rl and rc circuits - kntu homepage webftp clientwp.kntu.ac.ir/faradji/ec1/ec1_ch7.pdf · 7.0....

Chapter 7. Response of First-OrderRL and RC Circuits

By: FARHAD FARADJI, Ph.D.Assistant Professor,

Electrical Engineering,K.N. Toosi University of Technology

http://wp.kntu.ac.ir/faradji/ElectricCircuits1.htm

Reference: ELECTRIC CIRCUITS, J.W. Nilsson, S.A. Riedel, 10th edition, 2015.

Chapter Contents7.0. Introduction

7.1. The Natural Response of an RL Circuit

7.2. The Natural Response of an RC Circuit

7.3. The Step Response of RL and RC Circuits

7.4. A General Solution for Step and Natural Responses

7.5. Sequential Switching

7.6. The Integrating Amplifier

Electric Circuits 1 Chapter 7. Response of First-Order RL and RC Circuits 2

7.0. Introduction

7.1. The Natural Response of an RL Circcuuit

7.2. The Natural Response of an RRRRCCCCCC CCCCCCCCiiiiirrrrrccccccuuuuuuuiiiiiitttt

7.3. The Step Response of RL and RRRRRRCCCCCCC CCCCCCiiiiiirrrrrrrcccccccuuuuuuuuiiiiiitttttttsss

7.4. A General Solution for Step andddd NNNNNNNNNNNNNNNaaaaaaaaaaaattttttttttuuuuurral Responses

7.5. Sequential SSwwwwittttccchhhhiiinnnnggggggg

7.6. The Integrattinng AAAmmmmmmmmmppppppppplllliiiffffiiieeeeeeeeerrrrr


7.0. Introductiono Inductors and capacitors are able to store energy.

o We now determine currents and voltages that arisewhen energy is either released or acquired

by an inductor or capacitorin response to an abrupt change in a dc voltage or current source.

o In this chapter, we will focus on circuitsthat consist only of

sources,resistors, andeither (but not both) inductors or capacitors.

o Such circuits are calledRL (resistor-inductor) orRC (resistor-capacitor) circuits.


o Inductors and capacitors are able to store energy.

o We now determine currents and volttaaggges that arisewhen energy is either releasssseeeeedddddddd ooooooorrrrrr aaaaaaaccccccqqqquuuired

by an inductor or capacitooooooorrrrrin response to an abruupppptttttt ccccccchhhhhhhaaaaaaannnnnnggggggeeee in a dc voltage or current source.

o In this chapter, we willl fffffffffoooooooccccccccccccuuuuuuuuuuussssssss ooooooooooonnnnnnnnnnnnn ccccccccciiiiiiirrrrrrrrrrccccuuuuuuuiiiiiitttttttttttsssssssssthat consisssstttttttttt ooooooonnnnnnnnnllllllllyyyyy ooooooooooofffffff

sources,resistors, andeither (but not both) inductors or capacitors.

o Such circuits are calledRL (resistor-inductor) orRC (resistor-capacitor) circuits.


7.0. Introductiono Our analysis of RL and RC circuits will be divided into 3 phases.

o In 1st phase,we consider currents and voltages that arise

when stored energy in an L or C issuddenly released to a resistive network.

o This happens when L or C is abruptly disconnected from its dc source.

o We can reduce circuit to one of 2 equivalent forms shown.

o Currents and voltages that arise in this configuration are referred to asnatural response of circuit.

o Since nature of circuit itself,not external sources,

determines its behavior.


o Our analysis of RL and RC circuits will be divided into 3 phases.

o In 1st phase,we consider currents and voooollllttttttaaaaaaagggggggeeeeeeesssssss ttttttthhhhhhaaaattt arise

when stored energy in annnnnnn LLLL oooorrrr CCCC iiiiiisssssssuddenly released to aaaa rrrrrreeeeeeeesssssssiiiiiisssssssttttttttiiiiivvvvvveeeeee nnnetwork.

o This happens when L ooorrrrrrrrr CCCCCCCCCCC iiiiiiissssssssssss aaaaaaaaaaaabbbbbbbbbbbrrrrrrrrrrrruuuuuuuuuppppppppppttttttttllllyyyyyyyy ddddddddddiiiiiiisssssssssssssccccccccccccooooooooooonnnnnnnnnnnnnnneeeeeeeeeeecccccccccctttttttteeeeeeeedddd from its dc source.

o We can reduccceeeeeee ccccccccccciiiiiirrrrrrrrcccccccccccccuuuuuuuuuiiitttt tttttttttttoooooooooooo oooooooooooonnnnnnnneeeeeeeee oooooooooooffffffffffff 222222222222222 eeeeeeeeeeeeeqqqqqqqqqquuuuuuuuuuuuuiiiiiiiivvvvvvvvvvvvvaaaaaaaaaaallllleeeeeeeeeeeeennnnnnnnnnnttttttt ffffffffooooooooooooorrrrrrrrrmmmmmmmmmmmmmssssssssssssss sssssssssssshhhhhhhhhhhooooooooooooooowwwwwwwwwwwwwwnnnnnnnnnnn..

o Currents and voltageeeeeessssss tttttttttthhhhhhhaaaatttttt aaaaaarrrrrriiiisssssssssseeeeeeeee iiiiiiinnnn tttttttttthhhhhhhhhhiiiiiiissssssss ccccccooooooooonnnnnnnnnffffffffffffiiiiiiggggguuuuuurrrrrraaaaaaaaaattttttttttiiiiiioooooonnnnnnnn are referred to asnatural response of cccccccciiiiiirrrrrrrrrccccccccccuuuuuuuiiiiiiiittttttttt.

o Since nature of circuit itself,not external sources,

determines its behavior.


7.0. Introductiono In 2nd phase,

we consider currents and voltages that arisewhen energy is being acquired by an L or C

due to sudden application of a dc voltage or current source.

o This response is referred to asstep response.

o Process for finding both natural and step responses is the same.

o In 3rd phase,we develop a general method

that can be used to find response of RL and RC circuitsto any abrupt change in a dc voltage or current source.


o In 2nd phase,we consider currents and voltages that arise

when energy is being acqqquuuuuiiiiiiirrrrrrreeeeeeeeddddddd bbbbbbbyyyyyyy aaaannn L or Cdue to sudden applicaaaatttttttiiiiiiiooooooonnnnnn ooooffffff aaaaaaa dddddddcc voltage or current source.

o This response is referred to asstep response.

o Process for finnnnddddddddddddddiiiiiinnnnnnnnnngggggggggggg bbbboooooooooottttttthhhhhhhhhhhh nnnnnnnnnnnnaaaaaaaaaattttttttuuuuuuuuurrrrrrrrrrraaaaaaaaalllll aaaaaaaaaaaaaannnnnnnnnnnddddddddddd sssssssssssttttttttttteeeeeeeeeeeeppppppppppp rrrrrrrrrreeeeeeeeeeeesssssssspppppppppppoooooooooooonnnnnnnnnsssssssssseeeeeeeeeeeeeeessssssssss iiiiiiisssssssssss tttttttttttttthhhhhhhhhhheeeeeeeeeee sssssaaame.

o In 3rd phase,we develop a generaaaalllllllllll mmmmmmmmmmmmmmeeeeeeeeeeeettttttttttttthhhhhhhhhhhooooooooooddddddddddddd

that can be used to find response of RL and RC circuitsto any abrupt change in a dc voltage or current source.


7.0. Introductiono Figure shows 4 possibilities for

general configuration of RL and RC circuits.

o When there are no independent sources in circuit:Thevenin voltage or Norton current is 0.Circuit reduces to one of those shown below.We have a natural-response problem.


o Figure shows 4 possibilities forgeneral configuration of RL and RC circuits.

o When there are no independennnntttt sssssssooooooouuuuuurrrrrrrcccccceeeeeessss iiin circuit:Thevenin voltage or Nortonnnnnn ccccccuuuuurrrrrrrreeeennnntttttt iiiiiiissssss 0.Circuit reduces to one of thoooosssseeeeeeee sssssshhhhhhhhooooooowwwwwwnnnn below.We have a natural-rrrreeeeeeeeeesssssssssppppoooonnnnnnsssseeeeeeeeee pppprrrrrroooobbbbblllleeeemmmmmm..


