EE3301 Electrical Network Analysis - PowerPoint Presentation

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EE3301 Electrical Network Analysis

Dr. Jeanne Pitz

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Electrical Network Analysis

Analysis and design of RC, RL, and RLC electrical networksSinusoidal steady state analysis of passive networks using phasor representationmesh and nodal analyses Introduction to the concept of impulse response and frequency analysis using the Laplace transform. Prerequisites: MATH 2420 ( integral & differential eq)PHYS 2326 (electricity & magnetism) . Corequisite: EE 3101 (ENA lab)

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Course Logistics

Homework will be assigned but NOT collected.There will be a quiz every Wed. closed book and notes. It will cover the material that has already covered and the homework. No quiz on test days.3 Tests are scheduled during the semester.They are NOT comprehensive. The date will not be changed but what’s covered may be modified.The grade will be based on 4 tests The 4th will be the percentage based on the sum of the quizzes 10 best quizzes.

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Course Logistics: Schedule

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Chapter 1

OverviewInternational unitsVoltage and Current Definitions“Ideal” circuit elementPower and energy

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homework

Read chapter 1 esp. pages 10-18Work problems:1.14 a-d1.18 a-c1-25 a-e1-26 Answers are in the back of the book.

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Circuit language

Voltage is measured across two pointsCurrent is measured through an elementScale factor prefix smaller than 1Milli (m) – 10-3Micro (m) - 10-6Nano (n) – 10-9Pico (p) – 10-12Scale factor prefix larger than 1Kilo( K) – 103Mega( M) – 106Giga (G)– 109Terra (T) – 1012

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Definitions in circuit theory

Voltage is the energy per unit charge of separating a pos and neg charge, that is the potential difference between two pointsCurrent is the rate of change of chargeCurrent and voltage sign passive convention

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+

-

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Conventions in circuit theory

Voltage is joules per coulomb, the energy required to move a positive charge of 1colomb through the element.Power is i(t)*v(t)

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definitions

Power is the time rate of delivering or absorbing energyIn terms of voltage and current:

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Other facts

Current and voltage are VECTOR quantities having both magnitude and directionIf you compute power with the passive sign conventionPositive power is absorbedNegative power is suppliedEnergy is neither created or destroyed; it must be conserved.The charge on an electron is 1.6022 X 10 -19 coulombs

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+

-

v(t)

Power sign conventions

P is pos = vi if the current is entering the positive terminal.

See fig 1.6 for other cases

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i(t)

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Problem 1-14 pg 20 a, b

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+

-

v(t)

i(t)

a.

V=125V

I= 10 A

=1250 W

B->A

Supplied by B

Absorbed by A

b.

V=5VI= -240AW= 1200WA->B.Supplied by AAbsorbed by B

A

B

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Prob 1-18 a.

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Find the power at 625 s =0.625msWork out the exponents:1600*0.625ms=1000ms = 1.00s400*0.625 ms = 0.25s1000*0.625 ms = 625ms = .625 s

 

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Prob 1-18 b

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E

 

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Solution in book

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Chapter 2

Voltage and current sourcesIndependent, dependentElectrical resistanceOhm’s lawResistive CircuitsCircuit modelsKirchhoff's lawAnalysis of independent vs. dependent current sources

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homework

Read pages 24-48Make sure you understand the examplesLook it the following homework problems2.2, 2.4, 2.5, 2.7, 2.112.13, 2.1, 2.18 2.19, 2.22, 2.28 2.34, 2.35, 2.36, 2.37

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Ideal Sources

Ideal Voltage sources: can supply as much current as the circuit requires.Ideal Current sources : supplies current regardless of the voltage across it.Independent Sources: don’t rely on any other circuit parametersDependent Sources : depend on some parameter in the circuit.

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Sources: voltage

DC Independent Voltage source:Constant value regardless of what is attached. The sign indicates it’s polarity (positive on top)

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+

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5V

+

-

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Sources: current

DC Current source : independent arrow show it’s direction Constant value regardless of what is attached. Current flows around the loop.

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10A

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Dependent Voltage Sources

Dependent DC Voltage source : dependent on some parameter in the circuit.

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+-

5



V

+

-



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Dependent current Sources

Dependent DC Current source : dependent on some parameter in the circuit.

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a A

+

-

a

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Assessment problem 2.1

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a. What value of

ib makes this circuit valid:b. What power is associated with the 8A source:

 

a,

-2

P8A 216 W Current entering + node

 

b,

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Networks in Electrical Engineering

A network is an interconnection of components such as sources, resistors inductors and capacitors.

In studying first, DC (direct current) circuits composed only of resistors and sources, we can learn the basics of electricity, that will be useful with more complex circuits.In Direct current circuits the voltage or current is assumed to be constant overtime. For example the 4V supply never varies with time. Similarly for the current sources they are considered 2A and 4A for all time.

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Resistors and Ohm’s Law

Resistance is the property of a component to impede the flow of current.Such a component is called a “resistor”.The symbol for a resistor with resistance R is shown below.Ohms law relates the current through the resistor to the voltage across it. V=I*R

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R

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Circuit Elements

Resistors resist the flow of current Obey ohm’s lawVoltage and current relationshipsV=ir ohms law. R is usually considered a constant.but can be temperature dependent

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+

-

v

i

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Resistors: Current and Voltage conventions

The conventions for signs are as follows.

