Resolucionario do livro, Exercícios de Eletrônica de Potência

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CHAPTER 1 SOLUTIONS

(1-1)

(1-2)

(1-3)

Time

0s 5us 10us 15us V(V2:-)

-20V

0V

20V

40V

(3.150u,-1.052)

(3.150u,-1.052)

96.46n,23.94)

(1-4)

T i m e

0 s 2 u s 4 u s 6 u s 8 u s 1 0 u s 1 2 u s 1 4 u s 1 6 u s V ( V 2 : - )

• 5 V

0 V

5 V

1 0 V

1 5 V

2 0 V

2 5 V

( 3. 8 3 3 3 u , - 1. 0 5 1 7 )

( 8 0 0. 0 0 0 n , 2 3. 9 2 4 )

   

   

6 10 14

0 0 6 10

T^ ms^ ms^ ms

ms ms

W p t dt dt dt dt J

or W PT ms mJ

   

---

2-6)

   

   

dc avg

avg

avg

P V I

a I A P W

b I A P W

---

2-7)

a)

   

           

 

2

0

25sin 377.

25sin 377 1.0sin 377 25sin 377 12.5 1 cos 754.

R

T

R

v t i t R t V

p t v t i t t t t t W

P p t dt W T



b)

 

       

         

   

 

3

0

10 10 377 1.0 cos 377 3.77 cos 377.

3.77 cos 377 1.0sin 377 sin 754 1.89 sin 754. 2

L

L

T

L

di t v t L t t V dt

p t v t i t t t t t W

P p t dt T

   



c)

         

  0

12 1.0sin 377 12sin 377.

T

dc

p t v t i t t t W

P p t dt T



---

2-8) Resistor:

2

1/60 1/60 1/ 2

0 0 0 0

8 24sin 2 60.

8 24sin 2 60 2 6sin 2 60

16 96sin 2 60 144sin 2 60.

16 96sin 2 60 144sin 2 60 1/ 60

T

v t i t R t V

p t v t i t t t

t t W

P p t dt dt t dt t T

W



 

 

 

Inductor: PL^ 0.

dc source: Pdc^ ^ I^ avg Vdc^ ^  2   6 ^12 W.

---

2-9) a) With the heater on,

2 2 2

sin 120 2 12.5 2 sin 3000sin

max 3000.

m m m

m m

V I

P W I

p t V I t t t

p t W

b) P = 1500(5/12) = 625 W.

c) W = PT = (625 W)(12 s) = 7500 J. (or 1500(5) = 7500 W.)

---

2-10)

0 3

t

L L

L

i t v t dt d t t ms L

i ms A



a)

W  Li   J

b) All stored energy is absorbed by R: WR = 0.648 J_._

c)

16.. 40

R R

S R

W

P W

T ms

P P W

d) No change in power supplied by the source: 16.2 W.

---

2-13)

a) The zener diode breaks down when the transistor turns off to maintain inductor current. b) Switch closed: 0 < t < 20 ms.

 

   

L L

L L

L

di t v V L dt

di v A s dt L

at t ms i A

Switch open, zener on:

12 20 8.

L

L L

v V

di v A s dt L

t to return to zero

i t ms

Therefore, inductor current returns to zero at 20 + 30 = 50 ms. iL = 0 for 50 ms < t < 70 ms.

c)

Time

0s 10ms 20ms 30ms 40ms 50ms 60ms 70ms W(D1)

0W

40mW

80mW

SEL>>

Zener inst. power

W(L1)

-40mW

0W

40mW

Inductor inst. power

d)

      0

L T

Z Z

P

P p t dt W T



---

2-14) a) The zener diode breaks down when the transistor turns off to maintain inductor current.

b) Switch closed: 0 < t < 15 ms.

 

   

L L

L L

L

di t v V L dt

di v A s dt L

at t ms i A

Switch open, zener on:

20 30 10.

