 2.26
The z -parameter matrix for this network is
1. 5 1. 5
1.5 4.5
(B) =
(A) =
G
4. 5 1. 5
1.5 4.5G
1.5 4.5
(C) =
1.5 1.5 G
4.5 1.5
(D) =
1.5 4.5G
MCQ 2.27
The h -parameter matrix for this network is
−3 3
−3 −1
(B) =
(A) =
G
− 1 0.67
3 0.67 G
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Chap 2
Networks
3 3
(C) =
1 0.67 G
3
1
(D) =
− 3 − 0.67 G
GATE 2007
ONE MARK
MCQ 2.28
An independent voltage source in series with an impedance
Zs = Rs + jXs delivers a maximum average power to a load impedance
ZL when
(B) ZL = Rs
(A) ZL = Rs + jXs
(C) ZL = jXs
(D) ZL = Rs − jXs
MCQ 2.29
The RC circuit shown in the figure is
(A) a low-pass filter
(B) a high-pass filter
(C) a band-pass filter
(D) a band-reject filter
GATE 2007
TWO MARKS
MCQ 2.30
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 B1 is
B2
(A) 4
(B) 1
(C) 1/2
(D) 1/4
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Chap 2
Networks
MCQ 2.31
For the circuit shown in the figure, the Thevenin voltage and resistance
looking into X − Y are
(A)
4
3
V, 2 Ω
(B) 4 V, 23 Ω
(C)
4
3
V, 23 Ω
(D) 4 V, 2 Ω
MCQ 2.32
In the circuit shown, vC is 0 volts at t = 0 sec. For t > 0 , the capacitor
current iC (t), where t is in seconds is given by
(A) 0.50 exp (− 25t) mA
(B) 0.25 exp (− 25t) mA
(C) 0.50 exp (− 12.5t) mA
(D) 0.25 exp (− 6.25t) mA
MCQ 2.33
In the ac network shown in the figure, the phasor voltage VAB (in
Volts) is
(A) 0
(B) 5+30c
(C) 12.5+30c
(D) 17+30c
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Chap 2
Networks
GATE 2006
TWO MARKS
MCQ 2.34
A two-port network is represented by ABCD parameters given by
V1
A B V2
= I G = =C D G=− I G
1
2
If port-2 is terminated by RL , the input impedance seen at port-1 is
given by
(B) ARL + C
(A) A + BRL
C + DRL
BRL + D
(C) DRL + A
BRL + C
(D) B + ARL
D + CRL
MCQ 2.35
In the two port network shown in the figure below, Z12 and Z21 and
respectively
(A) re and βr0
(B) 0 and − βr0
(C) 0 and βro
(D) re and − βr0
MCQ 2.36
The first and the last critical frequencies (singularities) of a driving
point impedance function of a passive network having two kinds of
elements, are a pole and a zero respectively. The above property will
be satisfied by
(B) RC network only
(A) RL network only
(C) LC network only
(D) RC as well as RL networks
MCQ 2.37
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
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Chap 2
Networks
(A) 0.5 A
(B) 2.0 A
(C) 1.0 A
(D) 0.0 A
MCQ 2.38
In the figure shown below, assume that all the capacitors are initially
uncharged. If vi (t) = 10u (t) Volts, vo (t) is given by
(A) 8e -t/0.004 Volts
(B) 8 (1 − e -t/0.004) Volts
(C) 8u (t) Volts
(D) 8 Volts
MCQ 2.39
A negative resistance Rneg is connected to a passive network N having
driving point impedance as shown below. For Z2 (s) to be positive
real,
(A) Rneg # Re Z1 (jω), 6ω
(B) Rneg # Z1 (jω) , 6ω
(C) Rneg # Im Z1 (jω), 6ω
(D) Rneg # +Z1 (jω), 6ω
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Chap 2
Networks
GATE 2005
ONE MARK
MCQ 2.40
The condition on R, L and C such that the step response y (t) in the
figure has no oscillations, is
(A) R $ 1
2
L
C
(B) R $
L
C
(C) R $ 2
L
C
(D) R =
1
LC
MCQ 2.41
The ABCD parameters of an ideal n: 1 transformer shown in the
figure are
n 0
>0 x H
The value of x will be
(A) n
(B) 1
n
(C) n2
(D) 12
n
MCQ 2.42
In a series RLC circuit, R = 2 kΩ , L = 1 H, and C = 1 μF The
400
resonant frequency is
(A) 2 # 10 4 Hz
(B) 1 # 10 4 Hz
π
(C) 10 4 Hz
(D) 2π # 10 4 Hz
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Chap 2
Networks
MCQ 2.43
The maximum power that can be transferred to the load resistor RL
from the voltage source in the figure is
(A) 1 W
(B) 10 W
(C) 0.25 W
(D) 0.5 W
MCQ 2.44
The first and the last critical frequency of an RC -driving point
impedance function must respectively be
(A) a zero and a pole
(B) a zero and a zero
(C) a pole and a pole
(D) a pole and a zero
GATE 2005
TWO MARKS
MCQ 2.45
For the circuit shown in the figure, the instantaneous current i1 (t) is
(A) 10 3 90c A
2
(B) 10 3 − 90c A
2
(C) 5 60c A
(D) 5 − 60c A
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Chap 2
Networks
MCQ 2.46
Impedance Z as shown in the given figure is
(A) j29 Ω
(B) j9 Ω
(C) j19 Ω
(D) j39 Ω
MCQ 2.47
For the circuit shown in the figure, Thevenin’s voltage and Thevenin’s
equivalent resistance at terminals a − b is
(A) 5 V and 2 Ω
(B) 7.5 V and 2.5 Ω
(C) 4 V and 2 Ω
(D) 3 V and 2.5 Ω
MCQ 2.48
If R1 = R2 = R4 = R and R3 = 1.1R in the bridge circuit shown in the
figure, then the reading in the ideal voltmeter connected between a
and b is
(A) 0.238 V
(B) 0.138 V
(C) − 0.238 V
(D) 1 V
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Chap 2
Networks
MCQ 2.49
The h parameters of the circuit shown in the figure are
0. 1 0. 1
(A) =
− 0. 1 0. 3 G
10 − 1
(B) =
1 0.05G
30 20
(C) =
20 20G
10 1
(D) =
− 1 0.05G
MCQ 2.50
A square pulse of 3 volts amplitude is applied to C − R circuit shown
in the figure. The capacitor is initially uncharged. The output voltage
V2 at time t = 2 sec is
(A) 3 V
(B) − 3 V
(C) 4 V
(D) − 4 V
GATE 2004
ONE MARK
MCQ 2.51
Consider the network graph shown in the figure. Which one of the
following is NOT a ‘tree’ of this graph ?
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Chap 2
Networks
(A) a
(B) b
(C) c
(D) d
MCQ 2.52
The equivalent inductance measured between the terminals 1 and 2
for the circuit shown in the figure is
(A) L1 + L2 + M
(B) L1 + L2 − M
(C) L1 + L2 + 2M
(D)L1 + L2 − 2M
MCQ 2.53
The circuit shown in the figure, with R = 1 Ω, L = 1 H and C = 3 F
3
4
has input voltage v (t) = sin 2t . The resulting current i (t) is
(A) 5 sin (2t + 53.1c)
(B) 5 sin (2t − 53.1c)
(C) 25 sin (2t + 53.1c)
(D) 25 sin (2t − 53.1c)
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Chap 2
Networks
MCQ 2.54
For the circuit shown in the figure, the time constant RC = 1 ms.
