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Representation of the two vectors v1and v2.In this diagram, v1 leads v2 by 100o 30o 130o, although itcould also be argued that v2 leads v1 by 230o.Generally we express the phase difference by an angle lessthan or equal to 180o in magnitude.

Euler’s identity t 2 fA cos t A cos 2 f tIn Euler expression,A cos t Real (A e j t )A sin t Im( A e j t )Any sinusoidal function canbe expressed as in Eulerform.4-6

Applying Euler’s IdentityThe complex forcing function Vm e j ( t ) produces the complexresponse Im e j ( t ).

The sinusoidal forcing function Vm cos ( t θ) produces thesteady-state response Im cos ( t φ).The imaginary sinusoidal input j Vm sin ( t θ) produces theimaginary sinusoidal output response j Im sin ( t φ).

Re(Vm e j ( t ) ) Re(Im e j ( t ))Im(Vm e j ( t ) ) Im(Im e j ( t ))

Phasor DefinitionTime function : v1 t V1 cos ωt θ1 Phasor: V1 V1 1V1 Re(ej ( t 1 )V1 Re(ej ( 1 ))) by dropping t

A phasor diagram showing the sum ofV1 6 j8 V and V2 3 – j4 V,V1 V2 9 j4 V VsVs Ae j θA [9 2 4 2]1/2θ tan -1 (4/9)Vs 9.85 24.0o V.

Phasors AdditionStep 1: Determine the phase for each term.Step 2: Add the phase's using complex arithmetic.Step 3: Convert the Rectangular form to polar form.Step 4: Write the result as a time function.

Conversion of rectangular topolar form v1 t 20 cos( t 45 ) v2 (t ) 10 cos( t 30 ) V1 20 45V2 10 30

Vs V1 V2 20 45 10 30 14.14 j14.14 8.660 j 5 23.06 j19.14 29.97 39.7Vs Ae j 19.14A 23.06 ( 19.14) 29.96, tan 39.7 23.0622 1v s t 29.97 cos t 39.7

Phase relation ship

COMPLEX IMPEDANCEVL j L I LZ L j L L 90VL Z L I L

(a)(b)In the phasor domain,(a) a resistor R is represented by animpedance of the same value;(b) a capacitor C is represented by animpedance 1/j C;(c) an inductor L is represented by animpedance j L.(c)

V Vej tdVd Ve j tI C C j CVedtdtV1I j C V ZcIj Cj tZc is defined as the impedance of a capacitorThe impedance of a capacitor is 1/j C.As I j C V, if v V cos t , then i CV cos ( t 90 )

As I j C V, if v VM cos t , then i CVM cos ( t 90 )I M CVM

I Ie j tdId Ie j tV L L j LIedtdtVV j L I j L Z LIZL is the impedance of an inductor.The impedance of a inductor is j LAs V j C I, if i I cos t , then v LI cos ( t 90 )or i I cos ( t 90 ), and v LI cos t.j t

As V j CI, if i IM cos t, thenv LIM cos( t 90 )ori IM cos( t 90 ), and v LIM cos t, VM LIM

Complex Impedance in PhasorNotationVC ZC I C111 ZC j 90j C C CVL Z L I LZ L j L L 90 VR RI R

Kirchhoff’s Laws in Phasor FormWe can apply KVL directly to phasors. Thesum of the phasor voltages equals zero forany closed path.The sum of the phasor currents entering anode must equal the sum of the phasorcurrents leaving.

Ztotal 100 j(150 50) 100 j100 141.4 45 V100 30 I 0.707 30 45Ztotal 141.4 45 0.707 15 VR 100 I 70.7 15 , vR (t ) 70.7 cos(500t 15 )VL j150 I 150 90 0.707 15 106.05 90 15 106.05 75 , vL (t ) 106.05cos(500t 75 )VC j50 I 50 90 0.707 15 35.35 90 15 35.35 105 , vL (t ) 35.35cos(500t 105 )

11Z RC 1 / R 1 / Zc 1 / 100 1 /( j100)11j j 0.01As j0.012 j100 j100 j jZ RC11 0 70.71 450.01 j 0.01 0.01414 45 50 j50

