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RL and RC circuits firstorder response
Electric circuits
ENT 161/4
RL and RC circuit original response
 A first-order
circuit is characterized by
a first-order differential equation. This
circuit contain resistor and capacitor
or inductor in one close circuit.
 The
natural response of a circuit
refers to the behaviour ( in terms of
voltages and currents) of the circuit
itself, with no external sources of
excitation.
 RL circuit: circuit that have resistor
and inductor.
 RC circuit: circuit that have resistor
and capacitor.
Natural response RC circuit
Consider these three condition :
1. At initially, t=0 -, switch doesn’t
change for some time
2. At initial, t=0 +, switch doesn’t
change for some time
3. At final condition, t→∞, switch
doesn’t change for some time

Known t ≤ 0, v(t) = V0.
du
1
Therefore t ≥ 0:

dv
u
RC
ic  iR  0
v (t ) 1
1 t
du  
dv
dv(t ) v(t )


V0 u
0
C

0
RC
dt
R
1
dv(t ) v(t )
ln v(t )  ln V0  
(t  0)

0
RC
dt
RC
dv(t )
v(t )
 v(t ) 
t voltage



ln
dt
RC
 V    RC
 0 
dv(t )
1

dt
v(t )
RC
v(t )  V0 e
 t RC

For t > 0,
v(t )  V0 e
 t RC
V0
v(t )
iR (t )  
 e
R
R
t
RC
1
1
2
2
W (t )  C v(t )   C V0 e
2
2
 2t
RC
Natural response RC circuit graph
v(t )  V0
 V0 e
t0
 t RC
t0
 This
show that the voltage response
of the RC circuit is an exponential
decay of the initial voltage. constant, τ
= RC
v(t )  V0 e
 t
 Constant
τ define how fast voltage
reach stable condition :
Natural response RL circuit
Consider these three condition :
1. At initially, t=0 -, switch doesn’t
change for some time
2. At initial, t=0 +, switch doesn’t
change for some time
3. At final condition, t→∞, switch
doesn’t change for some time
 Known
at t ≤ 0, i(t) = I0
Therefore t > 0,
i (t ) 1
v(t )  R i (t )  0
R t
du    dv

i (0) u
0
di (t )
L
L
 R i (t )  0
dt
R
ln i (t )  ln i (0)   (t  0)
di (t )
L
  R i (t )
L
dt
 i (t ) 
R Current
di (t )
R
  t
ln 
  dt
i (t )
L
L
 i (0) 
du
R
  dv
u
L
i (t )  i (0) e
t R L
 For
t > 0,
i(t )  I 0 e
t R L
v(t )  i (t ) R
  RI 0 e
t R L
1
2
w(t )  Li (t ) 
2
1
2  2t R L
 LI 0 e
2
EXAMPLE
Switch in circuit for some time before
open at t=0. Calculate
a) IL (t) at t ≥ 0
b) I0 (t) at t ≥ 0+
c) V0 (t) at t ≥ 0+
d) Total energy percentage that stored
in inductor 2H that absorb by 10Ω
resistor.
Answer
a)
Switch close for some time until t=0,
known voltage at inductor should be
zero at t = 0-. Therefore, initial current
at inductor was 20A at t = 0-. Thus iL
(0+) also become 20A, because
immidiate changes for current didn’t
exist in inductor.

Equivalent resistance from inductor
and constant time
Req  2  40 10  10
L
2


 0.2 saat
Req 10

Therefore, current iL (t)

i L (t )  i(0 ) e
 20 e
5 t
 t
A
t0
b)
Current at resistor 40Ω could be
calculate by using current divider
law,
 10 
i0  i L 

 10  40 
 This
current was at t ≥ 0+ because
i0 = 0 at t = 0-. Inductor will
become close circuit when switch
open immediately and produce
changes immediately at current i0.
Therefore,
i0 (t )  4e
5t
A
t0

c)
V0 could be calculate by using
Ohm’s Law,
V0 (t )  40i0
5t
 160e V
t0

d)
Total power absorb by 10Ω
resistor
2
V0
p10 (t ) 
10
10t
 2560 e W
t0

 Total
energy absorb by 10Ω resistor

W10 (t )   2560e
0
 256 J
10t
dt
 Initial
energy stored at 2H
inductor
1 2
W (0)  L i (0)
2
1
 2 400   400 J
2

Therefore, energy percentage
that absorb by 10Ω resistor
256
100  64%
400
Step response RC circuit