The input voltage is vi (t) = 2 sin 103 t . The output voltage vo (t) is
equal to
(A) sin (103 t − 45c)
(B) sin (103 t + 45c)
(C) sin (103 t − 53c)
(D) sin (103 t + 53c)
MCQ 2.55
For the R − L circuit shown in the figure, the input voltage vi (t) = u (t)
. The current i (t) is
GATE 2004
TWO MARKS
MCQ 2.56
For the lattice shown in the figure, Za = j2 Ω and Zb = 2 Ω . The values
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Chap 2
Networks
z11 z12
of the open circuit impedance parameters 6 z @ = =
are
z21 z22 G
1−j 1+j
(A) =
1 + j 1 + jG
1−j 1+j
(B) =
−1 + j 1 − j G
1+j 1+j
(C) =
1 − j 1 − jG
1 + j −1 + j
(D) =
−1 + j 1 + j G
MCQ 2.57
The circuit shown in the figure has initial current iL (0−) = 1 A
through the inductor and an initial voltage vC (0−) =− 1 V across the
capacitor. For input v (t) = u (t), the Laplace transform of the current
i (t) for t $ 0 is
(A)
s
s2 + s + 1
(B)
s+2
s2 + s + 1
(C)
s−2
s +s+1
(D)
1
s +s+1
2
2
MCQ 2.58
The transfer function H (s) =
H (s) =
Vo (s)
of an RLC circuit is given by
Vi (s)
106
s + 20s + 106
2
The Quality factor (Q-factor) of this circuit is
(A) 25
(B) 50
(C) 100
(D) 5000
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Chap 2
Networks
MCQ 2.59
For the circuit shown in the figure, the initial conditions are zero. Its
V (s)
transfer function H (s) = c
is
Vi (s)
(A)
1
s + 106 s + 106
(B)
106
s + 103 s + 106
(C)
103
s2 + 103 s + 106
(D)
106
s2 + 106 s + 106
2
2
MCQ 2.60
Consider the following statements S1 and S2
S1 : At the resonant frequency the impedance of a series RLC circuit
is zero.
S2 : In a parallel GLC circuit, increasing the conductance G results
in increase in its Q factor.
Which one of the following is correct?
(A) S1 is FALSE and S2 is TRUE
(B) Both S1 and S2 are TRUE
(C) S1 is TRUE and S2 is FALSE
(D) Both S1 and S2 are FALSE
GATE 2003
ONE MARK
MCQ 2.61
The minimum number of equations required to analyze the circuit
shown in the figure is
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Chap 2
Networks
(A) 3
(B) 4
(C) 6
(D) 7
MCQ 2.62
A source of angular frequency 1 rad/sec has a source impedance
consisting of 1 Ω resistance in series with 1 H inductance. The load
that will obtain the maximum power transfer is
(A) 1 Ω resistance
(B) 1 Ω resistance in parallel with 1 H inductance
(C) 1 Ω resistance in series with 1 F capacitor
(D) 1 Ω resistance in parallel with 1 F capacitor
MCQ 2.63
A series RLC circuit has a resonance frequency of 1 kHz and a
quality factor Q = 100 . If each of R, L and C is doubled from its
original value, the new Q of the circuit is
(A) 25
(B) 50
(C) 100
(D) 200
MCQ 2.64
The differential equation for the current i (t) in the circuit of the
figure is
2
(A) 2 d 2i + 2 di + i (t) = sin t
dt
dt
2
(B) d 2i + 2 di + 2i (t) = cos t
dt
dt
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Chap 2
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2
(C) 2 d 2i + 2 di + i (t) = cos t
dt
dt
2
(D) d 2i + 2 di + 2i (t) = sin t
dt
dt
GATE 2003
TWO MARKS
MCQ 2.65
Twelve 1 Ω resistance are used as edges to form a cube. The resistance
between two diagonally opposite corners of the cube is
(A) 5 Ω
6
(B) 1 Ω
(C) 6 Ω
5
(D) 3 Ω
2
MCQ 2.66
The current flowing through the resistance R in the circuit in the
figure has the form P cos 4t where P is
(A) (0.18 + j0.72)
(B) (0.46 + j1.90)
(C) − (0.18 + j1.90)
(D) − (0.192 + j0.144)
The circuit for Q. 2.66 & 2.67 is given below.
Assume that the switch S is in position 1 for a long time and thrown
to position 2 at t = 0 .
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MCQ 2.67
At t = 0+ , the current i1 is
(A) − V
2R
(C) − V
4R
(B) − V
R
(D) zero
MCQ 2.68
I1 (s) and I2 (s) are the Laplace transforms of i1 (t) and i2 (t) respectively.
The equations for the loop currents I1 (s) and I2 (s) for the circuit
shown in the figure, after the switch is brought from position 1 to
position 2 at t = 0 , are
V
R + Ls + Cs1 − Ls I1 (s)
s
(A) >
=
R + Cs1 H=I2 (s)G = 0 G
− Ls
R + Ls + Cs1 − Ls I1 (s)
− Vs
(B) >
=
G
G
=
=
H
R + Cs1 I2 (s)
− Ls
0
R + Ls + Cs1
− Ls
I1 (s)
− Vs
(C) >
=
G
G
=
=
H
R + Ls + Cs1 I2 (s)
− Ls
0
V
R + Ls + Cs1
− Cs
I1 (s)
s
(D) >
== G
G
1 H=
R + Ls + Cs I2 (s)
− Ls
0
MCQ 2.69
The driving point impedance Z (s) of a network has the pole-zero
locations as shown in the figure. If Z (0) = 3 , then Z (s) is
(A)
3 (s + 3)
s + 2s + 3
(B)
2 (s + 3)
s + 2s + 2
(C)
3 (s + 3)
s + 2s + 2
(D)
2 (s − 3)
s − 2s − 3
2
2
2
2
MCQ 2.70
An input voltage v (t) = 10 2 cos (t + 10c) + 10 5 cos (2t + 10c)
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Chap 2
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V is applied to a series combination of resistance R = 1 Ω and an
inductance L = 1 H. The resulting steady-state current i (t) in ampere
is
(A) 10 cos (t + 55c) + 10 cos (2t + 10c + tan−1 2)
(B) 10 cos (t + 55c) + 10
3
2
cos (2t + 55c)
(C) 10 cos (t − 35c) + 10 cos (2t + 10c − tan−1 2)
(D) 10 cos (t − 35c) +
3
2
cos (2t − 35c)
MCQ 2.71
The impedance parameters z11 and z12 of the two-port network in the
figure are
(A) z11 = 2.75 Ω and z12 = 0.25 Ω
(B) z11 = 3 Ω and z12 = 0.5 Ω
(C) z11 = 3 Ω and z12 = 0.25 Ω
(D) z11 = 2.25 Ω and z12 = 0.5 Ω
GATE 2002
ONE MARK
MCQ 2.72
The dependent current source shown in the figure
(A) delivers 80 W
(B) absorbs 80 W
(C) delivers 40 W
(D) absorbs 40 W
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Chap 2
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MCQ 2.73
In the figure, the switch was closed for a long time before opening at
t = 0 . The voltage vx at t = 0+ is
(A) 25 V
(B) 50 V
(C) − 50 V
(D) 0 V
GATE 2002
TWO MARKS
MCQ 2.74
In the network of the fig, the maximum power is delivered to RL if
its value is
(A) 16 Ω
(B) 40 Ω
3
(C) 60 Ω
(D) 20 Ω
MCQ 2.75
If the 3-phase balanced source in the figure delivers 1500 W at a leading
power factor 0.844 then the value of ZL (in ohm) is approximately
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Chap 2
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(A) 90+32.44c
(B) 80+32.44c
(C) 80+ − 32.44c
(D) 90+ − 32.44c
GATE 2001
ONE MARK
MCQ 2.76
The Voltage e0 in the figure is
(A) 2 V
(B) 4/3 V
(C) 4 V
(D) 8 V
MCQ 2.77
If each branch of Delta circuit has impedance
of the equivalent Wye circuit has impedance
(B) 3Z
(A) Z
3
(C) 3 3 Z
(D) Z
3
3 Z , then each branch
MCQ 2.78
The admittance parameter Y12 in the 2-port network in Figure is
(A) − 0.02 mho
(B) 0.1 mho
(C) − 0.05 mho
(D) 0.05 mho
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Chap 2
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GATE 2001
TWO MARKS
MCQ 2.79
The voltage e0 in the figure is
(A) 48 V
(B) 24 V
(C) 36 V
(D) 28 V
MCQ 2.80
When the angular frequency ω in the figure is varied 0 to 3, the
locus of the current phasor I2 is given by
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Chap 2
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MCQ 2.81
In the figure, the value of the load resistor RL which maximizes the
power delivered to it is
(A) 14.14 Ω
(B) 10 Ω
(C) 200 Ω
(D) 28.28 Ω
MCQ 2.82
The z parameters z11 and z21 for the 2-port network in the figure are
(A) z11 = 6 Ω; z21 = 16 Ω
