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ESc102: Introduction to Electronics
Circuit Fundamentals:
quickA Recap
A R HarishDept. of Electrical Engineering
IIT Kanpur
26, 27 July 2010
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Concepts
Charge, Current, Voltage, Power, and Energy Ohms Law
KCL KVL
2
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Electrical Current
The time rate of flow ofelectrical charge The units are amperes (A), which areequivalent tocoulombs per second (C/s)
Andr-Marie Ampre1775-1836
Current has a magnitude and a direction 3
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Direct Current (DC) & Alternating Current (AC)
When current is constant with time, we say that wehave direct current,abbr
eviat
ed as DC.
On the otherhand, a current that varies with time, reversing direction
periodically, is called alternating current, abbreviated as AC
~
4
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Voltage
Voltage difference is a Source ofcurrent flow
Units ofVoltage: Volts (V)
Alessandro Giuseppe AntonioAnastasio Volta 1745-1827
1 2V V V !
5
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Electrical Systems ar e made of Voltage sources, wir es and a variety of electrical elements
Resistor Capacitor Inductor
Transformer
Diode
Transistor
6
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Connection of several circuit elements in closed paths by conductors
Electrical Circuit
Before we learn how to analyze and design circuits, we must become familiar withsome basiccircuit elements.
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Resistance
( ) ( )v t R i t ! v
Theconstant, R, is called the resistance of the component and ismeasuredin units of Ohm ()
Ohms law
esistor Symbol
:
8
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Conductance
( ) ( )v t R i t ! v ( )( ) ( )v ti t G v t R! ! v
G = 1/R is called conductance and its unitis Siemens (S)
ErnstWerner von Siemens1816-1892
9
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vR
ii
Gv
!
!
10
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11
Sec
Laboratory
(2 pm to 5 pm)
Tutorial
(8 am to
8:55 am)
Tutor
Day Venue Venue Name Email Phone
F1 Mon CL 102B TB 101 Dr. U. Das utpal 7150
F2 Mon CL 102B TB 102 Dr. G. Sharma govind 7922
F3 Thurs CL 102B TB 103 Dr. K. Vasudevan vasu 7109F4 Thurs CL 102B TB 104 Dr. P. Sircar sircar 7063
F5 Tue CL 102B TB 105 Dr. J. Akhtar mjakhtar 6523
F6 Tue CL 102B TB106 Dr. Y.N. Singh ynsingh 7944
F7 W
ed CL 102B TB 107
Dr. S. Gupta sumana73
10F8 Wed CL 102B TB 108 Dr. A. Biswas abiswas 7319
F9
-- -- TB 109Dr. A.K. Chaturvedi
(Tut only)akc 7613
Fri CL 102B--
Mr. A. Roy (Lab
only) aroy6
168
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12
CL 102B
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Expectations
Expectation % responses
Practical Examples, Applications,
Implementation
30
Learn the basics/concepts 27
Enjoy, fun learning, interesting 15
Knowledge 9
Good Grades 6
Informal Learning/Innovative Teaching/Healthy
student teacher interaction
6
Easy/less academic load3
13
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Any electrical element which obeys ohms lawcan be modeled as a resistor
14
Can we model an electric bulb as a resistor?
( ) ( )v t i t ! vOhms law
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Electric Bulb
15
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Electric Bulb
Even t
houg
hch
aracte
ristics ar
enon-lin
ear, ov
er a
certain
range, the bulb can be thought of as a resistor 16
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Power and Energy
X
12V
0V
Q Thecharge loses energy = Q x 12 Joules
This energy is taken from the voltagesource and delivered to thecircuitelement
A
charg
eof 1
coulomb r
eceiv
es or d
eliv
ers an
en
ergy of
1 joule in moving through a voltage of 1 volt.
dq
dwv !
17
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2
1
( ) ( )t
t
d w P t w p t d t
d t! !
Power:
dq
dwv !
)()()( titvdtdq
dqdw
dtdwt !!!
18
dt
dqi !
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X
V1
V2
Power
I
1 2( ) P V V I! v
IfV1 > V2 then P is positive and it means that power is beingdelivered to theelectrical element X
IfV1 < V2 then P is negative and it means that power is being
extra
cted from t
heelec
trical
elem
ent X.
X is a source of power ! 19
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Note on the direction ofcurrent
X
2A
-2AX
Similarly for the voltage between two points
20
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X
12V
6V
Examples
1AP = ?
1 2( )
(1 2 6 ) 1 6
P V V I
W
! v
! v !
X
12V
6V
1AP = ?
1 2( )
(1 2 6 ) 1 6
P V V I
W
! v
! v !
21
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X
6V
12V
1A
P = ?
1 2( )
( 6 1 2 ) 1 6
P V V I
W
! v
! v !
22
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There is only one battery in thecircuit. Can you find whichelement is abattery?
A battery is a source of power, so P is negative
Answer is C
Is energy conserved in this circuit? 23
2.5 V 2.5 V
2.5 V 2.5 V5 V
2 A
1 A1 A
+ ++
++
- -
---
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Power dissipated in a Resistor
R+
-
v
i v i R! v
P v i! v
2vP
R
!
2 P i R! v
vi
R!
24
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Fuse
Engineering Analysis
Real-life System
Abstract Model
Mathematicalproblem
Fuse
26
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Nodes and loops
Node: A point where 2 or morecircuit elements areconnected.
2
3
4
1
27
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A loop is formed by tracing a closed path through circuitelements without passing through any intermediate node moret
han on
ceR2
R3R4
R1I
This is not a valid loop !
28
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Kirchhoff's Current Law (KCL)
Sum of
curr
ents
ent
ering a nod
eis
equal to sum of
curr
ents
leaving a node
1 2 3i i i !
29
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Kirchhoff's Current Law (KCL)
Net current entering a node is zero1
0
N
ji !Current entering a node is considered negative and current
l
eaving a nod
eis
consid
er
ed as positiv
e
30
0321 ! iii
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3 4i i!
4ai A! 2bi A! 31
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8c
i ! 32
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Example
33
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The sum ofcurrents entering/leaving a closed surface is zero.
KCL: More general formulation
i1
i2
i3 i4
R2R3
R4VS
R1IX
34
04321
! iiii
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Series Circuit
Two elements ar e connected in series if there is no other
element connected t o the nodejoining them
A, B and C are in series
Theelements have the samecurrent going through them
a b ci i i! !35
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A and B are
in serie
sE, F and G are in series
36
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The algebraic sum of the voltages equals zero for anyclosed path (loop) in an electrical circuit
Kirchhoff's Voltage Law (KVL)
37
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Example
38
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KVL and Conservation ofEnergy
X
V1
V2
Q
Thecharge loses energy = Q x (V1-V2) Joules
Energy gained Energy lost
KVL: lawofconservation ofEnergy39
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Parallel Circuits
Two elements areconnected in parallel if bothends of oneelement areconnected directly to corresponding ends ofthe other
A and B areconnected in parallel
D, E and F areconnected in parallel
40
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The voltage across parallel elements areequal (both magnitudeand polarity)
41
cbavvv !!
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Example
3 5 0 8c cv v V ! !
( 1 0 ) 0 2c e e
v v v V ! !
42
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UseKVL , KCL and Ohms law to solve the given problem
43
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+
5V
-
+
5V
-
1A
+
v2=5V
-
0.5A
Use ohms law : v = I x R
i1 = ?
Apply KCL at the indicated node
1 15 1 2 . 5v i V! v !
1 2 1 2 5 5 1 7 5xv v v V! ! !
+ v1 -
44
Aii 5.20115.0 11 !!