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

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. _____________________________________________________________________________________ v2  t  [170sin  377t  ]2 2-2) a) p  t   v  t  i  t     2890sin 2 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 0  t  50 ms  p  t   v  t  i  t    40 50 ms  t  70 ms  0 70 ms  t  100 ms  b) T 70 ms 1 1 P   v  t  i  t  dt  100 ms 50ms 40 dt  8.0 W . T 0 c) T 70 ms W   p  t  dt   40 dt  800 mJ .; or W  PT   8W   100 ms   800 mJ . 0 50 ms _____________________________________________________________________________________ 2-5) a)  70 W . 0  t  6 ms  50 W . 6 ms  t  10 ms  p t  v t i t    40 W . 10 ms  t  14 ms  0 14 ms  t  20 ms b) 1   T 6 ms 10 ms 14 ms 1 P T  p  t  dt  0 20 ms    0 70 dt    50  dt   6 ms 10 ms 40 dt  19 W .  c) Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart T  6 ms 10 ms 14 ms  W   p  t  dt    70 dt    50  dt   40 dt   0.38 J .; 0  0 6 ms 10 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) di  t  vL  t   L  10  10   377   1.0  cos 377t  3.77 cos 377t V . 3 dt 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 1 p  t  dt  0 T 0 PL  c) p  t   v  t  i  t    12   1.0sin 377t   12sin 377t W . T 1 Pdc  T  p  t  dt  0 0 _____________________________________________________________________________________ 2 W 2 2 b) All stored energy is absorbed by R: WR = 0.648 J .1 0 90 d   900t 0  t  4 ms.5 2 sin 2t  3000sin 2t max  p  t    3000 W . b) P = 1500(5/12) = 625 W.6   0. Inductor: PL  0.6 A.2 W . P Vm I m  1500 W .  I m   1500   2   12. c) WR 0.eu/solution-manual-power-electronics-1st-edition-hart 2-8) Resistor: v  t   i  t  R  8  24sin 2 60t V .648 PR    16. L iL  4 ms    900   4   10  3  3.) _____________________________________________________________________________________ 2-10) t 1 1 iL  t    vL  t  dt  0. (or 1500(5) = 7500 W. d) No change in power supplied by the source: 16. p  t   v  t  i  t    8  24sin 2 60t   2  6sin 2 60t   16  96sin 2 60t  144sin 2 2 60t W . c) W = PT = (625 W)(12 s) = 7500 J.5 2 2 120 2    p  t   Vm I m sin 2t  120 2 12. Full file at http://testbank360. _____________________________________________________________________________________ .2 W. _____________________________________________________________________________________ 2-9) a) With the heater on.2 W . 1   T 1/60 1/60 1/60 1 P  p  t  dt    16 dt   96sin 2 60t dt   144sin 2 2 60t T 0 1/ 60  0 0 0   16  72  88 W . a) 1 2 1 Li   0. dc source: Pdc  I avgVdc   2   6   12 W .648 J. T 40 ms PS  PR  16.1  3. 9 ms L 10 mH    1.2 A. inst.7 ms 1. L0 15.7 ms. Full file at http://testbank360.0KW W(L1) 1. L 8.010 0 14 d   1400t A..0KW Source inst.0KW Ind.0KW 0s 20ms 40ms 60ms 80ms 100ms -W(Vcc) Time _____________________________________________________________________________________ .7 ms _____________________________________________________________________________________ 2-12) a) i(t) = 1800t for 0 < t < 4 ms i(4 ms) = 7.49  1400ton ton  11. or i    15. power 0W -1.62  R 5 1.296 J.9 ms.2  W Li .010 t t 1 1 i  t    v    d  0.eu/solution-manual-power-electronics-1st-edition-hart 2-11) a) 1 2 2W 2  1.. b) 10A 5A Inductor current SEL>> 0A I(L1) 10A Source current 0A -10A -I(Vcc) 1. Allowing five time constants. power (supplied) 0W -1. R   5.11. 