11
11
(B) z11 = 6 Ω; z21 = 4 Ω
11
11
(C) z11 = 6 Ω; z21 =− 16 Ω
11
11
(D) z11 = 4 Ω; z21 = 4 Ω
11
11
GATE 2000
ONE MARK
MCQ 2.83
The circuit of the figure represents a
(A) Low pass filter
(B) High pass filter
(C) band pass filter
(D) band reject filter
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Chap 2
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MCQ 2.84
In the circuit of the figure, the voltage v (t) is
(A) eat − ebt
(B) eat + ebt
(C) aeat − bebt
(D) aeat + bebt
MCQ 2.85
In the circuit of the figure, the value of the voltage source E is
(A) − 16 V
(B) 4 V
(C) − 6 V
(D) 16 V
GATE 2000
TWO MARKS
MCQ 2.86
Use the data of the figure (a). The current i in the circuit of the
figure (b)
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Chap 2
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(A) − 2 A
(B) 2 A
(C) − 4 A
(D) 4 A
GATE 1999
ONE MARK
MCQ 2.87
Identify which of the following is NOT a tree of the graph shown in
the given figure is
(A) begh
(B) defg
(C) abfg
(D) aegh
MCQ 2.88
A 2-port network is shown in the given figure. The parameter h21 for
this network can be given by
(A) − 1/2
(B) + 1/2
(C) − 3/2
(D) + 3/2
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Chap 2
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GATE 1999
TWO MARK
MCQ 2.89
The Thevenin equivalent voltage VTH appearing between the terminals
A and B of the network shown in the given figure is given by
(A) j16 (3 − j4)
(B) j16 (3 + j4)
(C) 16 (3 + j4)
(D) 16 (3 − j4)
MCQ 2.90
The value of R (in ohms) required for maximum power transfer in
the network shown in the given figure is
(A) 2
(B) 4
(C) 8
(D) 16
MCQ 2.91
A Delta-connected network with its Wye-equivalent is shown in the
given figure. The resistance 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
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Chap 2
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GATE 1998
ONE MARK
MCQ 2.92
A network has 7 nodes and 5 independent loops. The number of
branches in the network is
(A) 13
(B) 12
(C) 11
(D) 10
MCQ 2.93
The nodal method of circuit analysis is based on
(A) KVL and Ohm’s law
(B) KCL and Ohm’s law
(C) KCL and KVL
(D) KCL, KVL and Ohm’s law
MCQ 2.94
Superposition theorem is NOT applicable to networks containing
(A) nonlinear elements
(B) dependent voltage sources
(C) dependent current sources
(D) transformers
MCQ 2.95
The parallel RLC circuit shown in the figure is in resonance. In this
circuit
(A) IR < 1 mA
(B) IR + IL > 1 mA
(C) IR + IC < 1 mA
(D) IR + IC > 1 mA
MCQ 2.96
0 − 1/2
The short-circuit admittance matrix a two-port network is >
1/2 0 H
The two-port network is
(A) non-reciprocal and passive
(B) non-reciprocal and active
(C) reciprocal and passive
(D) reciprocal and acitve
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Chap 2
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MCQ 2.97
The voltage across the terminals a and b in the figure is
(A) 0.5 V
(B) 3.0 V
(C) 3.5 V
(D) 4.0 V
MCQ 2.98
A high-Q quartz crystal exhibits series resonance at the frequency ωs
and parallel resonance at the frequency ωp . Then
(A) ωs is very close to, but less than ωp
(B) ωs << ωp
(C) ωs is very close to, but greater than ωp
(D) ωs >> ωp
GATE 1997
ONE MARK
MCQ 2.99
The current i4 in the circuit of the figure is equal to
(A) 12 A
(B) − 12 A
(C) 4 A
(D) None or these
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Chap 2
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MCQ 2.100
The voltage V in the figure equal to
(A) 3 V
(B) − 3 V
(C) 5 V
(D) None of these
MCQ 2.101
The voltage V in the figure is always equal to
(A) 9 V
(B) 5 V
(C) 1 V
(D) None of the above
MCQ 2.102
The voltage V in the figure is
(A) 10 V
(B) 15 V
(C) 5 V
(D) None of the above
MCQ 2.103
In the circuit of the figure is the energy absorbed by the 4 Ω resistor
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Chap 2
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in the time interval (0, 3) is
(A) 36 Joules
(B) 16 Joules
(C) 256 Joules
(D) None of the above
MCQ 2.104
In the circuit of the figure the equivalent impedance seen across
terminals a, b, is
(A) b 16 l Ω
3
(B) b 8 l Ω
3
(C) b 8 + 12j l Ω
3
(D) None of the above
GATE 1996
ONE MARK
MCQ 2.105
In the given figure, A1, A2 and A3 are ideal ammeters. If A2 and A3
read 3 A and 4 A respectively, then A1 should read
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Chap 2
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(A) 1 A
(B) 5 A
(C) 7 A
(D) None of these
MCQ 2.106
The number of independent loops for a network with n nodes and b
branches is
(A) n − 1
(B) b − n
(C) b − n + 1
(D) independent of the number of nodes
GATE 1996
TWO MARKS
MCQ 2.107
The voltages VC1, VC2, and VC3 across the capacitors in the circuit in
the given figure, under steady state, are respectively.
(A) 80 V, 32 V, 48 V
(B) 80 V, 48 V, 32 V
(C) 20 V, 8 V, 12 V
(D) 20 V, 12 V, 8 V
***********
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Chap 2
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SOLUTIONS
SOL 2.1
Replacing P − Q by short circuit as shown below we have
Using current divider rule the current Isc is
ISC =
25
(16 0 ) = (6.4 − j4.8) A
25 + 15 + j30
Hence (A) is correct option.
SOL 2.2
Power transferred to RL will be maximum when RL is equal to the
thevenin resistance. We determine thevenin resistance by killing all
source as follows :
RTH = 10 # 10 + 10 = 15 Ω
10 + 10
Hence (C) is correct option.
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SOL 2.3
The given circuit is shown below
For parallel combination of R and C equivalent impedance is
R$ 1
jω C
R
Zp =
=
1
1
+
j
ωRC
R+
jω C
Transfer function can be written as
R
1 + jωRC
Vout = Z p
=
Vin
R
Zs + Zp
R+ 1 +
jωC 1 + jωRC
jωRC
=
jωRC + (1 + jωRC) 2
j
=
j + (1 + j) 2
j
Vout =
=1
3
Vin
(1 + j) 2 + j
Thus
Here ω = 1
RC
V
v out = b p l cos (t/RC)
3
Hence (A) is correct option.
SOL 2.4
From star delta conversion we have
Thus
R1 =
Ra Rb
6.6
=
= 2Ω
Ra + Rb + Rc 6 + 6 + 6
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Here
R1 = R 2 = R 3 = 2 Ω
Replacing in circuit we have the circuit shown below :
Now the total impedance of circuit is
Z =
Current
(2 + j4) (2 − j4)
+2 = 7Ω
(2 + j4) (2 − j4)
I = 14+0c = 2+0c
7
Hence (B) is correct option.
SOL 2.5
From given admittance matrix we get
I1 = 0.1V1 − 0.01V2 and
I2 = 0.01V1 + 0.1V2
Now, applying KVL in outer loop;
V2 =− 100I2
or
I2 =− 0.01V2
From eq (2) and eq (3) we have
− 0.01V2 = 0.01V1 + 0.1V2
− 0.11V2 = 0.01V1
V2 = − 1
11
V1
...(1)
...(2)
...(3)
Hence (D) is correct option.
SOL 2.6
Here we take the current flow direction as positive.
At t = 0− voltage across capacitor is
−3
Q
VC (0−) =− =− 2.5 # 10−6 =− 50 V
C
50 # 10
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Thus VC (0+) =− 50 V
In steady state capacitor behave as open circuit thus
V (3) = 100 V
Now,
VC (t) = VC (3) + (VC (0+) − VC (3)) e−t/RC
−t
= 100 + (− 50 − 100) e 10 # 50 # 10
−6
= 100 − 150e− (2 # 10 t)
ic (t) = C dV
dt
3
Now
= 50 # 10−6 # 150 # 2 # 103 e−2 # 10 t A
3
= 15e−2 # 10 t
3
ic (t) = 15 exp (− 2 # 103 t) A
Hence (A) is correct option.
