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
AC Circuits Transient Analysis
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
AC Circuits Transient Analysis
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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)(
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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-)
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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
−
=)(
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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)(
We use integration by substitution with:
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
We substitute K into Eq19:
AC Circuits Transient Analysis
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( ) 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
−
−=)(
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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
−
−−= )()(
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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
−
−=)(
AC Circuits Transient Analysis
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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
+
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We solve the differential equation:
RCtItI
dtd C
C)()( −=
Eq25: dtRCtI
tId
C
C 1)()(
−=
Integrate both sides of Eq25:
∫∫ −= dtRCtI
tId
C
C 1)()(
21)( KRC
tKtIC +−
=+ ln
Combine the constants of integration such that K3 = K2 – K1:
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
−
−=)(
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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
−
=)(
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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
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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) +
-
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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)(
AC Circuits Transient Analysis
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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
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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
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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
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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
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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 −=)()(
Integrate both sides of Eq39:
∫∫ −= dtLR
tvtvd
L
L
)()(
21)( KtLRKtvL +
−=+ ln
Combine the constants of integration such that K3 = K2 – K1:
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 ==+
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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)(
We use integration by substitution with:
dtLRdut
LRu −
=−
= ,
Giving:
KueREduue
REdt
LRt
LR
eREtIL +−=−=
−−
−= ∫∫)(
Eq44: Kt
LR
eREtIL +
−
−=)(
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We solve Eq44 for K using the initial condition 0)0( =−LI :
Eq45: REKKe
RE
=→+−= 00
We substitute K into Eq44:
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
−
−−=)(
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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 /)( −=
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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
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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 +−
=−
)(
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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
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Giving the same equation as found earlier:
RLwith
t
eREtI L =
−
−= tt1)(