# solution-manual-power-electronics-1st-edition-hart.doc

March 13, 2018 | Author: Zuhaili Zawawi | Category: Root Mean Square, Inductor, Ac Power, Electrical Resistance And Conductance, Electrical Equipment

#### Description

Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart

CHAPTER 2 SOLUTIONS 2/21/10

2-1) Square waves and triangular waves for voltage and current are two examples. _____________________________________________________________________________________ 2-2)

a) p  t   v  t  i  t  

v2  t 

R

[170sin  377t  ]2 10

 2890sin 2 377t W .

b) peak power = 2890 W. c) P = 2890/2 = 1445 W. _____________________________________________________________________________________ 2-3) v(t) = 5sin2πt V. a) 4sin2πt A.; p(t) = v(t)i(t) = 20 sin22πt W.; P = 10 W. b) 3sin4πt A.; p(t) = 15sin(2πt)sin(4πt) W.; P = 0 _____________________________________________________________________________________ 2-4)

a)

0  t  50 ms 50 ms  t  70 ms 70 ms  t  100 ms

 0  p  t   v  t  i  t    40  0  b)

70 ms

T

1 1 P   v  t  i  t  dt  40 dt  8.0 W . T 0 100 ms 50ms c) T

70 ms

0

50 ms

W   p  t  dt 

40 dt  800 mJ .; or W  PT   8W   100 ms   800 mJ .

_____________________________________________________________________________________ 2-5)

a)

 70 W .  50 W .  p t  v t i t    40 W .  0

0  t  6 ms 6 ms  t  10 ms 10 ms  t  14 ms 14 ms  t  20 ms

b)

P c)

1 T

T

 p  t  dt  0

1   20 ms 

6 ms

 0

10 ms

70 dt 

 40 dt  19 W .  10 ms

14 ms

  50  dt  

6 ms

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T

6 ms

W   p  t  dt   0

10 ms

70 dt 

 50  dt 

 40 dt   0.38 J .;  10 ms  14 ms

 0 6 ms or W  PT   19   20 ms   380 mJ .

_____________________________________________________________________________________ 2-6)

P  Vdc I avg a ) I avg  2 A., P   12   2   24 W . b) I avg  3.1 A., P   12   3.1  37.2 W . _____________________________________________________________________________________ 2-7) a)

vR  t   i  t  R  25sin 377t V .

p  t   v  t  i  t    25sin 377t   1.0sin 377t   25sin 2 377t  12.5  1  cos 754t  W . T

1 PR   p  t  dt  12.5 W . T 0 b)

vL  t   L

di  t  dt

 10  10 

3

 377   1.0  cos 377t  3.77 cos 377t V .

pL  t   v  t  i  t    3.77 cos 377t   1.0sin 377t  

 3.77   1.0  sin 754t  1.89 sin 754t W . 2

T

PL 

1 p  t  dt  0 T 0

c)

p  t   v  t  i  t    12   1.0sin 377t   12sin 377t W . Pdc 

1 T

T

 p  t  dt  0 0

_____________________________________________________________________________________

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2-8)

Resistor:

v  t   i  t  R  8  24sin 2 60t V .

p  t   v  t  i  t    8  24sin 2 60t   2  6sin 2 60t   16  96sin 2 60t  144sin 2 2 60t W . 1 1  p  t  dt    T 0 1/ 60  T

P

1/60

1/60

16 dt 

0

96sin 2 60t dt 

0

1/60

 144sin 0

2

 2 60t 

 16  72  88 W . Inductor: PL  0.

dc source: Pdc  I avgVdc   2   6   12 W . _____________________________________________________________________________________ 2-9)

a) With the heater on,

P

Vm I m  1500   2   12.5 2  1500 W .  I m  2 120 2



p  t   Vm I m sin 2t  120 2 12.5 2 sin 2t  3000sin 2t max  p  t    3000 W . b) P = 1500(5/12) = 625 W. c) W = PT = (625 W)(12 s) = 7500 J. (or 1500(5) = 7500 W.) _____________________________________________________________________________________ 2-10)

iL  t  

t

1 1 vL  t  dt  90 d   900t  L 0.1 0 iL  4 ms    900   4   10 

3

0  t  4 ms.