SOL 2.7
Given circuit is as shown below
By writing node equation at input port
I1 = V1 + V1 − V2 = 4V1 − 2V2
0.5
0.5
By writing node equation at output port
I2 = V2 + V2 − V1 =− 2V1 + 4V2
0.5
0.5
...(1)
...(2)
From (1) and (2), we have admittance matrix
4 −2
Y =>
− 2 4H
Hence (A) is correct option.
SOL 2.8
A parallel RLC circuit is shown below :
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Input impedance Z in =
1
1 + 1 + jω C
R jω L
At resonance
1 = ωC
ωL
So,
Z in = 1 = R
1/R
(maximum at resonance)
Thus (D) is not true.
Furthermore bandwidth is ωB i.e ωB \ 1 and is independent of L,
R
Hence statements A, B, C, are true.
Hence (D) is correct option.
SOL 2.9
Let the current i (t) = A + Be−t/τ
τ " Time constant
When the switch S is open for a long time before t < 0 , the circuit is
At t = 0 , inductor current does not change simultaneously, So the
circuit is
Current is resistor (AB)
i (0) = 0.75 = 0.375 A
2
Similarly for steady state the circuit is as shown below
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i (3) = 15 = 0.5 A
3
−3
τ = L = 15 # 10
= 10−3 sec
Req
10 + (10 || 10)
t
i (t) = A + Be− 1 # 10 = A + Be−100t
Now
i (0) = A + B = 0.375
and
i (3) = A = 0.5
So,
B = 0.375 − 0.5 =− 0.125
Hence
i (t) = 0.5 − 0.125e−1000 t A
Hence (A) is correct option.
−3
SOL 2.10
Circuit is redrawn as shown below
Where,
Z1 = jωL = j # 103 # 20 # 10−3 = 20j
Z2 = R || XC
1
XC = 1 =
=− 20j
jωC
j # 103 # 50 # 10−6
1 (− 20j)
Z2 =
R = 1Ω
1 − 20j
Voltage across Z2
VZ =
2
Z2 : 20 0 =
Z1 + Z 2
=c
− 20j
c 1 − 20j m
20j
c 20j − 1 − 20j m
: 20
(− 20j)
: 20 =− j
20j + 400 − 20j m
Current in resistor R is
j
V
I = Z =− =− j A
1
R
2
Hence (A) is correct option.
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SOL 2.11
The circuit can be redrawn as
Applying nodal analysis
VA − 10 + 1 + VA − 0 = 0
2
2
Current,
2VA − 10 + 2 = 0 = V4 = 4 V
I1 = 10 − 4 = 3 A
2
Current from voltage source is
I 2 = I1 − 3 = 0
Since current through voltage source is zero, therefore power delivered
is zero.
Hence (A) is correct option.
SOL 2.12
Circuit is as shown below
Since 60 V source is absorbing power. So, in 60 V source current flows
from + to - ve direction
So,
I + I1 = 12
I = 12 − I1
I is always less then 12 A So, only option (A) satisfies this conditions.
Hence (A) is correct option.
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SOL 2.13
For given network we have
(RL XC ) Vi
V0 =
R + (RL XC )
RL
V0 (s)
RL
+
1
sRL C
=
=
RL
Vi (s)
R + RRL sC + RL
R+
1 + sRL C
RL
1
=
=
R
R + RRL sC + RL
1+
+ RsC
RL
But we have been given
V (s)
1
=
T .F. = 0
2 + sCR
Vi (s)
Comparing, we get
1 + R = 2 & RL = R
RL
Hence (C) is correct option.
SOL 2.14
The energy delivered in 10 minutes is
E =
#0 VIdt = I #0Vdt
t
t
= I # Area
= 2 # 1 (10 + 12) # 600 = 13.2 kJ
2
Hence (C) is correct option.
SOL 2.15
From given circuit the load current is
20+0c
=
= 20+0c
IL = V
Zs + ZL
(1 + 2j) + (7 + 4j) 8 + 6j
where φ = tan - 1 3
= 1 (8 − 6j) = 20+0c = 2+ − φ
10+φ
5
4
The voltage across load is
VL = IL ZL
The reactive power consumed by load is
Pr = VL IL* = IL ZL # IL* = ZL IL 2
2
= (7 # 4j) 20+0c = (7 + 4j) = 28 + 16j
8 + 6j
Thus average power is 28 and reactive power is 16.
Hence (B) is correct option.
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SOL 2.16
At t = 0− , the circuit is as shown in fig below :
V (0−) = 100 V
V (0+) = 100 V
Thus
At t = 0+ , the circuit is as shown below
I (0+) = 100 = 20 mA
5k
At steady state i.e. at t = 3 is I (3)= 0
Now
i (t) = I (0+) e−
Ceq =
t
RCeq
u (t)
(0.5μ + 0.3μ) 0.2μ
= 0.16 μ F
0.5μ + 0.3μ + 0.2μ
1
1 =
= 1250
RCeq
5 # 103 # 0.16 # 10−6
i (t) = 20e−1250t u (t) mA
Hence (B) is correct option.
SOL 2.17
For Pmax the load resistance RL must be equal to thevenin resistance
Req i.e. RL = Req . The open circuit and short circuit is as shown below
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The open circuit voltage is
Voc = 100 V
From fig
I1 = 100 = 12.5 A
8
Vx =− 4 # 12.5 =− 50 V
I2 = 100 + Vx = 100 − 50 = 12.5 A
4
4
Isc = I1 + I2 = 25 A
Rth = Voc = 100 = 4 Ω
Isc
25
Thus for maximum power transfer RL = Req = 4 Ω
Hence (C) is correct option.
SOL 2.18
Steady state all transient effect die out and inductor act as short
circuits and forced response acts only. It doesn’t depend on initial
current state. From the given time domain behavior we get that
circuit has only R and L in series with V0 . Thus at steady state
i (t) " i (3) = V0
R
Hence (A) is correct option.
SOL 2.19
The given graph is
There can be four possible tree of this graph which are as follows:
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There can be 6 different possible cut-set.
Hence (C) is correct option.
SOL 2.20
Initially i (0−) = 0 therefore due to inductor i (0+) = 0 . Thus all
current Is will flow in resistor R and voltage across resistor will be
Is Rs . The voltage across inductor will be equal to voltage across Rs
as no current flow through R.
Thus
but
Thus
vL (0+) = Is Rs
di (0+)
+
vL (0 ) = L
dt
di (0+)
vL (0+) Is Rs
=
=
L
L
dt
Hence (B) is correct option.
SOL 2.21
Killing all current source and voltage sources we have,
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Zth = (1 + s) ( s1 + 1)
(1 + s)( s1 + 1)
[ s1 + 1 + 1 + s]
=
=
(1 + s) + ( s1 + 1)
s + s1 + 1 + 1
or
Zth = 1
Alternative :
Here at DC source capacitor act as open circuit and inductor act as
short circuit. Thus we can directly calculate thevenin Impedance as
1Ω
Hence (A) is correct option.
SOL 2.22
Z (s) = R 1 sL = 2
sC
s +
We have been given
Z (s) = 2 0.2s
s + 0.1s + 2
Comparing with given we get
1 = 0.2 or C = 5 F
C
1 = 0.1 or R = 2 Ω
RC
1 = 2 or L = 0.1 H
LC
s
C
s
RC
+
1
LC
Hence (D) is correct option.
SOL 2.23
Voltage across capacitor is
t
Vc = 1 idt
C 0
#
Here C = 1 F and i = 1 A. Therefore
#
t
#
t
Vc = dt
0
For 0 < t < T , capacitor will be charged from 0 V
Vc = dt = t
0
At t = T, Vc = T Volts
For T < t < 2T , capacitor will be discharged from T volts as
#
t
Vc = T − dt = 2T − t
T
At t = 2T, Vc = 0 volts
For 2T < t < 3T , capacitor will be charged from 0 V
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#
t
Vc = dt = t − 2T
2T
At t = 3T, Vc = T Volts
For 3T < t < 4T , capacitor will be discharged from T Volts
#
t
Vc = T − dt = 4T − t
3T
At t = 4T, Vc = 0 Volts
For 4T < t < 5T , capacitor will be charged from 0 V
#
t
Vc = dt = t − 4T
4T
At t = 5T, Vc = T Volts
Thus the output waveform is
Only option C satisfy this waveform.
