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
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart

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
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-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 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.
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
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
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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 
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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
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
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:
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
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.
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
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  
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
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)
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
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
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
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.
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
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
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
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
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
2.0KW
Instantaneous Power
Resistor
1.0KW
Inductor
0W
Vdc
-1.0KW
0s
W(R1)
W(L1)
5ms
W(V1)
10ms
15ms
20ms
Time
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
2-34)
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
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
_____________________________________________________________________________________
2-35) See Problem 2-10.
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
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.
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
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
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
_____________________________________________________________________________________
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.
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
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.
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
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
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
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
_____________________________________________________________________________________
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.
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
1.0A
(100.000m,577.350m)
0A
-1.0A
0s
20ms
-I(R1)
40ms
RMS(I(R1))
60ms
80ms
100ms
Time
_____________________________________________________________________________________