# 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$\mu$F 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.