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Transcript
1
Kulshreshtha, D. C.
@ McGraw-Hill Education
Basic Electrical Engineering
McGraw-Hill Education © 2010
BASIC ELECTRICAL ENGINEERING
D. C. KULSHRESHTHA,
PowerPoint Slides
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this PowerPoint slide may be displayed, reproduced or distributed in
any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for
their individual course preparation. If you are a student using this PowerPoint slide, you are using it without permission.
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2
Kulshreshtha, D. C.
@ McGraw-Hill Education
McGraw-Hill Education © 2010
Basic Electrical Engineering
Chapter 3
Network Analysis–Part I

D.C. Kulshreshtha
Next
Thought of The DAY
Whatever
THE MIND OF MAN
can CONCIEVE and BELIEVE,
it can
ACHIEVE.
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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3
Topics to be Discussed







Electric Circuit,
The Resistance
Parameter,
The Capacitance
Parameter,
The Inductance
Parameter,
Energy Sources-Classification
Ideal Voltage Source.
Ideal Current Source.
Friday, May 12, 2017





Series and Parallel
Combinations.
Practical Voltage
Source.
Practical Current
Source.
Source Transformation.
Kirchhoff’s Laws.
Ch. 3 Network Analysis- Part I
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4
Electric Circuit

It is a closed path, composed of active and
passive elements.

Active Element : It supplies energy to the
circuit.
Passive Element : It receives energy and then
1) either converts it to heat,
as in a Resistance (R).
2) or stores it in
(a) Electric Field, as in a Capacitor (C).
(b) Magnetic Field, as in an Inductor (L).

Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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5
Energy Sources

Classification
Independent Source
Voltage Source
DC Source
Ideal Source
Friday, May 12, 2017
Or
Or
Or
Or
Dependent Source
Current Source
AC Source
Practical Source
Ch. 3 Network Analysis- Part I
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6
Independent Ideal Voltage Source
Source
Load
Note that the source determines the voltage, but the
current is determined by the load.
The source has zero internal resistance.
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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7




The voltage source is said to be idle if the output
terminals are open such that i = 0.
When turned off (killed or made inactive), so that v =
0, it is equivalent to a short circuit.
Reference Marks : One terminal is marked plus and the
other minus. (Oversimplification; one mark can be
omitted.)
When actual polarity is opposite to the reference
marks, the voltage is a negative number.
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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8
Independent Ideal Current Source
Source
Load
Note that the source determines the current, but the
voltage is determined by the load.
The Source has infinite internal resistance (Ri).
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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9




The current source is said to be idle if the output
terminals are shorted together, such that v = 0.
When turned off (killed or made inactive), so that i = 0,
it is equivalent to an open circuit.
Reference Marks : An arrow is put.
When actual direction of current is opposite to the
reference (arrow) direction, the current is a negative
number.
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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10
Do you observe duality ?
The roles for the current and voltage are
interchanged in the two sources.
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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11
Practical Voltage Source
It is represented by an ideal voltage source in series
with an internal resistance (RSV).
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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12
Practical Current Source
It is modelled as an ideal current source in parallel
with an internal resistance (RSI).
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Ch. 3 Network Analysis- Part I
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13
Source Transformation


A practical current source can be converted
into its equivalent practical voltage source,
and vice versa.
This conversion is valid only for the external
load connected across the terminals of the
source.
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Ch. 3 Network Analysis- Part I
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14
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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15
Equivalence between
Voltage Source and Current Source

Two sources would be equivalent if they produce
identical values of VL and IL, when they are
connected across the same load.
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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16
Series and Parallel Combinations




What would be the net emf of the combination
if two ideal voltage sources of 2 V and 4 V are
connected in series so as to aid each other?
Click
Ans. 6 V
What would be the net emf of the combination
if two ideal voltage sources of 4 V and 4 V are
connected in parallel ? 4 V or 8 V ?
Click
Ans. Obviously, it should be 4 V
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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17





What would be the net emf of the combination if two
ideal voltage sources of 2 V and 4 V are connected in
parallel ? 2 V or 4 V or 3 V ?
Click
The question seems to be quite tricky!
Click
Ans. The question is wrong. The question contradicts
itself.
Ideal Voltage Sources in parallel are permissible only
when each has the same terminal voltage at every
instant of time.
What is its dual ?
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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18
Ideal Voltage Sources
Connected in Series
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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19
Ideal Current Sources
Connected in Parallel
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Ch. 3 Network Analysis- Part I
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20
Practical Current Sources
Connected in Series
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Ch. 3 Network Analysis- Part I
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21
Practical Voltage Sources
Connected in Parallel
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Ch. 3 Network Analysis- Part I
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22
Example 1 : Reduce the network shown in figure
to its simplest possible form by using source
transformation.
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Ch. 3 Network Analysis- Part I
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Solution
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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Example 2

