1. A document describing RC and RL circuits is provided. RC circuits are analyzed using Kirchhoff's laws. The time constant τ is defined as RC. For an RC circuit with an initial voltage V0, the voltage v(t) is given by v(t) = V0e-t/τ. 2. For an RL circuit with an initial current I0, the current i(t) is given by i(t) = I0e-t/τ, where the time constant τ is L/R. Kirchhoff's laws are again used to analyze the RL circuit. The voltage v(t) across the inductor is given by v(t) = RI0

1. RC/RL
Department of Electronic Engineering
National Taipei University of Technology

2. •
•
• RC RL
•
Department of Electronic Engineering, NTUT2/32

3. i
+
−
C
+
−
v
dv
i C
dt
=
L
vL
−+
i
L
di
v L
dt
=
( ) ( ) ( )0
0
1 t
t
i t v t dt i t
L
= +∫( ) ( ) ( )0
0
1 t
t
v t i t dt v t
C
= +∫
Department of Electronic Engineering, NTUT3/32

4. RC RL
RCv(t)
+
−
ic iR
+
−
v t( ) L R
−
+
vR
i t( )
KCL 0c Ri i+ =
( ) ( ) ( )
( )
1
0 0
dv t v t dv t
C v t
dt R dt RC
+ = ⇒ + =
( )
( )
( )
( )0 0
di t di t R
L Ri t i t
dt dt L
+ = ⇒ + =
KVL
( )
( ) ( )
dy t
ay t f t
dt
+ =
( )
( ) ( )
dy t
ay t f t
dt
+ =
• RC RL
Department of Electronic Engineering, NTUT4/32

5. RC RL ( )
Vs C
+
−
v
vR
+ −
it = 0 R
t =0 iL
+ −vL
RVs
vs(t) C
+
−
v
vR
+ −
iR iL
+ −vL
Rvs(t)
( ) 0f t ≠
Department of Electronic Engineering, NTUT5/32

6. • RC RL
y v(t) i(t) a
f(t) f(t)
•
c y(0)
t
c e-at
eat
( )
( ) ( )
dy t
ay t f t
dt
+ =
( )
( ) ( )at at at
dy t
e e ay t e f t
dt
+ =
( )
( )
( )
( )
( ) ( )at at at at at
dy t dy td
e y t e e ay t e ay t e f t
dt dt dt
 
  = + = + =  
 
( )
( ) ( ) ( )
at
at at at
d e y t
dt e f t dt e y t e f t dt c
dt
   = ⇒ = +∫ ∫ ∫
( ) ( )at at at
y t e e f t dt ce− −
= +∫
Department of Electronic Engineering, NTUT6/32

7. •
(Complete response)
( ) ( )at at at
y t e e f t dt ce− −
= +∫
( ) ( )at at at
y t e e f t dt ce− −
= +∫
yp (Particular solution)
(Forced response)
(Steady-state response)
(Homogeneous solution)
(Natural response) t
(Transient response)
t
A Be τ
−
= +
f ny y= +
( )at at
fy A e e f t dt−
= = ∫t
at
ny Be ceτ
−
−
= =
1
a
τ =
Department of Electronic Engineering, NTUT7/32

8. • f(t)
(Natural response)
f (t) = 0
f (t) = b
( )
( ) ( ) ( ) ( )at at at
dy t
ay t f t y t e e f t dt ce
dt
− −
+ = ⇒ = +∫
( ) ( )0at at
y t ce y e− −
= =
( ) atb
y t ce
a
−
= + ( )0
b
y c
a
= + ( )0
b
c y
a
= −
( ) ( )0 atb b
y t y e
a a
− 
= + −  
(y (0) = 0) f (t) = b
( ) atb
y t ce
a
−
= + ( )0 0y =
b
c
a
= −
( ) atb b
y t e
a a
−
= −
Department of Electronic Engineering, NTUT8/32

9. 1
• 5-1
(a) f (t) = 0 y(0) = 10 ( )
(b) f (t) = 5 y(0) = 10 ( )
(c) f (t) = 5 y(0) = 0 ( = )
(a)
(b)
(c)
( )
( ) ( )5
dy t
y t f t
dt
+ =
( ) ( )at at at
y t e e f t dt ce− −
= +∫
( ) 5t
y t ce−
=
( )0 10y c= =
( ) 5
10 t
y t e−
∴ =
( ) 5 5 5 5
5 1t t t t
y t e e dt ce ce− − −
= × × + = +∫
( )0 10 1 9y c c= = + ⇒ =
( ) 5
1 9 t
y t e−
∴ = +
( ) 5 5 5 5
5 1t t t t
y t e e dt ce ce− − −
= × × + = +∫
( )0 0 1 1y c c= = + ⇒ = −
( ) 5
1 t
y t e−
∴ = −
Department of Electronic Engineering, NTUT9/32

