Tài liệu Bài giải mạch P8 doc

Since it is in series with the +10V source, together they represent a direct short at t = 0+.. This means that the entire 2A from the current source flows through the capacitor and not

At t = 0+, i(0+) = i(0-) = 2A, v(0+) = v(0-) = 12V

(b) For t > 0, we have the equivalent circuit shown in Figure (b)

vL = Ldi/dt or di/dt = vL/L Applying KVL at t = 0+, we obtain,

vL(0+) – v(0+) + 10i(0+) = 0

vL(0+) – 12 + 20 = 0, or vL(0+) = -8 Hence, di(0+)/dt = -8/2 = -4 A/s

Hence, diR(0+)/dt = (-1/45)dvC(0+)/dt = -278/45

diR(0+)/dt = -6.1778 A/s

Also, iR = iC + iL

diR(0+)/dt = diC(0+)/dt + diL(0+)/dt -6.1788 = diC(0+)/dt + 0, or diC(0+)/dt = -6.1788 A/s

(c) As t approaches infinity, we have the equivalent circuit in Figure

(a) At t = 0+, since the inductor current and capacitor voltage cannot change abruptly,

the inductor current must still be equal to 0A, the capacitor has a voltage equal to –10V Since it is in series with the +10V source, together they represent a direct

short at t = 0+ This means that the entire 2A from the current source flows through the capacitor and not the resistor Therefore, vR(0+) = 0 V

(b) At t = 0+, vL(0+) = 0, therefore LdiL(0+)/dt = vL(0+) = 0, thus, diL/dt = 0A/s,

iC(0+) = 2 A, this means that dvC(0+)/dt = 2/C = 8 V/s Now for the value of

dvR(0+)/dt Since vR = vC + 10, then dvR(0+)/dt = dvC(0+)/dt + 0 = 8 V/s

(c) As t approaches infinity, we end up with the equivalent circuit shown in Figure (b)

i(0-) = 40/(3 + 5) = 5A, and v(0-) = 5i(0-) = 25V

Hence, i(0+) = i(0-) = 5A

(a) For t < 0, 4u(t) = 0 so that the circuit is not active (all initial conditions = 0)

