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1、Chapter 4TransientsElectrical Engineering and Electronics II Electrical Engineering and Electronics II 4 Course Hours4.Transients4.1 Introduction4.2 Initial state and steady state4.3 First-order RC Circuits4.4 First-order RL CircuitstENew steady statetransientCOld steady statesteady stateKRE+_Switch
2、 K is closed4.1 IntroductionNew steady statesteady stateRUs+_Conception of steady state and transient stateWhen t=0，uc(0)=0When t=, uc()=UsOld steady state Why the transient response happens?No transientIResistive circuitt = 0ER+_IKResistor is a energy-consumption element, current is proportional to
3、 voltage, no transient response will happen even if changing sourcelEnergy can not change instantly because of accumulating or decaying period.Charging or dischargingChange graduallyElectric field energyEKR+_CuCEtMagnetic field energyChange graduallyKRE+_t=0iLtE/RlEnergy can not change instantly bec
4、ause of accumulating or decaying period.lThe causes of transients1. Energy storage elements -inductors and capacitors change gradually；2.Changing circuit, such as switching source.lTransientsThe time-varying currents and voltages resulting from the sudden application of sources, usually due to switc
5、hing. By writing circuit equations, we obtain integro-differential equations. These equations can be converted to pure differential equations by differentiating w.r.t time.The study of transients require us to solve differential equations . 4.2 Initial state and steady state Assume changing circuit
6、when t=0, then t=0 is end point of old steady state; t=0+ is the start point of transient state.From t=0to t=0+,iL、uC can not change suddenly.t t=0=0t tt=0-t=0+The law of changing circuit(换路定则换路定则)Exa 4.2Exa 4.2SoluSolu(1)Brofore switch(1)Brofore switchWe know thatWe know thatBy the law ofBy the law
7、 of circuit changing circuit changingKnownsKnowns：before circuit before circuit changingchanging，C C、L L have no have no energy; Findenergy; Find：the the currents and voltages of all currents and voltages of all elementselements。S S(a)(a)C CU R R2 2R R1 1t t=0=0+-L LlHow to get initial valueExample
8、4.2, changing momentchanging moment，capacitance shortcapacitance short, changing moment, changing moment，inductance openinductance openiC 、uL alter suddenly(2) By the circuit at t=0+, Find other unknownsS SC CU R R2 2R R1 1t=0t=0+-L L(a) circuit(a) circuitiL(0+ )U iC (0+ )uC (0+)uL(0+)_u2(0+)u1(0+)i
9、1(0+ )R R2 2R1+_+-(b) (b) t = 0+ circiutlHow to get initial valueExercise 1: Assuming old circuit is in DC steady state before switch K is closed. how to get uC(0+),iR(0+)?iRR14k12 VKt=08kR22mFuC Solution:When t=0-, capacitor is considered as open circuit, we get equivalent circuit.R14k12VuC(0)8kt=0
10、-R14k12VuC(0)8kiRR14k12VKt=08kR22mFuC substituting voltage source for uC(0+)iR(0+) 8kR2+ uC(0+)t=0lHow to get initial valueExercise 2: Given by Exercise 2: Given by R R1 1=4, =4, R R2 2=6, =6, R R3 3=3, =3, C C=0.1F, =0.1F, L L=1mH, =1mH, U US S=36V, switch S is closed for a long time. =36V, switch
11、S is closed for a long time. Open the switch S wOpen the switch S whenhen t=0, how to get the initial values t=0, how to get the initial values of all elements?of all elements?lHow to get initial valueAnswer: Uc(0)=12V, iL(0)=4A,iR(0)=2AConclusions 1. At the moment of changing circuit1. At the momen
12、t of changing circuit，u uC C、 i iL L can not can not alter suddenly, but alter suddenly, but i ic c, u, uL L can change suddenly . can change suddenly . 3. 3. Before circuits changedBefore circuits changed, if , if uC(0(0-) ) 0 0, the, the capacitancecapacitance can be replaced by an can be replaced
13、 by an ideal voltage ideal voltage sourcesource with u with uc c(0(0+ +) ) at at t t=0=0+ +; if if iL(0(0-) ) 0 0, the , the inductinductanceance can be replaced by an can be replaced by an ideal current sourceideal current source with i with iL L(0(0+ +) ) at t=0at t=0+ +. .2. Before 2. Before circ
14、uitcircuits changed, if energy-storage elements s changed, if energy-storage elements have no energy, have no energy, just afterjust after circuitcircuits s changchanged (t=0 ed (t=0 + +) ) the the capacitancecapacitance and and inductance inductance are viewed as are viewed as short circuitsshort c
