Network Theorems

Question 1
In the given circuit, for maximum power to be delivered to R_{L}, its value should be ______________ \Omega.
(Round off to 2 decimal places.)

A
2.14
B
3.32
C
1.42
D
4.12
GATE EE 2021   Electric Circuits
Question 1 Explanation: 


=\frac{-j}{\omega C}=\frac{-j}{1000 \times 0.5 \times 10^{-3}}=-j 2 \Omega


Z_{\text {in }}=2 \| j 2=\frac{j 4}{2+j 2}=\frac{j 2}{1+j 1}
For maximum power transfer,
R_{L}=\left|Z_{\mathrm{TH}}\right|=\frac{2}{\sqrt{2}}=\sqrt{2}=1.414 \Omega
Question 2
A signal generator having a source resistance of 50\:\Omega is set to generate a \text{1 kHz} sinewave. Open circuit terminal voltage is \text{10 V} peak-to-peak. Connecting a capacitor across the terminals reduces the voltage to \text{8 V} peak-to-peak. The value of this capacitor is _____________\mu F. (Round off to 2 decimal places.)
A
4.25
B
6.32
C
1.25
D
2.38
GATE EE 2021   Electric Circuits
Question 2 Explanation: 
\begin{aligned} \frac{V_{0}(j \omega)}{V_{i}(j \omega)} &=\frac{1}{1+j \omega R C} \\ \frac{8}{10} \angle \theta &=\frac{1}{1+j \omega R C} \\ 1+j \omega R C &=1.25 \angle \theta \\ \omega R C &=\sqrt{1.25^{2}-1^{2}} \\ \omega R C &=0.75 \\ C &=\frac{0.75}{2000 \pi \times 50}=2.38 \mu\mathrm{F} \end{aligned}
Question 3
In the given circuit, for voltage V_y to be zero, the value of \beta should be _________. (Round off to 2 decimal places).

A
-4.58
B
-3.25
C
4.65
D
3.45
GATE EE 2021   Electric Circuits
Question 3 Explanation: 
\begin{aligned} \frac{V_{x}-6}{1}+\frac{V_{x}}{4}+\frac{V_{x}-V_{y}}{2} & =0 \\ 4 V_{x}-24+V_{x}+2 V_{x}-2 V_{y} & =0 \\ 7 V_{x}-2 V_{y} & =24 \\ \text { If } V_{y}=0 \qquad\qquad 7 V_{x} & =24 \\ \Rightarrow\qquad\qquad V_{x} & =\frac{24}{7} \\ \Rightarrow\qquad\qquad \frac{V_{y}-V_{x}}{2}+\frac{V_{y}-\beta V_{x}}{3} & =2 \\ 3 V_{y}-3 V_{x}+2 V_{y}-2 \beta V_{x} & =12 \\ 5 V_{y}-(3+2 \beta) V_{x} & =12 \\ V_{y} & =0\\ -(3+2 \beta) \frac{24}{7} &=12 \\ (3+2 \beta) &=\frac{-7}{2} \\ 2 \beta &=\frac{-7}{2}-3=\frac{-7-6}{2}=\frac{-13}{2} \\ \beta &=\frac{-13}{4}=-3.25 \end{aligned}
Question 4
For the network shown, the equivalent Thevenin voltage and Thevenin impedance as seen across terminals 'ab' is

A
\text{10 V} in series with 12\:\Omega
B
\text{65 V} in series with 15\:\Omega
C
\text{50 V} in series with 2\:\Omega
D
\text{35 V} in series with 2\:\Omega
GATE EE 2021   Electric Circuits
Question 4 Explanation: 
Given circuit can be resolved as shown below,


V_{T H}=15+50=65 \mathrm{~V}


\begin{aligned} V_{x} &=2+3+10=15 \mathrm{~V} \\ R_{\mathrm{TH}} &=\frac{V_{x}}{1}=15 \Omega \end{aligned}
Question 5
A benchtop dc power supply acts as an ideal 4 A current source as long as its terminal voltage is below 10 V. Beyond this point, it begins to behave as an ideal 10 V voltage source for all load currents going down to 0 A. When connected to an ideal rheostat, find the load resistance value at which maximum power is transferred, and the corresponding load voltage and current.
A
Short, \infty A, 10 V
B
Open, 4 A, 0 V
C
2.5 \Omega, 4 A, 10 V
D
2.5 \Omega, 4 A, 5 V
GATE EE 2020   Electric Circuits
Question 5 Explanation: 



Maximum power transistor of VI product is maximum. If draw the the curve, it intersect (10, 4) that will give maximum power. The terminal voltage is 10 V (Load voltage) and current is 4 A (Load current). Load resistance is \frac{10}{4}=2.5\Omega .
Question 6
The Thevenin equivalent voltage, V_{TH}, in V (rounded off to 2 decimal places) of the network shown below, is _______ .
A
5
B
7
C
14
D
6
GATE EE 2020   Electric Circuits
Question 6 Explanation: 
Only voltage source 4V is there and current source 5A is open circuited

From the above circuit,
V_{TH1}=4V

Case-II:
Only current source 5A is there and voltage source 4V is short circuited.

