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Esercitazioni E Esercizi, Matematica

Invia: 16 aprile 2013
Estratto
Engineering 43

Thevenin/Norton

AC Power
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Thevenin’s Equivalence Theorem
LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A

 vO _

i

a
b

LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART B

 Resistance to Impedance Analogy

vO  VO

RTH
 

 vO _

i

a
LINEAR CIRCUIT

i I
VTH  VTH RTH  ZTH
PART B

vTH
PART A

b

Thevenin Equivalent Circuit for PART A
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AC Thevenin Analysis
• Same Circuit, Find IO by Thevenin • Take as “Part B” Load the 1Ω Resistor Thru Which IO Flows  The Thevenin Ckt  Now Use Src-Xform

Vx  2 A0  (1  j ) Vx  (2  j 2)V

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AC Thevenin Analysis cont.
• The Xformed Circuit • Another Source Xform  Now the Combined 8+j2 Source and all The Impedance are IN SERIES
• Thus VOC Determined by V-Divider
8 2j

 Find ZTH by Zeroing the 8+j2V-Source

VOC

1 j 10  j 6  (8  2 j )  (1  j )  (1  j ) 2
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ReDraw the Voltage Divider
Z2 Z1

 VO 

• The V-Divider Formula:
VOC

VO  VS Z2 Z1  Z2 

1 j 1 j 8  8 j  2 j  2 j2  8  2 j   8  2 j   (1  j )  (1  j ) 2 2

VOC

8  8 j  2 j  2 j 2 8  2   8 j  2 j  10  6 j     53j 2 2 2
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AC Thevenin Analysis
• Find ZTH by Zeroing the 8+j2 V-Source
 So The Thevenin Equivalent Circuit

ZTH  (1  j ) || (1  j ) ZTH ZTH

1  j 1  j   1  j   1  j 
1 j2 2    1 11 2

 IO is Now Simple

VOC VTH IO   ZTH  1 1  1 53j IO  ( A) 2
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Norton’s Equivalence Theorem
LINEAR CIRCUIT May contain independent and dependent sources with their controlling variables PART A

 vO _

i

a
b

LINEAR CIRCUIT May contain independent and dependent sources with their cont
rolling variables PART B

 Resistance to Impedance Analogy

vO  VO



i

a
LINEAR CIRCUIT

i I
iN  I N
RN  Z N
PART B

iN

RN
PART A

vO _

b

Norton Equivalent Circuit for PART A
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AC Norton Analysis
• Same Circuit, Find IO by Norton • Take as the “Part B” Load the 1Ω Resistor Thru Which IO Flows  Shorting the Load Yields Isc= IN

 Possible techniques to Find ISC:
• Loops or Nodes • Source Transformation • SuperPosition

 Choose Nodes

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AC Norton Analysis cont
• Short-Ckt Node On The Norton Circuit
I SC 60 8  2 j  20   ( A) 1 j 1 j

 As Before Deactivate Srcs to Find ZN=ZTH

 Use ISC=IN, and ZN to Find IO
• See Next Slide

ZTH  (1  j ) || (1  j )  1
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AC Norton Analysis cont.2
• The Norton Equivalent Circuit with Load Reattached
 Find IO By I-Divider

I O  I SC

I SC 4  j 1   ZN 1 11 1 j

 Same as all the Others

4  j 1 j IO  1 j 1 j IO


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4  j    4 j  j 2   5  j3 
12  j 2 2

Outline – AC Steady State Power
• Instantaneous Power Concept
– For The Special Case Of Steady State Sinusoidal Signals

• Average Power Concept
– Power Absorbed Or Supplied During an Integer Number of Complete Cycles

• Maximum Average Power Transfer
– When The Circuit is in Sinusoidal Steady State
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Outline – AC SS Power cont.
• Power Factor
– A Measure Of The Angle Between the CURRENT and VOLTAGE Phasors

• Power Factor Correction
– Improve Power Transfer To a Load By “Aligning” the I & V Phasors

• Single Phase Three-Wire Circuits
– Typical HouseHold Power Distribution

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Instantaneous Power
• Consider in the Time Domain a Voltage Source Supplying Current to an Impedance Load
 Now Recall From Chp1 The Eqn for Power

p  vi
 Then at any Instant for Time-Varying Sinusoidal Signals

pt   vt
  it   VM I M cos t  v  cos t   i 
 In the General Case  Now Use Trig ID
cos x cos y  1 cos( x  y)  cos( x  y) 2

i(t )  I M cos t   i 

v(t )  VM cos t  v 

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Instantaneous Power cont.
• Rewriting the Power Relation for Sinusoids

VM I M cos( v  i )  cos(2 t   v  i ) p(t )  2 VM I M VM I M p(t )  cos( v   i )  cos(2 t   v   i ) 2 2
• The First Term is a CONSTANT, or DC value; i.e. There is No Time Dependence • The Second Term is a Sinusoid of TWICE the Frequency of the Driving Source

 Examine the TWO Terms of the Power Equation

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Example
• For the Single Loop Ckt
 Use Phasors To find I

V 4V60 I   2 A30 Z 230
 To Obtain the Time Domain current Take the Real Part of the Phasor Current

 Assume

v(t )  4V cos  t  60 i(t )  Re230  Re2e30e jt  or V  4V60 it   2 Amp  cos t  30 and Z  230
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Example cont
• Thus for This case
 In the Power Equation
p(t )  vt it   4W cos60  30  cos2 t  60  30

 Or
4V cos t  60

1.732  j

p(t )  3.46W  4W cos2 t  90
 See Next Slide for Plots
• v(t) • i(t) • p(t) = v(t)•i(t)

 The Amplitudes and Phase Angles

VM  4V

I M 2A

 v  60

 i  30

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Sinusoidal Power Example 9.1
8

p(t) Calculated by p(t)=v(t)•i(t)
6

Max-p = 7.46W

4

v(t) or i(t) or p(t)

Avg-p = 3.46W

2

0

-2

NEGATIVE POWER – Inductive Load can Release stored Energy to the Circuit
v(t) (V) i(t) (V) P(t) (V)

-4 0.000

0.003

0.006

0.009

0.012

0.015

0.018

0.021

0.024

0.027

0.030

file =Sinusoid_Lead-Lag_Plot_0311.xls

Time (S)

p = 0 if either i or v are
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