solution-manual-power-electronics-1st-edition-hart.doc
Short Description
Download solution-manual-power-electronics-1st-edition-hart.doc...
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 50ms 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 2t 120 2 12.5 2 sin 2t 3000sin 2t 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
_____________________________________________________________________________________ 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
_____________________________________________________________________________________
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
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
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
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
Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart
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
_____________________________________________________________________________________
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 jC 0.01 j 0.0189 V1
I1 4 0 187 62.1 Y1 0.01 j 0.0189
2 4 60 : Y2 1/R jC 0.01 j 0.0377 V2
I2 60 153 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 106
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
_____________________________________________________________________________________
View more...
Comments