The step response of a circuit is its
behaviour when the excitation is the step
function, which may be a voltage or a
current source.
Consider these three condition :
1. At initially, t=0 -, switch doesn’t
change for some time
2. At initial, t=0 +, switch doesn’t
change for some time
3. At final condition, t→∞, switch
doesn’t change for some time
 Known
For
at t ≤ 0, v(t)=V0
t > 0,
Vs  v(t )  Ri (t )
1
du

dv 
RC
u  Vs
dv(t )  t  ln v(t )  V   ln V  V 
s
0
s
Vs  v(t )  RC
RC
dt
 v(t )  Vs  voltan
t
1
dv(t )


 ln 
dt 
RC
V

V
0
s 

RC
Vs  v(t )
 t RC
v(t )  Vs  V0  Vs e
1
dv(t )

dt 
 t
RC
v(t )  Vs
 Vs  V0  Vs e

Current for step response RC circuit
dv
i (t )  C
dt
 1
 t 
 C   (V0  Vs )e 
 

1
 t
  V0  Vs e
R
 Vs V0   t 
   e
R R

i (t )  i(0 )e
 t

Then, for t >0
V  Vs  V0  Vs e
 V f  Vn
Where
V f V s
Vn  V0  Vs e
 t
 t
 Vf
= Force voltage or known as
steady-state response
 Vn = known as transient response
is the circuit’s temporary
response that will die out with
time.
Step response RC circuit graph
force
total
Natural
Step Response RL circuit
Consider these three condition :
1. At initially, t=0 -, switch doesn’t
change for some time
2. At initial, t=0 +, switch doesn’t
change for some time
3. At final condition, t→∞, switch
doesn’t change for some time
i(t)=I0 at t ≤ 0.
 For t > 0,
 known
Vs  Ri (t )  v(t )
di (t )
Vs  Ri (t )  L
dt
Vs
L di (t )
 i (t ) 
R
R dt
R
di (t )
dt  V
s
L
R  i (t )
R
di
 dt 
L
i  Vs R
R
du
 dv 
L
u  Vs R
i (t )
R t
du
  dv  
I 0 u  Vs
L 0
R
R
 t  ln i (t )  Vs R   ln I 0  Vs R 
L
Current
Vs


R
i (t )  R

 t  ln 
Vs
L
 I0  R 
i(t ) 
Vs
R
 I 0 
Vs
R
e
t R L
 Finally,
i(t )  I 0

Vs
R
 I 0 
Vs
R
e
di (t )
v(t )  L
dt
t R L
 Vs  R I 0 e
t 0
t R L
t 0
t0
t0
Question
Switch in those circuit was at x
position for some time. At t=0,
switch move to position y
immediately. Calculate,
(a) Vc(t) at t ≥ 0
(b) V0 (t) at t ≥ 0+
(c) i0 (t) at t ≥ 0+
(d) Total energy absorb by 60kΩ
resistor.
Answer (a)
 VC (0)=100V

Constant for circuit
  (0.5  10 )(80  10 )  40ms
6
3
 equivalent resistor = 80kΩ.
 Then,
VC(t) for t ≥ 0:
VC (t )  100e
25t
V
t0
Answer (b)
 V0
(t) could be calculate by using
voltage divider law.
48
V0 (t )  VC (t )
80
 25t
 60 e V
t0

Answer (c)
 current i0
(t) can be calculated by
using ohm’s law
V0 (t )
 25t

i0 (t ) 

e
mA
t

0
3
60 10
Answer (d)

Power absorb by 60kΩ
resistor

p60k (t )  i0 (t ) 60 10
2
 60e
50t
mW
3

t0

 Total
energy



W60k   i0 (t ) 60  10 dt
2
0
 1.2mJ
3
Second-order RLC circuit



1.
2.
RLC circuit : circuit that contain
resistor, inductor and capacitor
Second-order response : response
from RLC circuit
Type of RLC circuit:
RLC series circuit
RLC parallel circuit
Original response for parallel
RLC circuit
 Take
total current flows out from
node
V 1 t
dv
  vd  I 0  C
0
0
R L
dt
 differential
of t,
2
1 dv v
d v
 C 2  0
R dt L
dt
2
d v 1 dv v



0
2
dt
RC dt LC
v  Ae
 Take
st
As st
A st
As e 
e 
e 0
RC
LC
2
st
s
1 
 2
Ae  s 

0
RC
LC



st
characteristic equation
 Characteristic
equation known as
zero :
s
1 
 2

s 
0
RC LC 

 The
root of the characteristic
equation are
2
1
1
 1 
s1  
 
 
2 RC
 2 RC  LC
2
1
1
 1 
s2  
 
 
2 RC
 2 RC  LC
 Response
for RLC parallel circuit
v  A1 e  A2 e
s1t
s2t
 The
root of the characteristic equation
are
s1      0
2
s2      0
2
2
2
 where:
1