7.0. Introductiono RL and RC circuits are also known as 1st order circuits.

o Their voltages and currents are described by1st order differential equations.

o No matter how complex a circuit may appear.

o If circuit can be reduced toa Thevenin or Norton equivalent

connected to an equivalent L or C,it is a 1st order circuit.

o If multiple L’s or C’s exist in original circuit,they must be interconnected so that

they can be replaced bya single equivalent element.


o RL and RC circuits are also known as 1st order circuits.

o Their voltages and currents are descrribbed by1st order differential equatiioooonnnnnnnssssssss...

o No matter how complex a circuuuuuiiiiiittttttt mmmmmaaaaaayyyyyyyy aaaaaapppppppppppppear.

o If circuit can be reduced toa Thevenin or Norttttoooooooooonnnnnnnnnn eeeeeeeeeeeeqqqqqqqqqqqqqquuuuuuuuuuuuiiiiiiivvvvvvvvvvvvvvaaaaaaaaaaalllllllllleeeeeeeeeeennnntttttttt

conneccctttteeeeeeeeeeeedddddddddd tttttttttoooooooooooooooo aaaannnnnnnnnnnn eeeeeeeeeeeqqqqqqqqqqqqqquuuuuuuuuuuuiiiiiiivvvvaaaaaaaaaaaalllllleeeeeeeeeeeeennnnnnnnnnntttttttttttttt LLLLLLLLLLL ooooooooooooorrrrrrrrrrr CCCCCCCCCCCCC,,,,,CCit is a 1st ordddddddddeeeeeeeeeeeeerrrrrrrrrrrrr ccccccciiirrrrrrrrccccccccccuuuuuuuuuuuuuiiiiittttttttt.......

o If multiple L’s or C’sCC existttt iiiinnnnnnnnnnn ooooooorrrrrrriiiiiiiigggggggggiiiiiiiinnnnnnnnnaaaaaaaaallllllll ccciiiiirrrrrrccccuuuuuuuiiiiiiittttt,,they must be interconnected so that

they can be replaced bya single equivalent element.










7.0. Introduction








7.1. The Natural Response of an RL Circuito We assume that

independent current sourcegenerates a dc current of Is A.

o Switch has been in a closed positionfor a long time.

o We define phrase a long timemore accurately later in this section.

o For now, it means thatall currents and voltages have reached a constant value.

o Only constant (dc) currents can exist just prior to switch's being opened.

o L appears as a short circuit (Ldi/dt = 0)prior to release of stored energy.


o We assume thatindependent current source

generates a dc current of IIsssssI AAAAAAAA.....

o Switch has been in a closed posssssiiiiittttttttiiiioooonnnnfor a long time.

o We define phrase a lonnnggggggggggg tttttttttttiiiiiiiimmmmmmmmmmmmmeeeeeeeeeemore accuuurrrrrraaaaaaaaaaaaaaattteeeeeeeeeeeellllllyyyyyy llllaaaaaaaaaaaatttttttttteeeeeeeeerrrrrrrrrrrrr iiiiinnnnnnnnnnn ttttttttthhhhhhhhhhhhiiiiiissssssssss sssssssssssssseeeeeeeeeeeeccccccccccccttttttttiiiiiiooooooooooonnnnnnnnnnnn........

o For now, it means thaaaaaaaaaatttttttttttttall currents and voltaaaaaaaaaaagggggggggggggeeeeeeeeeeeeeesssssssss hhhhhhhhhhhhhaaaaaaaaaavvvvvvvvvveeeeeeeeeeee rrrrrrrrrrrrreeeeeaaaaaaaacccccchhhhhhhhhhhhheeeeeeeeeeeeedddddddddddd aaaaaaaaaa ccccccccccccooooooooooooonnnnnnnnsstant value.

o Only constant (dc) currents can exist just prior to switch's being opened.

o L appears as a short circuit (Ldi/dt = 0)prior to release of stored energy.


7.1. The Natural Response of an RL CircuitBecause L appears as a short circuit,

voltage across it is 0;there can be no current

in either R0 or R;all source current Is appears in L.

Finding natural response requiresfinding voltage and current at terminals of R

after switch has been opened, i.e., aftersource has been disconnected andL begins releasing energy.

If we let t = 0 denote the instantwhen switch is opened,

the problem becomes one of finding v(t) and i(t) for t 0.Electric Circuits 1 Chapter 7. Response of First-Order RL and RC Circuits 10

Because L appears as a short circuit,voltage across it is 0;there can be no current

in either R0 or R;all source current IsII appearssss iiiinnnnnnn LLLLLLL....

Finding natural responsssssssseeeeee rrreeeeqqqqqqqqquuuuiiiirrrrrreeeessssfinding voolltttttaaaaaaaaaaaaggggggggeeeeeee aaaannnnnnnnndddddddd cccuuuurrrrrrrrrrrrreeeeeeeennnnnnnnnnttttt aaaaaaaaaatttt tttteeeerrrmmmmmmmmmmmiiiinnnaaaalllsssssss oooooooooofffffffff RRRRRRRRR

after swiiitttccchhhh hhhaaaaaaaaassssssssss bbbbeeeeeeeeeeeeennnnnn oooooooppppppppppeeeeennnnnnnnnneeeeeddddddddd,,,,,,, iiii..eeeee..,,,,,,,, aaaaaaafffffffffffftteeerrrrrrrrsource has bbeennnnn dddddddddiiiiissssssccccccccccoooooooonnnnnnnnnnnnnnnneeeeeeeeeeeccccccttttttteeeeeeeeeeedddddddd aaaaaaaaaannnnnnnnndddddddL begins releasiiing energy.

If we let t = 0 denote the instantwhen switch is opened,

the problem becomes one of finding v(t)t and i(t)t for t 0.Electric Circuits 1 Chapter 7. Response of First-Order RL and RC Circuits 10

7.1. The Natural Response of an RL CircuitDeriving Expression for Current

o Equation is an ordinary 1st order differential equationwith constant coefficients.


Deriving Expression for Current

o Equation is an ordinary 1st ordeeeerrrrrr dddddddiiiiifffffffffffffeeeeeeeerrrrrrreeeeeeeennnnnntttttttial equationwith constant coefficients.



o An instantaneous change of currentcannot occur in L.

o In the 1st instant after switch has been opened,current in L remains unchanged.

o If we use0- to denote time just prior to switching, and0+ for time immediately following switching.

o Initial current in L is oriented in same direction as direction of i.



o An instantaneous change of currrrrrrrrrrrreeeeeennnnnnttttcannot occur in L.

o In the 1st instant after switch hass bbbbbbbbeeeeeeeeeeeeeeeennnnnnnn oooppppened,,current in LL rrreeemmmaaainnnsssss uuuuuuuunnnnnnnnccccccchhhhhhhhhhhaaaaaaaaannnnnnnggggggeeeeeedddddddd..

o If we use0- to denote timee jjjuuussssttttt pppppppprrrriiiioooooorrrr ttttooooo sssssssswwwwiiiittttttttttttccccchhhhiiiiiinnnnngggggggg,,,, aaaannnndddd0+ for time immediatttelly fffollllllowiing swiiittchhhhiing.

o Initial current in L is oriented in same direction as direction of i.



o i(t)starts from an initial value I0 anddecreases exponentially toward 0

as t increases.Electric Circuits 1 Chapter 7. Response of First-Order RL and RC Circuits 13


o i(t)tstarts from an initial value I0 anddecreases exponentially toward 0

as t increases.Electric Circuits 1 Chapter 7. Response of First-Order RL and RC Circuits 13


v is defined only for t > 0, not at t = 0.A step change occurs in v at t = 0.For t < 0, di/dt = 0, so v = Ldi/dt = 0.

v at t = 0 is unknown.



v is defined only for t > 0, nooootttttt aaaaaaattttttt tttttt ====== 00000.A step change occuurrrrrrrrrsssss iiiiinnnn vvvvvvvvv aaaatttttttttt tttt ====== 0000000...For t < 0, ddddiiiiiiiiiiii/////////////ddddddddddttttttt ==== 000000000000,,,, ssssoooo vvvvvvvvvv ======= LLLLLLLdddddddddiiiii////////////dddddttttt ===== 0000....

v at t = 0 is unknown.