6 Ω 4A

Since the 4-ohm resistor is in parallel with the capacitor,

i(0+) = vC(0+)/4 = 0/4 = 0 A

Also, since the 6-ohm resistor is in series with the inductor,

v(0+) = 6iL(0+) = 0V

Chapter 8, Solution 7

s2 + 4s + 4 = 0, thus s1,2 =

2

4x44

4± 2 −

−

= -2, repeated roots v(t) = [(A + Bt)e-2t], v(0) = 1 = A

dv/dt = [Be-2t] + [-2(A + Bt)e-2t] dv(0)/dt = -1 = B – 2A = B – 2 or B = 1

6± 2 −

−

= -3, repeated roots i(t) = [(A + Bt)e-3t], i(0) = 0 = A

di/dt = [Be-3t] + [-3(Bt)e-3t] di(0)/dt = 4 = B

Therefore, i(t) = [4te -3t ] A

Chapter 8, Solution 9

s2 + 10s + 25 = 0, thus s1,2 =

2

1010

−

= -5, repeated roots i(t) = [(A + Bt)e-5t], i(0) = 10 = A

di/dt = [Be-5t] + [-5(A + Bt)e-5t] di(0)/dt = 0 = B – 5A = B – 50 or B = 50

Therefore, i(t) = [(10 + 50t)e -5t ] A

Chapter 8, Solution 10

s2 + 5s + 4 = 0, thus s1,2 =

2

1625

Therefore, v(t) = (–(10/3)e -4t + (10/3)e -t ) V

Chapter 8, Solution 11

s2 + 2s + 1 = 0, thus s1,2 =

2

44

−

= -1, repeated roots v(t) = [(A + Bt)e-t], v(0) = 10 = A

dv/dt = [Be-t] + [-(A + Bt)e-t] dv(0)/dt = 0 = B – A = B – 10 or B = 10

Therefore, v(t) = [(10 + 10t)e -t ] V

Chapter 8, Solution 12

(a) Overdamped when C > 4L/(R2) = 4x0.6/400 = 6x10-3, or C > 6 mF

(b) Critically damped when C = 6 mF

(c) Underdamped when C < 6mF

1 = 5

For critical damping, ωo = α = Ro/(2L) = 5

04.0

1 = 5

ωo = α leads to critical damping i(t) = [(A + Bt)e-5t], i(0) = 2 = A

v = Ldi/dt = 2{[Be-5t] + [-5(A + Bt)e-5t]}

04.0

1 = 5

ωo = α leads to critical damping i(t) = [(A + Bt)e-5t], i(0) = 2 = A

v = Ldi/dt = 2{[Be-5t] + [-5(A + Bt)e-5t]}

v(0) = 6 = 2B – 10A = 2B – 20 or B = 13

Therefore, i(t) = [(2 + 13t)e -5t ] A

13

− = 20

ωo = α leads to critical damping i(t) = [(A + Bt)e-20t], i(0) = 0 = A di/dt = {[Be-20t] + [-20(Bt)e-20t]}, but di(0)/dt = -(1/L)[Ri(0) + vC(0)] = -(1/2.5)[0 + 24]

Hence, B = -9.6 or i(t) = [-9.6te -20t ] A

Chapter 8, Solution 17

.iswhich,204

12

10L

2

R

1025

141

1LC

1

240)

600(4)VRI(L

1dt

0 0

−

=+

2 1

2 1

t 32 37 2 t 68 2 1

2 o 2

ee

928.6)

t

(

i

A928.6Atoleads

This

240A

32.37A68.2dt

)0(di,AA0)

0

(

i

eAe

A)

t

(

i

32.37,68.23102030020

−α

±α

−

=

getwe,60dt)t(iC

1)t(v,

v(t) = (60 + 64.53e -2.68t – 4.6412e -37.32t ) V

Chapter 8, Solution 18

When the switch is off, we have a source-free parallel RLC circuit

5.02

1,

2125.0

1

=

RC x

LC

ω

936.125.04case

d

Io(0) = i(0) = initial inductor current = 20/5 = 4A

Vo(0) = v(0) = initial capacitor voltage = 0 V

)936.1sin936

.1cos()

sincos

v(0) =0 = A1

)936.1cos936.1936.1sin936.1()

936.1sin936

.1cos)(

5.0

2 1

.15.041

)40()(

)

0

(

2 2

dt

Thus,

t e

For t > 0, we have a series RLC circuit as shown in Figure (b) with R = 0 = α

ωo =

LC

1 = 4

1 = 0.5 = ωd

i(t) = [Acos0.5t + Bsin0.5t], i(0) = 12 = A

v = -Ldi/dt, and -v/L = di/dt = 0.5[-12sin0.5t + Bcos0.5t],

which leads to -v(0)/L = 0 = B Hence, i(t) = 12cos0.5t A and v = 0.5 However, v = -Ldi/dt = -4(0.5)[-12sin0.5t] = 24sin0.5t V

α = R/(2L) = 2/(2x0.5) = 2

ωo = 1/ LC=1/ 0 x1 4 =2 2Since α is less than ωo, we have an under-damped response

2482 2 o

d = ω −α = − =ω

i(t) = (Acos2t + Bsin2t)e-2t

i(0) = 6 = A

di/dt = -2(6cos2t + Bsin2t)e-2t + (-2x6sin2t + 2Bcos2t)e-αtdi(0)/dt = -12 + 2B = -(1/L)[Ri(0) + vC(0)] = -2[12 – 12] = 0

