l2 circuit fundamentals v2

8/8/2019 L2 Circuit Fundamentals v2

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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 !

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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

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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

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vR

ii

Gv

!

!

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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

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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

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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 !

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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

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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 !

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X

6V

12V

1A

P = ?

1 2( )

( 6 1 2 ) 1 6

P V V I

W

! v

! v !

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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!

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Fuse

Engineering Analysis

Real-life System

Abstract Model

Mathematicalproblem

Fuse

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Nodes and loops

Node: A point where 2 or morecircuit elements areconnected.

2

3

4

1

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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 !

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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 !

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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

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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

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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)

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Example

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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

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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 ! !

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UseKVL , KCL and Ohms law to solve the given problem

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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 !!

l2 circuit fundamentals v2

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