Z RCVc Vs( voltage division)Z L Z RC 70.71 4570.71 45 10 90 10 90 j100 50 j5050 j50 70.71 45 10 45 10 13570.71 45 vc (t ) 10 cos (1000t 135 ) 10 cos 1000t V

I VsZ L Z RC10 90 10 90 j100 50 j5050 j5010 90 0.414 13570.71 45 i (t ) 0.414 cos (1000t 135 )

VCIR R10 135 0.1 135 100iR (t ) 0.1 cos (1000t 135 )VC 10 135 10 135 IC 0.1 45Zc j100100 90 iR (t ) 0.1 cos (1000t 45 ) A

Solve by nodal analysisV1 V1 V2 2 90 j 2 eq(1)10 j5V2 V2 V1 1.5 0 1.5eq(2)j10 j5

V1 V1 V2V2 V2 V1 2 90 j 2 eq(1) 1.5 0 1.510 j5j10 j5From eq (1)1 1 jj10.1V1 j0.2V1 j0.2V2 j 2 As j, jj j j 1 j(0.1 j0.2)V1 j0.2V2 j 2From eq (2) j0.2V1 j0.1V2 1.5SolvingV1 by eq(1) 2 eq(2)(0.1 j0.2)V1 3 j 23 j23.6 33.69 V1 16.1 33.69 63.430.1 j0.2 0.2236 63.43 16.1 29.74 v1 16.1cos(100t 29.74 ) Veq(2)

Vs - j10, ZL j L j(0.5 500) j250Use mesh analysis,

Vs V R V Z 0 ( j10 ) I 250 I ( j 250 ) 0 j1010 90 I 0 . 028 90 45 250 j 250353 .33 45I 0 . 028 135 i 0 . 028 cos( 500 t 135 ) AV L I Z L ( 0 . 028 135 ) 250 90 7 45 v L ( t ) 7 cos( 500 t 45 ) VV R I R ( 0 . 028 135 ) 250 7 135 v R ( t ) 7 cos( 500 t 135 )

AC Power CalculationsP Vrms I rms cos PF cos v iQ Vrms I rms sin

apparent power Vrms I rms2P Q Vrms I rms 2P IQ I2rms2rmsRX2P Q V2RrmsRV2XrmsX

THÉVENIN EQUIVALENTCIRCUITS

The Thévenin voltage is equal to the open-circuitphasor voltage of the original circuit.Vt VocWe can find the Thévenin impedance byzeroing the independent sources anddetermining the impedance looking into thecircuit terminals.

The Thévenin impedance equals the open-circuitvoltage divided by the short-circuit current.Voc VtZ t I scI scI n I sc

Maximum Power Transfer If the load can take on any complex value,maximum power transfer is attained for a loadimpedance equal to the complex conjugate ofthe Thévenin impedance. If the load is required to be a pureresistance, maximum power transfer isattained for a load resistance equal to themagnitude of the Thévenin impedance.

UNIT III–NETWORK THEOREMS (BOTH AC AND DC CIRCUITS)– Superposition theorem– The venin’s theorem- Norton’s theorem-Reciprocity theorem- Maximum power transfer theorem.1619

9.1 – Introduction This chapter introduces importantfundamental theorems of network analysis.They are the Superposition theorem Thévenin’s theorem Norton’s theorem Maximum power transfer theorem Substitution Theorem Millman’s theorem Reciprocity theorem

9.2 – Superposition Theorem Used to find the solution to networks with two ormore sources that are not in series or parallel. The current through, or voltage across, anelement in a network is equal to the algebraic sumof the currents or voltages producedindependently by each source. Since the effect of each source will bedetermined independently, the number ofnetworks to be analyzed will equal the number ofsources.

Superposition Theorem The total power delivered to a resistive elementmust be determined using the total current throughor the total voltage across the element and cannotbe determined by a simple sum of the powerlevels established by each source.

9.3 – Thévenin’s Theorem Any two-terminal dc network can bereplaced by an equivalent circuit consistingof a voltage source and a series resistor.

Thévenin’s Theorem Thévenin’s theorem can be used to: Analyze networks with sources that are not in series orparallel. Reduce the number of components required toestablish the same characteristics at the outputterminals. Investigate the effect of changing a particularcomponent on the behavior of a network without havingto analyze the entire network after each change.

