RC/RL

Department of Electronic EngineeringNational Taipei University of Technology

••• RC RL

•

Department of Electronic Engineering, NTUT2/32

i

+

−C

+

−

v

dvi C

dt=

L

vL−+

i

L

div L

dt=

( ) ( ) ( )0

0

1 t

ti t v t dt i t

L= +∫( ) ( ) ( )

00

1 t

tv t i t dt v t

C= +∫

Department of Electronic Engineering, NTUT3/32

RC RL

RCv(t)+

−

ic iR

+

−v t( ) L R

−

+vR

i t( )

KCL 0c Ri i+ =

( ) ( ) ( ) ( )10 0

dv t v t dv tC v t

dt R dt RC+ = ⇒ + =

( ) ( ) ( ) ( )0 0di t di t R

L Ri t i tdt dt L

+ = ⇒ + =

KVL

( ) ( ) ( )dy tay t f t

dt+ =

( ) ( ) ( )dy tay t f t

dt+ =

• RC RL

Department of Electronic Engineering, NTUT4/32

RC RL ( )

Vs C

+

−v

vR+ −

it = 0 Rt =0 iL

+ −vL

RVs

vs(t) C

+

−v

vR+ −

iR iL

+ −vL

Rvs(t)

( ) 0f t ≠

Department of Electronic Engineering, NTUT5/32

• RC RL

y v(t) i(t) af(t) f(t)

•

c y(0)

t

c e-at

eat

( ) ( ) ( )dy tay t f t

dt+ =

( ) ( ) ( )at at atdy te e ay t e f t

dt+ =

( ) ( ) ( ) ( ) ( ) ( )at at at at atdy t dy tde y t e e ay t e ay t e f t

dt dt dt

= + = + =

( )( ) ( ) ( )

at

at at atd e y t

dt e f t dt e y t e f t dt cdt

= ⇒ = +∫ ∫ ∫

( ) ( )at at aty t e e f t dt ce− −= +∫

Department of Electronic Engineering, NTUT6/32

•(Complete response)

( ) ( )at at aty t e e f t dt ce− −= +∫

( ) ( )at at aty 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 atfy A e e f t dt−= = ∫

tat

ny Be ceτ− −= =

1

aτ =

Department of Electronic Engineering, NTUT7/32

• f(t)

::::

::::

(Natural response)

f (t) = 0

f (t) = b

( ) ( ) ( ) ( ) ( )at at atdy tay t f t y t e e f t dt ce

dt− −+ = ⇒ = +∫

( ) ( )0at aty t ce y e− −= =

( ) atby t ce

a−= + ( )0

by c

a= + ( )0

bc y

a= −

( ) ( )0 atb by t y e

a a− = + −

:::: (y (0) = 0) f (t) = b

( ) atby t ce

a−= + ( )0 0y =

bc

a= −

( ) atb by t e

a a−= −

Department of Electronic Engineering, NTUT8/32

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)

( ) ( ) ( )5dy t

y t f tdt

+ =

( ) ( )at at aty t e e f t dt ce− −= +∫

( ) 5ty t ce−=

( )0 10y c= =

( ) 510 ty t e−∴ =

( ) 5 5 5 55 1t t t ty t e e dt ce ce− − −= × × + = +∫( )0 10 1 9y c c= = + ⇒ =

( ) 51 9 ty t e−∴ = +

( ) 5 5 5 55 1t t t ty t e e dt ce ce− − −= × × + = +∫

( )0 0 1 1y c c= = + ⇒ = −

( ) 51 ty t e−∴ = −

Department of Electronic Engineering, NTUT9/32

(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

(II)

Rth =10 Ω 1 F

+

−vc t( )

(c) t > 0

( )4 2 / /6 7 10thR = + + = Ω

( ) 100 150 100V

10 5cv − = × = +

t 0

vc 100V vc

(c)( ) ( )0 0 100Vc cv v+ −= =

Department of Electronic Engineering, NTUT11/32

RC (I)

