Electronics and Communication Engineering - Networks
Exam Duration: 45 Mins Total Questions : 30
Find the current I flowing in the given figure.
- (a)
1 A
- (b)
2 A
- (c)
3 A
- (d)
4 A
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
In the circuit of figure, the power absorbed by the load RL is
- (a)
2 W
- (b)
4 W
- (c)
6 W
- (d)
8 W
Which of the following is not a dual of the other?
- (a)
G \(\rightarrow \) R
- (b)
KCL \(\rightarrow \) KVL
- (c)
CCVS \(\rightarrow \) VCVS
- (d)
\(C\frac { d{ V }_{ C }(t) }{ dt } \rightarrow L\frac { dI_{ L }(t) }{ dt } \)
Which of the following truly represents the Thevenin's equivalent circuit when a voltage source of 24 V undergoes a voltage drop of 0.6 V due to a load current of 1 A?
- (a)
\({ V }_{ Th }=24\quad V,{ R }_{ Th }=0.6\Omega \)
- (b)
\({ V }_{ Th }=24\quad V,{ R }_{ Th }=24\Omega \)
- (c)
\({ V }_{ Th }=23.4\quad V,{ R }_{ Th }=0.6\Omega \)
- (d)
\({ V }_{ Th }=23.4\quad V,{ R }_{ Th }=23.4\Omega \)
Which of the following is not a condition for maximum power transfer across a load ZL<\({ \theta }_{ L }\)in an AC Thevenin equivalent circuit of voltage VTh< 0\(°\) and ZTh< \(\theta _{ Th }\)
- (a)
ZL = ZTh
- (b)
\({ \theta }_{ L }\) = -\(\theta \)Th
- (c)
\(\)\(\theta \)L=\(\theta \)Th
- (d)
All of these
In the circuit shown in figure the effective resistance faced by the voltage source is
- (a)
4 \(\Omega \)
- (b)
3 \(\Omega \)
- (c)
2 \(\Omega \)
- (d)
1 \(\Omega \)
The circuit is shown in figure below
The current ratio transfer function \(\frac { { I }_{ 0 } }{ { I }_{ s } } \)is
- (a)
\(\frac { s(s+4) }{ { s }^{ 2 }+3s+4 } \)
- (b)
\(\frac { s(2s+2) }{ (s+1)(2s+3) } \)
- (c)
\(\frac { { s }^{ 2 }+3s+4 }{ s(s+4) } \)
- (d)
\(\frac { (s+1)(s+3) }{ s(s+4) } \)
The natural response of an RLC circuit is described by the differential equation
\(\frac { { d }^{ 2 }V(t) }{ { dt }^{ 2 } } +\frac { 2dV(t) }{ dt } +V(t)=0,V(0)=10,\frac { dv(0) }{ dt } =0,the\quad V(t)\) is
- (a)
10 (1+t) e-t V
- (b)
10 (1-t) e-t V
- (c)
10 e-t V
- (d)
10 te-t V
In the circuit of figure for positive time constant the range of \(\alpha\) is
- (a)
\(\alpha\) > 3.33
- (b)
\(\alpha\) < 3.33
- (c)
\(\alpha\) < 0.03
- (d)
\(\alpha\) > 0.03
The resonant frequency of the given series circuit is
- (a)
\(\frac { 1 }{ 2\pi \sqrt { 3 } } \) Hz
- (b)
\(\frac { 1 }{ 4\pi \sqrt { 3 } } \)Hz
- (c)
\(\frac { 1 }{ 4\pi \sqrt { 2 } } \)Hz
- (d)
\(\frac { 1 }{ \pi \sqrt { 2 } } \)Hz
The short-circuit admittance marix of a two-port network is
\(\begin{bmatrix} 0 & -1/2 \\ 1/2 & 0 \end{bmatrix}\)
The two-port network is
- (a)
non - reciprocal and passive
- (b)
non-reciprocal and active
- (c)
reciprocal and passive
- (d)
reciprocal and active
Given that F(s) is the one-sided Laplace transform of f(t), the Laplace transform of \(\int _{ 0 }^{ t }{ f(\tau )d\tau } \) is
- (a)
sF(s) - f(0)
- (b)
\(\frac { 1 }{ s } F(s)\)
- (c)
\(\int _{ 0 }^{ s }{ f(\tau )d\tau } \)
- (d)
\(\frac { 1 }{ s } [F(s)-f(0)]\)
In the following circuit the voltage Va is
- (a)
- 11 V
- (b)
11 V
- (c)
3 V
- (d)
- 3 V
The natural frequency of the circuit is given by
- (a)
- 6
- (b)
- 8
- (c)
\(\frac { 15 }{ 2 } \)
- (d)
- 10
In the circuit of figure the voltage V(t) is
- (a)
eat - ebt
- (b)
eat - ebt
- (c)
aeat - bebt
- (d)
aeat + be bt
In the circuit shown below, if f = 50 Hz then, current I(t) is
- (a)
14.14 cos (52\(\pi \)t - 45\(°\))A
- (b)
-11.4 cos (100t + 52.15\(°\)) A
- (c)
1.414 cos (100\(\pi \)t + 45\(°\))A
- (d)
1.414 cos (100t + 45\(°\))A
In the given circuit below, voltage Va is
