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AC Circuits Transient Analysis

Enzo Paterno Page 1

AC circuits Transient Analysis

1st order networks (RC & RL)

2d order networks (RLC)

Author: Enzo Paterno EP Revision 1/07/15

MADE IN THE USA


Enzo Paterno Page 2

CONTENTS

1 First Order RC Circuit Transient Analysis ........................................................................ 3 1.1 RC Circuit Capacitor Charging Phase ....................................................................... 3 • Capacitor current IC (t) with initial condition 0)0( =−

CV ...................................... 3

• Capacitor current IC (t) with initial condition iC VV =− )0( ..................................... 5

• Capacitor voltage VC (t) with initial condition 0)0( =−CV .................................... 6

• Capacitor voltage VC (t) with initial condition iC VV =− )0( ................................... 8

• A Shortcut approach to find the voltage )(tVC : ...................................................... 10 • Summary – Charging phase behavior for an RC circuit: ......................................... 11

1.2 RC Circuit Capacitor Charging Discharging Phase ................................................. 12 • Capacitor current IC (t)............................................................................................. 12 • Capacitor voltage VC (t) ........................................................................................... 14 • Summary – Capacitor Charging / Discharging phases ............................................ 15

1.3 Capacitor Transient Phases – Taking another look .................................................. 16 1.4 Capacitor Transient Phases – General Remarks ...................................................... 18

2 First Order RL Circuit Transient Analysis ...................................................................... 21 2.1 Inductor Storage Phase ............................................................................................ 21 • Inductor voltage VL (t) with initial condition 0)0( =−

LI .................................... 21

• Inductor current IL (t) with initial condition 0)0( =−LI ...................................... 23

• Inductor current IL (t) with initial condition iL II =− )0( ...................................... 24 • Summary – Storage phase behavior for an RL circuit: ............................................ 25

2.2 Inductor Storing Phase – Taking another look ........................................................ 26 2.3 Taking another look – General Remarks ................................................................. 28


Enzo Paterno Page 3

1 First Order RC Circuit Transient Analysis Circuits containing only a single storage element are defined as first-order networks and result in a first-order differential equation (i.e. RC & RL circuits).

1.1 RC Circuit Capacitor Charging Phase

Capacitor current IC (t) with initial condition 0)0( =−CV

The RC Circuit analysis provides a 1st order Differential Equation when performing the transient analysis:

@t = 0, the switch is closed providing a current path in the circuit. Using Kirchhoff’s voltage law:

Eq1: )()( tVtVE CR +=

The voltage across a resistor, the voltage across a capacitor and the current through a capacitor is

defined as:

Eq2: RtItV CR )()( =

Eq3: ∫−∞=

=t

tCC dI

CtV tt )(1)(

Eq4: dt

tdVCtI CC

)()( =

We solve for IC (t) and thus, substitute Eq2 and Eq3 into Eq1 and divide by R on both sides

∫+= dttIC

RtIE CC )(1)(

Eq5: ∫+= dttIRC

tIRE

CC )(1)(

E

VR (t) VC (t)

IC (t) +

-


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Differentiate both sides of Eq5:

Eq6:

+=

∫ dttI

RCtI

dtd

RE

dtd

CC )(1)(

Rearrange and evaluate the derivatives of Eq6, giving a first-order linear differential equation:

( )

=+ ∫ R

EdtddttI

dtd

RCtI

dtd

CC )(1)(

Eq7: 0)()( =+RC

tItIdtd C

C

We solve differential equation Eq7 by first rearranging the equation:

RCtItI

dtd C

C)()( −=

Eq8: dtRCtI

tId

C

C 1)()(

−=

Integrate both sides of Eq8:

∫∫ −= dtRCtI

tId

C

C 1)()(

21)( KRC

tKtIC +−

=+ ln

Combine the constants of integration such that K3 = K2 – K1:

Eq9: 3)( KRC

ttIC +−

= ln

Solve Eq9 for )(tIC :

( ) 33)( KeRC

t

eK

RCt

etCI

−

=+

−

=

lne

Let K4 = 3Ke

Eq10: RCt

eKtCI

−

= 4)(


Enzo Paterno Page 5

We solve Eq10 for K4 using initial conditionREIC =+ )0( which occurs when 0)0()0( == +−

CC VV :

REeKIC ==+ 04)0(

Giving:

Eq11: REK =4

Substitute Eq11 into Eq10. We can see that when 0)0( =−CV , the current )(tIC through capacitor C is

given by:

Eq12: RCt

eREtCI

−

=)( We define RC to be the network time constant with t =RC [sec].

