Tài liệu Bài giải mạch P6 pptx

Chapter 6, Solution 1.

= = ( − + = −3t −3t 5 2e 6 e

dt

dv i C ) 10(1 - 3t)e-3t A

p = vi = 10(1-3t)e-3t

⋅ 2t e-3t

= 20t(1 - 3t)e-6t W

Chapter 6, Solution 2.

2 2

1 1 (40)(120) 2

1 Cv

2

1

w = =

w2 =

2 2

1 (40)(80) 2

1

2

Cv = 1

∆ = − = ( − ) = 2 2

w w1 w2 20 120 80 160 kW

Chapter 6, Solution 3.

i = C = − = −

5

280 160 40x10

dt

dv 3 480 mA

Chapter 6, Solution 4.

idt v(0) C

1

v t

o = + ∫

∫ 6sin 4tdt +1

2

1 =

= 1 - 0.75 cos 4t

Chapter 6, Solution 5.

v = ∫ + t

o

idt v(0) C

1

For 0 < t < 1, i = 4t,

− ∫ = t

o 6 4t

20x10

1

v dt + 0 = 100t2

kV

v(1) = 100 kV

For 1 < t < 2, i = 8 - 4t,

∫ = − + −

t

1 6 (8 4t)dt v(1) 20x10

1

v

= 100 (4t - t2

- 3) + 100 kV

Thus v (t) = 







− − < <

< <

100(4t t 2)kV, 1 t 2

100t kV, 0 t 1

2

2

Chapter 6, Solution 6.

6 30x10

dt

dv i C − = = x slope of the waveform.

For example, for 0 < t < 2,

3 2x10

10

dt

dv − =

i = 150mA

2x10

10 30x10 x

dt

dv

3

6 = = −

− C

Thus the current i is sketched below.

t (msec)

150

2 10 12

8

6

4

-150

i(t) (mA)

Chapter 6, Solution 7.

∫ ∫ = + = + −

−

t

o

3

o 3 4tx10 dt 10

50x10

1 idt v(t ) C

1

v

= +10 = 50

2t

2

0.04k2

+ 10 V

Chapter 6, Solution 8.

(a) t t ACe BCe

dt

dv C 100 600 100 600 − − i = = − − (1)

i(0) = 2 = −100AC − 600BC → 5 = −A − 6B (2)

v = v → = A + B + − (0 ) (0 ) 50 (3)

Solving (2) and (3) leads to

A=61, B=-11

(b) 4 10 2500 5 J

2

1 (0) 2

1 2 3 = = = − Energy Cv x x x

(c ) From (1),

100 61 4 10 600 11 4 10 24.4 26.4 A 3 100t 3 600t 100t 600t i x x x e x x x e e e − − − − − − = − − = − −

Chapter 6, Solution 9.

v(t) = ( ) ( ) ∫ − − − + = + t

o

t t 6 1 e dt 0 12 t e V

1 2

1

v(2) = 12(2 + e-2) = 25.62 V

p = iv = 12 (t + e-t) 6 (1-e-t) = 72(t-e-2t

)

p(2) = 72(2-e-4) = 142.68 W

Chapter 6, Solution 10

dt

dv

x

dt

dv i C 3 2 10− = =











< <

< <

< <

=

64 -16t, 3 t 4 s

16, 1 t 3 s

16 , 0 1 s

µ

µ

t t µ

v











< <

< <

< <

=

-16x10 , 3 t 4 s

0, 1 t 3 s

16 10 , 0 1 s

6

6

µ

µ

x t µ

dt

dv











< <

< <

< <

=

- 32 kA, 3 t 4 s

0, 1 t 3 s

32 kA, 0 1 s

( )

µ

µ

t µ

i t

Chapter 6, Solution 11.

v = ∫ + t

o

idt v(0) C

1

For 0 < t < 1,

∫ = = −

−

t

o

3

6 40x10 dt 10t

4x10

1 v kV

v(1) = 10 kV

For 1 < t < 2,

vdt v(1) 10kV

C

1

v t

1 = + = ∫

For 2 < t < 3,

∫ = − + −

−

t

2

3

6 ( 40x10 )dt v(2) 4x10

1

v

= -10t + 30kV

Thus

v(t) = 









− + < <

< <

⋅ < <

10t 30kV, 2 t 3

10kV, 1 t 2

10t kV, 0 t 1

Chapter 6, Solution 12.

= = π − π − 3x10 x60(4 )( sin 4

dt

dv i C 3 t)

= - 0.7e π sin 4πt A

P = vi = 60(-0.72)π cos 4π t sin 4π t = -21.6π sin 8π t W

W = ∫ ∫ = − π π t dt t

o

8

1

o pdt 21.6 sin 8

= π

π

π 8

8

.6

cos

21 1/ 8

o = -5.4J