- (a)
2 V
- (b)
(2 + 6j) V
- (c)
1 V
- (d)
-2 V
In the following circuit, the value of Z(s) is
- (a)
\(\frac { 3{ s }^{ 2 }+8s+7 }{ s(5s+6) } \)
- (b)
\(\frac { s(5s+6) }{ 3{ s }^{ 2 }+8s+7 } \)
- (c)
\(\frac { 3s^{ 2 }+7s+6 }{ s(5s+6) } \)
- (d)
\(\frac { s(5s+6) }{ 3{ s }^{ 2 }+7s+6 } \)
In the following circuit the value of open circuit voltage and Thevenin resistance at terminal ab are
- (a)
Voc = 100 V, RTh = 1800 \(\Omega\)
- (b)
Voc =0 V. TTh = 270 \(\Omega\)
- (c)
Voc = 100 V, RTh = 90 \(\Omega\)
- (d)
Voc = 0 V,RTh = 90 \(\Omega\)
Two series resonant filters are as shown in the figure. Let the 3 dB bandwidth of filter 1 be B1 and the of filter 2 be B2. The value of \(\frac { { B }_{ 1 } }{ { B }_{ 2 } } \)is
- (a)
4
- (b)
1
- (c)
\(\frac { 1 }{ 2 } \)
- (d)
\(\frac { 1 }{ 4 } \)
In the circuit shown below, the current excitation is Iin(t) = 4u (-t) A. The IL(t) for t\(\ge \)0 is
- (a)
4e-5000t A
- (b)
2e-5000t A
- (c)
6 - 4e-5000t A
- (d)
6e-5000t A
Both capacitors shown in figure are initially uncharged. At t = 2 ms the current \({ I }_{ C_{ 1 } }\) is
- (a)
0.25 mA
- (b)
1 mA
- (c)
0.75 mA
- (d)
3 mA
For the circuit shown in figure, Thevenin.s voltage and Thevenin's equivalent resistance at terminals ab is
- (a)
5 V and 2 \(\Omega\)
- (b)
7.5 V and 2.5 \(\Omega\)
- (c)
4 V and 2 \(\Omega\)
- (d)
3 V and 2.5 \(\Omega\)
In the graph shown below soild lines are twigs and dotted lines are links.
The fundamental loop matrix is
- (a)
\(\begin{bmatrix} 1 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & -1 & -1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0 \\ -1 & 0 & -1 & 0 & -1 & 0 & 0 & 0 & 1 \end{bmatrix}\)
- (b)
\(\begin{bmatrix} -1 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & -1 & -1 & -1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & -1 & 1 & -1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 \end{bmatrix}\)
- (c)
\(\begin{bmatrix} 1 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & -1 & 1 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 \end{bmatrix}\)
- (d)
\(\begin{bmatrix} 1 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 1 & -1 & 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0 \\ -1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 1 \end{bmatrix}\)
In the following circuit when R = 0 \(\Omega\), the current IR equals 10 A
The value of R, for which it absorbs maximum power, is
- (a)
4 \(\Omega\)
- (b)
3 \(\Omega\)
- (c)
2 \(\Omega\)
- (d)
None of these
The circuit is as shown below.
In the circuit shown below, the current I1(t) is
- (a)
2.36 cos (4t - 41.07 \(°\)) A
- (b)
2.36 cos (4t + 41.07\(°\)) A
- (c)
1.37 cos (4t - 41.07\(°\)) A
- (d)
2.36 cos (4t + 41.07 \(°\))A
In a balanced Y-connected three phase generator, Vab = 400 Vrms
If phase sequence is acb, then phase voltage are
- (a)
\(231<0°,231<120°,231<240°\)
- (b)
\(231<-30°,213<-150°,231<90°\)
- (c)
\(231<30°,231<150°,231<-90°\)
- (d)
\(231<60°,231<180°,231<-60°\)
Consider the circuit shown below. All initial conditions are zero.
If Iin(t) =4\(\delta \)(t) then Io(t) will be
- (a)
4\(\delta \)(t)-e-t u(t) A
- (b)
4\(\delta \) (t) - 4e-t u(t) A
- (c)
4e-t u(t) - 4\(\delta \) (t) A
- (d)
e-t u(t) -\(\delta \) (t) A
The circuit is as shown below. solve the problem and choose correct option.
The transfer function H1(s)=\(\frac { { I }_{ 0 }(s) }{ { V }_{ s }(s) } \) is
- (a)
s(s3 + 2s2 + 3s +1)-1
- (b)
(s3 + 3s2 + 2s + 1)-1
- (c)
(s3 + 2s2 + 3s + 2)-1
- (d)
s(s3 + 3s2 + 2s2 +2 )-1
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