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TRANSIENTS IN R-L AND R-C CIRCUITS
• One capacitor and one resistor• The source and resistor may be equivalent to a
circuit with many resistors and sources.
R+
-Cvs(t)
+
-
vc(t)
+ -vr(t)
FIRST-ORDER RC CIRCUITS
R
1
C
2
K
E
RvEi c
c
KVL around the loop: EvRi Cc
EvRdtdvC c
c
EAev RCt
C
Initial condition 0)0()0( CC vv
)1()1( t
RCt
C eEeEv
dtdvCi c
c t
eRE
Switch is thrown to 1
RCCalled time constant
Transient Response of RC Circuits
EA
FIRST-ORDER RC CIRCUITS
)1( t
C eEv
/tc eE
dtdv
0
0
t
ct
c
dtdv
EEdtdv
RCTime Constant
R
1
C
2
K
E
Time
0s 1ms 2ms 3ms 4ms 5ms 6ms 7ms 8ms 9ms 10msV(2)
0V
5V
10V
SEL>>
RC
R=2kC=0.1F
FIRST-ORDER RC CIRCUITS
Switch to 2
R
1
C
2
K
E
RCt
c Aev
Initial condition Evv CC )0()0(
0 Riv cc
0dtdvRCv c
c
// tRCtc EeEev
/tc e
REi
Transient Response of RC Circuits
cc
dvi Cdt
FIRST-ORDER RC CIRCUITS
RCTime Constant
R
1
C
2
K
E
R=2kC=0.1F
Time
0s 1.0ms 2.0ms 3.0ms 4.0ms 5.0ms 6.0ms 7.0ms 8.0msV(2)
0V
5V
10V
SEL>>
t
RCt
C EeEetv
)(
E
dtdv
t
C 0
0
t
C
dtdv
E
FIRST-ORDER RC CIRCUITS
Time
0s 0.5ms 1.0ms 1.5ms 2.0ms 2.5ms 3.0ms 3.5ms 4.0ms 4.5ms 5.0ms 5.5ms 6.0msV(2) V(1)
0V
2.0V
4.0V
6.0V
FIRST-ORDER RC CIRCUITS
Ideal Linear Inductor
i(t) +
-
v(t)
Therestof
thecircuit
Ldt
tdiLdt
dtv )()(
t
dxxvL
ti )(1)(
)()( titi LLdtdiLiivP
)(21)( 2 tLiLidipdttwL Energy stored:
• One inductor and one resistor• The source and resistor may be equivalent to a
circuit with many resistors and sources.
FIRST-ORDER RL CIRCUITS
Switch to 1
R
1
L
2
K
E
dtdiLvL
KVL around the loop: EviR L
iRdtdiLE
Initial condition 0)0()0(,0 iit
called time constant
RL /
Transient Response of RL Circuits
FIRST-ORDER RL CIRCUITS
/
/
/
1
)1(
)1()1(
ttLRt
LR
L
tR
ttLR
EeeLR
RELe
RE
dtdL
dtdiLv
eEiRv
eREe
REi
Time constant
• Indicate how fast i (t) will drop to zero.• It is the amount of time for i (t) to drop to zero if it
is dropping at the initial rate .0t
t
dtdi
FIRST-ORDER RL CIRCUITS
Switch to 2
tLR
Aei
dtLR
idi
iRdtdiL
0
Initial conditionRE
It 0,0
/ttLR
eREe
REi
Transient Response of RL Circuits
R
1
L
2
K
E
0
0
0
0
0
: 0:
1
ln
i t t
I
i t tI
t ti I i t
Rdi dti L
Ri tL
tLR
Iti
0
)(ln tLR
eIti
0)(
FIRST-ORDER RL CIRCUITS
FIRST-ORDER RL CIRCUITS
R
1
L
2
K
E
Transient Response of RL Circuits
Time
0s 1ms 2ms 3ms 4ms 5ms 6ms 7ms 8ms 9ms 10msI(L1)
0A
2.0mA
4.0mA
SEL>>
Time
0s 1ms 2ms 3ms 4ms 5ms 6ms 7ms 8ms 9ms 10msI(L1)
0A
2.0mA
4.0mA
SEL>>
Input energy to L
L export its energy , dissipated by R
Initial
Value （ t = 0）
Steady Value (t
)
time constant
RL Circuits
Source(0
state) Source-
free(0
input)
RCCircuits
Source(0
state) Source-
free(0
input)
00 iREiL
REi 0 0i
00 v Ev
Ev 0 0v
RL /
RL /
RC
RC
SUMMARY
The Time Constant
• For an RC circuit, = RC• For an RL circuit, = L/R• -1/ is the initial slope of an exponential with an
initial value of 1• Also, is the amount of time necessary for an
exponential to decay to 36.7% of its initial value
SUMMARY
SUMMARY
• How to determine initial conditions for a transient circuit. When a sudden change occurs, only two types of quantities will remain the same as before the change. – IL(t), inductor current– Vc(t), capacitor voltage
• Find these two types of the values before the change and use them as the initial conditions of the circuit after change.
About Calculation for The Initial Value
iC iL
i
t=0
+
_
1A
+
-vL(0+
)
vC(0+)=4V
i(0+)
iC(0+) iL(0+)
1 3/ / 2R R
20 8V 4V2 2Cv
8V0 2A2 2
i
40 2A 1A4 4Li
0 0C Cv v
0 0L Li i
Method 1
(Analyzing an RC circuit or RL circuit)
Simplify the circuit
2) Find Leq(Ceq), and = Leq/Req ( = CeqReq)
1) Thévenin Equivalent.(Draw out C or L)
Veq , Req
3) Substituting Leq(Ceq) and to the previous solution of differential equation for RC (RL) circuit .
Method 2
(Analyzing an RC circuit or RL circuit)
3) Find the particular solution.
1) KVL around the loop the differential equation
4) The total solution is the sum of the particular and homogeneous solutions.
2) Find the homogeneous solution.
Examples
Method 3 (step-by-step)
(Analyzing an RC circuit or RL circuit)
1) Draw the circuit for t = 0- and find v(0-) or i(0-)2) Use the continuity of the capacitor voltage, or inductor current, draw
the circuit for t = 0+ to find v(0+) or i(0+)3) Find v(), or i() at steady state4) Find the time constant t
– For an RC circuit, t = RC– For an RL circuit, t = L/R
5) The solution is: /)]()0([)()( teffftf
Given f(0+) ， thus A = f(0+) – f(∞)
t
effftf
)]()0([)()(
Initial Steady
t
Aeftf
)()(In general,
Examples
vC (0)= 0, Find vC (t) for t 0. i1
6k
R1
R2 3k +
_ E
C=1000PF pf
i2 i3 t=0
9V
Method 3:
0
3K0 0, 9V 3V6K 3K
t
c c c c
c c
v t v v v e
v v
Apply Thevenin theorem :
6
1
6
2 10
1 1 2K6K 3K
2K 1000pF 2 10
3 3 V
Th
Th
t
c
R
R C
v t e
s
Examples
vC
R2 3k
+
_ v
U
I
t=0
6V
C=1000PF
R1=10k
R1=20k
+ - P3.2 vC (0)= 0, Find vC (t) for t 0.
Apply Thevenin’s theorem :
s
0 0
10K6V 4.615V10K 3K
c
c
v
v
6
1
6
2.31 10
1 1 30 K10K 3K 13
30 K 1000pF 2.31 1013
4.615 4.615 V
Th
Th
t
c
R
R C
v t e
THANK YOU
