Network Theorems - PracticePaper

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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