In the given figure,
(a) If RL = 80 Ω, find current iL.
(b) Transform the practical current source into a
practical voltage source and find iL if RL = 80 Ω
again.
(c) Find the power drawn from the ideal source in
each case.
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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26
Solution :
20
(a) iL  (100) 
 20 mA
20  80
(b) iL 
2
 20 mA
80  20
Click
(c) vL  iL  RL  (20 mA )(80 )  1.6 V
Click
P  vL  I  (1.6 V)(100 mA )  160 mW
Click
In the second case,
P  iL  V  (20 mA )( 2 V)  40 mW
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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27
Benchmark Example 3

Take the benchmark example of the circuit
given in figure. Using source transformation,
determine the voltage v across 3-Ω resistor.
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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28
Solution :
Transforming the 4-A current source into a
voltage source,
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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29
Combining the two voltage sources,
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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30
Again transforming the voltage source into
current source,
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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31
Combining the two current sources we get Fig.
(e). Transforming this current source into
voltage source (Fig. f )
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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32
Combining the two resistances, we get Fig.
(g).
Finally, using voltage
divider, we get
3
v  5
 2.5 V
33
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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33
Kirchhoff’s Laws

(1) KCL : Algebraic sum of currents
meeting at a junction of conductors in a
circuit is zero.
 It is simply a restatement of the principle
of conservation of charge.
b
I
j 1
Friday, May 12, 2017
j
0
Ch. 3 Network Analysis- Part I
Next
34

(2) KVL :The algebraic sum of voltages
around a closed circuit or a loop is zero.

It is simply a restatement of the principle of
conservation of energy.
k
v
j 1
Friday, May 12, 2017
j
0
Ch. 3 Network Analysis- Part I
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35
Polarity of Voltages
Note that polarity of the voltage (emf) across a
battery does not depend upon the assumed
direction of current.
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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36
Applying KVL
1.
2.
3.
4.
Select a closed loop.
Mark the voltage polarity (+ and -) across
each element in the closed loop.
Go round the selected loop, and add up
all the voltages with + or – signs.
Any one of the following two rules can be
followed :
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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37
(i) Rule 1 : While travelling, if you meet a
voltage rise, write the voltage with positive sign ;
if you meet a voltage drop, write the voltage with
negative sign.
(ii) Rule 2 : While travelling, write the voltage
with positive sign if + is encountered first; write
the voltage with negative sign if – is encountered
first.
We shall be following Rule 1, as it has a strong
analogy with the physical height (altitude) of a
place.
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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38
Example 5 : Use KVL to find vR2 and vx.
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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39
For finding vR2, we write KVL eqn. going
around loop abgha clockwise :
Click
 36  vR 2  4  0  vR 2  32 V
If you choose to go around the loop
anticlockwise, you get
Click
 4  36  vR 2  0  vR 2  32 V
Giving the same result.
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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40
There are two ways to determine vx
1) We can consider this voltage as the voltage
across the gap from d to f. Writing KVL
(habcdfgh) :
Click
 4  36  12  14  vx  0
vx  6 V
2) Knowing vR2 , apply a short-cut (bcdfgb) :
Click
 12  14  vx  32  0
vx  6 V
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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41
Important Note about KVL



The assumed direction of current through a
resistor and the polarity of voltage across it
are always in conformity.
The end into which the current enters is
marked positive.
Passive-element sign convention.
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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42
Example 6 : Find the current supplied by
the 60-V source in the network.
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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43
Solution :


We need not find the currents I1, I2 and I3.
Instead, we reduce the network.
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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44
Example 7 : Determine the value of
current I.
2–3–I–4=0
Friday, May 12, 2017
or
I = -5 A
Ch. 3 Network Analysis- Part I
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45
Example 8
Using KCL and KVL, determine the currents ix
and iy in the network shown.
Friday, May 12, 2017
Ch. 3 Network Analysis- Part I
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46
Solution : Using KCL, the currents in other branches
are marked as shown. Writing KVL equations for the
loops 1, 2 and 3,
100  10 I1  5I x  0

5I x  0 I y  10 I1  100
5I x  50  2( I1  I x )  2 I y  0

7 I x  2 I y  2 I1  50
 2 I y  50  3( I1  I x  I y )  0

Friday, May 12, 2017
3I x  5I y  3I1  50
Ch. 3 Network Analysis- Part I
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Writing the above equations in matrix form,
5
7

3
0
2
5
10 

 2
 3
 I x  100 
  

 I y    50 ;
 I   50
 1
Using Calculator, we solve for Ix and Iy,
I x  3.87 A;
Friday, May 12, 2017
and
Click
Click
I y  0.51 A
Ch. 3 Network Analysis- Part I
Next
48
Review







Electric Circuit,
The Resistance
Parameter,
The Capacitance
Parameter,
The Inductance
Parameter,
Energy Sources-Classification
Ideal Voltage Source.
Ideal Current Source.
Friday, May 12, 2017






Series and Parallel
Combinations.
Practical Voltage
Source.
Practical Current
Source.
Source Transformation.
Dependent Sources.
Kirchhoff’s Laws.
Ch. 3 Network Analysis- Part I
Next
49