As t becomes infinite,energy dissipated in R approaches initial energy stored in L.



As t becomes infinite,energy dissipated in R approaches initial energy stored in L.


7.1. The Natural Response of an RL CircuitTime Constant


Time Constant



o Time constant is an important parameter for the 1st order circuits.

o Several of its characteristics is worthwhile.

o First, it is convenient to think of time elapsed after switchingin terms of multiples of .

o One time constantafter L has begun to release its stored energy to R,

current has been reduced to e-1, orapproximately 0.37 of its initial value.


Time Constant

o Time constant is an important parrrraaaaaaaaaaaaammmmmmmmmmmmmmmeeeeeeeeeeetttteer for the 1st order circuits.

o Several of its ccchhhhaaarrraaaacccttteeeerrrriiiissssttttttttiiiiiiiiiccccssss iiiiissss wwwwoooorrrtttthhhhhhhhwwwwhhhhhhhhhhhhhiiiiiilllllllleeee...

o First, it is conveniientttttttt ttttttttttoooo tttthhhhhhhhiiiiinnnnnnkkkkk ooooooooffff ttttttiiiiiiiimmmmmmmmmeeeeee eeeellllaaaaaapppppppppssssssseeeeeeeeeeeddd aaaaaaaaaafffftttteeeeeeeeerrrrrrr sssswiittttchhhhhhhhiiiingin terms of multipllllessss ooooooooooofffffff ......

o One time constantafter L has begun to release its stored energy to R,

current has been reduced to e-1, orapproximately 0.37 of its initial value.



o When elapsed time exceeds 5 ,current is less than 1% of its initial value.

o We sometimes say that after 5 ,currents and voltages have, for most practical purposes,

reached their final values.

o For single circuits (1st order circuits) with 1% error,phrase a long time implies that 5 or more have elapsed.


Time Constant

o When elapseddd ttttiimmmmeeee eeeexxxxcccceeeeeeedddddddddsss 55555555555 ,,,,,,,current is llleeesssssss tttthhhhhhhhaaaannnn 11111111%%%%%%%% oooofffff iiiiittttssss iiiinnnniiiiittttiiiiaaaalllll vvvvvvvaaaaallllllluuuuee.

o We sometimes say thhattt aaaaaaaffffffffffttttttteeeeeeerrrrrrrrrr 5555555 ,,,,,,,,currents and voltages hhave, fffor most practiicall purposes,

reached their final values.

o For single circuits (1st order circuits) with 1% error,phrase a long time implies that 5 or more have elapsed.



o Existence of current in RL circuit shown isa momentary event and referred to as transient response of circuit.

o Response that exists a long time after switching has taken place is calledsteady-state response.

o Phrase a long time then also meansthe time it takes circuit to reach its steady-state value.


Time Constant

o Existence of ccuuuuuuurrrrrrrrrrrrrrrreeeeeeeeennnnnnnntttt iiiinnnnnnnnnnnnn RRRRRRRRLLLLLL ccccccccciiiiiiiirrccccccccccuuuuuuuuuuiiiiiittttttttt sssssssssssshhhhhhooooooooowwwwwwwwwwnnnnnnnnn iiiiiiisssssa momentary eveeeeeeeeeeeeeennnnnnnnntttttttt aaaannnnnnnnnddddddddddd referred to as transieeeennnnnnnnnnnnttttttttttt rrrrrrrreeeeeeeeeeeeessssssspppppppppppppoooooooooooonnnnnnnnnnnnssssssssssseeee ooooooooooooooffffffffffff ccccccccccccciiiiiiirrrrrrrrrrrccccccccuuuuuuiiiiittttttttttt........

o Response that exists a long time after switching has taken place is calledsteady-state response.

o Phrase a long time then also meansthe time it takes circuit to reach its steady-state value.



o Any 1st order circuit is characterized, in part, by value of its .

o If we have no method for calculating of such a circuit,perhaps because we don't know values of its components,

we can determine its value from a plot of circuit's natural response.

o gives time required for i to reach its final valueif i continues to change at its initial rate:

o Assume thati continues to change at this rate:

o i would reach its final value of 0in seconds.


Time Constant

o Any 1st order circuit is characterized,, inn part, by value of its .

o If we have no method for calcuuuullllaaaaatttttttiiiiiiinnnnnngggg oooooooffff such a circuit,perhaps because we don't kkkkkknnnnnnnoooooowwwww vvvvvvaaaallllluuuuuueeeeeeesss of its components,

we can determine its valuueee ffffffffrrrrrrrrroooooooooommmmmmmm aaaa pplot of circuit's natural response.

o gives time required ffffooooooooorrrrrr iiiiiiiiiiiiii tttttttttttttoooooooooooo rrrrrrrrrrrreeeeeeeeeeeeaaaaaaaaaaacccccccccccchhhhhhhhhhh iiiittttttttsssssssss ffffffffffiiiiiiiiiiinnnnnnnnnnnnnnnnaaaaaaaaaalllllll vvvvvvvvvaaaaaaaallllluuuuuuuuuuuuueeeeeeeeeeeeeif i continuuuueeeeeeeeeeeessssssssss tttttttttttoooooooooooooo ccccchhhhhhhhhhaaaaaaaaaaaannnnnnnnnnnngggggggggggggeeeeeeeeee aaaaattttttttt iiiiiiiitttttttttttttsssssssssss iiiiiiiiiiiiinnnnnnnnnnnnniiiiiiittttttttiiiiiiiiiiiaaaaaaaaaaaallllllllll rrrrrrrrraaaaatttttttttttttteeeeeeeeeeee:::::

o Assume thati continues to changeeeee aaaaattttttttt ttttttttttthhhhhhhhhhhiiiiisssss rrrraaaaaaaatttttttttttteeeee::::

o i would reach its final value of 0in seconds.


7.1. The Natural Response of an RL Circuito Calculating natural response of

an RL circuit is summarized as follows:1. Find initial current, I0, through inductor.2. Find time constant of circuit, = L/R.3. Use equation I0e-t/ , to generate i(t) from I0 and .

o All other calculations of interestfollow from knowing i(t).


o Calculating natural response ofan RL circuit is summarized as follows:

1. Find initial current, I0II , thhhhhrrrrrrooooooouuuuuuggggggghhhhhhh iiiiinnnnddductor.2. Find time constant of ccccciiiiirrrrrrrcccccccuuuuuuiiiitttt,,,, ======= LLLLL/R.3. Use equation I0II e-t/t , toooo gggggggeeeeeeennnnnnneeeeeeerrrrrrraaaaaattttttteeeee iii(t)t from I0 and .

o All other calculations ooofffffffff iiinnnnnntttteeeerrrrrreeeessssssssssttttfollow frommmmmm kkkkkknnnnnnnnnnoooooowwwwwwwwwiiiiiiiinnnnnnngggggggg iiii(((((tttttttt)))))ttt ...



o Switch has been closed for a long time prior to t = 0.

o Voltage across L must be 0 at t = 0-.o iL(0-) = 20 Ao iL(0+) = 20 Ao We replace resistive circuit connected to terminals of L with a single

resistor of 10 :


o Switch has beeeeennnn cccclloooossseeeedddd fffffffooorrr aaa llllloooonnnnggggg tttttiiiimmmmeeee ppppppppprrrriiiiiioooorrrr ttttttoooo ttttttttt ==== 0000...

o Voltage across LLL mussssssssttttttttttt bbbbbeeeee 000000000 aaaaaattttt tttttttt ==== 00000000--..o iL(0-) = 20 Ao iL(0+) = 20 Ao We replace resistive circuit connected to terminals of L with a single

resistor of 10 :



Electric Circuits 1 Chapter 7. Response of First-Order RL and RC Circuits 24Electric Circuits 1 Chapter 7. Response of First-Order RL and RC Circuits 24



o This result is difference betweeninitially stored energy (320 J) andenergy trapped in inductors (32 J).