2 L 0.49 A.1 ms b) Energy stored in L must be transferred to the resistor in (20 .1) = 8. WLpeak = 1. Full file at http://testbank360.075 at t  20 ms. b) Switch closed: 0 < t < 20 ms.02   3.7 A/s dt L 0.2 A. power 40mW SEL>> 0W 0s 10ms 20ms 30ms 40ms 50ms 60ms 70ms W(D1) Time . iL   160   0. c) 40mW Inductor inst. power 0W -40mW W(L1) 80mW Zener inst.2 t    30 ms 106.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. diL  t  vL  12 V . iL = 0 for 50 ms < t < 70 ms. diL vL 8    106. Switch open.075 t to return to zero : i 3. inductor current returns to zero at 20 + 30 = 50 ms.7 Therefore.7 106.  L dt diL vL 12    160 A/s dt L 0. zener on: vL  12  20  8 V . T 1  1  0.050 t to return to zero : i 6.0 t    30 ms 200 200 Therefore.eu/solution-manual-power-electronics-1st-edition-hart d) PL  0.050 at t  15 ms.03  64  13.  L dt diL vL 20    400 A/s dt L 0. inductor current returns to zero at 15 + 30 = 45 ms.015   6. diL t  vL  20 V . power 100W SEL>> 0W 0s 20ms 40ms 60ms 80ms W(D1) Time .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. diL vL 10    200 A/s dt L 0. Full file at http://testbank360. iL = 0 for 45 ms < t < 75 ms.0 A. zener on: vL  20  30  10 V. iL   400   0. c) 200W Inductor inst.73 W . power 0W -200W W(L1) 200W Zener inst. Switch open. 1 PZ   pZ  t  dt   T 0 0. 1 PZ   pZ  t  dt   T 0 0. T 1  1  0.7  5.006 0.  20 0. Full file at http://testbank360.01 0.5  216 W . 2 2 2-16)   2 Neutral conductor: PN  I 2 R  12 3  0.7  8. I rms  I m D  4 0. 2-4 Vrms  Vm D  10 0.006 0.02 1    5 4 2 I rms  7 2 dt  dt  2 dt  27.5  2. 2-5  14 Vrms  Vm D  10    8.37 V .26 A. PN 72 RN    0.5   72 W . _____________________________________________________________________________________ 2-18) Re: Prob.167    2 2 IN 12 3 _____________________________________________________________________________________ 2-17) Re: Prob.eu/solution-manual-power-electronics-1st-edition-hart d) PL  0.83 A.03  180  36 W . 0. _____________________________________________________________________________________ Phase conductors: P  I R  12  0.075  2  _____________________________________________________________________________________ 2-15) Examples are square wave (Vrms = Vm) and a triangular wave (Vrms = Vm/√3).02 0 0.36 V . Ptotal  3  72   216  432 W .01 _____________________________________________________________________________________ . 5   2     2.01  j 0.1 Y2  0.58 V .0   1. Full file at http://testbank360.1  I rms  1.0189  2  4 60 : Y2  1/R  jC  0.  4  Resistor: .1  2 2  1500  175  118  1793 W .eu/solution-manual-power-electronics-1st-edition-hart 2-19) 2 2  5  3 Vrms  22       4.2 A.  2  2  V I P  V0 I 0   m m cos   n  n  n 1 2  5  2    3  1.01  j 0.1    75.0 W .01  j 0.0377   Vm I m P  V0 I 0   cos   n  n  n 1 2  300  5    187   4  cos  153  6  cos  62. 1  2 60 : Y1  1/R  jC  0.  2  2 2 2  2   1.01  j 0.5      cos  20     cos  115   7.0377 I2 60 V2    153  75.1   2.1 Y1  0.0189 I1 4 0 V1    187  62. _____________________________________________________________________________________ 2-21) dc Source:  50  12 Pdc  Vdc I avg  12   114 W .  2  2   2 2 Notethat  cos(4 60t  45) is cos  4 60t  135  _____________________________________________________________________________________ 2-20) dc : V0  3  100   300 V . 