Thus, B = 6 and i(t) = (6cos2t + 6sin2t)e -2t A

/13/1LC/1

±α

v(t) = [Ae-t + Be-9t], v(0) = 16 = A + B (1)

i = Cdv/dt = C[-Ae-t - 9Be-9t] i(0) = 0 = C[-A – 9B] or A = -9B (2) From (1) and (2), B = -2 and A = 18

Hence, v(t) = (18e -t – 2e -9t ) V

Chapter 8, Solution 22

α = 20 = 1/(2RC) or RC = 1/40 (1)

2 2 o

α = 1/(2RC) = 1/(2x5x10-3) = 100

ωo = 1/ LC =1/ 0 x10−3 = 100

ωo = α (critically damped) v(t) = [(A1 + A2t)e-100t] v(0) = 0 = A1

9843.1)16/1(42 2 o

ω

vo(t) = (A1cosωdt + A2sinωdt)e-αt

i(t) = Is + [(A1cos4t + A2sin4t)e-t], Is = 10/5 = 2

i(0) = 2 = = 2 + A1, or A1 = 0 di/dt = [(A2cos4t)e-t] + [(-A2sin4t)e-t] = 4 = 4A2, or A2 = 1

0 = dv(0)/dt = -2A1 +2A2 or A2 = A1 = -3

v(t) = [3 – 3(cos2t + sin2t)e -2t ] volts

Chapter 8, Solution 28

The characteristic equation is s2 + 6s + 8 with roots

2,42

323662

I t

i( )= + −2 + −4

5.112

8I s = → I s =

B A

i(0)=0 → 0=1.5+ + (1)

t

Ae dt

A dt

di

21

04

22)

0

(

++

i( )=1.5−2 − 2 +0.5 − 4 A

Chapter 8, Solution 29

(a) s2 + 4 = 0 which leads to s1,2 = ±j2 (an undamped circuit)

v(t) = Vs + Acos2t + Bsin2t 4Vs = 12 or Vs = 3 v(0) = 0 = 3 + A or A = -3 dv/dt = -2Asin2t + 2Bcos2t

dv(0)/dt = 2 = 2B or B = 1, therefore v(t) = (3 – 3cos2t + sin2t) V

(b) s2 + 5s + 4 = 0 which leads to s1,2 = -1, -4

i(t) = (Is + Ae-t + Be-4t) 4Is = 8 or Is = 2 i(0) = -1 = 2 + A + B, or A + B = -3 (1)

di/dt = -Ae-t - 4Be-4t

di(0)/dt = 0 = -A – 4B, or B = -A/4 (2) From (1) and (2) we get A = -4 and B = 1

i(t) = (2 – 4e -t + e -4t ) A

(c) s2 + 2s + 1 = 0, s1,2 = -1, -1

v(t) = [Vs + (A + Bt)e-t], Vs = 3

v(0) = 5 = 3 + A or A = 2 dv/dt = [-(A + Bt)e-t] + [Be-t] dv(0)/dt = -A + B = 1 or B = 2 + 1 = 3

v(t) = [3 + (2 + 3t)e -t ] V

Chapter 8, Solution 30

2 2 2

2 2

s =− =−α+ α −ω =− =−α− α −ω

L

R s

s

26502

13002

Hence,

mH8.1536502

α

LC s

2

F25.16)

45.632(

1

=

L C

α = R/(2L) = 6/2 = 3, ωo = 1/ LC =1/ 0.04

s = −3± 9−25 = −3± j4Thus, v(t) = Vf + [(Acos4t + Bsin4t)e-3t]

where Vf = final capacitor voltage = 50 V v(t) = 50 + [(Acos4t + Bsin4t)e-3t] v(0) = -12 = 50 + A which gives A = -62

i(0) = 0 = Cdv(0)/dt dv/dt = [-3(Acos4t + Bsin4t)e-3t] + [4(-Asin4t + Bcos4t)e-3t]