We assume that steady state is reached at t = 5 t

The Changes in IC (t) between time constants:

Capacitor current IC (t) with initial condition iC VV =− )0(

If the initial voltage across the capacitor is not zero, we define the initial condition as iC VV =− )0(

As a result, we getR

VEI iC

−=+ )0(

Rapid decay Slow decay

E

VR (t) VC (t)

IC (t) +

- Vi = VC (0-)


Enzo Paterno Page 6

We revisit Eq10 and solve for K4 using initial conditionR

VEI iC

−=+ )0( :

RVEeKI i

C−

==+ 04)0(

Giving:

Eq13: R

VEK i−=4

Substitute Eq13 into Eq10. We can see that when iC VV =− )0( , the current )(tIC through capacitor C is

given by:

Eq14: RCt

eR

VEtCIi

−−

=)(

Capacitor voltage VC (t) with initial condition 0)0( =−CV

We substitute RCt

eREtIC

−

=)( into ∫−∞=

=t

tCC dI

CtV tt )(1)( to derive the equation for the voltage

)(tV C across the capacitor when 0)0( =−CV :

∫∫∫−

=

−

== dtRCt

eRC

EdtRCt

eRE

CdttI

CtV CC

11)(1)(

We use integration by substitution with:

dtRC

duRC

tu 1 , −=

−=

Giving:

KuEeduueEdtRCt

eRC

EtVC +−=−=

−

= ∫∫1)(

Eq15: KRCt

EetVC +

−

−=)(


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We solve Eq15 for K using the initial condition 0)0( =−CV :

Eq16: EKKEe =→+−= 00

We substitute K into Eq15:

RCt

EeEtVC

−

−=)(

We can see that when 0)0( =−CV , the voltage )(tVC across capacitor C is given by:

Eq17:

−

−= RCt

eEtVC 1)(

The voltage across the resistor can be found by:

RCtR

RCtR

CR

EeEEtVeEEtV

tVEtV

−

−

+−=

−−=

−=

)()1()(

)()(

Thus with the initial condition 0)0( =−CV :

Eq18: RCt

eEtVR

−

=)(


Enzo Paterno Page 8

Capacitor voltage VC (t) with initial condition iC VV =− )0(

We substitute RCt

eR

VEtCI i

−−

=)( into ∫−∞=

=t

tCC dI

CtV tt )(1)( to derive the equation for the voltage

)(tV C across the capacitor when 0)0( =−CV :

( )∫∫∫−

−=

−−

== dtRCt

eRC

VEdtRCt

eR

VEC

dttIC

tV ii

CC11)(1)(


dtRC

duRC

tu 1 , −=

−=

Giving:

( ) ( ) ( ) KueVEduueVEdtRCt

eRC

VEtV iiiC +−−=−−=

−

−= ∫∫1)(

Eq19: ( ) KRCt

eVEtV iC +

−

−−=)(

We solve Eq19 for K using the initial condition iC VV =− )0( :

Eq20: ( ) EKKeVEV ii =→+−−= 0



Enzo Paterno Page 9

( ) RCt

eVEEtV iC

−

−−=)(

We can see that when iC VV =− )0( , the voltage )(tVC across capacitor C is given by:

Eq21: ( ) RCt

eVEEtV iC

−

−−=)(

One can see that Eq21 becomes Eq17 when 0)0( =−CV .

The voltage across the resistor with the initial condition iC VV =− )0( is found to be:

Eq22: ( ) RCt

eVEtV iR

−

−=)(


Enzo Paterno Page 10

A Shortcut approach to find the voltage )(tVC : Substitute:

RCt

eREtIC

−

=)( into RtIRtItV CRR )()()( ==

We get:

RCt

eEtVR

−

=)( Given that EtVtV CR =+ )()( We get:

RCt

eEEtVEtV RC

−

−=−= )()( Resulting in:

VC(0-) = 0

−

−= RCt

eEtVC 1)(

VC(0-) = Vi RCt

eVEEtV iC

−

−−= )()(



Summary – Charging phase behavior for an RC circuit:

E represents the final voltage the capacitor reaches and RC=t represents the circuit’s time constant.

When 0)0( =+CV

tt

eREtCI

−

=)(

−

−= tt

eEtVC 1)(

When iC VV =+ )0(

tt

eR

VEtCI i

−−

=)( ( ) tt

eVEEtV iC

−

−−=)(

Furthermore:

RCt

eEtVR

−

=)(

And

( ) RCt

eVEtV iR

−

−=)(



1.2 RC Circuit Capacitor Discharging Phase

We have so far analyzed the network below in its charging phase when the switch is placed in position 1.