67 0.375 A. 4  j  8 60   0.25 156. P  I rms 2 _____________________________________________________________________________________ 2-23)  Vm I m P  V0 I 0   cos   n  n  n 1 2 n Vn In Pn ∑Pn 0 20 5 100 100 1 20 5 50 150 2 10 1.  2  2 R   0.025  2 2  0. _____________________________________________________________________________________ . 2 I rms  0.2rms  I 2.641 I rms  9.9 W .0923 A.51  0.3125 0.9 Power including terms through n = 4 is 158.01 10 I2   0.25 3 6.641 A.025  3 I2   0.375   2     0.556 1. 16  j  6 60   0.781 158. 30 I1   3..5   2     9. Full file at http://testbank360.623 A.269 A.85 158. R 16 5 I1   0. _____________________________________________________________________________________ 2-22) P  I rms 2 R V0 6 I0   0.1 4 5 0.5 A.9 watts.83 A. 16  j  2 60   0.eu/solution-manual-power-electronics-1st-edition-hart P  I rms 2 R I rms  I 02  I1.269  0.01 2 2  3.25 6.  2  2 PR  I rms 2 R  386 W .0923  I rms  0.51 A.426   16   2.426 A. 4  j  4 60   0.2 rms I 0  9. 44 51.83 1.01 0.46 81.32 25.4 1.7 Through n = 4.0 26.00 0.17 6.7 A R 20 P0.1 4 12. b) THD = 10% → I9 = (0.0 22.0 1 50.66 2 63.05)(10) = 0.ϕn° Pn 0 50.20)(10) = 2 A. d) THD = 40% → I9 = (0. Full file at http://testbank360.05 0.00 29.625 63.99 PR = ∑ Pn ≈ 300 W. ∑Pn = 753 W.7 0. _____________________________________________________________________________________ 2-25)  Vm I m P  V0 I 0   cos   n  n  n 1 2 V V 50  36 I 0  0 dc   0.48 1.22 3 42.1 3 16.66 37. c) THD = 20% → I9 = (0.0000 10.0000 2.10)(10) = 1 A.00 20.8 W (dc component only ) 2 PVdc  I 0Vdc   0.32 0.71 1.6667 1.0 0 500.7   36   25. _____________________________________________________________________________________ 2-26) a) THD = 5% → I9 = (0.6 2 25.7  20  9.00 9.2 W PL  0 Resistor Average Power n Vn Zn In angle Pn 0 50.6 223.11 56.5 A.94 0.67 250.  2  2 .16 0.43 5.8 1 127.0000 10.83 65.5 45.26 2.24 1.87 4 31.40)(10) = 4 A. R  I 02 R   0.5000 0.33 5 25.3 5.eu/solution-manual-power-electronics-1st-edition-hart 2-24)  Vm I m P  V0 I 0   cos   n  n  n 1 2 n Vn In θn .31 1. _____________________________________________________________________________________ 2-27) a)  170  10  P   Pn     cos  30   0  0  736 W . eu/solution-manual-power-electronics-1st-edition-hart b) 2 2 2  10  6   3 I rms         8.  2 P 781 pf    0.rms 12/ 2 DF    0.62 A.53  53% 12/ 2 _____________________________________________________________________________________ 2-29) .88 I rms 9.  2   2  2  170 S  Vrms I rms    9.68 S 1156 c) I1.  2 P 736 pf    0.62 d) 2 2  5  4      2  2 THDI   0.  2  2  2  170 S  Vrms I rms    8.67  67% 10/ 2 _____________________________________________________________________________________ 2-28) a)  170  12  P   Pn     cos  40   0  0  781 W .  2  2 b) 2 2 2  12  5   4 I rms         9.51 d) 2 2  6  3      2  2 THDI   0.rms 10/ 2 DF    0.62  1156 VA. Full file at http://testbank360.831 I rms 8.719 S 1024 c) I1.51  1024 VA.51 A. 32 I1. ( graphically ) a) P  V1.49 A.662  2..rms   8.36 c) THDI    0.66 d) DF    89.38 ( graphically ) a) P  V1.eu/solution-manual-power-electronics-1st-edition-hart 8 4 I1.80  80% S Vrms I rms  240   10.rms cos  1  1    240   5.rms 6.38 e) crest factor    1.6 _____________________________________________________________________________________ .895  89.rms   5.rms I1.6 A.rms 2.3 e) crest factor    1.822  6. Full file at http://testbank360.36 A.rms cos  1  n    240   10.362  10.6 I1.72 I rms 10.rms   2.6% I rms 6. I 2.. I peak  10.32 A. rms 5. 