Electric and Electronic Lecture Presentation - Chapter02.ppt

7/29/2019 Electric and Electronic Lecture Presentation - Chapter02.ppt

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Fundamentals of Electric

Circuits


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An Electrical System

Source Control Load

Transmission system

1) Source to provide energy for the electrical system

can be voltage source or current source

2) Transmission systemconducts the energy from source to load

3) Control apparatusto control the flow of energy

4) Loadto absorb electrical energy supplied by source


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Figure 2.1,2.2

2-1

Ideal voltage sources

Various representations of an electrical system


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Figure 2.3, 2.4

2-2

Symbol for ideal current source

Symbols for dependent sources

By convention : +ve current flow out of a voltage

thru +ve terminal


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Some Definitions

Electrical networka collection of elements thru which current flows

Branchany portion of a circuit with two terminals connected to it. A branch may

consist of one or more circuit elements.

Nodethe junction of two or more branches


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Figure 2.5,

2.6

2-3

Definition of a branch

Definitions of node and supernode


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Figure 2.7, 2.8

Definition of a loop

2-4

Figure 2.7

Definition of a mesh

Figure 2.8


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Figure 2.11,

2.13

2-5

Demonstration of KCLIllustration of Kirchhoffs current lawFigure 2.11


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Electric Charge and Current

- An electricsystem transmits energy due to movement of electric charge.- So fundamental electric quantity is chargesmallest amount of charge- electron

- Electrical currenttime rate of charge of charge passing a predetermined area


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Example 1

Determine current given charge suppose that charge is given by

For current to flow, there must exist a closed circuit.

i = current flowing in closed circuit

Note : Current flowing from source to load is the same as current flowing from load to source- No current is lost around closed circuit

- This principle is known as Kirchhoffs Current Law (KCL)


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KCL state that the sum of current at a node must equal to zero, or

For node 1

Define : current entering a node asve

current leaving a node as +ve

Node 1KCL :


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Exercise


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Voltage and Kirchhoffs Voltage Law- change moving in an electric circuit given rise to a current.

- So it must take some work, or energy for the charge to move between two points in a circuit.

- Total work per unit charge associated with the motion of charge between two points is calledVOLTAGE

- Definition :

- The voltage, or potential difference, between two points in a circuit indicates the energy

required to move charge from one point to the other

- The direction or polarity of the voltage related to whether energy is being dissipated orgenerated

- as in case of current, energy in the system is not lost or the sum of

voltages associated with sources must equal the sum of the load voltages

The net voltage around a closed circuit is zero

KIRCHHOFFS VOLTAGE LAW (KVL)


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- Introduce reference (ground) voltage

- Voltage V2 is the difference between two node voltages Va and

V2 = VaVb

- Select any mode as the reference node, so all node voltages may be referenced to this

reference voltage.

- In figure, select node b as reference (normally assign 0V at reference)

So, Vb = 0V

V1 = 1.5V

V2 = Vab = VaVb = Va0 = Va

but Va and V1 are the same so V1 = V2


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Exercise 1Determine unknown voltage V2

Vs = 12V

V1 = 6V

V3 = 1V

KVL: -Vs + V1 + V2 + V3 = 0

V2 = 5V

V4 = 5V


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Exercise 2

Vs1 = 12V, Vs2 = -4V

V2 = 2V , V3 = 6V

V5 = 12V

Find V1 and V4 ?

KVL : - V2 + Vs2V4 = 0V4 = -V2 +Vs2

- Vs1 - V1 +V2 +V3 = 0

1

1


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Electrical Power

- Power is defined as work per unit time

- P = work = work x charge

time charge time

= voltage x current

P = VI Watts (W)

- Power is a signal signed quantitypositive power

negative power

- Positive Power = power dissipated P = vi (power dissipated)

- Negative Power = power generated P = vi (power generated)

i

+

v

i

+

v


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Resistance and Ohms Law

- When current flows thru a wire/other circuit elements, it encounters resistance

- This causes energy to be dissipated (heat)

- According to Ohms Law

V = IR

- Voltage is proportional to the current flowing thru it

R Ohms () 1 = 1 V

A

- The resistance of a material depends on Resistivity () ; the inverse called conductivity ()

- For a cylindrical resistance element, resistance is proportional to length of sample, l , and

inversely proportional to its crosssectional area A and conductivity

A

lR = l A


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- Convenient to define conductance of circuit element as the inverse of its resistance, used

symbol G

G = I Siemen (S)

RThus ohms law can be written as

I = GV (V = IR = I )

G

- For resistors, in addition to resistance in ohm, the max allowable power dissipation (power

rating) is specified. Exceeding this power rating could cause overheating and burn out.