Thévenin’s Theorem Procedure to determine the proper values of RTh andETh Preliminary1. Remove that portion of the network across which theThévenin equation circuit is to be found. In the figurebelow, this requires that the load resistor RL betemporarily removed from the network.

Thévenin’s Theorem2. Mark the terminals of the remaining two-terminal network. (Theimportance of this step will become obvious as we progressthrough some complex networks.)RTh:3. Calculate RTh by first setting all sources to zero (voltagesources are replaced by short circuits, and current sources byopen circuits) and then finding the resultant resistancebetween the two marked terminals. (If the internal resistanceof the voltage and/or current sources is included in the originalnetwork, it must remain when the sources are set to zero.)

Thévenin’s TheoremETh:4. Calculate ETh by first returning all sources to their originalposition and finding the open-circuit voltage between themarked terminals. (This step is invariably the one that will leadto the most confusion and errors. In all cases, keep in mindthat it is the open-circuit potential between the two terminalsmarked in step 2.)

Thévenin’s Theorem Conclusion:5.Draw the Thévenin equivalentcircuit with the portion of thecircuit previously removedreplacedbetweentheterminals of the equivalentcircuit. This step is indicatedby the placement of theresistor RL between theterminals of the Théveninequivalent circuit.Insert Figure 9.26(b)

Experimental Procedures Two popular experimental procedures for determiningthe parameters of the Thévenin equivalentnetwork: Direct Measurement of ETh and RTh For any physical network, the value of ETh can be determinedexperimentally by measuring the open-circuit voltage acrossthe load terminals. The value of RTh can then be determined by completing thenetwork with a variable resistance RL.

Thévenin’s Theorem Measuring VOC and ISC The Thévenin voltage is again determined bymeasuring the open-circuit voltage across the terminalsof interest; that is, ETh VOC. To determine RTh, a shortcircuit condition is established across the terminals ofinterest and the current through the short circuit (Isc) ismeasured with an ammeter. Using Ohm’s law:RTh Voc / Isc

Vth173

Rth174

Norton’s Theorem Norton’s theorem states the following: Any two-terminal linear bilateral dc network can bereplaced by an equivalent circuit consisting of acurrent and a parallel resistor. The steps leading to the proper values of IN andRN. Preliminary steps:1. Remove that portion of the network across which theNorton equivalent circuit is found.2. Mark the terminals of the remaining two-terminalnetwork.

Norton’s Theorem Finding RN:3. Calculate RN by first setting all sources to zero (voltagesources are replaced with short circuits, and currentsources with open circuits) and then finding theresultant resistance between the two marked terminals.(If the internal resistance of the voltage and/or currentsources is included in the original network, it mustremain when the sources are set to zero.) Since RN RTh the procedure and value obtained using theapproach described for Thévenin’s theorem willdetermine the proper value of RN.

Norton’s Theorem Finding IN :4. Calculate IN by first returning all the sources to theiroriginal position and then finding the short-circuitcurrent between the marked terminals. It is the samecurrent that would be measured by an ammeterplaced between the marked terminals. Conclusion:5. Draw the Norton equivalent circuit with the portion ofthe circuit previously removed replaced between theterminals of the equivalent circuit.

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9.5 – MMaximum PowerTransfer Theorem The maximum power transfer theoremstates the following:A load will receive maximum power from anetwork when its total resistive value isexactly equal to the Thévenin resistanceof the network applied to the load. Thatis,RL RTh

Maximum Power Transfer Theorem For loads connected directly to a dc voltage supply, maximum power will bedelivered to the load when the load resistance is equal to the internalresistance of the source; that is, when:RL Rint

Reciprocity Theorem The reciprocity theorem is applicable only tosingle-source networks and states the following: The current I in any branch of a network, due to asingle voltage source E anywhere in the network, willequal the current through the branch in which thesource was originally located if the source is placed inthe branch in which the current I was originallymeasured. The location of the voltage source and the resulting currentmay be interchanged without a change in current

UNIT IV - TRANSIENT RESPONSE FOR DC CIRCUITS–– Transient response of RL, RC and RLC–– Laplace transform for DC input with sinusoidal input.