o Leq for parallel inductorspredicting terminal behavior of parallel combination

has an initial energy of 288 J.Electric Circuits 1 Chapter 7. Response of First-Order RL and RC Circuits 28

oooooooooo TTTTTThhhhhhhhiiiiiiiisssssssss rrrrrrrrrrreeeeeeeeessssssssssssuuuuuuuulllllltttttttttt iiiiiiiisssssssssssss ddddddddddddddiiiiiiiifffffffffffffffffffeeeerence betweeniiiiiiiiiiiiinnnnnnnnnniiiiittttiiiiiiiiaaaaaaaaaaaalllllllllllyyy stored energy (320 J) andeeeennnneerrgy trapped in inductors (32 J).

ooo LLLLLLLLLLeq ffffffffooor parallel inductorspredicting terminal behavior of parallel combination

has an initial energy of 288 J.



o Energy stored in Leq representsamount of energy that will be delivered to resistive network at terminals of original inductors.


oooooooooooo EEEEEEEnnnnnnnneeeeeeeeeeerrrrrrrgggggggggggyyyyyyyyyy ssssssssstttttttttttttooooooooooorrrrrrrrreeeeeeeeeeeeeddddddddddddd iiiiiiiiiinnnnnnnnnn LLLLeq representsaaaaaaaaaaaaammmmmmmmmmooooooooooouuuuuuuuuuuunnnnt of energy that will be ddddeeeelliiivvveeeerred to resistive network at ttttttttttteeeeeerrrminals of original inductors.









7.0. Introduction








7.2. The Natural Response of an RC Circuito Natural response of an RC circuit

is developed from the circuit shown.

o Switch has been in position “a” for a long time,allowing loop made up of

dc voltage source Vg , R1 ,and Cto reach a steady-state condition.

o A capacitor behaves as an open circuitin presence of a constant voltage.

o Voltage source cannot sustain a current.

o Source voltage appears across capacitor terminals.


o Natural response of an RC circuitis developed from the circuit shown.

o Switch has been in position “a” fffooooooorrrrrrrr aaaaaaa lllllllooooooonnnnnngggg time,allowing loop made up of

dc voltage source VgVV , R1 ,aaaannnnnnddddddd CCCCCCCCto reach a steaaaaaadddddddyyyyyyyyyyy---sssstttttttaaaaaatttteeeeeeeeee ccccooooooonnnnnnndddddiiiitttiiioooooonnnn..

o A capacitor beeeehhhhhhhhhhhhhhaaaaavvvvvvvvvvvveeeeeeeeeeesss aaaaaaaaaaassssssssss aaaaaaaaaaannnnnnnnnnnn ooooooooooopppppppppppeeeeeeeeeeeennnnnnnnnn cccccccccccciiiiiiiiirrrrrrrrrrrrccccccccccuuuuuuuuiiiiiiiiitttttttttttttin presence of a cccccccccccoooooooooooonnnnnnnnnnnssssssssttttaaaaaaaaannnnnnnnnnnntttttttt vvvvvvvvvvvvvooooooooooooollllllltttttttttttttaaaaaaggggggggggggeeeeeeeeeee.....

o Voltage source cannot ssssuuuuuuuuuuuuusssssssssssssttttttttttttaaaaaaaaaaiiiiiiiiinnnnnnnnnnn aaaaaaaaaa ccccccccccuuuuuuuuuuuuurrrrrrrrrrreeeeeeeennnnnnnnntttttttttttt....

o Source voltage appears across capacitor terminals.


7.2. The Natural Response of an RC Circuito At t = 0, when switch is moved

from position “a” to position “b”,voltage on C is Vg.

o Because there can be no instantaneous changein voltage at terminals C,

problem reduces tosolving the circuit shown below.


o At t = 0, when switch is movedfrom position “a” to position “b”,

voltage on C is VgVV .

o Because there can be no instannnntttttttaaaaaannnneeeeoooouuuusssss cccccchhhhhangein voltage at terminals C,C

problem reducess tttttooooooooosolvinnnngggggggggggggg tttttttthhhhhhheeeeee ccccccccciiiiiirrrrrrccccccuuuiiitttt sssssshhhhhhhhhoooooooooowwwwwwwwwnnnn bbbbeeeellloooooooooowwwww..


7.2. The Natural Response of an RC CircuitDeriving the Expression for the Voltage

o We can easily find voltage v(t)by thinking in terms of node voltages:


Deriving the Expression for the Voltage

o We can easily find voltage v(t)tby thinking in terms of nodeee vvvvvooooooollllltttttttaaaaaagggggggeeeeeeessssss:::



o Natural response of an RC circuit isan exponential decay of initial voltage.

o Time constant RC governs rate of decay.



o Natural responnseee offf aaaaaaaaaaaannnn RRRRCCCCCCCCC ccccciiiirrrrrcccccccccccuuuuiiiitttttt iiissssssssan exponential decaaayyyyyyyyy ooooooooofffffff iiiiiinnnnnnnnniiiiiiittttttttiiiiiiaaaaaaaallllll vvvvvvvvvooooooolllllllllltttttttttaaaaaaagggggggggggggeeeeeeeeee....

o Time constant RC governs rate of decay.


7.2. The Natural Response of an RC Circuito Calculating natural response of

an RC circuit is summarized as follows:1. Find initial current, V0, across capacitor.2. Find time constant of circuit, = RC.3. Use equation V0e-t/ , to generate v(t) from V0 and .

o All other calculations of interestfollow from knowing v(t).


o Calculating natural response ofan RC circuit is summarized as follows:

1. Find initial current, V0VV , acccccccrrrrrroooooooossssssssssss ccccccaaaaaappppacitor.2. Find time constant of ccccciiiiirrrrrrrcccccccuuuuuuiiiitttt,,,, ======= RRRRRC.CC3. Use equation V0VV e-t/t , toooo gggggggeeeeeennnnnneeeeeeeerrrrraaaaaatttttteeee v(t)t from V0VV and .

o All other calculations ooofffffffff iiinnnnnntttteeeerrrrrreeeessssssssssttttfollow frommmmmm kkkkkknnnnnnnnnnoooooowwwwwwwwwiiiiiiiinnnnnnngggggggg vvvv((((((((((ttt))))))))))t ...



o Switch has been in position “x” for a long time.

o C will charge to 100 V and be positive at upper terminal.

o We can replace resistive network connected to C at t = 0+ withan equivalent resistance of 80 k .

o = (0.5 10-6)(80 103) = 40 msElectric Circuits 1 Chapter 7. Response of First-Order RL and RC Circuits 37

o Switch has been in pppositiioooonnnnn “““““““““““xxxx”””””””””” fffffffffooooorrrr aaaaa lllllllloooooooonnnnnnnggggg tttttttttttiiiiiiiimmmmmmmmmmmmmeeeeeee...

o C will charge too 1111100000000 VVVV aaaannnnddddd bbbbeee ppppoooossssiiiittttiiiivvvveeee aaaaatttt uuuupppppppppppeeeeerrrr ttterrrrmmmmmiiiinnnnaaaallll....

o We can replace resistive neeeetttttttwwwwwwwwwwoooorrrrrrkkkk ccccccccoooonnnnnnnnnnnnneeeeeecccccccttttttttttteeeeeeedddd tttttttttooooo CCCC aaaattttt tttt === 000++++ withan equivalent resistanccceee oofff 888000 kkk .

o = (0.5 10-6)(80 103) = 40 msElectric Circuits 1 Chapter 7. Response of First-Order RL and RC Circuits 37



o Once we know v(t),we can obtain i(t) from Ohm's law.

o After determining i(t)we can calculate v1(t) and v2(t).

o To find v(t),we replace series-connected C’s with a Ceq.

o Ceq has a capacitance of 4 F and is charged to a voltage of 20 V.

o Circuit reduces to one shown below.o = (4)(250) 10-3 = 1 s.


o Once we know v(t)t ,we can obtain i(t)t from Ohm's law.

oooo After determining i(t)twe can calculate v1(t)t and v2(t)t .

ooooo To find v(t)t ,we replace series-connected C’s with CCaaaa CCCCCCCCeqeqeqeqeqeqeeqeqqqCCC ...

