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https://www.book4me.xyz/solution-manual-power-electronics-hart/ CHAPTER 1 SOLUTIONS (1-1)

https://www.book4me.xyz/solution-manual-power-electronics-hart/ (1-2)

https://www.book4me.xyz/solution-manual-power-electronics-hart/ 25V

20V

15V

10V

5V

0V

-5V 0s

2us

4us

6us

8us

10us

12us

14us

16us

10us

12us

14us

16us

V(D1:2) Time 25V

(1.4333u,23.800) 20V

15V

10V

5V (4.0833u,-851.690m) 0V

-5V 0s

2us

4us

6us

8us

V(S1:4) Time

In part (b), the voltage across the current source is reduced from 24 V by the switch resistance and diode voltage drop.

https://www.book4me.xyz/solution-manual-power-electronics-hart/ (1-3)

40V

96.46n,23.94) 20V

(3.150u,-1.052)

0V

(3.150u,-1.052)

-20V 0s

5us

10us

V(V2:-) Time

15us

https://www.book4me.xyz/solution-manual-power-electronics-hart/ (1-4)

25V

20V

(800.000n,23.924)

15V

10V

5V (3.8333u,-1.0517) 0V

-5V 0s

2us

4us

6us

8us

V(V2:-) Time

10us

12us

14us

16us

https://www.book4me.xyz/solution-manual-power-electronics-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  

v 2  t  [170sin  377t ]2   2890sin2 377t W . R 10

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  p t   v t i t   40 0 

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

b) 70ms

P

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

c) T

70ms

0

50ms

W   p t dt 



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

_____________________________________________________________________________________ 2-5)

a)

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

70 W . 50 W .  p t   v t i t    40 W . 0 b)

1 P T c)

1   p t dt    0 20ms  T

6 ms

 0

10 ms

70 dt 

 40 dt  19 W . 10 ms  14 ms

  50 dt  

6 ms

https://www.book4me.xyz/solution-manual-power-electronics-hart/ 10 ms 14 ms T  6 ms  W   p  t  dt    70 dt    50  dt   40 dt   0.38 J .; 0 6ms 10ms  0 

or W  PT  19  20 ms  380 mJ. _____________________________________________________________________________________ 2-6)

P  Vdc I avg a) Iavg  2 A., P  12 2   24 W . b) Iavg  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 377t  12.5 1 cos 754t  W . 2

T

PR 

1 p  t  dt  12.5 W . T 0

b)

di t  3  10 10   377 1.0  cos377 t  3.77 cos377 t V . dt 3.77 1.0  pL t   v t  i t   3.77 cos377t 1.0sin 377 t   sin 754 t  1.89sin 754 t W. 2 T 1 PL   p t  dt  0 T0 vL  t   L

c)

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

Pdc 

1 p  t  dt  0 T 0

_____________________________________________________________________________________

https://www.book4me.xyz/solution-manual-power-electronics-hart/ 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 60t W . 2

1/60 1/60 1/60 T  1 1  2 P   p t  dt    16dt   96sin 2 60t dt   144sin 2 60t  T0 1/ 60  0 0 0 

 16  72  88 W . Inductor: PL  0.

dc source: Pdc  Iavg Vdc  2 6  12 W.

_____________________________________________________________________________________

2-9)

a) With the heater on,

P

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







p  t   VmI 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 90 d   900t v L  t  dt   0.1 0 L

0  t  4 ms .

iL  4 ms    900 4 10  3.6 A. 3

a)

1 1 2 W  Li 2   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. _____________________________________________________________________________________...