α = R/(2L) = 5/2 = 2.5 4

/1LC/1

v(t) = Vs + [A1e-4.95t + A2e-0.05t], v = 20

Since α is less than ωo, we have an underdamped response Therefore,

i(t) = A1cos8t + A2sin8t where i(0) = 0 = A1

di(0)/dt = (1/L)vL(0) = -(1/L)v(0) = -4x20 = -80 However, di/dt = 8A2cos8t, thus, di(0)/dt = -80 = 8A2 which leads to A2 = -10

Now we have i(t) = -10sin8t A

Chapter 8, Solution 35

At t = 0-, iL(0) = 0, v(0) = vC(0) = 8 V For t > 0, we have a series RLC circuit with a step input

α = R/(2L) = 2/2 = 1, ωo = 1/ LC = 1/ 1/5 = 5

o

2 −ω = − ±α

±α

−

v(t) = Vs + [(Acos2t + Bsin2t)e-t], Vs = 12

v(0) = 8 = 12 + A or A = -4, i(0) = Cdv(0)/dt = 0 But dv/dt = [-(Acos2t + Bsin2t)e-t] + [2(-Asin2t + Bcos2t)e-t]

0 = dv(0)/dt = -A + 2B or 2B = A = -4 and B = -2

v(t) = {12 – (4cos2t + 2sin2t)e -t V

Chapter 8, Solution 36

For t = 0-, 3u(t) = 0 Thus, i(0) = 0, and v(0) = 20 V

For t > 0, we have the series RLC circuit shown below

s1,2 = 2 0.8 j0.6

o

2 −ω = − ±α

±α

−v(t) = Vs + [(Acos0.6t + Bsin0.6t)e-0.8t]

Vs = 15 + 20 = 35V and v(0) = 20 = 35 + A or A = -15

i(0) = Cdv(0)/dt = 0 But dv/dt = [-0.8(Acos0.6t + Bsin0.6t)e-0.8t] + [0.6(-Asin0.6t + Bcos0.6t)e-0.8t]

0 = dv(0)/dt = -0.8A + 0.6B which leads to B = 0.8x(-15)/0.6 = -20

i(0) = i1 = 5A -10 – 6i2 + v(0) = 0

v(0) = 10 + 6x5/3 = 20 For t > 0, we have a series RLC circuit

R = 6||12 = 4

ωo = 1/ LC = 1/ (1/2)(1/8) = 4

α = R/(2L) = (4)/(2x(1/2)) = 4

α = ωo, therefore the circuit is critically damped

v(t) = Vs +[(A + Bt)e-4t], and Vs = 10

i(0) = 2A, i1(0) = 10(2)/(10 + 15) = 0.8 A

v(0) = 5i1(0) = 4V For t > 0, we have a source-free series RLC circuit

di(0)/dt = (1/L)[-Ri(0) + v(0)] = (4/3)(-4x2 + 4) = -16/3 = -5.333

Hence, -5.333 = -4.431A – 0.903B (2) From (1) and (2), A = 1 and B = 1

For t > 0, the circuit is shown in Figure (b)

R = 20||30 = 12 ohms

ωo = 1/ LC = 1/ (1/2)(1/4) = 8

α = R/(2L) = (12)/(0.5) = 24 Since α > ωo, we have an overdamped response

s1,2 = −α± α −ω2 =

o

2 -47.833, -0.167 Thus, v(t) = Vs + [Ae-47.833t + Be-0.167t], Vs = 30

v(0) = 24 = 30 + A + B or -6 = A + B (1)

i(0) = Cdv(0)/dt = 0 But, dv(0)/dt = -47.833A – 0.167B = 0

Since α = ωo, we have a critically damped response

v(t) = Vs + [(A + Bt)e-5t], Vs = 24 – 12 = 12V

v(0) = 0 = 12 + A or A = -12

i = Cdv/dt = C{[Be-5t] + [-5(A + Bt)e-5t]}

i(0) = 3 = C[-5A + B] = 0.02[60 + B] or B = 90 Thus, i(t) = 0.02{[90e-5t] + [-5(-12 + 90t)e-5t]}