• t > 0

KCL

( ) t > 0

( ) 00cv V− =

0c Ri i+ =

( ) ( )0c cdv t v t

Cdt R

+ =

( ) ( )10C

C

dv tv t

dt RC+ = ( )

t

RCcv t ce

−=

( ) 00cv V− = ( ) 00cv c V− = =

( ) ( ) 00 0c cv v v+ −= =

( ) 0

t

RCcv t V e

−= ( ) ( ) 0

tc RC

c

dv t Vi t C e

dt R

−= = − ( ) ( ) 0

tc RC

R

v t Vi t e

R R

−= =

RCvc(t)+

−

ic iR

Department of Electronic Engineering, NTUT12/32

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 → ∞

( ) ( ) 0dy t

ay tdt

+ =

RCτ =

R v i= C q v=

( ) 0f t →

τ 1

a

τ τ

τ

V0

vc(t)

0t

0.368V0

1 Kτ =2 2Kτ =

3 3Kτ =

K 2K 3K1τ 2τ 3τ

Department of Electronic Engineering, NTUT13/32

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+ −= =

( ) ( )0.110100 100 Vt

tcv t e e

− −= =

( ) ( ) ( ) ( )0.1 0.17 7100 70 V

7 6 // 2 4 10t t

cv t v t e e− −= = × =+ +

( ) ( ) ( ) ( )0.1 0.1100 0.1 10 Ac t tc

dv ti t C e e

dt− −= = × − = −

Department of Electronic Engineering, NTUT14/32

3 ( )

• v(0) = 4 (V) t > 0 i

KCL

t > 0

+−

18 F 3 Ω2i (V)

+

−v

i

6 Ω

2 10

6 8 3

v i dv v

dt

− + + =1

8

dv dvi C

dt dt= =

114 0 6 0

6 8 3

dvv

dv v dvdt vdt dt

−+ + = ⇒ + =

( ) ( ) ( )60 4 Vt

tv t v e eτ− −= =

( ) ( ) ( )6 614 6 3 A

8t tdv

i t C e edt

− −= = × × − = −

Department of Electronic Engineering, NTUT15/32

RC (I)

• v(0–)=V0

t = 0

• t > 0 v(t)

(b)

( )

( )

RC K( V0–Vs) (

RC )

( ) ( )0

t

RCs sv t V V V e

−= + −

sV

( )0

t

RCsV V e

−−

(a)

Vs

0 t

v(t)

( )0

t

RCsV V e

−−( )0 sV V−

(b)0 t

V0

Vs

v(t)

( )0

t

RCs sV V V e

−+ −

Vs C+

−v

vR+ −

it = 0 R

Department of Electronic Engineering, NTUT16/32

RC (II)

• RC

1. RC

2. ( )

3. ( ) τ

t

e τ−

×

Department of Electronic Engineering, NTUT17/32

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 IC I

dt R dt RC C+ = ⇒ + =

( )t t t t

RC RC RC RCI

v t e e dt ce RI ceC

− − − −= + = +∫

( ) ( ) 00 0v v v+ −= = ( ) 00v RI c V+ = + = 0c V RI= −

( ) ( )0 , 0t

RCv t RI V RI e t−

= + − >

Department of Electronic Engineering, NTUT18/32

5

• (a) v(0 ) = 15 (V) t > 0 v(t)

24 Ω 13F

+

−v40V

8 Ωi a

b(a) (b)

40V

a

bRth = 6 Ω

8 Ω

24 Ω

A. ( (b))

30(V) abVoc

( )2440 30 V

24 8× =

+

B. (c) 8 24

68 24thR

×= = Ω+

16 2 sec

3thR Cτ = × = × =

( ) 0.5215 30 15t

te e− −− = −

C. t > 0 ( ) ( )0.530 15 Vtv t e−= −

13F30V

a

b(c)

6 Ω

Department of Electronic Engineering, NTUT19/32

RC

• RCv(0) = 0 t > 0

• RC

Voc = ( )

Rth =

( ) ( )1t

tRCs s sv t V V e V e τ− −= − = −

( ) ( )1 tht R Cocv t V e−= −

a te−

Vs C

+

−v

vR+ −

it =0 R

Department of Electronic Engineering, NTUT20/32

6

• v(0) = 0 t 0 v(t) i(t)

(b)

(a)

+−36 V 12 Ω

+

−v t( )

6 Ω

18

F

i t( )

+−Voc 24= V

a

b

+

−v t( )

Rth = 4 Ω

(b)

18

F

( )1236 24 V

12 6ocV = × =+

6 / /12 4thR = = Ω

1 14 sec

8 2thR Cτ = = × =

( ) ( )( )224 1 Vtv t e−= −

( ) ( ) ( )236 6=2+4e Ati t v t −= −

Department of Electronic Engineering, NTUT21/32

•I0 ( )