15、ircuits and and open circuitopen circuit, respectively., respectively.lDC Steady State ResponseThe transient terms for currents/voltages decay to zero with time. Under Steady State :For capacitors with dc source, capacitance behaves as open circuits.For inductors with dc source, inductance behaves a
16、s short circuits.lDC Steady State ResponseThe steps in determining the steady state response for RLC circuits with dc sources are:1. Replace capacitances with open circuits.2. Replace inductances with short circuits.3. Solve the remaining circuit using methods in chapter 2.Example 4.1 Find steady-st
17、ate values of vx and ix in this circuit for t0.Answer: vx =5V, ix = 1A t0Exercise 4.3 Find steady-state values of labeled currents and voltages for t0.Answer: va =50V, ia = 2Ai1 = 2A, i2=1A, i3=1AHomeworkP4.2P4.6P4.8lEquivalent circuit of First-order circuit Two parts: one (equivalent) capacitor or
18、inductor; a two terminal network with resistance and sources.N NL LN NC CorlFirst-order circuit Only one (equivalent) capacitor or inductor is included in a linear circuit. 4.3 First-order RC CircuitslAccording to Thevenin LawNL LNC CorRULuLiL+- -RUCuCiC+- -4.3 First-order RC CircuitslDifferential e
19、quation of first-order RC circuitRULuLiL+- -RUCuCiC+- -Solution: lFirst-order RC CircuitsExample: to find the transient response after changing circuit when t=0. lFirst-order RC Circuitshomogeneous solutionparticular solutionlFirst-order RC Circuitslhomogeneous solutionlFirst-order RC CircuitsTheref
20、oreThen, the final solution islParticular solutionlFirst-order RC CircuitslThe solution of differential equationSubstituting the initial condition：lFirst-order RC Circuits Time constant reflects the length of transient period.t 2 3 4 5 6 7 e-t/ 36.8%13.5%5%1.8%0.3%0.25%0.09%After about five time con
21、stants, the transient response is over.After one time constants, the transient response is equal to 36.8 percent of its initial value.lThe curves versus timeThe initial slop intersects the final value at one time constant.Mounting curveDecaying curve Time constant reflects the length of transient pe
22、riod.lThe solution of differential equationTime constantSteady state valueInitial valuelFirst-order RC CircuitslSolution of other parameterslThree elements methodThree elements: 1.steady state value f(); 2.time constant ; 3. initial value f(0+).lFormula of Three element method：f()steady state valuet
23、ime constantf(0+)initial value=RC time constant of RC circuit= ？ time constant of RL circuitThree elements method解：解： 用三要素法求解用三要素法求解用三要素法求解用三要素法求解电路如图，电路如图，t=0时合上开关时合上开关S，合，合S前电路已处于前电路已处于稳态。试求电容电压稳态。试求电容电压 和电流和电流 。(1)(1)确定初始值确定初始值确定初始值确定初始值由由由由t t=0=0- -电路可求得电路可求得电路可求得电路可求得由换路定则由换路定则由换路定则由换路定则Applic
24、ation Examplet=0-等效电路等效电路9mA+-6k RS9mA6k 2 F3k t=0+-C R(2) (2) 确定稳态值确定稳态值确定稳态值确定稳态值由换路后电路求稳态值由换路后电路求稳态值(3) (3) 由换路后电路求由换路后电路求由换路后电路求由换路后电路求 时间常数时间常数时间常数时间常数 t 电路电路9mA+-6k R 3k t=0-等效电路等效电路9mA+-6k RR0的计算类似于戴维宁等效的计算类似于戴维宁等效电阻。即从储能元件两端看电阻。即从储能元件两端看进去的等效电阻。进去的等效电阻。三要素三要素三要素三要素u uC C 的变化曲线如图的变化曲线如图的变化曲线如
25、图的变化曲线如图18V54Vu uC C变化曲线变化曲线变化曲线变化曲线tO4.4 First-order RL CircuitslTime constant=RC=L/RRULuLiL+- -RUCuCiC+- -4.4 First-order RL CircuitsThree element methodlInitial value: t=0-t=0+ f(0+)lSteady state value: t = f()lTime constant : =RC =L/RlSubstituting three elementslDraw the curve versus timelStepsl
26、Limited Condition:1) first-order circuit2) DC sourceExample 4.2 Find voltage of v(t) and current i(t) in this circuit for t0.Answer:i(0)=0,i()=2AV(0)=100,V()=0Example 4.3 Find voltage of v(t) and current i(t) in this circuit for t0.Answer:i(0)=Vs/R1, i()=0AV(0)=Ldi/dt,V()=0Exercise 4.4 Find voltage
27、of v(t) and current iR(t) , iL(t) in this circuit for t0, assume that iL(0)=0.Answer:iL(0)=0,iL()=2A; iR(0)=2,iR()=0A; V(0)=20,V()=0Exercise 4.5 Find voltage of v(t) and current i(t), v(t) in this circuit for t0, assume that the switch has been closed for a very long time prior to t=0.Answer:HomeworkP4.9P4.10P4.12Main Contents1. Solve first-order RC or RL circuits.2. Understand the concepts of transient response and steady-state response. 3. Relate the transient response of first-order circuits to the time constant.4. Solve RLC circuits in dc steady-state conditions.