We now throw the switch to position 2. The capacitor begins to discharge at a rate controlled by the time

constant Ƭ = RC.

Capacitor current IC (t) Using Kirchhoff’s voltage law:

Eq23: )()(0 tVtV RC +=

We get;

∫+= dttIC

RtI CC )(1)(0

This gives;

Eq24: 0)()( =+RC

tItIdtd C

C

Remark:

• If the switch is moved to position 2 at after 5 Ƭ, then the capacitor would have charged fully to E. However, In general, if the switch is moved to position 2 before 5 Ƭ, then the initial capacitor voltage is Vi.

• The current reverses direction during the discharging phase

V(0-) = E

+



We solve the differential equation:

RCtItI

dtd C

C)()( −=

Eq25: dtRCtI

tId

C

C 1)()(

−=


∫∫ −= dtRCtI

tId

C

C 1)()(

21)( KRC

tKtIC +−

=+ ln


Eq26: 3)( KRC

ttIC +−

= ln

Solve Eq26 for )(tIC :

( ) 33)( KeRC

t

eK

RCt

etCI

−

=+

−

=

lne

Let K4 = 3Ke

Eq27: RCt

eKtCI

−

= 4)(

We use the initial conditionREIC −=+ )0( to give (Note: the minus sign denotes the fact that current is

now flowing in the opposite direction):

Eq28: REK −=4

The current )(tIC through capacitor C is given by:

Eq29: RCt

eREtCI

−

−=)(



Capacitor voltage VC (t) The voltage across the resistor is given by:

RRCt

eRERtIRtItV CRR

−

−=== )()()(

Eq30: RCt

EetVR

−

−=)(

We know that:

)()(0 tVtV CR +=

)()( tVtV RC −=

Eq31: RCt

EetVC

−

=)(



Summary – Capacitor Charging / Discharging phases

−

−= tt

eEtVC 1)( RCt

EetVC

−

=)(

tt

eREtCI

−

=)( RC

t

eREtCI

−

−=)(

RCt

eEtVR

−

=)( RC

t

EetVR

−

−=)(

Charging Phase Discharging Phase Assume VC (0-) = 0



1.3 Capacitor Transient Phases – Taking another look

@t = 0, the switch is closed providing a current path in the circuit. We recall Eq1, Eq2, and Eq4:

Eq1: )()( tVtVE CR += Eq1’: )()( tVRtIE CC +=

The voltage across a resistor and the current through a capacitor is defined as:

Eq2: RtItV CR )()( =

Eq4: dt

tdVCtI CC

)()( =

We solve for VC (t) thus, substitute Eq4 into Eq1’ and divide by R on both sides

EtVdt

tdVRC CC =+ )()(

This first order linear differential equation with constant coefficients has a general solution comprising of two parts:

)()()( )()( tVtVtV FCNCC +=

VC (N) is the complementary solution, (i.e. also called the natural response), whereby we set the right hand side of the differential equation to zero giving a homogeneous equation.

VC (F) is the particular solution, (i.e. also called the forced response), whereby the solution is based on the particular right hand side of the equation.

We compute the Natural response VC (N):

0)(1)(=+ tV

RCdttdV

CC

Let t

NC KetV α=)()(

E

VR (t) VC (t)

IC (t) +

-



RCt

NC

t

tt

tt

KetV

GivingRC

KeRC

KeRC

eK

KeRCdt

dKe

−=

−=

=

+

=+

=+

)(

;

1

01

01

01

)(

α

α

α

α

αα

αα

We compute the Forced response VC (F): Since the right hand side of the linear equation is a constant, then

AtV FC =)()(

EtVdt

tdVRC CC =+ )()(

EAdtdARC =+

Giving; A = E The solution to the differential equation becomes:

EKetV RCt

C +=−

)( In order to solve for K, we use the initial condition VC (0) = 0

EKEKe

−==+ 00

Substituting;

EEetV RCt

C +−=−

)(

RCwith

t

eEtVC =

−

−= tt1)(



1.4 Capacitor Transient Phases – General Remarks If voltages and currents in a 1st order RC circuit satisfy a differential equation of the form:

x(t) represents i(t) or v(t) Where f(t) is the forcing function (i.e., the independent sources driving the circuit). If x(t) = VC(t), then the solution of the differential equation at t > 0 takes the general form:

( ) CRevvvtv

vvK

vKKvK

CReKKtv

TH

t

CCC

CC

C

C

TH

t

C

C

=⇒∞−+∞=

∞−=

=+

∞=

=⇒+=

−+

+

+

−

t

t

t

t

)()0()()(

)()0(

)0()(

)(

2

21

1

21

For the series RC circuit: K1 = VC (∞) = E E + K2= VC (0+) = 0 K2 = -E Giving the same equation as found earlier:

RCwith

t

eEtVC =

−

−= tt1)(

)()()( tftxadt

tdx=+

K1 is referred as the steady state constant and is found when t = ∞

K2 is referred as the initial state constant and is found when t = 0+

RTH is the circuit Thevenin resistance. RTH is found at t = ∞.

NOTE: In order to find VC(0+), one needs to first find VC(0-). VC(0-) = VC(∞-) VC(0+) as VC(t) cannot change instantaneously



Example 1: For the circuit below, @ t = 0, the switch opens. Find vo(t) @ t > 0 Let:

R = 3 kΩ R1 = 4 kΩ R2 = 2 kΩ C = 100 uF

closed)(Switch 8243

2412)0()0(

open) isswitch since dischargedfully (Capacitor0)(

0,)(

21

1

21

=

Ω+Ω+ΩΩ+Ω

=−=+=+

=∞=

>+=−

kkkkkvvKK

vK

teKKtv

CC

C

t

Ct

6.0)10100)(106()( timedischarge theis

6321 =×Ω×=+= − FCRRt

t

0,8)( 6.0 >=−

tetvt

C

0,38

422)()( 6.0 >=

Ω+ΩΩ

=−

tekk

ktvtvt

CO



Example 2: For the circuit below, @ t = 0, the switch moves to position 2. Find i(t) @ t > 0.

1)position at (Switch 463

312)0()0(

2)position toconnected isswitch since dischargedfully (Capacitor0)(

3)()(

0,)(

21

1

21

=+

=−=+=+

=∞=

Ω=

>+=−

kkkvvKK

vK

ktvti

teKKtv

CC

C

C

t

Ct

2.0)10100)(102(

2)position at (Switch timedischarge theis

63

21

21 =×Ω×=+

= − FCRR

RRt

t

0,4)( 2.0 >=−

tetvt

C

0,3

4)( 2.0 >Ω

=−

tek

tit



2 First Order RL Circuit Transient Analysis

Circuits containing only a single storage element are defined as first-order networks and result in a first-

order differential equation (i.e. RC & RL circuits).

2.1 Inductor Storage Phase

Inductor voltage VL (t) with initial condition 0)0( =−LI

The RL Circuit provides a 1st order Differential Equation when performing the transient analysis:

Using Kirchhoff’s voltage law:

Eq32: )()( tVtVE LR +=

The voltage across a resistor, the current through an inductor and the voltage across an inductor is

defined as:

Eq33: RtItV LR )()( =

Eq34: ∫−∞=

=t

tL dv

LtI tt )(1)(

Eq35: dt

tdILtV LL

)()( =

Substitute Eq33 and Eq34 into Eq32 and divide by R on both sides

Eq36: )()( tvdttvLRE LL += ∫

Differentiate both sides of Eq36:

Eq37:

+= ∫ )()( tvdttv

LR

dtdE

dtd

LL

VL (t)

IL (t)

E



Rearrange and evaluate the derivatives of Eq37, giving a first order linear differential equation:

Eq38: 0)()( =+ tvdtdtv

LR

LL

We solve differential equation Eq38 by first rearranging the equation:

Eq39: dtLR

tvtvd

L

L −=)()(


∫∫ −= dtLR

tvtvd

L

L

)()(

21)( KtLRKtvL +

−=+ ln


Eq40: 3)( KtLRtvL +

−= ln

Solve Eq40 for )(tvL :

( ) 33)( Ke

tLR

eKt

LR

etLv

−

=+

−

=

lne

Let K4 = 3Ke

Eq41: t

LR

eKtLv

−

= 4)(

We solve Eq41 for K4 using initial condition EvL =+ )0( since the model for an inductor at t = 0 is an

open: 04)0( eKELv ==+



Giving:

Eq42: EK =4

Substitute Eq42 into Eq41 to get:

Eq43: t

LR

eEtLv

−

=)( We define RL to be the network time constant with t = R

L [sec].