ooo CCCCCCCeeqeqeqeqeqeC hhhhhhhhhaaaaaaassss aaaaaaaa ccccaaaappppppaaaaaacccciiiitttaaannncccce of 4 F and is ccccchhhhhhharrrrggggeeeeeddddddddddd ttttttttttoooo aaaaa vvvvoooooooooollllllttttttttttaaaaaaggge of 20 V.

ooooo CCCCCCCCCCCCiiiiiiirrrrccccuuuuiiiiiittttttttt rrrrrreeeeddddddddduuuuuces to one shown below.ooooooooo ====== ((((((((((((444444444444))))((250) 10-3 = 1 s.



o Initial Energy stored in C1 and C2:


o Initial Energy stored in C1 and C2:



o Energy delivered to R:


ooooo Energy delivered to R:



o Energy stored in Ceq is 800 J.o Because Ceq predicts terminal behavior of

original series-connected capacitors,energy stored in Ceq is energy delivered to R.


ooooooooooo EEEEEEEEEEnnnnnnnnnnneeeeeeeeeeerrrrrrrrgggggggggyyyyyyyyyyyyyy sssssssssssttttttttttooooooorrrred in CeqC is 800 J.ooooooooooo BBBBBBBBBBeeeeeeeeeeeccccccccccaaaaaauuuuuuuuusssssseeeeeeeeeee CCCCCCCCCCCeqeqeqeqeqeqeeeqeeCC ppppppppprrrrrrrrrreeeeeeeeeeeeeeddddddddddddiiiiiiiccccccccccctttttttttsssss terminal behavior of

ooooooorrrrrriiiiiigggggggggginnnnaaaaaaaallllll sssseeeeeeeerrrrrrriiiieeeesss-connected capacitors,eeeennnneeeerrrrggggyyy ssttttored in CeqC is energy ddddddddddeeeeelllliiiiivvvveeered to R.









7.0. Introduction








7.3. Step Response of RL and RC Circuitso We now find

currents and voltages generated in the 1st order RL or RC circuitswhen either dc voltage or current sources are suddenly applied.

o Response of a circuit tosudden application of a constant voltage or current source

is referred to as step response of circuit.

o We show how circuit respondswhen energy is being stored in L or C.


o We now findcurrents and voltages generated in the 1st order RL or RC circuits

when either dc voltage or cccccccuuuuuuurrrrrrrrrrrrrreeeeeeennnnnnntttttt ssssooources are suddenly applied.

o Response of a circuit tosudden application of a connnsssstttttttaaaaaaaannnnnnnttttttt vvvvvvvoooooollllttttaaage or current source

is referred to as sssstttttttttteeeeeeeepppp rrrreeeeeeeeeessssppppppppppoooonnnnnnsssseeeeeee ooooffff cccccciiiirrrcccuuuiiitttt......

o We show howww cccccccccccccciiiiiirrrrrrcccccccccuuuuuuuuuuiiitttt rrrrrrrrrrrreeeeeeeeeeesssssssssssppppppppppppppoooooooooonnnddddddddddddssssssswhen energy is bbbbbeeeeeeeeeeeeeeiiiiiiiiinnnnnnnnnnnggggggg ssssssssstttttttttooooooooooorrrrrrrrreeeeeeeeeeeeeddddddddddddd iiiiiiiinnnnnnn LLLLLL ooooooooooooorrrrrrrr CCCCCCCCCCCCCC....


7.3. Step Response of RL and RC CircuitsThe Step Response of an RL Circuit

o Energy stored in L when switch is closedis given in terms of

a nonzero initial current i(0).

o Task is to find expressions for i(t) and v(t)after switch has been closed.


The Step Response of an RL Circuit

o Energy stored in L when switch is clooseedis given in terms of

a nonzero initial currennnnnnntttttt iiiii(((((0000)))))....

o Task is to find expressions for i((((ttt))))))ttt aaaaaaaannnnnnnnnddddddddd vvvvvv(((((ttt)))ttafter switch has beeeeeeeeeeennnnnnn cccccccccccllllllloooooooooooosssssssssseeeeeeeeeddddddddddddd..



Algebraic sign of I0 is positive if I0 is in same direction as i.

Otherwise, I0 carries a negative sign.

When initial energy in L is 0, I0 = 0:



Algebraic sign of I0I is poooosssssssssssiiiitttttttttttttiiiiiiiiivvvvvvvvvvveeeeeeeeee iiiiiiffffffffff IIIIIIIIIIII0000II iiiiiisssssss iiiiiinnnnnn sssssssssssaaaaaaaaaammmmmmmmmmeeeeeeee dddddddddddddiiiiiiirrrrrrrreeection as i.

Otherwise, I0I carries a negative sign.

When initial energy in L is 0, I0II = 0:



o 1 after switch has been closed,current will have reached

approximately 63% of its final value:



o 1 after switch has been closedddddd,,,current will have reached

approximately 66663333333333%%%%%%%%%%%%%% ooooooooooooooffffffffffff iiiiiittttttttttttsssssssssss ffffffffffffiiiiiiinnnnnnnnnnnaaaaaaaallll vvvvvvvvvvvvaaaaaaaaaaallllluuuuuuuuuuueeeeeeeeeee::::::::::



o If current continuesto increase at its initial rate,

it reaches its final value at t = :



o If current continuesto increase at its initial rate,

it reaches its finaaaallllllll vvvvvvvvvvvaaaaaaaaaaaalllllllllllluuuuuuuuuuuueeeeeeeeeeeee aaaaaaaaaaaaaattttttttt ttttttttt ==== ::::::::



o Voltage across L is 0 before switch is closed (t < 0).

o Initial current is I0.

o L prevents an instantaneous change in current.

o Hence, current is I0 in instant after switch has been closed.

o Voltage drop across resistor is I0R.

o When switch is closed, v jumps to Vs - l0R and decays exponentially to 0.Electric Circuits 1 Chapter 7. Response of First-Order RL and RC Circuits 50


o Voltage acrossssss LLLLLLLLLLLL iiiiiisssssss 0000 bbbbbbbbbbbbeeeeeeeeeffffffoooooooorrrrrrreeeeeeeeeee ssssssssswwwwwwwwwwwiiiiiitttttttttccccccccccccchhhhhhhh iiissssssss cccccccccccllllllloooooooosssssseeeeeeeedddddddddddd (((((tttttttttt <<<<<<<<< 00000000000)))))))))))....

o Initial current is I0I .

o L prevents an instantaneeeeeeeeeoooooooooooouuuuuuuuuuussssssssss cccccccccchhhhhhhhhhaaaaaaaaannnnnnnnnnggggggggggggeeee iiinnnnnn ccccccccuuuuuuuuurrrrrrrrrrrrrrreeeeeeeeeennnnnnnnnnntttttttttt..

o Hence, current is I0I in instant after switch has been closed.

o Voltage drop across resistor is I0I R.

o When switch is closed, v jumps to VsVV - l0R and decays exponentially to 0.Electric Circuits 1 Chapter 7. Response of First-Order RL and RC Circuits 50


o When initial inductor current is 0:

o Initial current is 0.

o v jumps to Vs.

o v approaches 0 as t increases.

o Current is approaching constant value of Vs/R.Electric Circuits 1 Chapter 7. Response of First-Order RL and RC Circuits 51


o When initial iinnnnnnnddddddddddddddduuuuuuuucccccctttttttoooorrrrrrrrr cccccccccuuuuuuurrrrrrrrrrrrreeeeeeeeeeennnnnnnttttttt iiiiiisssssss 0000000000000:::::

o Initial current is 0.

o v jumps to VsVV .

o v approaches 0 as t increases.

o Current is approaching constant value of VsVV /R.Electric Circuits 1 Chapter 7. Response of First-Order RL and RC Circuits 51


o When I0 = 0:



o When I0 I = 0:


7.3. Step Response of RL and RC Circuits

o Switch has been in position a for a long time.o L is a short circuit across 8 A current source.o L carries an initial current of 8 A.o This current is oriented opposite to reference direction for i: I0 = -8 A.o When switch is in position b: final value of i will be 24/2 = 12 A.o = L/R = 200/2 = 100 ms.