Hence (C) is correct option.
SOL 2.24
Writing in transform domain we have
1
Vc (s)
= 1 s
= 2 1
Vs (s)
^ s + s + 1h
(s + s + 1)
Since Vs (t) = δ (t) " Vs (s) = 1 and
Vc (s) = 2 1
(s + s + 1)
3
2
Vc (s) = 2 =
G
3 (s + 12 ) 2 + 43
Taking inverse laplace transform we have
Vt = 2 e− sin c 3 t m
2
3
Hence (D) is correct option.
or
t
2
SOL 2.25
Let voltage across resistor be vR
VR (s)
= 1 1
= 2 s
VS (s)
( s + s + 1) (s + s + 1)
Since vs = δ (t) " Vs (s) = 1 we get
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s
s
=
(s2 + s + 1) (s + 12 ) 2 + 43
1
(s + 12 )
2
=
−
(s + 12 ) 2 + 43
(s + 12 ) 2 + 43
vR (t) = e− cos 3 t − 1 # 2 e− sin 3 t
2
2
2
3
VR (s) =
or
1
2
1
2
t
= e− 2 =cos
3 t − 1 sin 3 t
2
2 G
3
Hence (B) is correct option.
SOL 2.26
From the problem statement we have
z11 = v1
= 6 = 1.5Ω
i1 i = 0 4
z12 = v1
= 4.5 = 4.5Ω
i2 i = 0
1
z21 = v2
= 6 = 1.5Ω
i1 i = 0 4
z22 = v2
= 1.5 = 1.5Ω
i2 i = 0
1
2
1
2
2
Thus z -parameter matrix is
z11 z12
1.5 4.5
=z z G = =1.5 1.5 G
21 22
Hence (C) is correct option.
SOL 2.27
From the problem statement we
h12 = v1
=
v2 i = 0
h22 = i2
=
v2 i = 0
1
1
have
4. 5 = 3
1. 5
1 = 0.67
1.5
From z matrix, we have
v1 = z11 i1 + z12 i2
v2 = z21 i1 + z22 i2
If v2 = 0
i2 = − z21 = − 1.5 =− 1 = h
Then
21
i1
z22
1.5
or
i2 =− i1
Putting in equation for v1, we get
v1 = (z11 − z12) i1
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v1
= h11 = z11 − z12 = 1.5 − 4.5 =− 3
i1 v = 0
Hence h −parameter will be
h11 h12
−3 3
=h h G = =− 1 0.67 G
21 22
2
Hence (A) is correct option.
SOL 2.28
According to maximum Power Transform Theorem
ZL = Zs* = (Rs − jXs)
Hence (D) is correct option.
SOL 2.29
At ω " 3 , capacitor acts as short circuited and circuit acts as shown
in fig below
Here we get V0 = 0
Vi
At ω " 0 , capacitor acts as open circuited and circuit look like as
shown in fig below
Here we get also V0 = 0
Vi
So frequency response of the circuit is as shown in fig and circuit is
a Band pass filter.
Hence (C) is correct option.
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SOL 2.30
We know that bandwidth of series RLC circuit is R . Therefore
L
Bandwidth of filter 1 is B1 = R
L1
Bandwidth of filter 2 is B2 = R = R = 4R
L2
L1
L1 /4
Dividing above equation B1 = 1
B2
4
Hence (D) is correct option.
SOL 2.31
Here Vth is voltage across node also. Applying nodal analysis we get
Vth + Vth + Vth − 2i = 2
2
1
1
i = Vth = Vth
1
But from circuit
Therefore
Vth + Vth + Vth − 2Vth = 2
2
1
1
or
Vth = 4 volt
From the figure shown below it may be easily seen that the short
circuit current at terminal XY is isc = 2 A because i = 0 due to short
circuit of 1 Ω resistor and all current will pass through short circuit.
Therefore
Rth = Vth = 4 = 2 Ω
isc
2
Hence (D) is correct option.
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SOL 2.32
The voltage across capacitor is
At t = 0+ ,
Vc (0+) = 0
At t = 3 ,
VC (3) = 5 V
The equivalent resistance seen by capacitor as shown in fig is
Req = 20 20 = 10kΩ
Time constant of the circuit is
τ = Req C = 10k # 4μ = 0.04 s
Using direct formula
Vc (t) = VC (3) − [Vc (3) − Vc (0)] e−t/τ
= VC (3) (1 − e−t/τ) + VC (0) e−t/τ = 5 (1 − e−t/0.04)
or
Vc (t) = 5 (1 − e−25t)
dV (t)
Now
IC (t) = C C
dt
= 4 # 10−6 # (− 5 # 25e−25t) = 0.5e−25t mA
Hence (A) is correct option.
SOL 2.33
Impedance
= (5 − 3j) (5 + 3j) =
=
(5 − 3j) # (5 + 3j)
5 − 3j + 5 + 3j
(5) 2 − (3j) 2
= 25 + 9 = 3.4
10
10
VAB = Current # Impedance = 5+30c # 34 = 17+30c
Hence (D) is correct option.
SOL 2.34
The network is shown in figure below.
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Now
and
also
From (1) and (2)
Thus
V1 = AV2 − BI2
I1 = CV2 − DI2
V2 =− I2 RL
we get
V1 = AV2 − BI2
I1
CV2 − DI2
...(1)
...(2)
...(3)
Substituting value of V2 from (3) we get
Input Impedance Zin = − A # I2 RL − BI2
− C # I2 RL − DI2
or
Zin = ARL + B
CRL + D
Hence (D) is correct option.
SOL 2.35
The circuit is as shown below.
At input port
V1 = re I1
At output port V2 = r0 (I2 − βI1) =− r0 βI1 + r0 I2
Comparing standard equation
V1 = z11 I1 + z12 I2
V2 = z21 I1 + z22 I2
z12 = 0 and z21 =− r0 β
Hence (B) is correct option.
SOL 2.36
For series RC network input impedance is
Zins = 1 + R = 1 + sRC
sC
sC
Thus pole is at origin and zero is at − 1
RC
For parallel RC network input impedance is
1 R
sC
sC
=
Zin =
1 +R
1 + sRC
sC
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Thus pole is at − 1 and zero is at infinity.
RC
Hence (B) is correct option.
SOL 2.37
We know
v = Ldi
dt
Taking laplace transform we get
V (s) = sLI (s) − Li (0+)
As per given in question
− Li (0+) =− 1 mV
Thus
i (0+) = 1 mV = 0.5 A
2 mH
Hence (A) is correct option.
SOL 2.38
At initial all voltage are zero. So output is also zero.
Thus
v0 (0+) = 0
At steady state capacitor act as open circuit.
Thus,
v0 (3) = 4 # vi = 4 # 10 = 8
5
5
The equivalent resistance and capacitance can be calculate after
killing all source
Req = 1 4 = 0.8 kΩ
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Ceq = 4 1 = 5 μF
τ = Req Ceq = 0.8kΩ # 5μF = 4 ms
v0 (t) = v 0 (3) − [v 0 (3) − v 0 (0+)] e−t/τ
= 8 − (8 − 0) e−t/0.004
v0 (t) = 8 (1 − e−t/0.004) Volts
Hence (B) is correct option.
SOL 2.39
Here
Z2 (s) = Rneg + Z1 (s)
or
Z2 (s) = Rneg + Re Z1 (s) + j Im Z1 (s)
For Z2 (s) to be positive real, Re Z2 (s) $ 0
Thus
Rneg + Re Z1 (s) $ 0
or
Re Z1 (s) $− Rneg
But Rneg is negative quantity and − Rneg is positive quantity. Therefore
Re Z1 (s) $ Rneg
or
For all ω.
Rneg # Re Z1 (jω)
Hence (A) is correct option.