- Power dissipated in R

P = IV = I . IR

= IR

= VR


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ExampleDetermine the minimum resistor size that can be connected to a given battery w/o exceeding

the resistors 1/4W power rating

SolutionPower rating = 0.25W

Battery voltage = 1.5V

P = V

R0.25 (1.5)

R

R (1.5) = 9 So min R = 9

0.25

R (1/4 W)+1.5 V


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Exercise

Determine iB and power supplied by battery if: i1 = 0.2 mA

i2 = 0.4 mA

i3 = 1.2 mAVB = 3 V

VB+

iB


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Open and Short Circuit

Short circuit:- 1) R 0

2) V = 0 for any i3) Allow unimpeded current

Open Circuit :- 1) R

2) I = 0

+

V

i

8

+

V

R8

i = 0 for any V


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Series Resistors

-

Series Circuit

Def : Two or more circuit elements are said to be in series if the current flow from one

elements exclusively flows into the next element. In the example, to the battery, the

resistors appear an a single equivalent resistance,

REQ where REQ = R1 + R2

So, for series resistance

+

i R1

+V1 -

R2

+V2

-1.5 V

by KVL1.5 = V1 + V2

= iR1 + iR2

= i(R1 + R2)


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Voltage Divider- Closely tied to series resistors

- Source voltage divides among the resistors in series according to KVL

- We can write the voltage across each R

i R1

+V1 -

R2

+V2

-V

+

R3-V3 +

V = i (R1 + R2+R3)

V = i REQ

i = V

REQ

V1 = iR1 = R1 V

REQ

V2 = iR2 = R2 V

REQ

V3 = iR3 = R3 V

REQ


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Parallel Resistors

Def : Two or more circuit elements are said to be in parallel if the elements share the

same terminals. From KVL , it follows that the elements will have the samevoltage.

Ex:

KCL requires that

is = i1 + i2 + i3

ohms lawi1 = V , i2 = V, i3 = V

R1 R2 R3

so is = V 1 + 1 + 1 = V 1

R1 R2 R3 REQ

isR3

i1

R1 R2

i2 i3 +

V

-

R1 R2

i2 i3


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Where

or REQ =

- Note : for parallel combinations, normally indicate R1 // R2 // R3 .

1 = 1 + 1 + 1

REQ R1 R2 R3

1 = 1 + 1 + 1 + .... + 1REQ R1 R2 R3 RN

1 + 1 + + 1R1 R2 RN

1


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Current Divider

From circuit ,

ohms law: V = i REQ

i1 = i REQ = i

=

i1 = V, i2 = V, i3 = V

R1 R2 R3

11 + 1 + 1

R1 R2 R3

R1

R1

i1

R1

1 + 1 + 1

R1 R2 R3


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i2 = =

i3 = =

So :

i REQR2

i1

R2

1 + 1 + 1

R1 R2 R3

i REQR3

i1

R3

1 + 1 + 1

R1 R2 R3

1 + 1 +.. + 1

R1 R2 Rn

is

1

Rnin =


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Ex:

Determine i1 , R1= 10

R2= 2

R3= 20

Is = 4 A

-

V

+

R1 R2

i2 i3

R3

i1

Is

1 + 1 + 1

R1 R2 R3

Is

1

R1i1 =

1 + 1 + 1

10 2 20

(4)

1

10=

8

13= A


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Ex:

Determine V3

1 k

1 k

+V3

-5V

+

1 k

R1

R2//R3

+V3

-Vs+

Electric and Electronic Lecture Presentation - Chapter02.ppt

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