2 2 I rms  8.66  cos  0   1358 W .446  44.32 _____________________________________________________________________________________ 2-30) 12 9 I1.3 A. I 2.82 A.. I peak  18.6  cos  0   2036 W .492  6.32  I 2.66 A..49 d) DF    80% I rms 10.60  60% I rms 10.6 I peak 18. rms I1.64 I rms 6.5% S Vrms I rms  240   6.rms   6..82 c) THDI    0.32 I peak 10. rms 8. P P 1358 b) pf     0.6% I rms 6. P P 2036 b) pf     0.6  I 2.. 2 2 I rms  5. eu/solution-manual-power-electronics-1st-edition-hart 2-31) 5V: I = 0 (capacitor is an open circuit) 1 1 25cos(1000t ): Z  R  j L  j  2  j1000(.3 W _____________________________________________________________________________________ 2-32) PSpice shows that average power is 60 W and energy is 1. Use VPULSE and IPULSE for the sources.5cos(1000t ) A 2 10cos(2000t ): Z  2  j1.5 2 2  12.2 J. 2  j1.001)  j  2  j0 C 1000  1000  106 25 I cos(1000t )  12.28 A  2  2 2 PR  I rms R  9.3 W PL  0 PC  0 Psource  172.5  10 I10   4  37 A. .5  4 I rms       9. Full file at http://testbank360.282  2   172. 1.000m.60.000m.eu/solution-manual-power-electronics-1st-edition-hart Energy 2.0 (20.000) 0W Inst Power -400W W(I1) AVG(W(I1)) 20 0 SEL>> -20 0s 4ms 8ms 12ms 16ms 20ms I(I1) V(V1:+) Time _____________________________________________________________________________________ . Full file at http://testbank360.2000) 0 S(W(I1)) 400W Avg Power (20. 361u) Vdc -1.0KW Average Power (16.670m. the average power is zero (slightly different because of numerical solution).9998K) Resistor 1.0.189. Full file at http://testbank360.670m. For the inductor and dc source.eu/solution-manual-power-electronics-1st-edition-hart 2-33) Average power for the resistor is approximately 1000 W.0KW Inductor (16.670m.131u) 0W (16. 2.0KW 0s 5ms 10ms 15ms 20ms AVG(W(R1)) AVG(W(L1)) AVG(W(V1)) Time .-30. Full file at http://testbank360.0KW Instantaneous Power Resistor 1.0KW Inductor 0W Vdc -1.0KW 0s 5ms 10ms 15ms 20ms W(R1) W(L1) W(V1) Time _____________________________________________________________________________________ .eu/solution-manual-power-electronics-1st-edition-hart 2. Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart 2-34) . 2631) Current 0A SEL>> -10A 0s 4ms 8ms 12ms 16ms 20ms I(I1:+) RMS(I(I1)) Time _____________________________________________________________________________________ 2-35) See Problem 2-10.eu/solution-manual-power-electronics-1st-edition-hart Rms voltage is 8. Full file at http://testbank360.3666) 5V 0V V(V1:+) RMS(V(V1:+)) 10A (20.000m. 10V Voltage (20. Rms current is 5.8.2631 A. .5.000m.3666 V. The difference is absorbed by the switch and diode.007m) Resistor (40.022m.647.0 Inductor 2. _____________________________________________________________________________________ . meaning 16.eu/solution-manual-power-electronics-1st-edition-hart 0W (40.0000m.2 W supplied.0 (4.021m.2 W absorbed. The source power is -16.648.