10. (I)
•
• RC (a)
(a) RC
4 Ω 6 Ω
2 Ω 7 Ω
C = 1 F 150 V
+ −v
t =05 Ω
+
−
vc(t)
(b) t = 0
t = 0 t = 0
t = 0
(b)
vc ( )0− 150 V
+ −v
+
−
4 Ω 6 Ω
2 Ω 7 Ω 5 Ω
Department of Electronic Engineering, NTUT10/32

11. (II)
Rth
=10 Ω 1 F
+
−
vc t( )
(c) t > 0
( )4 2 / /6 7 10thR = + + = Ω  
( ) 10
0 150 100V
10 5
cv −  
= × = 
+ 
t 0
vc 100V vc
(c)
( ) ( )0 0 100Vc cv v+ −
= =
Department of Electronic Engineering, NTUT11/32

12. RC (I)
• t > 0
KCL
( ) t > 0
( ) 00cv V−
=
0c Ri i+ =
( ) ( )
0c cdv t v t
C
dt R
+ =
( )
( )
1
0C
C
dv t
v t
dt RC
+ = ( )
t
RC
cv t ce
−
=
( ) 00cv V−
= ( ) 00cv c V−
= =
( ) ( ) 00 0c cv v v+ −
= =
( ) 0
t
RC
cv t V e
−
= ( )
( ) 0
t
c RC
c
dv t V
i t C e
dt R
−
= = − ( )
( ) 0
t
c RC
R
v t V
i t e
R R
−
= =
RCvc(t)
+
−
ic iR
Department of Electronic Engineering, NTUT12/32

13. RC (II)
• RC
(a)
K t
R C
RC ( , )
• RC
(Time constant)
v(t) i(t)
(b)
K
0 t
f t( )
(a)
(b)
( )
t
f t Ke τ
−
= t → ∞
( )
( ) 0
dy t
ay t
dt
+ =
RCτ =
R v i= C q v=
( ) 0f t →
τ
1
a
τ τ
τ
V0
vc(t)
0
t
0.368V0
1 Kτ =
2 2Kτ =
3 3Kτ =
K 2K 3K
1τ 2τ 3τ
Department of Electronic Engineering, NTUT13/32

14. 2
• (a) t = 0 t = 0
t 0 vc(t) , v(t) ic(t)
(c) t ＞ 0
τ = RC
(a) RC
4 Ω 6 Ω
2 Ω 7 Ω
C = 1 F 150 V
+ +v
t =05 Ω
+
−
vc(t) Rth
=10 Ω 1 F
+
−
vc t( )
(c) t > 0
( ) ( ) ( )0 0 100 Vc cv v+ −
= =
( ) ( )0.110
100 100 V
t
t
cv t e e
−
−
= =
( )
( )
( ) ( )0.1 0.17 7
100 70 V
7 6// 2 4 10
t t
cv t v t e e− −
= = × =
+ +  
( )
( )
( ) ( )0.1 0.1
100 0.1 10 Ac t t
c
dv t
i t C e e
dt
− −
= = × − = −
Department of Electronic Engineering, NTUT14/32

15. 3 ( )
• v(0) = 4 (V) t > 0 i
KCL
t > 0
+
−
1
8
F 3 Ω2i (V)
+
−
v
i
6 Ω
2 1
0
6 8 3
v i dv v
dt
−
+ + =
1
8
dv dv
i C
dt dt
= =
1
14 0 6 0
6 8 3
dv
v
dv v dvdt v
dt dt
−
+ + = ⇒ + =
( ) ( ) ( )6
0 4 V
t
t
v t v e eτ
−
−
= =
( ) ( ) ( )6 61
4 6 3 A
8
t tdv
i t C e e
dt
− −
= = × × − = −
Department of Electronic Engineering, NTUT15/32

16. RC (I)
• v(0–)=V0
t = 0
• t > 0 v(t)
(b)
( )
( )
RC K
( V0–Vs) (
RC )
( ) ( )0
t
RC
s sv t V V V e
−
= + −
sV
( )0
t
RC
sV V e
−
−
(a)
Vs
0 t
v(t)
( )0
t
RC
sV V e
−
−( )0 sV V−
(b)0 t
V0
Vs
v(t)
( )0
t
RC
s sV V V e
−
+ −
Vs C
+
−
v
vR
+ −
it = 0 R
Department of Electronic Engineering, NTUT16/32