• RL

RL

0di dt = 0Lv L di dt= =

Department of Electronic Engineering, NTUT22/32

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 0di t di t R

L Ri t i tdt dt L

+ = ⇒ + =

( )R

tLi t ce

−=

( ) 00i c I− = =

( ) 0

Rt

Li t I e−

= ( ) 0

Rt

LL

div t L I Re

dt

−= = − ( ) 0

Rt

LRv t iR I Re

−= =

( )t

f t Ke τ−

=

Department of Electronic Engineering, NTUT23/32

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 Ω1 H

3 Ω1

i 1H158 Ω

a

b

15

845

4

t

ab

div e

dt

−= = − ( )

15 15

8 82 45 2 9

V3 2 4 5 2

t t

abv t t e e− −

= × = − × = −+

(3)

1

6A

2 a-b

( )0

30 306 A

2 3 5i I= = = =

+

( )( )

3 2 3 15

3 2 3 8thR× +

= = Ω+ +

( )15

86t

i t e−

= 8 sec

15th

L

Rτ = =

(1)

(2)

t > 0

Department of Electronic Engineering, NTUT24/32

8 ( )

•L/R i(0) = 2 A

v6A

a

b

2 Ω

32H

4 Ω

i

+ −v

34 2

2 6ab

di vv i i

dt = + = × −

3

2

di div L

dt dt= =

33 24 22 6

didi dti idt

+ = × −

6 0di

idt

+ = 62 Ati e−=

1

6τ = L

Rτ =

Department of Electronic Engineering, NTUT25/32

RL (I)

• RL

• i(0 ) = I0 t = 0

KVL t > 0

t = 0+

t > 0

t =0 iL

+ −vL

RVs

ss

di di R VL Ri V i

dt dt L L+ = ⇒ + =

( )R R R R

t t t ts sL L L L

V Vi t e e dt ce ce

L R

− − −= + = +∫

( ) ( ) 00 0i i I+ −= =

( ) 0 00 s sV Vi c I c I

R R+ = + = ⇒ = −

( ) 0

Rt

s s LV V

i t I eR R

− = + −

ss

VI

R= ( ) ( )0

Rt

Ls si t I I I e

−= + −

Department of Electronic Engineering, NTUT26/32

RL (II)

• i(t) Vs / R

•

• RL

RL

– × τ

( ) ( )0

Rt

Ls si t I I I e

−= + −

0

Rt

s LV

I eR

− −

t

e τ−

0

Rt

s LV

I eR

− −

VssV

R

sV R

Department of Electronic Engineering, NTUT27/32

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+−6 V

2 Ωa

2 H

b

i t( )

3 336 6(V)

6 3 3 3oc abV v = = × = + + ( )6 / /6 3 / /3 2thR = + = Ω

21

2th

L

Rτ = = = ( ) ( ) ( )3 2 3 3 At ti t e e− −= + − = −

( ) ( )36 63 A

6 6 / /3 3 6× =

+ +( )6 / /6 3 / /3 2+ = Ω ( ) ( ) ( )3 2 3 3 At ti t e e− −= + − = −

Department of Electronic Engineering, NTUT28/32

•

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 tK t

( ) = <= >

0 00

t t0R t t0

,,

u t t0( )− = <= >

0

1

0 t

(a)

(b)

( )u t

1

0 t0t

( )0u t t−

Department of Electronic Engineering, NTUT29/32

10

• i(t)

KCL

B. t ＞ 0 i t i i Aef nt( ) = + = + −3 2di

dt i+ =2 6

A. t ＜ 0

i(0) = 0 A = 0 i(t) = 0

didt 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

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 didt i+ =2 0 i t Ae t( ) = −2

t 0 i1 t i1 i1 Aef nt( ) = + = + −0.5 2

di1dt 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

• RC RL

yf yn

• yf ( ) ( )yf yf yn y(0)

B yn = Be –at

( ) ( ) ( )dy tay t f t

dt+ =

1

aτ

( )t

aty t A Be A Beτ− −= + = +

( ) ( )0 atf f f ny t y y y e y y− = + − = +

( ) ( )0 0 fB y A y y= − = −

Department of Electronic Engineering, NTUT32/32