o Switch has been in posssiiiiiiittttttttttiiiiioooooooooonnnnnn aaaaaaaa fffffffffoooooooooooorrrr aaaaaaaaaaa lllllllllloooooooooooonnnnnggggggggg ttttttttiiiiiiimmmmmmmmmeeeeeeeee..o L is a short circuit across 8 AAAAAAAAAAAAA ccccccccccccuuuuuuuurrrrrrrrrrrrrrrrrrreeeeeeeennnnnnnnnntttttttttttt ssssssssssoooooooooooouuurrrrrrrcccccceeeee..o L carries an initial current of 8 A.o This current is oriented opposite to reference direction for i: I0 = -8 A.o When switch is in position b: final value of i will be 24/2 = 12 A.o = L/R = 200/2 = 100 ms.



o In instant after switch has been moved to position b,L sustains a current of 8 A counterclockwise around newly formed closed path.

o This current causes a 16 V drop across 2 resistor.

o This voltage drop adds to drop across source, producing a 40 V drop across L.


o In instant after switch has bbbbbbbbbeeeeeeeeennnn mmmmoooovvvveeeedddddddddddd tttoooo ppppooossssssiiiiittttttttiioooonnnn bbbbbbbbb,L sustains a current of 8 A counterclockwise around newly formed closed path.

o This current causes a 16 V drop across 2 resistor.

o This voltage drop adds to drop across source, producing a 40 V drop across L.


7.3. Step Response of RL and RC CircuitsThe Step Response of an RC Circuit

o For mathematical convenience,we choose Norton equivalent of network

connected to equivalent capacitor.

o For RL circuit, we had:

o By comparing the equations, we have:


The Step Response of an RC Circuit

o For mathematical convenience,we choose Norton equivaleeennnntttttt oooooooofffffff nnnnnnneeeeeeettttttwwwwwooork

connected to equivalent cccccccaaaaaappppaaaacccciiiittttoooooorrrrrrr....

o For RL circuit,, wwwwwwwwwwwweeeeee hhhhhhaaaaddddddddd:::::

o By comparing the equations, we have:



o We obtained solutions byusing a mathematical analogy to

solution for step response of RL circuit.

o Let's see whether these solutionsfor RC circuit make sense in terms of

known circuit behavior.

Initial voltage across C is V0.Final voltage across C is ISR.

= RC.Solution for vC is valid for t > 0.



o We obtained solutions byusing a mathematical analoooggggyyyyyyy ttttttttooooooo

solution for step responseeeeeee ooooooffffff RRRRLLLL ccccciiiirrrrrcccccuit.

o Let's see whether these solutioonnssssfor RC circuit makee ssssssssssseeeeeeeeeennnnnnnnnnnssssssssssssseeeeeeeeeeee iiiiiiinnnnnnnnnnnnn ttttttteeeeeeeeerrrrrrrrrmmmmmmmsssssssssss ooooooooooffffffffff

known cccciiiiiiiiirrrrrrrrrrrccccuuuuuuuuuiiiiiiiittttt bbbbbbbbbeeeeeeeeeeeehhhhhhhhhaaaaaaaaaaaaavvvvvvvvvvviiiiiiiioooooorrrrrr........

Initial voltage across C is V0VV .Final voltage across C is ISI RSS .

= RC.CCSolution for vC is valid for t > 0.



Current in C at t = 0+ is Is – V0/R.This prediction makes sense.

Capacitor voltage cannot change instantaneously.Initial current in R is V0/R.Capacitor branch current changes instantaneously

from 0 at t = 0- to Is – V0/R at t = 0+.Capacitor current is 0 at t = + .



Current in C at t = 0000++++++++++ iiiiiiiiiisssssssssss IIIIIIIIIIIIIssssII ––––––––– VVVVVVVVVVVV0000VVVV //////////RRRRRRRRRR...This prediccccttttttttttttiiiiiiiioooooooooooonnnnnnn mmmmmmmmmaaaaaaaakkkkkkkkkkkeeeeeeeeeeessssss sssssssseeeeeennnnnnnnnnnssssssssseeeeeeeeeee...

Capacitor voltaaaaaaagggggggggggeeeeeee ccccaaaaaaannnnnnnnnnnnnnnnooooooootttttttt cccccccccchhhhhhaaaaaaannnnnnngggggggeeeeeee iiiiiiinnnnnnsssssssssttttttttttttaaaaaaaaannnnttttttttaaaaaannnnnnnnnnnneeeeeeeeoooooouuuuuusly.Initial current in RRRRRR iiiisssssss VVVVVVVVVVVV000VVVVV //////////RRRRRRRRRRR.Capacitor branch current changes instantaneously

from 0 at t = 0- to IsI – V0VV /R at t = 0+.Capacitor current is 0 at t = + .



o We find Norton equivalent with respect to terminals of C for t > 0:


o We find Norton equivaallllllleeeeeeennnnnnnnnntttt wwwwwwwwwwiiiitttthhhhhhh rrrreeeeeeeesssspppppppeeeeeeeeccccctttt ttttooooo tttteeeeerrrrrrrmmmmmiiiiiinnnnaaals of C for t > 0:



o We check consistency of solutions:


o We check consistency of solutions:










7.0. Introduction








7.4. Solution for Step & Natural Responseso To generalize solution of these 4 possible circuits,

we let x(t) represent unknown quantity.

o x(t) can have 4 possible values:i(t) or v(t) of an L or C.

o Constant K can be 0.

o Sources are constant voltages and/or currents.

o Final value of x or xf will be constant.

o Final value must satisfy above equation.

o When x reaches its final value, dx/dt = 0:


o To generalize solution of these 4 possible circuits,we let x(t)t represent unknown quantity.

o x(t)t can have 4 possible values:::i(t)t or v(t)t of an L or C.

o Constant K caaannnnnnnn bbbbbeeeeeeeeee 0000..

o Sources are constantttttttttttt vvvvvvvvvvvvoooooooooollllttttaaaaaaaaagggggggggggggeeeeeeeessssssssss aaaaaaaaannnnnnnnnnnnddddddddd////////////oooooooooorrrrrrrrrrrr ccccccccuuuuuuuuuuurrrrrrrrrrrrrreeeeeeeeeeennnnttttssssssssssss..

o Final value of x or xfx willlll bbbbbbbbbbbbbbeeeeeeeeeeeeee ccccccccccccooooooooooonnnnnnnnnnssssssssssttttttttttttaaaaaaaaaaaannnnnnnnnnntttt...

o Final value must satisfy above equation.

o When x reaches its final value, dx/dt = 0:


o To obtain a general solution,we use time t0 as lower limit and t as upper limit.

o t0 corresponds to time of switching or other change.

7.4. Solution for Step & Natural Responses


o To obtain a general solution,we use time t0 as lower limit and t as upper limit.

o t0 corresponds to time of switching or other change.




o Previously we assumed that t0 = 0.

o u and v are symbols of integration.


o Previously we assumed that t0 = 0.

o u and v are symbols of integration.

Calculating natural or step response of RL or RC circuits:

1. Identify variable of interest for circuit.For RC circuits, it is best to choose vC.For RL circuits, it is best to choose iL.

2. Determine variable initial value (its value at t0).If variable is vC or iL, it is not necessary

to distinguish between t = t0- and t = t0

+.They both are continuous variables.If another variable is chosen,

its initial value is defined at t = t0+.

3. Calculate variable final value (its value as t ).

4. Calculate for circuit.

5. Use:



Calculating natural or step response of RL or RC circuits::

1. Identify variable of interest for circuit.For RC circuits, it is best to chooooooooooossssssseeeeeee vvvvvCCCCCCC.For RL circuits, it is best to chhhhhhoooooooooooooosssssseeee iiiiLLLLii ....