We assume that steady state is reached at t = 5 t

Inductor current IL (t) with initial condition 0)0( =−LI

We substitute Eq43 into Eq34 to derive the equation for )(tI L when 0)0( =−LI :

∫∫∫−

=

−

== dtt

LR

eLEdt

tLR

EeL

dttvL

tI LL1)(1)(


dtLRdut

LRu −

=−

= ,

Giving:

KueREduue

REdt

LRt

LR

eREtIL +−=−=

−−

−= ∫∫)(

Eq44: Kt

LR

eREtIL +

−

−=)(



We solve Eq44 for K using the initial condition 0)0( =−LI :

Eq45: REKKe

RE

=→+−= 00


tLR

eRE

REtIL

−

−=)(

We can see that when 0)0( =−LI , the voltage )(tI L through the inductor L is given by:

Eq46: REI

tLR

eIt

LR

eREtI ffL =

−

−=

−

−= with 11)(

Inductor current IL (t) with initial condition iL II =− )0( With initial condition iL II =− )0( the current through the inductor becomes:

Eq47: ( )t

LR

eIIItI iffL

−

−−=)(



With initial condition iL II =− )0( the voltage across the inductor becomes:

For an ideal inductor, the winding resistance is negligible (i.e. 0Ω ) and as such:

The voltage across the resistor can be found by:

tLR

R

LR

EeEtV

tVEtV−

−=

−=

)(

)()(

Thus giving;

Eq48:

−

−=t

LR

eEtVR 1)(

Summary – Storage phase behavior for an RL circuit:

tLR

eEtLv

−

=)(

−

−=t

LR

eEtVR 1)(

( )t

LR

eIIItI iffL

−

−−=)(

LtRWiL eRIEtV /)()( −−=

LtRL EetV /)( −=



2.2 Inductor Storing Phase – Taking another look

@t = 0, the switch is closed providing a current path in the circuit. We recall Eq32, Eq33, and Eq35:

Eq32: )()( tVtVE LR += Eq32’: )()( tVRtIE LL +=

The voltage across a resistor and the current through an inductor is defined as:

Eq33: RtItV LR )()( =

Eq35: dt

tdILtV LL

)()( =

We solve for IL (t) thus, substitute Eq35 into Eq32’ and divide by R on both sides

EtRIdt

tdIL LL =+ )()(

This first order linear differential equation with constant coefficients has a general solution comprising of two parts:

)()()( )()( tItItI FLNLL +=

IL (N) is the complementary solution, (i.e. also called the natural response), whereby we set the right hand side of the differential equation to zero giving a homogeneous equation.

IL (F) is the particular solution, (i.e. also called the forced response), whereby the solution is based on the particular right hand side of the equation.

We compute the Natural solution IL (N):

0)()(=+ tI

LR

dttdI

LL

Let t

NL KetI α=)()(

VL (t)

IL (t)

E



tLR

NL

t

tt

tt

KetI

GivingLR

KeLR

KeLReK

KeLR

dtdKe

−=

−=

=

+

=+

=+

)(

;

0

0

0

)(

α

α

α

α

αα

αα

We compute the forced response IL (F): Since the right hand side of the linear equation is a constant, then

AtI FL =)()(

EtRIdt

tdIL LL =+ )()(

ERAdtdAL =+

Giving;

REA =

The solution to the differential equation becomes:

REKetI

tLR

L +=−

)(

In order to solve for K, we use the initial condition IL (0) = 0

REK

REKe

−=

=+ 00

Substituting;

REe

REtI

tLR

L +−

=−

)(



RLwith

t

eREtI L =

−

−= tt1)(

2.3 Taking another look – General Remarks If the voltages and currents in a 1st order RL circuit satisfy a differential equation of the form:

Where f(t) is the forcing function (i.e., the independent sources driving the circuit). The solution of the differential equation at t > 0 takes the general form:

( )TH

t

LLL

LL

L

L

TH

t

RLeIIItI

IIK

IKKIK

RLeKKtI

L

L

=⇒∞−+∞=

∞−=

=+

∞=

=⇒+=

−+

+

+

−

t

t

t

t

)()0()()(

)()0(

)0()(

)(

2

21

1

21

For the RL circuit: K1 = IL (∞) = + K2= IL (0+) = 0 K2 =

)()()( tftxadt

tdx=+

RE

RE

RE

−

K1 is referred as the steady state constant and is found when t = ∞

K2 is referred as the initial state constant and is found when t = 0+

RTH is the circuit Thevenin resistance



Giving the same equation as found earlier:

RLwith

t

eREtI L =

−

−= tt1)(

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