SOL 2.40
Transfer function is
1
1
Y (s)
1
sC
LC
=
= 2
=
U (s)
s LC + scR + 1
R + sL + 1
s2 + R s + 1
sC
L
LC
Comparing with s2 + 2ξωn s + ωn2 = 0 we have
Here
2ξωn = R ,
L
and
Thus
ωn =
1
LC
ξ = R
2L
LC = R
2
C
L
For no oscillations, ξ $ 1
R C $1
Thus
2
L
or
R $2
L
C
Hence (C) is correct option.
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SOL 2.41
For given transformer
I2 = V1 = n
I1
V2
1
or
I1 = I2 and V1 = nV2
n
Comparing with standard equation
V1 = AV2 + BI2
I1 = CV2 + DI2
A B
n 0
=C D G = = 0 1 G
n
Thus
x = 1
n
Hence (B) is correct option.
SOL 2.42
We have L = 1H and C = 1 # 10−6
400
Resonant frequency
f0 =
1
==
2π LC
2π
1
1 # 1 # 10 - 6
400
3
4
= 10 # 20 = 10 Hz
2π
π
Hence (B) is correct option.
SOL 2.43
Maximum power will be transferred when RL = Rs = 100Ω
In this case voltage across RL is 5 V, therefore
2
Pmax = V = 5 # 5 = 0.25 W
R
100
Hence (C) is correct option.
SOL 2.44
For stability poles and zero interlace on real axis. In RC series
network the driving point impedance is
Zins = R + 1 = 1 + sRC
Cs
sC
Here pole is at origin and zero is at s =− 1/RC , therefore first critical
frequency is a pole and last critical frequency is a zero.
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For RC parallel network the driving point impedance is
R 1
R
Cs =
Zinp =
1
1
+
sRC
R+
Cs
Here pole is s =− 1/RC and zero is at 3, therefore first critical
frequency is a pole and last critical frequency is a zero.
Hence (C) is correct option.
SOL 2.45
Applying KCL we get
i1 (t) + 5+0c = 10+60c
or
i1 (t) = 10+60c − 5+0c = 5 + 5 3j − 5
or
i1 (t) = 5 3 +90c = 10 3 +90c
2
Hence (A) is correct option.
SOL 2.46
If L1 = j5Ω and L3 = j2Ω the mutual induction is subtractive because
current enters from dotted terminal of j2Ω coil and exit from dotted
terminal of j5Ω. If L2 = j2Ω and L3 = j2Ω the mutual induction is
additive because current enters from dotted terminal of both coil.
Thus
Z = L1 − M13 + L2 + M23 + L3 − M31 + M32
= j5 + j10 + j2 + j10 + j2 − j10 + j10 = j9
Hence (B) is correct option.
SOL 2.47
Open circuit at terminal ab is shown below
Applying KCL at node we get
Vab + Vab − 10 = 1
5
5
or
Vab = 7.5 = Vth
Short circuit at terminal ab is shown below
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Short circuit current from terminal ab is
Isc = 1 + 10 = 3 A
5
Thus
Rth = Vth = 7.5 = 2.5 Ω
Isc
3
Here current source being in series with dependent voltage source
make it ineffective.
Hence (B) is correct option.
SOL 2.48
Here Va = 5 V because R1 = R2 and total voltage drop is 10 V.
Now
Vb = R3 # 10 = 1.1 # 10 = 5.238 V
R3 + R4
2.1
V = Va − Vb = 5 − 5.238 =− 0.238 V
Hence (C) is correct option.
SOL 2.49
For h parameters we have to write V1 and I2 in terms of I1 and V2 .
V1 = h11 I1 + h12 V2
I2 = h21 I1 + h22 V2
Applying KVL at input port
V1 = 10I1 + V2
Applying KCL at output port
V2 = I + I
1
2
20
or
I2 =− I1 + V2
20
Thus from above equation we get
h11 h12
10 1
=h h G = =− 1 0.05G
12 22
Hence (D) is correct option.
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SOL 2.50
Time constant RC = 0.1 # 10 - 6 # 103 = 10 - 4 sec
Since time constant RC is very small, so steady state will be reached
in 2 sec. At t = 2 sec the circuit is as shown in fig.
Vc = 3 V
V2 =− Vc =− 3 V
Hence (B) is correct option.
SOL 2.51
For a tree there must not be any loop. So a, c, and d don’t have any
loop. Only b has loop.
Hence (B) is correct option.
SOL 2.52
The sign of M is as per sign of L If current enters or exit the dotted
terminals of both coil. The sign of M is opposite of L If current
enters in dotted terminal of a coil and exit from the dotted terminal
of other coil.
Thus
Leq = L1 + L2 − 2M
Hence (D) is correct option.
SOL 2.53
Here ω = 2 and V = 1+0c
Y = 1 + jω C + 1
R
jω L
= 3 + j 2 # 3 + 1 1 = 3 + j4
j2 # 4
= 5+ tan - 1 4 = 5+53.11c
3
I = V * Y = (1+0c)( 5+53.1c) = 5+53.1c
Thus
i (t) = 5 sin (2t + 53.1c)
Hence (A) is correct option.
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SOL 2.54
vi (t) = 2 sin 103 t
Here ω = 103 rad and Vi = 2 +0c
1
jωC
1
Now
V0 =
.Vt =
V
+
ωCR i
1
1
j
R+
jω C
1
2 +0c
=
1 + j # 103 # 10 - 3
= 1 − 45c
v0 (t) = sin (103 t − 45c)
Hence (A) is correct option.
SOL 2.55
vi (t) = u (t)
Vi (s) = 1
s
Input voltage
Taking laplace transform
Impedance
or
Z (s) = s + 2
V (s)
1
I (s) = i
=
s + 2 s (s + 2)
I (s) = 1 ; 1 − 1 E
2 s s+2
Taking inverse laplace transform
i (t) = 1 (1 − e−2t) u (t)
2
At t = 0 ,
i (t) = 0
1
At t = 2 ,
i (t) = 0.31
At t = 3 ,
i (t) = 0.5
Graph (C) satisfies all these conditions.
Hence (C) is correct option.
SOL 2.56
We know that
where
V1 = z11 I1 + z12 I2
V2 = z11 I1 + z22 I2
z11 = V1
I1 I = 0
2
z21 = V2
I1
I1 = 0
Consider the given lattice network, when I2 = 0 . There is two similar
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path in the circuit for the current I1. So I = 1 I1
2
For z11 applying KVL at input port we get
V1 = I (Za + Zb)
Thus
V1 = 1 I1 (Za + Zb)
2
z11 = 1 (Za + Zb)
2
For Z21 applying KVL at output port we get
V2 = Za I1 − Zb I1
2
2
Thus
V2 = 1 I1 (Za − Zb)
2
z21 = 1 (Za − Zb)
2
For this circuit z11 = z22 and z12 = z21. Thus
V
R
S Za + Zb Za − Zb W
z11 z12
2
2
=z z G = SS Za − Zb Za + Zb WW
21 22
S 2
2 W
X
T
Here Za = 2j and Zb = 2Ω
z11 z12
1+j j−1
Thus
=z z G = = j − 1 1 + j G
21 22
Hence (D) is correct option.
SOL 2.57
Applying KVL,
v (t) = Ri (t) +
Ldi (t) 1
+
dt
C
#0
3
i (t) dt
Taking L.T. on both sides,
V (s) = RI (s) + LsI (s) − Li (0+) +
I (s) vc (0+)
+
sC
sC
v (t) = u (t) thus V (s) = 1
s
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1 = I (s) + sI (s) − 1 + I (s) − 1
s
s
s
2 + 1 = I (s) 6s2 + s + 1@
s
s
or
I (s) = 2 s + 2
s +s+1
Hence (B) is correct option.
Hence
SOL 2.58
Characteristics equation is
s2 + 20s + 106 = 0
Comparing with s2 + 2ξωn s + ωn2 = 0 we have
ωn = 106 = 103
2ξω = 20
Thus
2ξ = 203 = 0.02
10
Now
Q = 1 = 1 = 50
2ξ
0.02
Hence (B) is correct option.
