Network Theory

Question 1
For a 2-port network consisting of an ideal lossless transformer, the parameter S_{21} (rounded off to two decimal places) for a reference impedance of 10 \Omega, is _____.
A
1
B
0.5
C
1.5
D
0.8
GATE EC 2020      Two Port Networks
Question 1 Explanation: 
For ideal transformer of n : 1, the scattering matrix is
\begin{bmatrix} S_{11} &S_{12} \\ S_{21} & S_{22} \end{bmatrix}=\begin{bmatrix} \frac{n^{2}-1}{n^{2}+1} &\frac{2n}{n6{}} \\ & \end{bmatrix}
S_{21}=\frac{2n}{n^{2}+1}=\frac{2(2)}{2^2+1}=\frac{4}{5}=0.8
Question 2
The current I in the given network is
A
0 A
B
2.38 \angle -96.37^{\circ} A
C
2.38 \angle 143.63^{\circ} A
D
2.38 \angle -23.63^{\circ} A
GATE EC 2020      Basics of Network Analysis
Question 2 Explanation: 


I=-[I_{1}+I_{2}]
I=-\left [ \frac{120\angle -90^{\circ}}{80-j35}+\frac{120\angle -30^{\circ}}{80-j35} \right ]
I=2.38\angle 143.7^{\circ}
Question 3
The current in the RL-circuit shown below is i(t) = 10cos(5t-\pi /4)A. The value of the inductor (rounded off to two decimal places) is _________ H.
A
4.82
B
5.24
C
2.83
D
1.32
GATE EC 2020      Transient Analysis
Question 3 Explanation: 
Z=\frac{V}{I}=\frac{200\angle 0^{\circ}}{10\angle -45^{\circ}}=20\angle 45^{\circ}
Z=10\sqrt{2}+j10\sqrt{2}
X_{L}=10\sqrt{2}
\omega L=10\sqrt{2}
L=\frac{10\sqrt{2}}{5}=2.828\, H
Question 4
In the given circuit, the two-port network has the impedance matrix[Z] =\begin{bmatrix} 40 & 60\\ 60 & 120 \end{bmatrix}. The value of Z_L for which maximum power is transferred to the load is _________ \Omega.
A
48
B
38
C
40
D
54
GATE EC 2020      Two Port Networks
Question 4 Explanation: 
From maximum power transfer theorem
Z_{L}=Z_{th}
Z_{th}=Z_{22}-\frac{Z_{12}\times Z_{21}}{R_{S}+Z_{11}}
For given data,
Z_{th}=120-\frac{60\times 60}{10+40}=48\Omega
Z_{L}=48\Omega
Question 5
In the circuit shown below, the Thevenin voltage V_{TH} is
A
2.4 V
B
2.8 V
C
3.6 V
D
4.5 V
GATE EC 2020      Network Theorems
Question 5 Explanation: 
By applying the Source Transformation
Question 6
The RC circuit shown below has a variable resistance R(t) given by the following expression:

R(t)=R_0\left ( 1-\frac{t}{T} \right ) \; for \;0\leq t \lt T

where R_0=1\Omega, and C=1F. We are also given that T=3R_0C and the source voltage is V_s=1V. If the current at time t=0 is 1A, then the current I(t), in amperes, at time t=T/2 is ___________(rounded off to 2 decimal places).
A
0.25
B
0.12
C
0.68
D
0.72
GATE EC 2019      Transient Analysis
Question 6 Explanation: 
\begin{array}{c} T=3 R_{0} C=3 \mathrm{sec} \\ R(t)=\left(1-\frac{t}{3}\right) ; 0 \leq t \leq 3 \mathrm{sec} \end{array}


\begin{aligned} R(t) i(t)+\frac{1}{C} \int i(t) d t&=1 \\ \left(1-\frac{t}{3}\right) i(t)+\int i(t) d t&=1 \end{aligned}
Differentiating both sides, we get
\begin{aligned} \left(1-\frac{t}{3}\right) \frac{d i}{d t}-\frac{i}{3}+i &=0 \\ (3-t) \frac{d i}{d t}+2 i &=0 \\ \frac{d i}{i} &=-\frac{2}{(3-t)} d t \end{aligned}
Integrating on both sides, we get,
\begin{aligned} \ln (i)&=2 \ln (3-t)+\ln (c)\\ i(t)&=c(3-t)^{2} ; t \geq 0\\ \text{Given that,} i(0)&=1 A.\\ \text{So,}\quad c(3-0)^{2} &=1 \mathrm{A} \\ c &=\frac{1}{9} \mathrm{A} \\ i(t) &=\frac{1}{9}(3-t)^{2} \mathrm{A}\\ \text{At }t&=\frac{T}{2}=1.5 \mathrm{sec} \\ i(1.5)&=\frac{1}{9}(1.5)^{2}=0.25 A \end{aligned}
Question 7
In the circuit shown, if v(t)=2sin(1000t) volts, R=1k\Omega and R=C=1\mu F, then the steady-state current i(t), in milliamperes (mA), is
A
sin(1000t) + cos(1000t)
B
2sin(1000t) + 2cos(1000t)
C
3sin(1000t) + cos(1000t)
D
sin(1000t) + 3cos(1000t)
GATE EC 2019      Sinusoidal Steady State
Question 7 Explanation: 