 3.6 A.

a)

W

1 2 1 2 Li   0.1  3.6   0.648 J . 2 2

b) All stored energy is absorbed by R: WR = 0.648 J. c)

PR 

WR 0.648   16.2 W . T 40 ms PS  PR  16.2 W .

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

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2-11) a)

W

1 2 Li , or i  2

2  1.2  2W   15.49 A. L 0.010

t

t

1 1 i  t    v    d  14 d   1400t A. L0 0.010 0 15.49  1400ton ton  11.1 ms b) Energy stored in L must be transferred to the resistor in (20 - 11.1) = 8.9 ms. Allowing five time constants,



L 8.9 ms   1.7 ms.; R 5

R

L 10 mH   5.62  1.7 ms 1.7 ms

_____________________________________________________________________________________ 2-12) a) i(t) = 1800t for 0 < t < 4 ms i(4 ms) = 7.2 A.; WLpeak = 1.296 J. b) 10A 5A

Inductor current

SEL>> 0A I(L1) 10A Source current

0A -10A

-I(Vcc) 1.0KW Ind. inst. power 0W -1.0KW W(L1) 1.0KW Source inst. power (supplied) 0W -1.0KW 0s

20ms

40ms

60ms

80ms

100ms

-W(Vcc) Time

_____________________________________________________________________________________

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2-13) a) The zener diode breaks down when the transistor turns off to maintain inductor current. b) Switch closed: 0 < t < 20 ms.

vL  12 V .  L

diL  t  dt

diL vL 12    160 A/s dt L 0.075 at t  20 ms, iL   160   0.02   3.2 A. Switch open, zener on:

vL  12  20  8 V .

diL vL 8    106.7 A/s dt L 0.075 t to return to zero : i 3.2 t    30 ms 106.7 106.7

Therefore, inductor current returns to zero at 20 + 30 = 50 ms. iL = 0 for 50 ms < t < 70 ms. c) 40mW

Inductor inst. power 0W

-40mW W(L1) 80mW

Zener inst. power 40mW

SEL>> 0W 0s

10ms

20ms

30ms

40ms

W(D1) Time

50ms

60ms

70ms

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d)

PL  0. T

1 1  1 PZ   pZ  t  dt   0.03  64  13.73 W .  T 0 0.07  2  _____________________________________________________________________________________ 2-14)

a) The zener diode breaks down when the transistor turns off to maintain inductor current. b) Switch closed: 0 < t < 15 ms.

vL  20 V .  L

diL t  dt

diL vL 20    400 A/s dt L 0.050 at t  15 ms, iL   400   0.015   6.0 A.

Switch open, zener on:

vL  20  30  10 V.

diL vL 10    200 A/s dt L 0.050 t to return to zero : i 6.0 t    30 ms 200 200

Therefore, inductor current returns to zero at 15 + 30 = 45 ms. iL = 0 for 45 ms < t < 75 ms. c) 200W Inductor inst. power 0W

-200W W(L1) 200W Zener inst. power 100W

SEL>> 0W 0s

20ms

40ms

W(D1) Time

60ms

80ms

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d)

PL  0. T

1 1  1 PZ   pZ  t  dt   0.03  180  36 W .  T 0 0.075  2  _____________________________________________________________________________________ 2-15) Examples are square wave (Vrms = Vm) and a triangular wave (Vrms = Vm/√3). _____________________________________________________________________________________ 2-16)

2 2 Phase conductors: P  I R  12  0.5   72 W .

Neutral conductor: PN  I 2 R  12 3

Ptotal  3  72   216  432 W . RN 

PN 72  2 IN 12 3

2

2

 0.5  216 W .

 0.167 

_____________________________________________________________________________________ 2-17) Re: Prob. 2-4

Vrms  Vm D  10 0.7  8.37 V . I rms  I m D  4 0.5  2.83 A. _____________________________________________________________________________________ 2-18) Re: Prob. 2-5

 14   8.36 V .  20

Vrms  Vm D  10  I rms 

1 0.02

0.006

 0

0.01

7 2 dt 

  5

0.006

2

0.02

dt 

4

2

dt  27.7  5.26 A.