SOL 2.59
H (s) =
V0 (s)
Vi (s)
1
1
sC
=
= 2
1
s
LC
sCR + 1
+
R + sL +
sC
1
= 2 −2
s (10 # 10−4) + s (10−4 # 10 4) + 1
106
= −6 2 1
= 2
10 s + s + 1 s + 106 s + 106
Hence (D) is correct option.
SOL 2.60
Impedance of series RLC circuit at resonant frequency is minimum,
not zero. Actually imaginary part is zero.
Z = R + j ` ωL − 1 j
ωC
At resonance ωL − 1 = 0 and Z = R that is purely resistive. Thus
ωC
S1 is false
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Now quality factorQ = R
C
L
Q = 1
G
C
L
Since G = 1 ,
R
If G - then Q . provided C and L are constant. Thus S2 is also false.
Hence (D) is correct option.
SOL 2.61
Number of loops = b − n + 1
= minimum number of equation
Number of branches = b = 8
Number of nodes = n = 5
Minimum number of equation
= 8−5+1 = 4
Hence (B) is correct option.
SOL 2.62
For maximum power transfer
ZL = ZS* = Rs − jXs
Thus
ZL = 1 − 1j
Hence (C) is correct option.
SOL 2.63
Q = 1
R
L
C
When R, L and C are doubled,
2L = 1
Q' = 1
2R 2C
2R
Thus
Q' = 100 = 50
2
L =Q
C
2
Hence (B) is correct option.
SOL 2.64
Applying KVL we get,
di (t) 1
+
i (t) dt
dt
C
di (t)
sin t = 2i (t) + 2
+ i (t) dt
dt
#
sin t = Ri (t) + L
or
#
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Differentiating with respect to t , we get
2di (t) 2d2 i (t)
cos t =
+ i (t)
+
dt
dt2
Hence (C) is correct option.
SOL 2.65
For current i there is 3 similar path. So current will be divide in three
path
so, we get
Vab − b i # 1l − b i # 1l − b 1 # 1l = 0
3
6
3
Vab = R = 1 + 1 + 1 = 5 Ω
eq
i
6
3 6 3
Hence (A) is correct option.
SOL 2.66
Data are missing in question as L1 &L2 are not given.
SOL 2.67
At t = 0 - circuit is in steady state. So inductor act as short circuit
and capacitor act as open circuit.
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At t = 0 - ,
i1 (0 -) = i2 (0 -) = 0
vc (0 -) = V
At t = 0+ the circuit is as shown in fig. The voltage across capacitor
and current in inductor can’t be changed instantaneously. Thus
i1 = i2 =− V
2R
At t = 0+ ,
Hence (A) is correct option.
SOL 2.68
When switch is in position 2, as shown in fig in question, applying
KVL in loop (1),
RI1 (s) + V + 1 I1 (s) + sL [I1 (s) − I2 (s)] = 0
s
sC
I1 (s) 8R + 1 + sL B − I2 (s) sL = − V
s
sc
or
z11 I1 + z12 I2 = V1
Applying KVL in loop 2,
sL [I2 (s) − I1 (s)] + RI2 (s) + 1 I2 (s) = 0
sC
Z12 I1 + Z22 I2 = V2
or
− sLI1 (s) + 8R + sL + 1 BI2 (s) = 0
sc
Now comparing with
Z11 Z12 I1
V1
=Z Z G=I G = =V G
21
22
2
2
we get
R
− sL
SR + sL + 1
sC
S
R + sL + 1
− sL
SS
sC
T
Hence (C) is correct option.
V
W I1 (s)
−V
W=
=> s H
G
WW I2 (s)
0
X
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SOL 2.69
Zeros =− 3
Pole1 =− 1 + j
Pole 2 =− 1 − j
K (s + 3)
Z (s) =
(s + 1 + j)( s + 1 − j)
K (s + 3)
K (s + 3)
=
2
2
(s + 1) − j
(s + 1) 2 + 1
=
From problem statement Z (0) ω = 0 = 3
Thus 3K = 3 and we get K = 2
2
2 (s + 3)
Z (s) = 2
s + 2s + 2
Hence (B) is correct option.
SOL 2.70
v (t) = 10 2 cos (t + 10c) + 10 5 cos (2t + 10c)
1 4444 2 4444 3 1 4444
4 2 4444
43
v1
v2
Thus we get ω1 = 1 and ω2 = 2
Now
Z1 = R + jω1 L = 1 + j1
Z2 = R + jω2 L = 1 + j2
v (t) v (t)
i (t) = 1 + 2
Z1
Z2
=
10 2 cos (t + 10c) 10 5 cos (2t + 10c)
+
1+j
1 + j2
=
10 2 cos (t + 10c) 10 5 cos (2t + 10c)
+
12 + 22 + tan−1 1
12 + 22 tan−1 2
=
10 2 cos (t + 10c) 10 5 cos (2t + 10c)
+
2 + tan−1 45c
5 tan−1 2
i (t) = 10 cos (t − 35c) + 10 cos (2t + 10c − tan−1 2)
Hence (C) is correct option.
SOL 2.71
Using 3− Y conversion
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2 # 1 = 2 = 0. 5
2+1+1
4
R2 = 1 # 1 = 1 = 0.25
2+1+1
4
R3 = 2 # 1 = 0.5
2+1+1
R1 =
Now the circuit is as shown in figure below.
z11 = V1
I1
Now
I2 = 0
= 2 + 0.5 + 0.25 = 2.75
z12 = R3 = 0.25
Hence (A) is correct option.
SOL 2.72
Applying KCL at for node 2,
V2 + V2 − V1 = V1
5
5
5
or
V2 = V1 = 20 V
Voltage across dependent current source is 20 thus power delivered
by it is
PV2 # V1 = 20 # 20 = 80 W
5
5
It deliver power because current flows from its +ive terminals.
Hence (A) is correct option.
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SOL 2.73
When switch was closed, in steady state, iL (0 -) = 2.5 A
At t = 0+ , iL (0+) = iL (0 -) = 2.5 A and all this current of will pass
through 2 Ω resistor. Thus
Vx =− 2.5 # 20 =− 50 V
Hence (C) is correct option.
SOL 2.74
For maximum power delivered, RL must be equal to Rth across same
terminal.
Applying KCL at Node, we get
0.5I1 = Vth + I1
20
or
but
Vth + 10I1 = 0
I1 = Vth − 50
40
Thus Vth + Vth − 50 = 0
4
or
Vth = 10 V
For Isc the circuit is shown in figure below.
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but
Isc = 0.5I1 − I1 =− 0.5I1
I1 =− 50 =− 1.25 A
40
Isc =− 0.5 # − 12.5 = 0.625 A
Rth = Vth = 10 = 16 Ω
Isc
0.625
Hence (A) is correct option.
SOL 2.75
IP , VP " Phase current and Phase voltage
IL, VL " Line current and line voltage
Now
VP = c VL m and IP = IL
3
So,
Power = 3VP IL cos θ
1500 = 3 c VL m (IL) cos θ
3
also
IL = c VL m
3 ZL
1500 = 3 c VL mc VL m cos θ
3
3 ZL
2
(400) (.844)
ZL =
= 90 Ω
1500
As power factor is leading
So,
cos θ = 0.844 " θ = 32.44
As phase current leads phase voltage
ZL = 90+ − θ = 90+ − 32.44c
Hence (D) is correct option.
SOL 2.76
Applying KCL, we get
e0 − 12 + e0 + e0 = 0
4
4 2+2
or
Hence (C) is correct option.
e0 = 4 V
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SOL 2.77
The star delta circuit is shown as below
Here
and
Now
ZAB = ZBC = ZCA = 3 Z
ZAB ZCA
ZA =
ZAB + ZBC + ZCA
ZB =
ZAB ZBC
ZAB + ZBC + ZCA
ZC =
ZBC ZCA
ZAB + ZBC + ZCA
ZA = ZB = ZC =
3Z 3Z
= Z
3Z+ 3Z+ 3Z
3
Hence (A) is correct option.
SOL 2.78
y11 y12
y1 + y3 − y3
=y y G = = − y y + y G
21 22
3
2
3
y12 =− y3
y12 =− 1 =− 0.05 mho
20
Hence (C) is correct option.