2 RC
0 
1
LC
 summarize
Parameter
Terminology
s1, s2
characteristic
equation
α
0
frequency Neper
resonant radian
frequency
Value in natural
response
s1     2  0
2
s2     2  0
1

2 RC
0 
1
LC
2


1.
2.
3.
Roots solution s1 and s2 depend on
α and 0
Consider these cases saperately:
If 0 < α , voltage response was
overdamped
If 0 > α , voltage response was
underdamped
If 0 = α , voltage response was
critically damped
Overdamped voltage response

Solution for overdamped voltage
v  A1 e  A2 e
s1t
s2t
 constant A1
and A2 can be
determined from the initial conditions

v(0+) and dv(0 )
dt
 Known,

v(0 )  A1  A2

dv(0 )
 s1 A1  s2 A2
dt
 Here
v(0+) = V0 and initial value for
dv/dt was


dv(0 ) iC (0 )

dt
C
1.
2.
Solution for overdamped natural
response, v(t) :
Calculate characteristic equation, s1
and s2, using R, L and C value.
Calculate v(0+) and dv(0 )
dt
using circuit analysis.
3.
Calulate A1 and A2 by solve those
equation

v(0 )  A1  A2


dv(0 ) iC (0 )

 s1 A1  s2 A2
dt
C
4.
Insert s1, s2, A1 and A2 value to
calculate overdamped natural
response for t ≥ 0.
 Example
for overdamped natural
response for v(0) = 1V and
i(0) = 0
Underdamped voltage response
 At
0 > α2, root of the
characteristic equation was
complex number and those
response called underdamped.
 Therefore
s1     (0   )
2
   j 0  
2
   jd
s2    jd
 ωd
: damped radian frequency
2
2
 underdamped
voltage response
for RLC parallel circuit was
v(t )   B1 e
 t
 B2 e
cos d t
 t
sin d t
 constant
B1 and B2 was real
number.
Solve those two linear equation
to calculate B1 and B2,

v(0 )  V0  B1


dv(0 ) iC (0 )

 1B1  d B2
dt
C
Example for underdamped voltage
response for v(0) = 1V and i(0) = 0
Critically Damped Voltage
Response
 Second-order circuit was critically
damped when 0 = α . When
circuit was critically damped, two
characterictic root equation was
real and same,
1
s1  s2    
2 RC
 Solution
for voltage
v(t )  D1t e
t
 D2 e
t
•Linear equation to calculate D1 and
D2 value

v(0 )  V0  D2


dv(0 ) iC (0 )

 D1  D2
dt
C
Example for critically damped
voltage response at v(0) = 1V and
i(0) = 0
Step response RLC
parallel circuits
 From
Kirchhoff current law
iL  iR  iC  I
dv
v
I
iL   C
dt
R
di
 Known v  L
dt
Therefore
2
dv
d iL
L 2
dt
dt
 Have,
2
L diL
d iL
iL 
 LC 2  I
R dt
dt
2
d iL
1 diL iL
I



2
dt
RC dt LC LC
 There
are two solution to solve
the equation, direct approach and
indirect approach.
Indirect approach
 From
Kirchhoff’s current law:
1 t
v
dv
vd



C

I

0
L
R
dt
 Differential
2
v 1 dv
d v

C 2  0
L R dt
dt
2
d v
1 dv
v



0
2
dt
RC dt LC
 Depend
on characteristic
equation root :
v  A1 e  A2 e
s1t
v   B1 e
t
 B2 e
v  D1t e
cos d t
t
t
s2t
sin d t
 D2 e
t
 Insert
in Kirchhoff’s current law eq:


iL  I  A1 e  A2 e

s1t
s2t
 t
iL  I  B1 e cos d t
 t
 B2 e sin d t
  t
  t
iL  I  D1 t e  D2 e
Direct approach
 It’s
simple to calculate constant for
the equation






A1 , A2 , B1 , B2 , D1 , D2
directly by using initial value
response function.
 Constant
of the equation could be
calculate from
iL (0)
and
diL (0)
dt
 The
solution for a second-order
differential equation with a
constant forcing function equals
the forced response plus a
response funtion identical in form
to the natural response.
 If
and Vf represent the final value
of the response function. The final
value may be zero,
 function of the same form
i  If 

as the natural response 
 function of the same form
v  Vf  

as the natural response 
Natural response for RLC Series
circuit
 The
procedures for finding the natural
or step responses of a series RLC
circuit are the same as those used to
find the natural or step responses of a
parallel RLS circuit, because both
circuits are described by differential
equations that have the same form.
RLC series circuit
 Summing
the voltages around the
closed path in the circuit,
di 1 t
Ri  L   i d  V0  0
dt C 0
 differential
2
di
d i i
R L 2  0
dt
dt
C
2
d i R di
i