i(t) = {(3 – 9t)e -5t } A

Chapter 8, Solution 41

At t = 0-, the switch is open i(0) = 0, and

v(0) = 5x100/(20 + 5 + 5) = 50/3 For t > 0, we have a series RLC circuit shown in Figure (a) After source

transformation, it becomes that shown in Figure (b)

where ωd = 4.583 and Vs = 20 v(0) = 50/3 = 20 + A or A = -10/3

i(t) = Cdv/dt = C(-2) [(Acosωdt + Bsinωdt)e-2t] + Cωd[(-Asinωdt + Bcosωdt)e-2t]

i(0) = 0 = -2A + ωdB

B = 2A/ωd = -20/(3x4.583) = -1.455 i(t) = C{[(0cosωdt + (-2B - ωdA)sinωdt)]e-2t}

= (1/25){[(2.91 + 15.2767) sinωdt)]e-2t}

i(t) = {0.7275sin(4.583t)e -2t } A

Chapter 8, Solution 42

For t = 0-, we have the equivalent circuit as shown in Figure (a)

i(0) = i(0) = 0, and v(0) = 4 – 12 = -8V

v(0) = -8 = -12 + A or A = 4

i = Cdv/dt, or i/C = dv/dt = [-3(Acos4t + Bsin4t)e-3t] + [4(-Asin4t + Bcos4t)e-3t]

i(0) = -3A + 4B or B = 3

v(t) = {-12 + [(4cos4t + 3sin4t)e -3t ]} A

11

100

11

,50012

x L

R

o

ωα

→

α

s1,2 = −α ± α −ω2 =

o

2 -0.5 ± j1.323 Thus, i(t) = Is + [(Acos1.323t + Bsin1.323t)e-0.5t], Is = 4

i(0) = 1 = 4 + A or A = -3

v = vC = vL = Ldi(0)/dt = 0 di/dt = [1.323(-Asin1.323t + Bcos1.323t)e-0.5t] + [-0.5(Acos1.323t + Bsin1.323t)e-0.5t]

di(0)/dt = 0 = 1.323B – 0.5A or B = 0.5(-3)/1.323 = -1.134

Thus, i(t) = {4 – [(3cos1.323t + 1.134sin1.323t)e -0.5t ]} A

Chapter 8, Solution 46

For t = 0-, u(t) = 0, so that v(0) = 0 and i(0) = 0

For t > 0, we have a parallel RLC circuit with a step input, as shown below

5µF 8mH

s1,2 = −α± α −ω2 ≅

o

2 -50 ± j5,000 Thus, i(t) = Is + [(Acos5,000t + Bsin5,000t)e-50t], Is = 6mA

i(0) = 0 = 6 + A or A = -6mA v(0) = 0 = Ldi(0)/dt di/dt = [5,000(-Asin5,000t + Bcos5,000t)e-50t] + [-50(Acos5,000t + Bsin5,000t)e-50t]

di(0)/dt = 0 = 5,000B – 50A or B = 0.01(-6) = -0.06mA Thus, i(t) = {6 – [(6cos5,000t + 0.06sin5,000t)e -50t ]} mA

Since α = ωo, we have a critically damped response

s1,2 = -10 Thus, i(t) = Is + [(A + Bt)e-10t], Is = 3

i(0) = 1 = 3 + A or A = -2

vo = Ldi/dt = [Be-10t] + [-10(A + Bt)e-10t]

vo(0) = 0 = B – 10A or B = -20 Thus, vo(t) = (200te -10t ) V

Chapter 8, Solution 48

For t = 0-, we obtain i(0) = -6/(1 + 2) = -2 and v(0) = 2x1 = 2

For t > 0, the voltage is short-circuited and we have a source-free parallel RLC circuit

