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Engineering 43 Thevenin/Norton AC Power Docsity.com 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 Docsity.com 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 Docsity.com 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 Docsity.com 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 Docsity.com 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 Docsity.com 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 Docsity.com 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 Docsity.com 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 Docsity.com 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  Docsity.com 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 Docsity.com 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 Docsity.com 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  Docsity.com 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 Docsity.com 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 Docsity.com 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 Docsity.com 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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