17. RC (II)
• RC
1. RC
2. ( )
3. ( ) τ
t
e τ
−
×
Department of Electronic Engineering, NTUT17/32

18. 4
• t = 0 v(0 ) = V0
t 0 v(t)
v(t)
t = 0+
KCL t 0
I
+
−
vC
t =0a
b
R
iR iC
dv v dv v I
C I
dt R dt RC C
+ = ⇒ + =
( )
t t t t
RC RC RC RC
I
v t e e dt ce RI ce
C
− − − −
= + = +∫
( ) ( ) 00 0v v v+ −
= = ( ) 00v RI c V+
= + = 0c V RI= −
( ) ( )0 , 0
t
RC
v t RI V RI e t
−
= + − >
Department of Electronic Engineering, NTUT18/32

19. 5
• (a) v(0 ) = 15 (V) t > 0 v(t)
24 Ω
1
3F
+
−
v40V
8 Ωi a
b
(a) (b)
40V
a
b
Rth
= 6 Ω
8 Ω
24 Ω
A. ( (b))
30(V) ab
Voc
( )
24
40 30 V
24 8
× =
+
B. (c)
8 24
6
8 24
thR
×
= = Ω
+
1
6 2 sec
3
thR Cτ = × = × =
( ) 0.52
15 30 15
t
t
e e
−
−
− = −
C. t > 0 ( ) ( )0.5
30 15 Vt
v t e−
= −
1
3F30V
a
b
(c)
6 Ω
Department of Electronic Engineering, NTUT19/32

20. RC
• RC
v(0) = 0 t > 0
•
RC
Voc = ( )
Rth =
( ) ( )1
t
tRC
s s sv t V V e V e τ
−
−
= − = −
( ) ( )1 tht R C
ocv t V e−
= −
a t
e
−
Vs C
+
−
v
vR
+ −
it =0 R
Department of Electronic Engineering, NTUT20/32

21. 6
• v(0) = 0 t 0 v(t) i(t)
(b)
(a)
+
−
36 V 12 Ω
+
−
v t( )
6 Ω
1
8
F
i t( )
+
−Voc 24= V
a
b
+
−
v t( )
Rth = 4 Ω
(b)
1
8
F
( )
12
36 24 V
12 6
ocV = × =
+
6 / /12 4thR = = Ω
1 1
4 sec
8 2
thR Cτ = = × =
( ) ( )( )2
24 1 Vt
v t e−
= −
( ) ( ) ( )2
36 6=2+4e At
i t v t −
= −  
Department of Electronic Engineering, NTUT21/32

22. •
I0 ( )
•
RL
RL
0di dt = 0Lv Ldi dt= =
Department of Electronic Engineering, NTUT22/32

23. RL
• RL i (0 ) = I0
t > 0
• RL τ
L/R τ ( L )
i(0+) = i(0 ) = I0 t > 0
i(0 ) = I0
+
−
v t( ) L R
−
+
vR
i t( )
( )
( )
( )
( )0 0
di t di t R
L Ri t i t
dt dt L
+ = ⇒ + =
( )
R
t
L
i t ce
−
=
( ) 00i c I−
= =
( ) 0
R
t
L
i t I e
−
= ( ) 0
R
t
L
L
di
v t L I Re
dt
−
= = − ( ) 0
R
t
L
Rv t iR I Re
−
= =
( )
t
f t Ke τ
−
=
Department of Electronic Engineering, NTUT23/32

24. 7
• 1 (
1 ) t = 0
2 t > 0 v(t) i(t)
2 Ω
t =0
+
−
30V
2
+
−
v t( )
3 Ωa
b
i t( ) 2 Ω1H
3 Ω1
i 1H
15
8 Ω
a
b
15
8
45
4
t
ab
di
v e
dt
−
= = − ( )
15 15
8 8
2 45 2 9
V
3 2 4 5 2
t t
abv t t e e
− −
= × = − × = −
+
(3)
1
6A
2 a-b
( )0
30 30
6 A
2 3 5
i I= = = =
+
( )
( )
3 2 3 15
3 2 3 8
thR
× +
= = Ω
+ +
( )
15
8
6
t
i t e
−
=
8
sec
15th
L
R
τ = =
(1)
(2)
t > 0
Department of Electronic Engineering, NTUT24/32

25. 8 ( )
•
L/R i(0) = 2 A
v
6A
a
b
2 Ω
3
2H
4 Ω
i
+ −v
3
4 2
2 6
ab
di v
v i i
dt
 
= + = × − 
 
3
2
di di
v L
dt dt
= =
3
3 24 2
2 6
di
di dti i
dt
 
 
+ = × − 
 
 
6 0
di
i
dt
+ = 6
2 At
i e−
=
1
6
τ =
L
R
τ =
Department of Electronic Engineering, NTUT25/32