α = 1/(2RC) = (1)/(2x1x0.25) = 2

ωo = 1/ LC = 1/ x0.25 = 2 Since α = ωo, we have a critically damped response

s1,2 = -2 Thus, i(t) = [(A + Bt)e-2t], i(0) = -2 = A

v = Ldi/dt = [Be-2t] + [-2(-2 + Bt)e-2t]

vo(0) = 2 = B + 4 or B = -2

Thus, i(t) = [(-2 - 2t)e -2t ] A and v(t) = [(2 + 4t)e -2t ] V

Chapter 8, Solution 49

For t = 0-, i(0) = 3 + 12/4 = 6 and v(0) = 0

For t > 0, we have a parallel RLC circuit with a step input

α = 1/(2RC) = (1)/(2x5x0.05) = 2

ωo = 1/ LC = 1/ x0.05 = 2 Since α = ωo, we have a critically damped response

s1,2 = -2 Thus, i(t) = Is + [(A + Bt)e-2t], Is = 3

s1,2 = −α± α −ω2 =

o

2 -10, -2.5

Thus, i(t) = Is + [Ae-10t] + [Be-2.5t], Is = 9

i(0) = 3 = 9 + A + B or A + B = -6 di/dt = [-10Ae-10t] + [-2.5Be-2.5t], v(0) = 0 = Ldi(0)/dt or di(0)/dt = 0 = -10A – 2.5B or B = -4A

Thus, A = 2 and B = -8 Clearly, i(t) = { 9 + [2e -10t ] + [-8e -2.5t ]} A Chapter 8, Solution 51

Let i = inductor current and v = capacitor voltage

At t = 0, v(0) = 0 and i(0) = io For t > 0, we have a parallel, source-free LC circuit (R = ∞)

α = 1/(2RC) = 0 and ωo = 1/ LC which leads to s1,2 = ± jωo

v = Acosωot + Bsinωot, v(0) = 0 A

iC = Cdv/dt = -i dv/dt = ωoBsinωot = -i/C dv(0)/dt = ωoB = -io/C therefore B = io/(ωoC)

v(t) = -(i o /(ω o C))sinω o t V where ω o = LC Chapter 8, Solution 52

o

d

1575.264300

50)575.264(

12

1

x C R

α

0 = R2C2dvo/dt + R1(C2dvo/dt – C1dv1/dt) (4) Applying KVL to the outer loop produces,

vs = v1 + i2R2 + vo = v1 + R2C2dvo/dt + vo, which leads to

v1 = vs – vo – R2C2dvo/dt (5) Substituting (5) into (4) leads to,

14

1

2 2

2

=++

→

+

++

=

dt

dv dt

v d dt

v d dt

dv dt

dv v v

199.125.10

35.2

t Be

t Ae

v= − 1 25tcos1.199 + − 1 25tsin1.199

v(0) = 2=A Let w=1.199

)cossin

()sincos

(25

1 Ae 1 25 wt Be 1 25 wt w Ae 1 25 wt Be 1 25 wt dt

085.2199.1

225.125

.10)0(

=

=

→

+

−

=

dt dv

V 199.1sin085

.2199.1cos

Which leads to s2 + 7.25s = 0 = s(s + 7.25) or s1,2 = 0, -7.25

v(t) = A + Be-7.25t (3) v(0) = 4 = A + B (4)

For t < 0, i(0) = 0 and v(0) = 0

For t > 0, the circuit is as shown below

6dio/dt + 0.25d2io/dt2 + 25(io + i) = 0 (1)

For the smaller loop, 4 + 25∫(i+io)dt = 0

Taking the derivative, 25(i + io) = 0 or i = -io (2) From (1) and (2) 6dio/dt + 0.25d2io/dt2 = 0

This leads to, 0.25s2 + 6s = 0 or s1,2 = 0, -24