0.01

_____________________________________________________________________________________

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2-19) 2

 5     2 

Vrms  22  

3  2

2

2

 4.58 V . 2

 2   1.1  I rms  1.5       2.2 A.  2  2  V I P  V0 I 0   m m cos   n  n  2 n 1 2

 5  2    3  1.1   cos  20     cos  115   7.0 W . 2  2  2   2 Notethat  cos(4 60t  45) is cos  4 60t  135 

  2.0   1.5   

_____________________________________________________________________________________ 2-20)

dc : V0  3  100   300 V .

1  2 60 : Y1  1/R  jC  0.01  j 0.0189 V1 

I1 4 0   187  62.1 Y1  0.01  j 0.0189 

2  4 60 : Y2  1/R  jC  0.01  j 0.0377 V2 

I2 60   153  75.1 Y2  0.01  j 0.0377  

Vm I m cos   n  n  2 n 1

P  V0 I 0    300  5  

 187   4  cos 2

 62.1  

 153  6  cos 2

 75.1 

 1500  175  118  1793 W . _____________________________________________________________________________________ 2-21)

dc Source:

 50  12  114 W . 4  

Pdc  Vdc I avg  12 

Resistor:

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2 P  I rms R

I rms  I 02  I1,2rms  I 2,2 rms I 0  9.5 A. I1 

30  3.51 A. 4  j  4 60   0.01

I2 

10  0.641 A. 4  j  8 60   0.01 2

I rms

 3.51  0.641   9.5      2  2  2

2

 9.83 A.

2 PR  I rms R  386 W .

_____________________________________________________________________________________ 2-22) 2 P  I rms R

V0 6   0.375 A. R 16 5 I1   0.269 A. 16  j  2 60   0.025  I0 

I2 

3  0.0923 A. 16  j  6 60   0.025  2

I rms

 0.269  0.0923   0.375      2  2  2

2

 0.426 A.

2 I rms  0.623 A.; P  I rms R   0.426   16   2.9 W . 2

_____________________________________________________________________________________ 2-23) 

Vm I m cos   n  n  2 n 1

P  V0 I 0  

n Vn In Pn ∑Pn 0 20 5 100 100 1 20 5 50 150 2 10 1.25 6.25 156.25 3 6.67 0.556 1.85 158.1 4 5 0.3125 0.781 158.9 Power including terms through n = 4 is 158.9 watts. _____________________________________________________________________________________

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2-24) 

Vm I m cos   n  n  2 n 1

P  V0 I 0  

n Vn In θn - ϕn° Pn 0 50.0000 10.0 0 500.0 1 50.0000 10.0 26.6 223.6 2 25.0000 2.5 45.0 22.1 3 16.6667 1.11 56.3 5.1 4 12.5000 0.625 63.4 1.7 Through n = 4, ∑Pn = 753 W. _____________________________________________________________________________________ 2-25) 

Vm I m cos   n  n  2 n 1 V V 50  36 I 0  0 dc   0.7 A R 20 2 P0, R  I 02 R   0.7  20  9.8 W (dc component only ) P  V0 I 0  

PVdc  I 0Vdc   0.7   36   25.2 W PL  0 Resistor Average Power

n Vn Zn In angle Pn 0 50.00 20.00 0.7 0.00 9.8 1 127.32 25.43 5.01 0.67 250.66 2 63.66 37.24 1.71 1.00 29.22 3 42.44 51.16 0.83 1.17 6.87 4 31.83 65.94 0.48 1.26 2.33 5 25.46 81.05 0.31 1.32 0.99 PR = ∑ Pn ≈ 300 W. _____________________________________________________________________________________ 2-26)

a) b) c) d)

THD = 5% → I9 = (0.05)(10) = 0.5 A. THD = 10% → I9 = (0.10)(10) = 1 A. THD = 20% → I9 = (0.20)(10) = 2 A. THD = 40% → I9 = (0.40)(10) = 4 A.