2. Determine variable initial valueeee ((((iiiiitttttttssssss vvvvvvvvaaaaaaallllluuuuuuueeee aaat t0).If variable is vC or iLi , it is not neeccccccceeeeeeeessssssssssssssaaaaaaarrryyy

to distinggguish bettwwwwwwwwwweeeeeeeeeeeeeeeeeeeeennnnnnnnnnn ttttttttt ========= tttttt0000----- aaaaaannnnnnnnndddddddddd tttttttttttt ============ tttttttt0000

+++++++.They both aaaarrrrrrrrrrrreeeeeeeeee ccccccccccoooooooooooonnnnnnnttttiiiiiiiinnnnnnnnuuuuuuuuuuuoooooooooooouuuuuuuusss vvvvvvvvvaaaaaaaaaarrrrrrrriiiiiiiaaaaaaaaaaaaabbbbbbbbbbbblllllleeeeeeeeeeesssssssssss....If another variableeeeeeeeee iiiiiiiiiiissssssss cccchhhhhhhhhoooooooossssssssssssseeeeeeennnnnnnnnnn,,,,

its initial value is deeeeffffffffffffiiiiiiiiiiinnnnnnnnnnnnnneeeeeeeeeeeedddddddddddddd aaaaaaaaaatttttttttttt ttttttttt ============= tttttt0000++++++++...

3. Calculate variable final value (its value as t ).

4. Calculate for circuit.

5. Use:


Calculating natural or step response of RL or RC circuits:

o You can then find equationsfor other circuit variables using:

circuit analysis techniques orby repeating preceding steps

for other variables.



Calculating natural or step response of RL or RC circuits::

o You can then find equationsfor other circuit variables ussssiinnnnnnngggggggg:::::::

circuit analysis techniqueeeeeessssssss oooooorrrrby repeating preceding stttteeeeeeppppppppssssss

for other variaabbbbllleeessss....


o Switch has been in position a for a long time.

o C looks like an open circuit.

o After switch has been in position b for a long time,C will look like an open circuit in terms of 90 V source.

o Final value of vC is +90 V.



vC = v60 =o Switch has been in positiooonnnnnnnn aaaaaaaaaaa fffffffffffoooooooooorrrrrr aaaaaaaaaaaa llllllllloooooooonnnnnnnnnngggg ttttttiiiiimmmmmmmmmmmmeeeeeeeeeee.....

o C looks like an ooooppppppppppppppeeeeennnnnnnnnnn ccccccccciiiirrrccccuuuuuuuuuuuuiiiiiiiiiiiittttttttt.....

o After switch has been in poooossssssssssssiiiiiiiiiiitttttttttttttiiiiiiiiiooooooooooonnnnnnnnnnnnn bbbbbbbbbbbb fffffffffffffoooooooooooorrrrrrrrrrr aaaaaaa llllloooooooonnnnnnnnnnngggggggggg tttttttiiiiiiiiimmmmmmmmmmmmeeeeeeeeeeeee,,,,,,,C will look like an open circuit in terms of 90 V source.

o Final value of vC is +90 V.


vvvvvvvCCCCCCCCCC =========== vvvv666666660000000 =========

o doesn't change for i(t).o We need to find only initial and final values for i.o When obtaining initial value, we must get i(0+).o Current in capacitor can change instantaneously.o This current is equal to current in resistor:o Final value of i(t) is 0.

o Alternative solution is i(t) = CdvC(t)/dt.



o doesn't change for i(t)t ..o We need to findddd oooooooooonnnnnnnnnnnnllllllyyyyyyyyyyy iiiiiinnnniiiittttttttttttiiiiiiiaaaaaaaaalllllllllll aaaaaaaaaaannnnnnndddd ffffffffffiiiiiinnnnnnnnnnnnaaaaaaaalllllllll vvvvvvvvvaaaaaaaaaalllllllluuuuuuuuuuueeeeeeeeeessssssss fffffffffoooooooorrrrrrrrrrr iiiiiiiii.o When obtaining initial vvvvvvvvaaaaaaaaaaallllllllllluuuuuueeee,,,, wwwwwwwwwwweeeeeeeee mmmmmmmmmmmuuuuuuuuuuuuusssssssstttt gggggggggeeeeeeetttttttttt iiiiiiii(((((((((0000000000000++++++++)))))))))).....o Current in capacitor can chhhhaaaaaaaaaaannnnnnnnnnnnnggggggggggeeeeeeeeeee iiiiiiiinnnnnnnnnssssssssssttttttttttaaaaaaaaaaannnnnnnnnnnntttttttttttaaaannnnnnneeeeeeooooooooooouuuuuuuuuusssssssslllllllllyyyyyyy...o This current is equal to current in resistor:o Final value of i(t)t is 0.

o Alternative solution is i(t) = t CdvC(t)/t dt.Electric Circuits 1 Chapter 7. Response of First-Order RL and RC Circuits 70


o Magnetically coupled coilscan be replaced by a single L

having an inductance of:

o By hypothesis, initial value of io is 0.

o Final value of io will be 120/7.5 or 16 A.

o of circuit is 1.5/7.5 or 0.2 s.


See Problems 6.40 and 6.41.

o Magnetically coupled coilscan be replaced by a single LLLLLLL

having an inductance of:

o By hypothesisss,,, iiiiiiiinnnnnniiiiiiitttttttiiiiiaaaalll vvvvvvvvvvvvaaaaaaallluuuueeee oooooofffffffff iiiiiiiooooooooo iiiiiiiissss 0000..

o Final value of io will bbbbbbbbbbbeeeeeeeeeeee 111111112222220000000/////////////7777777777...55555555555 oooooooooooorrrrr 111111666666666666 AAAAAAAAAAAAAAAAA....

o of circuit is 1.5/7.5 or 0000000000000..22222222222222 ssssssssssss...


See Problems 6.40 and 6.41.









7.0. Introduction








o Whenever switching occurs more than once in a circuit,we have sequential switching.

o For example,a single 2-position switch may be switched back and forth, ormultiple switches may be opened or closed in sequence.

o Time reference for all switchings cannot be t = 0.

o Wedetermine v(t) and i(t) for a given position of switch or switches andthen use these solutions to determine initial conditions

for next position of switch or switches.



o Whenever switching occurs more than once in a circuit,we have sequential switching.

o For example,a single 2-position switch mmmmaaaaaaayyyyyy bbbbbeeee sssswwwwwwwiiiiiiitttttttcccched back and forth, ormultiple switches may be ooppppeeeeeennnnnnneeeeeeedddddddd oooooorrrrrr ccclosed in sequence.

o Time reference for all sswwwwwwwwwwwiiiiiiiitttttttttttccccccccccccchhhhhhhhhhiiiiiinnnnnnnnnnnggggggggggggggssssssss ccccccccccaaaannnnnnnnnnnnnnnnnnnooooooooottttttttttttt bbbbbbbbbbeeeeeeeee tttttt ========== 000000000.....

o Wedetermine v(t)t annnnnnnnnnddddddddddddd iiiiiiii((((ttttttttt))))))))))tttt fffffffffoooooooooorrrrrrrrrrr aaaaaaaaaaaaa ggggggiiiivvvvvvvvvvveeeeeeeeeeeennnnnnnnnn pppppppppppppooooooooooosssssssssssiiiiiiiiiitttttttttiiiooooooooooonnnnnnnn ooooooooooooffffffffff ssssssssssssswwwwwwwitch or switches andthen use these solutiiiiooooooooooooonnnnnnnnnnnnnnssssssssss tttttttttttttooooooooooo dddddddddddddeeeeeeeeeettttttttttttteeeeeeeeeeeerrrrmmmmmmmmiiiiiiiinnnnnnnnnnnneeeeeeeee iiiiiiiinnnnnnnnnniiiiiiiiiiittttttttttttiiiiiiiiiaaaaaaaaaaaallllll conditions

for next position of switch or switches.


o With sequential switching problems,obtaining initial value x(t0) is important.

o Anything but inductive currents and capacitive voltagescan change instantaneously at time of switching.

o Solving first for inductive currents and capacitive voltagesis even more pertinent.

o Drawing circuit pertaining to each time interval is often helpful.



o With sequential switching problems,obtaining initial value x(t0) is important.

o Anything but inductive currentttssss aaaaaaannnnnnnnddddddd cccccccaaaaaaappppppaaaacitive voltagescan change instantaneouslyyyyyyy aaaaaattttt ttttiiiimmmmeeee oooooooofffff switching.