L

L L

v V

di v A s dt L

t to return to zero

i t ms

Therefore, inductor current returns to zero at 15 + 30 = 45 ms.

iL = 0 for 45 ms < t < 75 ms.

c)

Time

0s 20ms 40ms 60ms 80ms W(D1)

0W

100W

200W

SEL>>

Zener inst. power

W(L1)

-200W

0W

200W

Inductor inst. power

2-19)

 

       

 

2 2 2

2 2 2

0 0 1

cos 2

2.0 1.5 cos 20 cos 115 7.. 2 2 2 2

cos(4 60 45 ) cos 4 60 135

rms

rms

m m n n n

V V

I A

V I

P V I

W

Notethat t is t







---

2-20)

 

 

 

 

 

     

   

0

1 1

1 1 1

2 2

2 2 2

0 0 1

cos 2

300 5 cos 62.1 cos 7 2 2

m m n n n

dc V V

Y R j C j

I

V

Y j

Y R j C j

I

V

Y j

V I

P V I







 5.1

1500 175 118 1793 W.

---

2-21) dc Source:

P dc Vdc I avg W

^ ^ 

Resistor:

   

   

2

2 2 2 0 1, 2,

0

1

2

2 2 2

2

rms

rms rms rms

rms

R rms

P I R

I I I I

I A

I A

j

I A

j

I A

P I R W

---

2-22)

   

   

   

2

0 0

1

2

2 2 2

2 2

rms

rms

rms rms

P I R

V

I A

R

I A

j

I A

j

I A

I A P I R W

---

2-23)

0 0 ^  1

cos

m m n n n

V I

P V I  





  (^)  

n Vn In Pn ∑Pn

0 20 5 100 100

1 20 5 50 150

2 10 1.25 6.25 156.

3 6.67 0.556 1.85 158.

4 5 0.3125 0.781 158.

Power including terms through n = 4 is 158.9 watts.

---

b)

2 2 2 10 6 3

8.. 2 2 2

rms

rms rms

I A

S V I VA

P

pf S

c)

1, 10/^2

rms

rms

I

DF

I

d)

2 2 6 3

THD I

---

2-28) a)

 

cos 40 0 0 781. 2 2

P P n W



b)

2 2 2 12 5 4

9.. 2 2 2

rms

rms rms

I A

S V I VA

P

pf S

c)

1, 12/^2

rms

rms

I

DF

I

d)

2 2 5 4

THD I

---

2-29)

1, 2,

2 2

rms rms

rms peak

I A I A

I A I graphically

a) P^ ^ V 1,^ rms I 1,^ rms cos^   1 ^  1  ^  240   5.66 cos 0  ^ ^1358 W.

b)    

rms rms^240 6.

P P

pf S V I

c)

2, 2.^

rms I rms

I

THD

I

d)

1, 5.^

rms

rms

I

DF

I

e)

peak

rms

I

crest factor I

---

2-30)

1, 2,

2 2

rms rms

rms peak

I A I A

I A I A graphically

a) P^ ^ V 1,^ rms I 1,^ rms cos^   1 ^  n  ^  240   10.6 cos 0  ^ ^2036 W.

b)    

rms rms^240 10.

P P

pf S V I

c)

2, 6.^

rms I rms

I

THD

I

d)

1, 8.^

rms

rms

I

DF

I

e)

peak

rms

I

crest factor I

---

Time

0s 4ms 8ms 12ms 16ms 20ms I(I1) V(V1:+)

0

20

SEL>>

W(I1) AVG(W(I1))

-400W

0W

400W

Inst Power

Avg Power (20.000m,60.000)

S(W(I1))

0

Energy (20.000m,1.2000)

---

2-33) Average power for the resistor is approximately 1000 W. For the inductor and dc source,

the average power is zero (slightly different because of numerical solution).

Time

0s 5ms 10ms 15ms 20ms AVG(W(R1)) AVG(W(L1)) AVG(W(V1))

-1.0KW

0W

1.0KW

2.0KW

Average Power

Vdc

Inductor

Resistor

(16.670m,-30.131u)

(16.670m,189.361u)

(16.670m,0.9998K)