Basics of Network Analysis

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Question 1

Consider the circuit shown in the figure. The current

I

flowing through the

10\Omega

resistor is _________.

A	1A
B	0A
C	0.1A
D	-0.1A

GATE EC 2022 Network Theory

Question 1 Explanation:

Here, there is no any return closed path for Current (I) . Hence I=0
Current always flow in loop.

Question 2

The current

I

in the circuit shown is ________

A	$1.25 \times 10^{-3}A$
B	$0.75 \times 10^{-3}A$
C	$-0.5 \times 10^{-3}A$
D	$1.16 \times 10^{-3}A$

GATE EC 2022 Network Theory

Question 2 Explanation:

Applying Nodal equation at Node-A

\begin{aligned} \frac{V_A}{2k}+\frac{V_A-5}{2k}&=10^{-3}\\ \Rightarrow 2V_A-5&=2k \times 10^{-3}\\ V_A&=3.5V\\ Again,&\\ I&=\frac{5-V_A}{2k}\\ &=\frac{5-3.5}{2k}\\ &=0.75 \times 10^{-3}A \end{aligned}

Question 3

Consider the circuit shown in the figure.

The value of

v_{0}

(rounded off to one decimal place) is _________ V.

A	0.5
B	0.8
C	2
D	1

GATE EC 2021 Network Theory

Question 3 Explanation:

Write KVL equation first loop,

\begin{aligned} V_{o}-4+1\left(V_{o}+2\right) &=0 \\ V_{o} &=1 \mathrm{~V} \end{aligned}

Question 4

Consider the circuit shown in the figure.

The current I flowing through the

7\:\Omega

resistor between P and Q (rounded off to one decimal place) is ________ A.

A	0.2
B	0.5
C	0.8
D	1.4

GATE EC 2021 Network Theory

Question 4 Explanation:

Redraw the circuit,

\begin{aligned} 3 \Omega \;||\; 6 \Omega&=2 \Omega \\ 2 \Omega \;||\; 2 \Omega&=1 \Omega \\ \qquad I&=5 \times \frac{1}{10}=0.5 \mathrm{~A} \end{aligned}

Question 5

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 Network Theory

Question 5 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 6

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 Network Theory

Question 6 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 7

A connection is made consisting of resistance A in series with a parallel combination of resistances B and C. Three resistors of value 10

\Omega

, 5

\Omega

, 2

\Omega

are provided. Consider all possible permutations of the given resistors into the positions A, B, C, and identify the configurations with maximum possible overall resistance, and also the ones with minimum possible overall resistance. The ratio of maximum to minimum values of the resistances (up to second decimal place) is ____________.