o Solving first for inductive curreenntttsssss aaaaaaaaaaannnnnnnnnddddddddd ccccaappacitive voltagesis even more pertinnneeeeeeeeeeennnnnnnnnnnttttttttttt....

o Drawing circuuuiiiittttttttttttt pppppppppppppeeeeeeeeeeerrrrrrrrrrrttttttttaaaaiiiiinnnnnnnnnnnniiiiiinnnnnnnnnnnnnngggggggggggg ttttttttoooooooooo eeeeeeeeeeeaaaaaaaaaaaaccccccccccccccchhhhhhhhhhhhhh ttttttttttttiiiiiiimmmmmmmmmmmmmeeeeeeeeeeeee iiiiiiinnnnnnnnntttttttttttteeeeeeeeeeerrrrrvvvvvvvvaaaaaaaaaaaallllll iiiiiiiiissssssssss oooooooooooofffffffffffftttttttttteeeeeeeeeennnnnnnnnnnnn hhhhhhhhhhhheeeeeeeeeelllpful.



o At instant switch is moved to position b, initial voltage on C is 0.

o If switch were to remain in position b, C would eventually charge to 400 V.

o when switch is in position b is 10 ms.


o At instant switchh is movvvvvvvvveeeeeeeeddddd ttttoooooooo ppppppppppooooossssssiiiiiiiittttiiiiooooonnnnnnnnnn bbbbbbbbb,,,,,, iiiinnnniiiitttttiiiiiaaaaalllllll vvvvvvvvvvooooooooolttttaaaaaaaaggggggggggeeee oooooooooonnnnnnn CCC iis 000.

o If switch were to remain in pppppppppppoooooooooooosssssssiiiiitttttttttttiiiiiiiioooooooooonnnnnnnnnn bbbbbbbbbbb,,,,,,,, CCCCCCCCCC wwwwwwwwwwooooooooooouuuuuuuuuuuuulllllllldddddddddddd eeeeeeeeevvvvvvveeeeeeeennnnnnnnnnnnttttttttttuuuuuually charge to 400 V.

o when switch is in position b is 10 ms.



o Switch remains in position b for only 15 ms.

o This expression is valid for 0 < t < 15 ms.

o When switch is moved to position c, initial voltage on C is 310.75 V.


o Switch remains in position bbbbbbbbbb ffffffffffffooooooooooorrrrrrrr oooooooooooonnnnnnnnnnllllllyyyyyyyyyyyyyy 1111111111115555555555 mmmmmmmmmmmmsssssssss.......

o This expression is valid for 0 < t < 15 ms.

o When switch is moved to position c, initial voltage on C is 310.75 V.


o With switch in position c, final value of C voltage is 0.

o is 5 ms.

o t0 = 15 ms



o With switch in posiiittiiionnnnn cccccccccc,,,,,, ffffffiiiinnnnaaaaaaallll vvvvvaaaaaallllllluuuueeeee ooooooooofffffff CCCCCC vvvvoooolllltttttaaaaaaagggggggeeeeeeeeeeee issss 00000000000.....

o is 5 ms.

o t0 = 15 msElectric Circuits 1 Chapter 7. Response of First-Order RL and RC Circuits 78

o C voltage will equal 200 V at 2 different times:



o C voltage will equal 222000000000000000 VVVVV aaaattttttt 22222 ddddddiiiifffffffffffffffeeeerrrreeeeeeeeeennnntttttt tttttiiiimmmmeeeeeessss::::










7.0. Introduction








o Output voltage isproportional to integral of input voltage.

o We assume that op amp is ideal.

o t0 represents instant in time

when we begin integration.

o vo(t0) is value of output voltage at that time: vo(t0) = vCf(t0)



o Output voltage isproportional to integral of input volltage.

o We assume that op amp is ideaaallll...

o t0 represents instant in time

when we begin integration.

o vo(t0) is value of outpuuuuttttttttttt vvvvvvvvvvvoooooooooooooolllllllllllltttttttttttaaaaaaaaagggggggggeeeeeeeeeee aaaaaaaaaaaatttttttttt tttthhhhhhaaaaaaaaaaaaatttttttttt tttttttttttiiiiiiiiiimmmmmmmmmmmmmmeeeeeeeeeee:::::: vvvvvvvvvvvvvoooo((((((((((((ttttttttttt0000)))) = vCfff(ff t0)


o Output voltage equalsinitial value of voltage on Cf

plus an inverted (minus sign),scaled (1/RsCf) replica of integral of input voltage.

o If no energy is stored in Cf when integration starts,output voltage is proportional to integral of input voltage only if opamp operates within its linear range (if it doesn't saturate).



o Output voltaggeeee eeeeqqqqqquuuuaaalllssssinitial valuuueee oooofffff vvvvoooollllttttaaagggeeee oooonnnn CCCCCCCCCfffffCCCC

plus an inverteeddddddddd (((mmmmmiiinnnnuuuussss ssssiiiiggggnnn)))))))),,,,scaled (1/RsCfC )ff rrrreepppllliiiccaaa oooffff iiinnttteegggrrraaall oooffff iiiinnnpppuut voltage.

o If no energy is stored in CfCC when integration starts,output voltage is proportional to integral of input voltage only if opamp operates within its linear range (if it doesn't saturate).


o During this interval, vo is decreasing, and op amp eventually saturates at -6 V:



vO(0.009) = -5 + 9 = 4 Vo During this interval, vo is decreasing, and op amp eventually saturates at -6 V:


vO(0.009) = -5 + 9 = 4 V

o Integrating amplifier can perform integration function very well,but only within specified limits

that avoid saturating op amp.

o Op amp saturates due toaccumulation of charge on feedback capacitor.

o We can prevent from saturatingby placing an R in parallel with Cf.



o Integrating amplifier caaaannnnnnnnnnn pppppppppppppeeeeeeeeeeeeeerrrrrrrrrrrrfffffffffffoooooooooooorrrrrrrrrrrmmmmmmmmmmmm iiiiiiinnnntttttteeeeeeeeeeegggggggggggrrrrrrrrrrrrraaaaaaaaaaaaaatttttttttiiiiiiiiooooooooooooooonnnnnnnnn ffffffffffffuuuuuuuuuuuunnnnnnnnnnnncccccttttion very well,but only wwwwiiiitttttttttttthhhhhhhhhhhiiiiiiiinnnnnnnnnn ssssssssppppeeeeeeeeeeeecccccccccccciiiiiiiiiiiifffffffffffffiiiiiiiiiieeeeeeeeeedddddd lllllliiiiiiiiimmmmmmmmmmmmmiiiiiiiiiiiitttttttttttttssssssssssss

that avoid satuuuuuuuuuurrrrrrrrrrrraaaaaaaaaaattttttttiiiinnnnnnnngggggggg ooooooooopppppppppp aaaaaaaaaaaaammmmmmmmmmppppppppppp.

o Op amp saturates due tooooaccumulation of charge on feedback capacitor.

o We can prevent from saturatingby placing an R in parallel with CfCC .


o We can convert integrating amplifier to

a differentiating amplifier

by interchanging Rs and Cf:

o We can design both integrating- and differentiating-amplifier circuits

by using an L instead of a C.

o Fabrication of capacitors for integrated-circuit devices is much easier.

o Inductors are rarely used in integrating amplifiers.



o We can convert integrating amplifier to

a differentiating amplifier

by interchanging Rs and CCCCffffffCCCCC ::::

o We can designnn bbbbbbbbboooottthhhhhhhhhhh iiiinnnntttteeegggrrraaaatttiiiinnnngggggg-- aaaannnnddddddd dddddddddiiiiiffffffffffffffffffeeeeerrrreeeennttttiiiiiiiaaaatttttiiiiiinnnngggggg--aaaammmppppppppppllllliiiiiiiiiffffffiiiier circuits

by using an L insteaddd oooooooooofffff aaaa CCCC....

o Fabrication of capacitors for integrated-circuit devices is much easier.

o Inductors are rarely used in integrating amplifiers.


rl and rc circuits - kntu homepage webftp clientwp.kntu.ac.ir/faradji/ec1/ec1_ch7.pdf · 7.0....

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