\begin{aligned} \text{Here,}X_{C}&=\frac{1}{\omega C}=\frac{1}{10^{3} \times 10^{-6}}=\frac{1}{10^{-3}}\\ x_{C} &=10^{3} \Omega \\ R &=10^{3} \Omega \quad \text { (Given) } \\ v(t) &=2 \sin 1000 t \vee=2 \angle 0^{\circ} \mathrm{V} \end{aligned}
Redrawing the given network, we get,


As the bridge is balanced, it can be redrawn as


\begin{aligned} \therefore \quad Y_{e q} &=Y_{1}+Y_{2} \\ &=\frac{3}{2} R+\frac{1}{-2 j X_{C}} \\ &=\frac{3}{2} \times 10^{-3}+j \frac{1}{2} \times 10^{-3} \\ \therefore \quad i(t) &=(t) \times Y_{\mathrm{eq}}=2 \angle 0^{\circ}\left[\frac{3}{2}+j \frac{1}{2}\right] \mathrm{m} \mathrm{A} \\ &=(3+j 1) \mathrm{mA} \\ &=3 \sin (1000 t)+\cos (1000 t) \mathrm{mA} \end{aligned}
Question 8
Consider the two-port resistive network shown in the figure. When an excitation of 5 V is applied across Port 1, and Port 2 is shorted, the current through the short circuit at Port 2 is measured to be 1 A (see (a) in the figure).
Now, if an excitation of 5 V is applied across Port 2, and Port 1 is shorted (see(b) in the figure), what is the current through the short circuit at Port 1?
A
0.5 A
B
1.0 A
C
2.0 A
D
2.5 A
GATE EC 2019      Network Theorems
Question 8 Explanation: 
According to reciprocity theorem,
In a linear bilateral single source network the ratio of response to excitation remains the same even after their positions get interchanged.
\therefore \quad \frac{I}{5}=\frac{1}{5} \Rightarrow I=1 \mathrm{A}
Question 9
Consider the network shown below with R_{1} = 1\Omega , R_{2} = 2\Omega \; and \; R_{3} = 3\Omega . The network is connected to a constant voltage source of 11V.

The magnitude of the current (in amperes, accurate to two decimal places) through the source is _______.
A
5
B
8.0
C
8.7
D
9.5
GATE EC 2018      Basics of Network Analysis
Question 9 Explanation: 


As the network is symmetric,
V_{A}= V_{B} and V_{C}= V_{D}
So current throught R_{2} resistor is zero and V_{A}=V_{B} and V_{C}=V_{D},electrically the circuit can reduced as,


Total resistance,
\begin{aligned} R_{T} &=2\left(R_{1} \| R_{1}\right)+\left(R_{1}\left\|R_{1}\right\| R_{3} \| R_{3}\right) \\ &=R_{1}+\left(\frac{R_{1}}{2} \| \frac{R_{3}}{2}\right) \end{aligned}
Given that, R_{1}=1 \Omega and R_{3}=3 \Omega
\begin{aligned} \text{So,}\qquad R_{T} &=1+\left(\frac{1}{2} \| \frac{3}{2}\right) \Omega=1+\frac{3 / 2}{4}=\frac{11}{8} \Omega \\ I &=\frac{11 \mathrm{V}}{R_{T}}=\frac{11}{(11 / 8)}=8 \mathrm{A} \end{aligned}
Question 10
For the circuit given in the figure, the magnitude of the loop current (in amperes, correct to three decimal places) 0.5 second after closing the switch is _______.
A
0.3
B
0.4
C
0.5
D
0.2
GATE EC 2018      Transient Analysis
Question 10 Explanation: 
\begin{aligned} \text { Loop current, } i(t)&=\frac{1}{1+1}\left(1-e^{-t / \tau}\right) \mathrm{A} ; t \gt 0 \\ \tau&=\frac{L}{R_{e q}}=\frac{1}{1+1}=\frac{1}{2} \mathrm{sec} \\ i(t)&=\frac{1}{2}\left(1-e^{-2 t}\right) \mathrm{A} ; t \gt 0\\ \text{At }t=0.5 \mathrm{sec} \\ i(t)&=\frac{1}{2}\left(1-e^{-1}\right) \mathrm{A}=0.316 \mathrm{A} \end{aligned}


There are 10 questions to complete.
Like this FREE website? Please share it among all your friends and join the campaign of FREE Education to ALL.