# Basics of Network Analysis

 Question 1
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 1 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 2
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 C 8.7 D 9.5
GATE EC 2018   Network Theory
Question 2 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 3
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 ____________.
 A 0.466 B 2.14 C 16.758 D 6.098
GATE EC 2017-SET-2   Network Theory
Question 3 Explanation:
The connection of resistors is as shown below:

Given resistor values are :$10 \Omega, 5 \Omega, 2 \Omega$
The maximum resistance possible is,
\begin{aligned} R_{T(\max )} &=10 \Omega+(5 \Omega \| 2 \Omega) \\ &=\left(10+\frac{10}{7}\right) \Omega=\frac{80}{7} \Omega \end{aligned}
The minimum resistance possible is,
\begin{aligned} R_{T(\min )} &=2 \Omega+(10 \Omega \| 5 \Omega) \\ &=\left(2+\frac{10}{3}\right) \Omega=\frac{16}{3} \Omega \\ \frac{R_{T(\max )}}{R_{T(\min )}} &=\frac{80 / 7}{16 / 3}=\frac{15}{7}=2.143 \end{aligned}
 Question 4
In the figure shown, the current i (in ampere) is __________
 A 0 B -1 C -2 D -4
GATE EC 2016-SET-3   Network Theory
Question 4 Explanation:

Using KCL at $V_{1}$
$\frac{V_{1}}{1}+\frac{V_{1}-8}{1}+\frac{V_{1}-8}{1}+\frac{V_{1}}{1}=0$
or $V_{1}=4 V$
considering KVL, we get

 Question 5
In the circuit shown in the figure, the magnitude of the current (in amperes) through $R_{2}$ is ___
 A 2 B 3 C 4 D 5
GATE EC 2016-SET-2   Network Theory
Question 5 Explanation:

Using KVL in the outer loop
$60-5\left(0.16 V_{x}\right)-\frac{V_{x}}{5} \times 3-V_{x}=0$
$\text{or}\quad v_{x}=25 \mathrm{V}$
$\therefore$ The current flowing through
$R_{2}=\frac{V_{x}}{5}=\frac{25}{5}=5 \mathrm{A}$
 Question 6
In the given circuit, each resistor has a value equal to 1 $\Omega$.

What is the equivalent resistance across the terminals a and b?
 A 1/6 $\Omega$ B 1/3 $\Omega$ C 9/20 $\Omega$ D 8/15 $\Omega$
GATE EC 2016-SET-2   Network Theory
Question 6 Explanation:
$R_{eq}\Rightarrow$

By using delta to star conversion

Again by star to delta conversion

\begin{aligned} R_{\mathrm{ab}} &=\{4 \| 1)+(4 \| 1)\}\|\{1 \| 4)\} \\ &=\left(\frac{4}{5}+\frac{4}{5}\right) \| \frac{4}{5}=\frac{8}{15} \Omega \end{aligned}
 Question 7
An AC voltage source V = 10 sin(t) volts is applied to the following network. Assume that $R_{1}=3k \Omega$, $R_{2}=6k \Omega$ and $R_{3}=9k \Omega$, and that the diode is ideal.

RMS current $I_{rms}$ (in mA) through the diode is ________
 A 0.5 B 1 C 1.5 D 2
GATE EC 2016-SET-1   Network Theory
Question 7 Explanation:

The equivalent resistance across terminal ah ( outer loop) is

\begin{aligned} V &=\frac{I}{3} \times 3 \mathrm{k} \Omega+\frac{I}{6} \times 6 \mathrm{k} \Omega+\frac{I}{3} \times 9 \mathrm{k} \Omega \\ V &=5 I \\ \text{or}\quad \frac{V}{I} &=5 \mathrm{k} \Omega \end{aligned}
For half wave rectifier
\begin{aligned} I_{\mathrm{rms}} &=\frac{I_{m}}{(2)}=\frac{10 \sin t}{5 \mathrm{k} \Omega}=2 \sin t \mathrm{m} \mathrm{A} \\ \therefore \quad I_{\mathrm{rms}} &=\frac{I_{m}}{2}=1 \mathrm{m} \mathrm{A} \end{aligned}
 Question 8
In the circuit shown, the voltage $V_{X}$ (in Volts) is _______.
 A 7 B 8 C 9 D 10
GATE EC 2015-SET-3   Network Theory
Question 8 Explanation:

\begin{aligned} \frac{V_{x}}{20}+\frac{V_{x}-V_{y}}{10}+0.5 V_{x}&=5 \mathrm{A} \\ V_{x}\left[\frac{1}{20}+\frac{1}{10}+0.5\right]&=5+\frac{V_{y}}{10} \\ 13 \mathrm{V}_{x}&=100+2 \mathrm{V}_{y} \quad\ldots(i)\\ \text { and also, } V_{y}&=0.25 \mathrm{V}_{x} \quad\dots(ii) \end{aligned}
Solving equations (i) and (ii), we have
\begin{aligned} 52 V_{y} &=100+2 V_{y} \\ 50 V_{y} &=100 \Rightarrow V_{y}=2 \mathrm{V} \\ V_{x} &=4 V_{y}=8 \mathrm{V} \end{aligned}
 Question 9
In the given circuit, the values of $V_{1} \; and \; V_{2}$ respectively are

 A 5 V, 25 V B 10 V,30 V C 15 V,30 V D 0 V , 20V
GATE EC 2015-SET-1   Network Theory
Question 9 Explanation:

Current flowing through both the parallel $4 \Omega$ will
be I.
$\text { So, } V_{2}=4(I+I+2 I)+4 I \quad \text { by KVL }$
$I+I+2 I=5 \quad \text{by KCL}$
\begin{aligned} I &=\frac{5}{4} A \\ V_{2} &=4 \times 5+\frac{4 \times 5}{4}=25 \mathrm{V} \\ V_{1} &=4 I=\frac{4 \times 5}{4}=5 \mathrm{V} \end{aligned}
 Question 10
In the network shown in the figure, all resistors are identical with R = 300 $\Omega$. The resistance $R_{ab}$ (in $\Omega$) of the network is ______.

 A 95 B 100 C 110 D 120
GATE EC 2015-SET-1   Network Theory
Question 10 Explanation:
Modifying the given circuit

\begin{aligned} R_{a b} &=\left(\frac{1}{2 R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{2 R}\right)^{-1} \\ &=\frac{R}{3}=\frac{300}{3}=100 \Omega \end{aligned}
There are 10 questions to complete.