0
2
dt
L dt LC
 Characteristic
equation for RLC
series circuit
R
1
s  s
0
L
LC
2
 Characteristic
equation root
2
s1, 2
R
1
 R 

   
2L
 2 L  LC
@
s1, 2      0
2
2
 Neper
frequency (α) for RLC
series circuit
R

rad / s
2L
and resonant radian frequency was,
0 
1
rad / s
LC
Current response
2
Overdamped
0
 
2
Underdamped
0  
2
critically damped
0  
2
2
2
 Three
kind of solution
i(t )  A1 e  A2 e
s1t
i(t )  B1e
t
 B2e
i(t )  D1t e
cos d t
t
 t
s2t
sin d t
 D2 e
 t
Step response for RLC series
circuit
 The
procedures for finding the
step responses of series RLC
circuit are the same as those
used to find the step response of
a parallel RLC circuit.
RLC series circuit
 Using
Kirchhoff’s voltage law,
di
v  Ri  L  vC
dt
 Current
known as,
dvC
iC
dt
 Differential
for current
2
di
d vC
C 2
dt
dt
 Insert
in Voltage current law
equation
2
d vC R dvC vC
V



2
dt
L dt
LC LC
 Three
solution that possibly for vC


vC  V f  A1 e  A2 e

s1t
vC  V f  B1 e

 B2 e
t
t
s2t
cos d t
sin d t
  t
  t
vC  V f  D1 t e  D2 e
Contoh 1
1.
2.
3.
4.
Tenaga awal yang disimpan oleh litar
berikut adalah sifar. Pada t = 0, satu
punca arus DC 24mA diberikan kepada
litar. Nilai untuk perintang adalah 400Ω.
Apakah nilai awal untuk iL?
di L
Apakah nilai awal untuk
?
dt
Apakah punca-punca persamaan ciri?
Apakah ungkapan numerik untuk iL(t)
pada t ≥ 0?
Jawapan
1.
Tiada tenaga yang disimpan
dalam litar sebaik sahaja punca
arus digunakan, maka arus awal
bagi induktor adalah sifar.
Induktor mencegah perubahan
yang serta-merta pada arus
induktor, oleh itu iL (0)=0 sebaik
sahaja suis dibuka.
2.
Nilai awal voltan kapasitor adalah
sifar sebelum suis dibuka, oleh itu
ia akan sifar sebaik sahaja suis
dibuka. Didapati:
di L
vL
dt

maka
di L (0 )
0
dt
3.
Dari elemen-elemen dalam litar,
diperolehi
0
12
2
1
10
8


 16  10
LC (25)(25)
9
1
10


2 RC (2)( 400)( 25)
 5 10 rad / s
4
  25  10
2
8
 Oleh
kerana , maka punca-punca
persamaan ciri adalah nyata
s1  5  10  3  10
4
4
 20 000 rad / s
s 2  5  10  3  10
4
 80 000 rad / s
4
4.
sambutan arus induktor adalah
overdamped dan persamaan
penyelesaian adalah


i L  I f  A1 e  A2 e
s1t
s2t
 Dua
persamaan serentak:


i L (0)  I f  A1  A2  0
di L (0)


 s1 A1  s 2 A2  0
dt

A1  32mA

A2  8mA

Penyelesaian numerik:
 24  32e
iL (t )  
80000t

8
e

untuk
t0
20000t

mA


Contoh 2

Tiada tenaga disimpan dalam
inductor 100mH atau
kapasitor 0.4µF apabila suis
di dalam litar berikut ditutup.
Dapatkan vC (t) untuk t ≥ 0.
Jawapan
 Punca-punca
persamaan ciri:
2
280
10
 280 
s1  
 
 
0.10.4
0 .2
 0. 2 
  1400  j 4800 rad / s
s 2   1400  j 4800 rad / s
6
 Punca-punca
adalah kompleks,
maka sambutan voltan adalah
underdamped. Oleh itu, diperolehi
voltan vC :

1400t
vC  48  B1 e
cos 4800t
 1400t
 B2 e
sin 4800t
t0
 Pada
awalnya, tiada tenaga
tersimpan dalam litar, maka:

vC (0)  0  48  B1

dvC (0 )


 0  4800 B2  1400 B1
dt


 Selesaikan B untuk B dan
2
1

B1  48V

B2  14V
 penyelesaian
untuk vC (t)
 48  48 e
cos 4800t 
V
vC (t )  
1400t

sin 4800t
  14 e

untuk
t0
1400t