SOL 2.79
We apply source conversion the circuit as shown in fig below.
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Now applying nodal analysis we have
e0 − 80 + e0 + e0 − 16 = 0
10 + 2 12
6
or
4e0 = 112
e0 = 112 = 28 V
4
Hence (D) is correct option.
SOL 2.80
At ω = 0
and at ω = 3,
jωC
I2 = Em +01c = Em +0c
1 + jωCR2
R2 + jωC
+90c
+I2 =
+ tan−1 ωCR2
E m ωC
I2 =
+ (90c − tan−1 ωCR2)
2 2
2
1 + ω C R2
I2 = 0
I2 = Em
R2
Only fig. given in option (A) satisfies both conditions.
Hence (A) is correct option.
SOL 2.81
Xs = ωL = 10 Ω
For maximum power transfer
RL = Rs2 + Xs2 =
Hence (A) is correct option.
102 + 102 = 14.14 Ω
SOL 2.82
Applying KVL in LHS loop
E1 = 2I1 + 4 (I1 + I2) − 10E1
or
E1 = 6I1 + 4I2
11
11
Thus z11 = 6
11
Applying KVL in RHS loop
E2 = 4 (I1 + I2) − 10E1
= 4 (I1 + I2) − 10 c 6I1 + 4I2 m
11
11
=− 16I1 + 4I2
11
11
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Thus z21 =− 16
11
Hence (C) is correct option.
SOL 2.83
At ω = 0 , circuit act as shown in figure below.
V0 = RL
Vs
RL + Rs
(finite value)
At w = 3 , circuit act as shown in figure below:
V0 = RL
Vs
RL + Rs
At resonant frequency ω =
V0 = 0 .
(finite value)
1 circuit acts as shown in fig and
LC
Thus it is a band reject filter.
Hence (D) is correct option.
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SOL 2.84
Applying KCL we get
iL = eat + ebt
Now
V (t) = vL = L diL = L d [eat + ebt] = aeat + bebt
dt
dt
Hence (D) is correct option.
SOL 2.85
Going from 10 V to 0 V
10 + 5 + E + 1 = 0
or
E =− 16 V
Hence (A) is correct option.
SOL 2.86
This is a reciprocal and linear network. So we can apply reciprocity
theorem which states “Two loops A & B of a network N and if an
ideal voltage source E in loop A produces a current I in loop B ,
then interchanging positions an identical source in loop B produces
the same current in loop A. Since network is linear, principle of
homogeneity may be applied and when volt source is doubled, current
also doubles.
Now applying reciprocity theorem
i = 2 A for 10 V
V = 10 V, i = 2 A
V =− 20 V, i =− 4 A
Hence (C) is correct option.
SOL 2.87
Tree is the set of those branch which does not make any loop and
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connects all the nodes.
abfg is not a tree because it contains a loop l node (4) is not connected
Hence (C) is correct option.
SOL 2.88
For a 2-port network the parameter h21 is defined as
h21 = I2
I1 V = 0 (short circuit)
2
Applying node equation at node a we get
Va − V1 + Va − 0 + Va − 0 = 0
R
R
R
3Va = V1
& Va = V1
3
Now
V1 − V1
V
3 = 2V1
1 − Va
I1 =
=
R
R
3R
and
0 − V1 − V
−
0
V
3 =
a
1
I2 =
=
R
R
3R
Thus
I2
I1
V2 = 0
= h21 =
− V1 /3R − 1
=
2
2V1 /3R
Hence (A) is correct option.
SOL 2.89
Applying node equation at node A
Vth − 100 (1 + j0) Vth − 0
=0
+
3
4j
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or
or
By simplifying
4jVth − 4j100 + 3Vth = 0
Vth (3 + 4j) = 4j100
4j100
Vth =
3 + 4j
Vth =
4j100
3 − 4j
3 + 4j # 3 − 4j
Vth = 16j (3 − j4)
Hence (A) is correct option.
SOL 2.90
For maximum power transfer RL should be equal to RTh at same
terminal.
so, equivalent Resistor of the circuit is
Req = 5Ω 20Ω + 4Ω
Req = 5.20 + 4 = 4 + 4 = 8 Ω
5 + 20
Hence (C) is correct option.
SOL 2.91
Delta to star conversion
Rab Rac
= 5 # 30 = 150 = 3 Ω
R1 =
50
Rab + Rac + Rbc 5 + 30 + 15
R2 =
Rab Rbc
= 5 # 15 = 1.5 Ω
Rab + Rac + Rbc 5 + 30 + 15
R3 =
Rac Rbc
= 15 # 30 = 9 Ω
Rab + Rac + Rbc 5 + 30 + 15
Hence (D) is correct option.
SOL 2.92
No. of branches = n + l − 1 = 7 + 5 − 1 = 11
Hence (C) is correct option.
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SOL 2.93
In nodal method we sum up all the currents coming & going at
the node So it is based on KCL. Furthermore we use ohms law to
determine current in individual branch. Thus it is also based on ohms
law.
Hence (B) is correct option.
SOL 2.94
Superposition theorem is applicable to only linear circuits.
Hence (A) is correct option.
SOL 2.95
Hence (B) is correct option.
SOL 2.96
For reciprocal network y12 = y21 but here y12 =− 12 ! y21 = 12 . Thus
circuit is non reciprocal. Furthermore only reciprocal circuit are
passive circuit.
Hence (B) is correct option.
SOL 2.97
Taking b as reference node and applying KCL at a we get
Vab − 1 + Vab = 3
2
2
or
or
Vab − 1 + Vab = 6
Vab = 6 + 1 = 3.5 V
2
Hence (C) is correct option.
SOL 2.98
Hence (A) is correct option.
SOL 2.99
The given figure is shown below.
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Applying KCL at node a we have
I = i 0 + i1 = 7 + 5 = 12 A
Applying KCL at node f
I =− i 4
so
i 4 =− 12 amp
Hence (B) is correct option.
SOL 2.100
so
V = 3 − 0 = 3 volt
Hence (A) is correct option.
SOL 2.101
Can not determined V without knowing the elements in box.
Hence (D) is correct option.
SOL 2.102
The voltage V is the voltage across voltage source and that is 10 V.
Hence (A) is correct option.
SOL 2.103
Voltage across capacitor
−t
VC (t) = VC (3) + (VC (0) − VC (3)) e RC
Here VC (3) = 10 V and (VC (0) = 6 V. Thus
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−t
Now
−t
−t
VC (t) = 10 + (6 − 10) e RC = 10 − 4e RC = 10 − 4e 8
VR (t) = 10 − VC (t)
−t
−t
= 10 − 10 + 4e RC = 4e RC
Energy absorbed by resistor
−t
2
−t
3 V R (t)
3 16e 4
3
E #
= #
= # 4e 4 = 16 J
4
R
0
0
0
Hence (B) is correct option.
SOL 2.104
It is a balanced whetstone bridge
R1 R 3
b R2 = R 4 l
so equivalent circuit is
Zeq = (4Ω 8Ω) = 4 # 8 = 8
3
4+8
Hence (B) is correct option.
SOL 2.105
Current in A2 ,
I2 = 3 amp
Inductor current can be defined as I2 =− 3j
Current in A 3 ,
I3 = 4
Total current
I1 = I 2 + I 3
I1 = 4 − 3j
I = (4) 2 + (3) 2 = 5 amp
Hence (B) is correct option.
SOL 2.106
For a tree we have (n − 1) branches. Links are the branches which
from a loop, when connect two nodes of tree.
so if total no. of branches = b
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No. of links = b − (n − 1) = b − n + 1
Total no. of links in equal to total no. of independent loops.
Hence (C) is correct option.
SOL 2.107
In the steady state condition all capacitors behaves as open circuit &
Inductors behaves as short circuits as shown below :
Thus voltage across capacitor C1 is
VC = 100 # 40 = 80 V
10 + 40
1
Now the circuit faced by capacitor C2 and C 3 can be drawn as below :
Voltage across capacitor C2 and C 3 are
VC = 80 C 3 = 80 # 3 = 48 volt
5
C2 + C3
2
VC = 80
3
C2 = 80 2 = 32 volt
#5
C2 + C3
Hence (B) is correct option.
***********
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