_____________________________________________________________________________________ 2-27)

a)

 170  10    cos  30   0  0  736 W .  2  2

P   Pn  

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b) 2

I rms

 10       2 

2

6   3    2  2

2

 8.51 A.

 170  8.51  1024 VA.  2 P 736 pf    0.719 S 1024

S  Vrms I rms  

c)

DF 

I1,rms I rms

10/ 2  0.831 8.51

d) 2

 6  3      2  2 THDI  10/ 2

2

 0.67  67%

_____________________________________________________________________________________ 2-28)

a)

 170  12    cos  40   0  0  781 W .  2  2

P   Pn   b)

2

I rms

 12       2 

2

5   4     2  2

2

 9.62 A.

 170  9.62  1156 VA.  2 P 781 pf    0.68 S 1156

S  Vrms I rms  

c)

DF 

I1,rms I rms

12/ 2  0.88 9.62

d) 2

 5  4      2  2 THDI  12/ 2

2

 0.53  53%

_____________________________________________________________________________________ 2-29)

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8  5.66 A.; 2

I1,rms 

I 2,rms 

4  2.82 A.; 2

I rms  5.662  2.822  6.32 A.; I peak  10.38 ( graphically ) a) P  V1,rms I1,rms cos  1  1    240   5.66  cos  0   1358 W . b) pf 

P P 1358    0.895  89.5% S Vrms I rms  240   6.32 

c) THDI 

d) DF 

I 2,rms I rms

I1, rms I rms

2.82  0.446  44.6% 6.32

5.66  89.6% 6.32

e) crest factor 

I peak I rms

10.38  1.64 6.32

_____________________________________________________________________________________ 2-30)

I1,rms 

12  8.49 A.; 2

I 2,rms 

9  6.36 A.; 2

I rms  8.492  6.362  10.6 A.; I peak  18.3 A. ( graphically ) a) P  V1, rms I1,rms cos  1  n    240   10.6  cos  0   2036 W . b) pf 

P P 2036    0.80  80% S Vrms I rms  240   10.6 

c) THDI 

d) DF 

I 2,rms I rms

I1, rms I rms

6.36  0.60  60% 10.6

8.49  80% 10.6

e) crest factor 

I peak I rms

18.3  1.72 10.6

_____________________________________________________________________________________

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2-31) 5V: I = 0 (capacitor is an open circuit) 25cos(1000t ): Z  R  j L  j I

1 1  2  j1000(.001)  j  2  j0 C 1000  1000  106

25 cos(1000t )  12.5cos(1000t ) A 2

10cos(2000t ): Z  2  j1.5  I10 

10  4  37 A. 2  j1.5 2

 12.5     2 

I rms  

2

4   9.28 A 2

2 PR  I rms R  9.282  2   172.3 W

PL  0 PC  0 Psource  172.3 W _____________________________________________________________________________________

2-32) PSpice shows that average power is 60 W and energy is 1.2 J. Use VPULSE and IPULSE for the sources.

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Energy (20.000m,1.2000)

2.0 0 S(W(I1)) 400W

Avg Power (20.000m,60.000)

0W

Inst Power

-400W W(I1)

AVG(W(I1))

I(I1)

4ms V(V1:+)

20 0 SEL>> -20 0s

8ms

12ms

16ms

20ms

Time

_____________________________________________________________________________________

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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).

2.0KW Average Power

Resistor

1.0KW

Inductor

(16.670m,0.9998K)

(16.670m,-30.131u)

0W Vdc

(16.670m,189.361u)

-1.0KW 0s AVG(W(R1))

5ms AVG(W(L1))

10ms AVG(W(V1)) Time

15ms

20ms

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2.0KW

Instantaneous Power

Resistor 1.0KW

Inductor

0W Vdc

-1.0KW 0s W(R1)

W(L1)

5ms W(V1)

10ms

15ms

20ms

Time

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2-34)

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Rms voltage is 8.3666 V. Rms current is 5.2631 A.

10V Voltage

(20.000m,8.3666)

5V

0V V(V1:+)

RMS(V(V1:+))

10A (20.000m,5.2631) Current

0A

SEL>> -10A 0s I(I1:+)

4ms RMS(I(I1))

8ms

12ms

16ms

20ms

Time

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2-35) See Problem 2-10.