Solution to First Order Differential EquationConsider the general Equationdx(t ) x(t ) K s f (t )dtLet the initial condition be x(t 0) x( 0 ), then we solve the differentialequation:dx(t ) x(t ) K s f (t )dtThe complete solution consists of two parts: the homogeneous solution (natural solution) the particular solution (forced solution)

The Natural ResponseConsider the general Equationdx(t ) x(t ) K s f (t )dtSetting the excitation f (t) equal to zero,dx N (t )dx N (t )x N (t ) dx N (t )dt x N (t ) 0 or , dtdt x N (t ) dx N (t )dt , x N (t ) x N (t ) e t / It is called the natural response.

The Forced ResponseConsider the general Equationdx(t ) x(t ) K s f (t )dtSetting the excitation f (t) equal to F, a constant for t 0dxF (t ) xF (t ) K S FdtxF (t ) K S F for t 0 It is called the forced response.

The Complete ResponseSolve for ,Consider the general Equationdx(t ) x(t ) K s f (t )dtThe complete response is: the natural response the forced responsefor t 0x(t 0) x(0) x( ) x(0) x( )The Complete solution:x x N (t ) x F (t ) e t / K S F e t / x ( )x (t ) [ x(0) x( )] e t / x ( )[ x(0) x( )] e t / called transient responsex ( ) called steady state response

TRANSIENT RESPONSEFigure 5.1

A general model of the transientCircuit with switched DC excitationFigure5.2,5.3analysis problem

For a circuit containing energy storage elementFigure 5.5, 5.6

(a) Circuit at t 0(b) Same circuit a long time after the switch is closedFigure5.9, 5.10The capacitor acts as open circuit for the steady state condition(a long time after the switch is closed).

(a) Circuit for t 0(b) Same circuit a long time before the switch is openedThe inductor acts as short circuit for the steady state condition(a long time after the switch is closed).

Reason for transient response The voltage across a capacitor cannot bechanged instantaneously. VC (0 ) VC (0 ) The current across an inductor cannot bechanged instantaneously. I L (0 ) I L (0 )

ExampleFigure5.12, 5.135-6

Transients Analysis1. Solve first-order RC or RL circuits.2. Understand the concepts of transientresponse and steady-state response.3. Relate the transient response of firstordercircuits to the time constant.

TransientsThe solution of the differential equationrepresents are response of the circuit. Itis called natural response.The response must eventually die out,and therefore referred to as transientresponse.(source free response)

Discharge of a Capacitance through aResistance i 0,iciRiC iR 0dvC t vC t C 0dtRSolving the above equationwith the initial conditionVc(0) Vi

Discharge of a Capacitance through aResistancedvC t vC t C 0dtRdvC t RC vC t 0dtvC t KestststRCKse Ke 0 1s RC t RCvC t Ke vC (0 ) Vi Ke KvC t Vi e0 / RC t RC

vC t Vi e tRCExponential decay waveformRC is called the time constant.At time constant, the voltage is36.8%of the initial voltage.vC t Vi (1 e t RC )Exponential rising waveformRC is called the time constant.At time constant, the voltage is 63.2%of the initial voltage.

RC CIRCUITfor t 0-, i(t) 0u(t) is voltage-step function Vu(t)

RC CIRCUITVu(t)iR iCvu (t ) vCdvCiR , iC CRdtdvCRC vC V , v u (t ) V for t 0dtSolving the differential equation

Complete ResponseComplete response natural response forced response Natural response (source free response)is due to the initial condition Forced response is the due to theexternal excitation.

Figure5.17,5.18a). Complete, transient and steady stateresponseb). Complete, natural, and forced responsesof the circuit5-8

Circuit Analysis for RC CircuitApply KCLiR iCvs v RdvCiR , iC CRdtdvC11 vR vsdtRCRCvs is the source applied.

Solution to First Order Differential EquationConsider the general Equationdx(t ) x(t ) K s f (t )dtLet the initial condition be x(t 0) x( 0 ), then we solve the differentialequation:dx(t ) x(t ) K s f (t )dtThe comple

UNIT I - BASIC CIRCUIT CONCEPTS - Circuit elements - Kirchhoff's Law - V-I Relationship of R,L and C - Independent and Dependent sources - Simple Resistive circuits - Networks reduction - Voltage division - current source transformation. - Analysis of circuit using mesh current and nodal voltage methods. 1 Methods of Analysis