Tài liệu Bài giải mạch P6 pptx

Under dc conditions, the circuit becomes that shown below: 4C 1eq... Because this is a totally capacitive circuit, we can combine all the capacitors using the property that capacitors in

2

1Cv2

1

1 (40)(80)2

480 mA

Chapter 6, Solution 4

)0(vidtC

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

1

v dt + 0 = 100t2 kV v(1) = 100 kV

1v

1t0,

kVt100

2 2

Chapter 6, Solution 6

6

10x

30

dt

dv

C

i= = − x slope of the waveform

For example, for 0 < t < 2,

3

10x2

10dt

10x10x

3 3

10x50

1)

t(vidtC

(a) ACe t BCe t

AC

i(0)=2=−100 −600 → 5=− −6 (2)

B A v

.2410

41160010

461

s3t116,

s10

,16

µµ

µ

t t

v

s10,1016

6

6

µµ

µ

t x

s3t10,

s10

,kA32)

(

µµ

µ

t t

i

Chapter 6, Solution 11

v = ∫t +

oidt v(0)C

3

10x

1

v(1) = 10 kV For 1 < t < 2,

kV10)1(vvdtC

3

6 ( 40x10 )dt v(2)10

x

1v

= -10t + 30kV Thus

2t1,

kV10

1t0,

kVt10

Chapter 6, Solution 12

π

−π

=

= x10− x60(4 )( sin4dt

dvC

o 21.6 sin8pdt

π

π88

6.cos

o = -5.4J

Under dc conditions, the circuit becomes that shown below:

4C

1eq

w20 = 2 = x20x10− 6x1002 =

2

1Cv

Cv

2 1

80

80

++

(b) Ceq = 5 + [6 || (4 + 2)] = 5 + (6 || 6) = 5 + 3 = 8F

(c) 3F in series with 6F = (3 x 6)/9 = 6F

13

16

12

1C

1eq

=++

Ceq = 1F

For the capacitors in parallel = 15 + 5 + 40 = 60 µF

1 eqC

Hence

10

160

130

120

1C

1eq

=++

20

8

6µF in series with 6µF = 3µF 3µF in parallel with 2µF = 5µF 5µF in series with 5µF = 2.5µF Hence Ceq = 2.5µF

120

160

1C

1

1

eq

=++

eq

Thus

Ceq = 10 + 40 = 50 µF

(a) 3µF is in series with 6µF 3x6/(9) = 2µF

1CC

vs = v1 + v2 = 2

1

2 1 2 2 1

C

CCvvC

=

2 1

1

CC

C+

=v

2 1

2

CC

Cv

1C

QC

Q =

2

2 1 2 2 2

C

CCQQC

=+

or

Q2 =

2 1

2CC

C+

s 2 1

1

CC

CQ

1

CC

C+

=

2 1

2

CC

C+

=i

2 2

C1

= 393.8mJ

(a)

20

720

110

15

1C

1C

1C

1C

1

3 2 1 eq

=++

=++

12

2 2

eqvC

40

130

130

110

140

110

1

403 + + =

Ca = 5µF

30210

1300

1400

1300

1400

75.23x

in series with C =

5

C3

2

C52

C3Cx

2C

C

1C2

1C2

1C

1eq

=+

=

Ceq = C

vo = ∫t +

oidt i(0)C

1

For 0 < t < 1, i = 60t mA,

kVt100tdt6010

x

10

o 6

3

−

vo(1) = 10kV For 1< t < 2, i = 120 - 60t mA,

10

t + = [40t – 10t2]1 10kV

1t0,

kVt10)

3t1,

mA20

1t0,tmA20)t(

is

Ceq = 4 + 6 = 10µF

)0(vidtC

1

o eq

3

t2010

x10

10v

kV1t

kV1t2

1t0,

kVt)t(v

2 2

dt

dv10xdt

dvC

1 1

3t1,

mA12

1t0,

tmA12

dt

dv10x4dt

dvC

2 1

3t1,

mA8

1t0,

tmA8

Chapter 6, Solution 32

(a) Ceq = (12x60)/72 = 10 µF

13001250

501250

)0(30

1012

0

0

2 1

2 6

e x

v

230250

20250

)0(30

1060

2 2

2 6

e x

v

(b) At t=0.5s,

03.138230

250,

15.8401300

w µF

J 1905.0)03.138(10202

w µF

J 381.0)03.138(10402

w

Because this is a totally capacitive circuit, we can combine all the capacitors using the property that capacitors in parallel can be combined by just adding their values and we combine capacitors in series by adding their reciprocals

3F + 2F = 5F 1/5 +1/5 = 2/5 or 2.5F The voltage will divide equally across the two 5F capacitors Therefore, we get:

VTh = 7.5 V, CTh = 2.5 F

Chapter 6, Solution 34

i = 6e-t/2

2 /

2

1)6(10x10dt

6.0

10x60t/

VL

3

200 mH

Chapter 6, Solution 36

V)t2sin)(

2)(

12(10x4

1dt

diL

diL

∫ =∫

Jt200cos200

6

9 11/200

o

−

= =−48(cosπ−1)mJ= 96 mJ

Chapter 6, Solution 38

(e 2te )dt10

x40dt

diL

v= = − 3 − 2 t − − 2 t

= 40(1−2t)e− 2 tmV t>0

1idt

diL

v= → = ∫0t +

1dt)4t2t3(10x200

1

i= −3∫0t 2 + + +

1)t4tt

40

-ms3t110t,

-20

ms10

ms3t1,10x10

-ms10

,10

200

ms3t1V,

200

-ms10

t 2 t

2

1)0(ivdtL

1)0(ivdtL

w = L Li ( )

2

1)t(Li2

1idt

2 t

10x80

t o

5

1tivdtL

1

= 0.25t2 kA For 1 < t < 2, v = -10 + 5t

1

3 ( 10 5t)dt i(1)10

x10

1i

1t0,

kAt25.0)

t

(

i

2 2

= (3)

24

4

iL 2A, vc = 0V

L2

1

C2

10)

2R

R10Ri

vc L

+

=

=

2 6

2

c

c

)2R(

R100x10x80Cv

2

1

L

)2R(

100x10x2Li

2 6

)2R(

100x10x)2Rx(

R100x10

−

+

−

+ vC1

=+

=

=

64

30i

160

[5(8 12) 6(8 4)]

8106

L

L Lx L

L L L

5.02

5.025

.0//

++

=+

5L

L3

5LxL

3

2L

L

85

dt

di4vv

i = i1 + i2 i2 = i – i1 (3)

3

vdt

diordt

di3

and

0dt

di5dt

di2

di2

di7dt

di5dt

di5dt

di2

dt

di73

51

35dt

vs = 1+ 2

2 1

sLL

vdt

dt

di

L2

2 =v

,vLL

L

2 1

1

2 1

2

LL

Lv

diLv

2

1 1 2

2 1

i = +

2 1

2 1 2

1

2 1 s

LL

LLvL

vL

vdt

didt

didt

=+

=+

LLL

1vdtL

1

2 1

2 1 1 1

2 1

LLL+

=

dt

diLL

LLL

1vdtL

1

2 1

2 1 2 2

2 1

LL

L+

Chapter 6, Solution 60

8

155//

=

eq L

di L

2 2

0

A5.15.05

.12)

15(5

12)0()

s

2030

diL

diL

8.0

2

1 t t

(a) 40mH

80

60202560//

eq

x i

dt t v L

i dt

3

3

)0()1(

1.0)0(12

1040

10)

0()(1

Using current division,

i i i

i

i

4

1,

.0)0(75.0)

-A )08667.01

,2

30

,2

2 1

1

t dt

di dt

,4

42

,0

20

,4

2

2

t t

t dt

di dt

di

L

v

v1

v2

34/12)

A 93)]

()0([)()

(b) -12 + 4i(0) + v=0, i.e v=12 – 4i(0) = 36 V

(c) At steady state, the inductor becomes a short circuit so that v= 0 V

(a) = 2 = 2 =

1 1

2

1iL2

1

= − 2 =

20 (20)( 2)2

dt

diL

v=

ovdt i(0)L

1i = 1 − ∫t12 dt + 0 mA

o

10x60

i 50(1 - cos 4t) mA

mH2440

mV)t4cos1)(

50(dt

d10x24dt

diL

= 4.8 sin 4t mV

= - 5t + 30 mV

Thus vo(t) is as shown below:

2 5

6

5

7 5

5 0

1

=C

1dtvCR

1dtvC

1 1

o

−+

For the given problem, C = 2µF,

R1C = 1 R1 = 1/(C) = 1006/(2) = 500 kΩ

R2C = 1/(4) R2 = 1/(4C) = 500kΩ/(4) = 125 kΩ

R3C = 1/(10) R3 = 1/(10C) = 50 kΩ

10xx10x10

10x.010x20

−+

R

At node b,

dt

dvCR

vvR

v2

Combining (1) and (2),

dt

dv2

RCv

v

o o

dv2.0dt

dvRC

3t1,V2

1t0,V2

2

Chapter 6, Solution 75

,dt

dvRC

Chapter 6, Solution 76

,dt

5t0,10dt

−

=

−

110

o i

Thus vi is obtained from vo as shown below:

–dvo(t)/dt

– vo(t) (V)

4 -4

4 -4

2

vdt

dv2t2sin

Chapter 6, Solution 79

We can write the equation as

)(4)(t y t

− +

−+

− +

+

−

sin2t

−+

vo

R

From the given circuit,

dt

dvk200

k1000v

k5000

k1000)

t(dt

v

o 2

o 2

Ω

Ω

−Ω

dv5dt

vd

o

o 2

o

2

=++

Chapter 6, Solution 81

We can write the equation as

)(25

2

2

t f v

qI

150(36 20)

1t0,4i

1t0,L4dt

diLv

1t0,mV5v

1

CC

=++

11

For the parallel-connected strings,

=µ

C10C10

s

(b) vT = 8 x 100V = 800V

2 T

2

1vC

2

1