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0W (40.022m,-16.200) Source Power -100W

SEL>> -200W AVG(W(V1)) 4.0 Inductor 2.0

(4.0000m,648.007m)

Resistor (40.021m,647.946

0 0s I(L1)

10ms S(W(L1))

20ms

30ms

40ms

S(W(R1)) Time

The inductor peak energy is 649 mJ, matching the resistor absorbed energy. The source power is -16.2 W absorbed, meaning 16.2 W supplied. b) If the diode and switch parameters are changed, the inductor peak energy is 635 mJ, and the resistor absorbed energy is 620 mJ. The difference is absorbed by the switch and diode. _____________________________________________________________________________________

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2-36)

The inductor current reaches a maximum value of 3.4 A with the resistances in the circuit: I = 75/ (20+1+1) = 3.4 A. 4.0A

Inductor Current

2.0A

SEL>> 0A I(L1) 4.0A Source Current 0A

-4.0A 0s

20ms

40ms

60ms

80ms

-I(V1) Time

Quantity Inductor resistor average power Switch average power Diode average power Source average power

Probe Expression AVG(W(R1))

Result 77.1 W

AVG(W(S1)) AVG(W(D1)) AVG(W(Vcc))

3.86 W each 81 mW each -85.0 W

100ms

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2-37) a) Power absorbed by the inductor is zero. Power absorbed by the Zener diode is 13.8 W.

4.0A

2.0A Inductor Current

0A I(L1) 4.0A

2.0A

SEL>> 0A 0s

Zener Diode Current

10ms -I(D1)

20ms

30ms

40ms

50ms

60ms

70ms

Time

b) Power in the inductor is zero, but power in the 1.5Ω resistor is 1.76 W. Power absorbed by the Zener diode is 6.35 W. Power absorbed by the switch is 333 mW. _____________________________________________________________________________________

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3-38) See Problem 3-37 for the circuit diagram. a) Power absorbed by the Zener diode is 36.1 W. Power absorbed by the inductor is zero. 10A

5A

Inductor Current

SEL>> 0A I(L1) 10A

5A

Zener Diode Current

0A 0s

20ms

40ms

60ms

80ms

-I(D1) Time

b) Power in the inductor is zero, but power in the 1.5Ω resistor is 4.4 W. Power absorbed by the Zener diode is 14.2 W. Power absorbed by the switch is 784 mW.

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2-39)

40A

Total Current 20A

0A

-20A 0s I(I1)

Quantity Power rms current Apparent power S Power factor

4ms I(I2)

I(I3)

8ms I(I4)

12ms -I(V1) Time

Probe Expression AVG(W(V1)) RMS(I(V1)) RMS(V(I1:+))* RMS(I(V1)) AVG(W(V1)) / (RMS(V(I1:+))* RMS(I(V1)))

16ms

20ms

Result 650 W 14 A 990 VA 0.66

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2-40) DESIRED QUANTITY

ORIGINAL RESULT

NEW VALUES

Inductor Current Energy Stored in Inductor Average Switch Power Average Source Power (absorbed) Average Diode Power AVG(W(D1)) 0.464 W. Average Inductor Power Average Inductor Voltage Average Resistor Power Energy Absorbed by Resistor Energy Absorbed by Diode Energy Absorbed by Inductor rms Resistor Current

max = 4.5 A. max = 2.025 J 0.010 W. -20.3 W. 0.464 W.

4.39 A 1.93 L 0.66 W -19.9 W .449 W

0 0 19.9 W. 1.99 J. .046 J. 0 0.998 A.

0 0 18.8 W 1.88 J .045 J 0 0.970 A

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2-41) Use the part VPULSE or IPULSE (shown). Here, the period is 100 ms, and the rise times chosen are 20 ms, 50 ms, and 80 ms. The fall times are the period minus the rise times. Each rms value is 0.57735, which is identical to 1/√3.

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1.0A (100.000m,577.350m)

0A

-1.0A 0s

20ms -I(R1)

40ms RMS(I(R1))

60ms

80ms

100ms

Time

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