From the above circuit,
V_{TH2}=2 \times 5=10V
By applying superposition theorem,
V_{TH}=V_{TH1}+V_{TH2}=10+4=14V
Question 7
x_R\; and \; x_A are, respectively, the rms and average values of x(t) = x(t - T), and similarly, y_R\; and \; y_A are, respectively, the rms and average values of y(t) = kx(t). k, \;T are independent of t. Which of the following is true?
A
y_A=kx_A;\; y_R=kx_R
B
y_A=kx_A;\; y_R\neq kx_R
C
y_A \neq kx_A;\; y_R= kx_R
D
y_A \neq kx_A;\; y_R\neq kx_R
GATE EE 2020   Electric Circuits
Question 7 Explanation: 
Given that,
\begin{aligned} x(t)&=x(t-T) \; \text{ i.e. periodic signal}\\ \text{Average of } x(t)&=x_{A}\\ \text{Rms of }x(t)&=x_{R} \\ \text{Average of }y(t)&=y_{A}\\ \text{Rms of } y(t)&=y_{R} \\ y(t)&=k_{x}(t_{0}) \; \; ....(i) \\ &\text{Using equation(i),}\\ \text{Average of }y(t)&=k \times \text{ Average of }x(t) \\ y_{A}&=kx_{A}\\ \text{Power of }y(t)&=|k|^{2}\text{ Power of }x(t) \\ Rms^{2} \text{ of } y(t)&=|k|^{2}\; Rms^{2}\text{ of }x(t) \\ y_{R}^{2}=|k|^{2}\cdot x_{R}^{2} \\ y_{R}=|k|x_{R}\end{aligned}
Question 8
The current flowing in the circuit shown below in amperes is _____
A
0
B
1
C
2
D
4
GATE EE 2019   Electric Circuits
Question 8 Explanation: 


By Millman'e theorem,
E=\frac{\frac{200}{50}+\frac{160}{40}-\frac{100}{25}-\frac{80}{20}}{\frac{1}{50}+\frac{1}{40}+\frac{1}{25}+\frac{1}{20}}=0V
\frac{1}{R}=\frac{1}{50}+\frac{1}{40}+\frac{1}{25}+\frac{1}{20}
Simplified circuit,
\therefore \;\;I=0A
Question 9
In the circuit shown below, the value of capacitor C required for maximum power to be transferred to the load is
A
1nF
B
1\muF
C
1mF
D
10mF
GATE EE 2017-SET-2   Electric Circuits
Question 9 Explanation: 


The frequency at which the load is resistive and it is equal to 0.5\Omega i.e. The load is resistive means, the imaginary part of load is equal to zero and real part is equal to 0.5\Omega.
Z_{load}=\frac{1 \times \frac{1}{Cs}}{1 + \frac{1}{Cs}}+Ls
\;\;=\frac{1}{1+Cs}+Ls
\;\; =\frac{(1-Cs)}{1-C^2s^2}+Ls
Put, s=j\omega;
Z_{load}=\frac{1-j\omega C}{1+C^2\omega ^2}+j\omega L
\;\;\;=\frac{1}{1+C^2\omega ^2}+j\omega \left ( L-\frac{C}{(1+C^2\omega ^2)} \right )
Real part of the
Z_{load}=\frac{1}{1+C^2\omega ^2}=0.5
Putting, \omega =100 r/s
we get, C=10mF
Question 10
For the network given in figure below, the Thevenin's voltage V_{ab} is
A
-1.5 V
B
-0.5 V
C
0.5V
D
1.5V
GATE EE 2017-SET-2   Electric Circuits
Question 10 Explanation: 
Consider the following circuit,

After rearrangement, we get,

From circuit using KCL,
\frac{V_{ab}+30}{15}+\frac{V_{ab}-8}{5}=0
V_{ab}+30+3V_{ab}-24=0
V_{ab}=-1.5V
There are 10 questions to complete.