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R

-

+

V

r

=

I

r

* R

V

r

I

r

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Single loop analysis

Ohm’s law : the voltage across a resistor is directly proportional to the current flowing through it.For simple resistors R is constant sov=i*RIn DC analysis the voltage (or current ) is constant so on a plot of I vs. V, the slope is 1/R

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i

v

1/R

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Conductance

If we use the reciprocal of ohms we get Conductance which is measured in “siemens” S. some literature uses “mho” G= 1/RMho :

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

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Prob 2.11 v=i*r =>r=v/i

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Find slope from 2 points: (54,2m) (108,4m): m=(4-2)mA/(108-54)V=1/R

R=26K

Slope m =

Δ

y/

Δ

x; on this graph y is current in mA and x is V

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Adding resistors series vs parallel

Series share the same currentRT = R1 + R2Parallel share the same voltage1/RT= 1/R1 +1/R2RT= R1*R2/(R1 + R2)

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R1

R2

R

1

R

2

+

-

-

I

V

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Circuit Models

Some terminologyBranch – a portion of the circuit containing a single element and the nodes at either end of the element.Node – a point connecting two or more elements.Loop – a closed path in which no node is encountered more than once (except the starting point).

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Loops, nodes, branches

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2A

- +

6V

- 8V +

* -

Vx

4A

-

6V

+

+

4V

-

-

2V

+

6A

branch

node

Closed loop

node

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Kirchhoff’s laws

Current lawSum of currents entering or leaving a node is 0, taking direction into accountVoltage lawSum of voltages around a closed loop is 0 taking polarity into account

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Circuit convention rules

Sum of voltages around a closed loop is 0.Sum of currents at a node is 0.Power passive sign convention: Arrange I and v so magnitude are positive then if the sign of power is positive it being absorbed; if the sign of power is negative it being supplied.

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+

-

v(t)

i(t)

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Problem 1-22

proceed in a loop; Add the voltage if you reach the + terminal first; subtract if you reach the – terminal first. 0 = -8V + Vx - 6V - 4VVx= 18V

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2A

- +

6V

- 8V +

+ -

Vx

4A

-

6V

+

+

4V

-

-

2V

+

6A

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Analysis with dependent sources

Write the equations using the “parameter” specified.Then another equation setting the parameterSolve the equations simultaneously

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Example

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Example assessment 2.9

Solve for current i1 write a kvl loop equation around the outside loop.Voltage v: kvl around the outside loop, solve for v

 

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Power calculations: assessment 2.9 c,d

componentVoltage(v)Current(A)Power(W)absorbeddelivered5V supply5V2512512554K resis1.35V2533.733.751V supply1V2525256K resis4.65V7753603.753603.75Dep src-2V750150015001.8K resis1.35V7501012.51012.58V875060006000power61506150

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Voltage and current division

Voltage divides across resistors in series in proportion to their values.Vt = V1 + V2 = I R1 + I R2Current divides through resistors in proportion to their values. It = I1 + I2 =V/R1 + V/R2

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Chapter 3

Resistors in seriesResistors in parallelVoltage divider circuitsCurrent divider circuitsMeasuring voltageMeasuring currentMeasuring resistanceWheatstone bridges(skip section 3.7)

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Adding Resistors in Series

Two resistors connected at a single nodeWrite Kirchhoff's law around the loop:

 

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Adding Resistors in Parallel

Parrallel connected elements have the same voltage

Write Kirchhoff's current law at node a.

 

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Adding Resistors in Parallel

 

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Example Assessment Problem 3.1

Find the voltage vFind the power delivered by the current sourceFind the power dissipated by the 10  resistor.

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Solution

a. solve for Req then v = 5A*Reqb. solve for power delivered by the 5A source

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Solution

Find the power dissipated by the 10  resistorTo find V1 across the 10 + 6 resistors we must first determine the current in 30 and 7.2 then find current in the 10 branch

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V1

 

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Solution

Now we have the current in the branch we can find voltage in the 10  and power is v*i.

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Voltage Divider

Resistors can used to develop voltage level from the source voltage.

 

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Similarly for V2:

 

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Current Divider

The current source is is divided at the node:Is=i1 + i2

 

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Assessment problem 3.3 pg. 64

Find the value of R that will cause 4A of current to flow through the 80.How much power will R dissipate?How much power will the Current source supply

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Solution to Assessment 3.3 a

Find the value of R that will cause 4A of current to flow through the 80.

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Write an equation that includes R showing how the current divides between the two branches:

 

 

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Solution: Assessment 3.3 b

How much power will R dissipate?

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Find the current in the branch

iRFind voltage at VR then power is vR*IR

 

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Solution Assessment 3.3 c

How much power will the Current source supply?Find the voltage Vs then total power is Vs*20A

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Vs

V

R

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Solving circuits using voltage division

Generalizing voltage and current division to help solve and analyze circuitsSeries connectionsSame “i” goes through all the resistors

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Solving circuits using current division

Next consider the circuit shown. Parallel connections:All resistors have the same voltage drop

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Resistance MeasurementsWheatstone bridge

A Wheatstone bridge is a “balanced circuit used to measure the resistance of Rx by varying R3 until no current is measured in the meter igThen the unknown resistance is:

 

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Wheatstone bridge cont’d

If ig =0 then:Points a and b are at the same potential since no current is flowing in the meter (ig=0)

 

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Wheatstone bridge cont’d

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Now divide equation 1 by equation 2. (1) (2)

 

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Assessment problem 3.7

R1=100 R2=1000 R3=150 What is Rx?

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Assessment problem 3.7 b

How much power is dissipated by he circuitNow the power in each resistor is R*i2

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Power Calculations

P100= 100*(20mA)2=40mWP150= 150* (20mA)2=60mWP1000=1000*(2mA)2=4mWP1500=1500*(2mA)2=6mWSo yes all resistors are well within the 250mW max limits

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Skip delta and wye configurations

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Backup

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Content chapter 1

Outline: introduction Components: Resistors. Inductors CapacitorsParametersVoltagecurrentPassive sign conventionPower and energy

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Steps to check

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Steps to check

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