-16. the inductor peak energy is 635 mJ.200) Source Power -100W SEL>> -200W AVG(W(V1)) 4. and the resistor absorbed energy is 620 mJ.946 0 0s 10ms 20ms 30ms 40ms I(L1) S(W(L1)) S(W(R1)) Time The inductor peak energy is 649 mJ. matching the resistor absorbed energy. b) If the diode and switch parameters are changed. Full file at http://testbank360. Full file at http://testbank360.1 W power Switch average power AVG(W(S1)) 3.0A Inductor Current SEL>> 0A I(L1) 4.0A Source Current 0A -4.eu/solution-manual-power-electronics-1st-edition-hart 2-36) The inductor current reaches a maximum value of 3.4 A.0A 2. 4.0 W .0A 0s 20ms 40ms 60ms 80ms 100ms -I(V1) Time Quantity Probe Expression Result Inductor resistor average AVG(W(R1)) 77.86 W each Diode average power AVG(W(D1)) 81 mW each Source average power AVG(W(Vcc)) -85.4 A with the resistances in the circuit: I = 75/ (20+1+1) = 3. 35 W.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 13.5Ω resistor is 1.0A 2.0A 2.76 W. _____________________________________________________________________________________ .8 W.0A Zener Diode Current SEL>> 0A 0s 10ms 20ms 30ms 40ms 50ms 60ms 70ms -I(D1) Time b) Power in the inductor is zero. Full file at http://testbank360. 4. Power absorbed by the Zener diode is 6. but power in the 1.0A Inductor Current 0A I(L1) 4. Power absorbed by the switch is 333 mW. Full file at http://testbank360. a) Power absorbed by the Zener diode is 36. but power in the 1.2 W.4 W. Power absorbed by the Zener diode is 14.5Ω resistor is 4. 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. . Power absorbed by the switch is 784 mW.eu/solution-manual-power-electronics-1st-edition-hart 3-38) See Problem 3-37 for the circuit diagram.1 W. Full file at http://testbank360.eu/solution-manual-power-electronics-1st-edition-hart 2-39) 40A Total Current 20A 0A -20A 0s 4ms 8ms 12ms 16ms 20ms I(I1) I(I2) I(I3) I(I4) -I(V1) Time Quantity Probe Expression Result Power AVG(W(V1)) 650 W rms current RMS(I(V1)) 14 A Apparent power S RMS(V(I1:+))* RMS(I(V1)) 990 VA Power factor AVG(W(V1)) / (RMS(V(I1:+))* RMS(I(V1))) 0.66 _____________________________________________________________________________________ . 88 J Energy Absorbed by Diode . 0. 50 ms.010 W. Full file at http://testbank360. 1. 18. Average Inductor Power 0 0 Average Inductor Voltage 0 0 Average Resistor Power 19.39 A Energy Stored in Inductor max = 2.998 A.5 A.99 J. Each rms value is 0.8 W Energy Absorbed by Resistor 1. and 80 ms. the period is 100 ms.66 W Average Source Power (absorbed) -20. Here. . The fall times are the period minus the rise times.93 L Average Switch Power 0.464 W.9 W Average Diode Power 0. which is identical to 1/√3. .025 J 1. -19.970 A _____________________________________________________________________________________ 2-41) Use the part VPULSE or IPULSE (shown).9 W.3 W.464 W. 0. .eu/solution-manual-power-electronics-1st-edition-hart 2-40) DESIRED QUANTITY ORIGINAL RESULT NEW VALUES Inductor Current max = 4. and the rise times chosen are 20 ms.57735.449 W AVG(W(D1)) 0. 4.045 J Energy Absorbed by Inductor 0 0 rms Resistor Current 0.046 J. 350m) 0A -1.eu/solution-manual-power-electronics-1st-edition-hart 1.0A (100.577.0A 0s 20ms 40ms 60ms 80ms 100ms -I(R1) RMS(I(R1)) Time _____________________________________________________________________________________ .000m. Full file at http://testbank360.