26. RL (I)
• RL
• i(0 ) = I0 t = 0
KVL t > 0
t = 0+
t > 0
t =0 iL
+ −vL
RVs
s
s
di di R V
L Ri V i
dt dt L L
+ = ⇒ + =
( )
R R R R
t t t t
s sL L L L
V V
i t e e dt ce ce
L R
− − −
= + = +∫
( ) ( ) 00 0i i I+ −
= =
( ) 0 00 s sV V
i c I c I
R R
+
= + = ⇒ = −
( ) 0
R
t
s s L
V V
i t I e
R R
− 
= + − 
 
s
s
V
I
R
= ( ) ( )0
R
t
L
s si t I I I e
−
= + −
Department of Electronic Engineering, NTUT26/32

27. RL (II)
• i(t) Vs / R
•
• RL
RL
– τ
( ) ( )0
R
t
L
s si t I I I e
−
= + −
0
R
t
s L
V
I e
R
− 
− 
 
t
e τ
−
0
R
t
s L
V
I e
R
− 
− 
 
Vs
sV
R
sV R
Department of Electronic Engineering, NTUT27/32

28. 9
• i(0 ) = 2 A t > 0 i(t)
A. t > 0 a-b
( )
B.
a-b
+
−
3 Ω
6 Ω 3 Ω2 H
b
a
it =0
6 Ω
36V
+
−
6V
2 Ω
a
2 H
b
i t( )
3 3
36 6(V)
6 3 3 3
oc abV v
 
= = × = 
+ + 
( )6 / /6 3 / /3 2thR = + = Ω  
2
1
2th
L
R
τ = = = ( ) ( ) ( )3 2 3 3 At t
i t e e− −
= + − = −
( )
( )
36 6
3 A
6 6 / /3 3 6
× =
+ +
( )6 / /6 3 / /3 2+ = Ω   ( ) ( ) ( )3 2 3 3 At t
i t e e− −
= + − = −
Department of Electronic Engineering, NTUT28/32

29. •
K K = 1 u(t)
(Unit step function)
us(t) (a)
t = 0 0 1
• t – t0 t
R (b) u(t – t0)
t0 t0
u t t
K t
( ) = <
= >
0 0
0
t t0
R t t0
,
,
u t t0( )− = <
= >
0
1
0
t
(a)
(b)
( )u t
1
0 t0
t
( )0u t t−
Department of Electronic Engineering, NTUT29/32

30. 10
• i(t)
KCL
B. t ＞ 0 i t i i Aef n
t
( ) = + = + −
3 2di
dt i+ =2 6
A. t ＜ 0
i(0) = 0 A = 0 i(t) = 0
di
dt i+ =2 0 i t Ae t
( ) = −2
C. t i(t) = 0 , t ＜ 0
= 3 – 3e-2t , t ＞ 0 i t e u tt
( ) ( ) ( )( )= − −
3 1 2
A
i(0) = 0 = 3 + A = 0 , A = -3 i t e t
( ) ( )( )= − −
3 1 2
A
+
−
6u t( ) V
2 Ω
1H
i
Department of Electronic Engineering, NTUT30/32

31. 11
• i(t)
A. u(t) (b)
B. –u(t – 2) (c)
i2(t) i2 t( ) = – 0.5 [1 – e-2(t – 2)] , t ＞ 2
i1 t( ) = 0.5 (1 – e-2t)(A) , 0 ＜ t ＜ 2i t( ) =
i1 t( ) += i2 t( ) = – 0.5 [1 – e-2(t – 2)] (A) , t ＞ 20.5 (1 – e-2t)
i1(0) = 0 A = 0 i1(t) = 0
t 0
di
dt i+ =2 0 i t Ae t( ) = −2
t 0 i1 t i1 i1 Aef n
t
( ) = + = + −
0.5 2
di1
dt
i1+ =2 0
i1(0) = 0.5 + A = 0 A = - 0.5⇒
0.5 (1 – e-2t) (A) , t ＞ 0i1 t( ) =
+
−[u(t)−u(t−2)] V
2 Ω i
1H
+
−
v t( )
u(t)
i1(t) –u(t – 2)
i2(t) i(t) = i1(t) + i2(t)
+
−
2 Ω
1H
(c)
i2(t)
−u(t−2) V
(a)
+
−
u t( )
2 Ω
1H
(b)
i1(t)
Department of Electronic Engineering, NTUT31/32

32. • RC RL
yf yn
• yf ( ) ( )
yf yf yn y(0)
B yn = Be –at
( )
( ) ( )
dy t
ay t f t
dt
+ =
1
a
τ
( )
t
at
y t A Be A Beτ
−
−
= + = +
( ) ( )0 at
f f f ny t y y y e y y−
 = + − = + 
( ) ( )0 0 fB y A y y= − = −
Department of Electronic Engineering, NTUT32/32