Power Electronics Circuits Devices and Applications 4th Edition Rashid Solutions Manual Full download: People also search: power electronics circuits devices & applications (4th edition) pdf power electronics circuits devices and applications 3rd edition pdf power electronics circuits devices and applications 4th edition pdf free download power electronics by mh rashid 4th edition pdf power electronics: devices, circuits, and applications power electronics by mh rashid 2nd edition pdf power electronics rashid 4th edition power electronics circuits devices and applications solution manual. Power electronics circuits devices and applications 4th edition rashid solutions manual • 1. 10 2 Power Electronics Circuits Devices and Applications 4th Edition Rashid SOLUTIONS MANUAL Full download: and-applications-4th-edition-rashid-solutions-manual/ Prob 2.1 trr:= 5⋅1 − 6 0 (a) Eq. (2.10) Chapter 2 – Diodes Circuits di_dt:= 80⋅ 6 10 2 QRR:= 0.5⋅di_dt⋅trr (b) Eq. (2-11) 6 3 QRR⋅10 = 1 × 10 μC IRR:= 2⋅QRR⋅di_dt IRR = 400 A Prob 2.2 − 6 6 A trr:= 5⋅10 trr s Ifall_rate:= 800⋅10 s SF:= 0.5 ta:= 1 + SF − 6 ta = 3.333 × 10 (a) tb:= SF⋅ta IRR:= Ifall_rate⋅ta Using Eq.

(2-7), 1 tb = 1.667 × − 6 3 IRR = 2.667 × 10 • 2 QRR:= ⋅(Ifall_rate⋅ta)⋅(ta + tb) 1 6 QRR⋅10 = 6.667 × 10 3 μC QRR:= ⋅IRR⋅(ta + tb) 6 3 QRR⋅10 = 6.667 × 10 Using Eq. (2-6), μC (b) IRR:= Ifall_rate⋅ta 3 IRR = 2.667 × 10 • 0 10 Chargestorage Prob 2.3 trr:= 5⋅1 − 6 trr SF:= 0.5 ta:= 1 + SF − 6 ta = 3.333 × 10 tb:= SF⋅ta 1 tb = 1.667 × − 6 m:= ⋅ta⋅trr m 8.333 10 12 2 QRR(x):= m⋅x = × − Plot of the charge storage verus di/dt 0.008 0.006 QRR (x) 0.004 0.002 0 2.10 8 4.10 8 6.10 8 8.10 8 1.10 9 • x di/dt • 10 Chargestorage IRR(x):= ta⋅x 4000 Plot of the charge storage verus di/dt 3000 IRR (x) 2000 1000 0 1.10 8 2.10 8 3.10 8 4.10 8 5.10 8 6.10 8 7.10 8 8.10 8 9.10 8 1.10 9 x di/dt Prob 2.5 VT:= 25.8⋅ − 3 VD2:= 1.6 Using Eq. Download darmowy program wirtualny dysk software reviews. (2-3), VD1:= 1.2 ID2:= 1500 ID1:= 100 (a) VD2 −VD1 η:= ⎛ID2⎞ VT⋅ln⎜ I ⎟ η = 5.725 (b) x:= ⎝ VD1 η⋅V D1⎠ x = 8.124 • T Using Eq. (2-3), ⎛ID1⎞ IS:= ID1 x e IS = 0.03 VT⋅η⋅ln⎜ ⎝ IS ⎟ = 1.2 ⎠ • Prob 2-7 0 VD1:= 2200 − 3 VD2:= 2200 − 3 3 R1:= 100⋅10 IS1:= 20⋅10 (a) IS2:= 35⋅10 IR1:= VD1 R1 IR1 = 0.022 Using Eq. (2-13), (b) IR2:= IS1 + IR1 − IS2 VD2 IR2 = 7 × 1 − 3 5 R2:= IR2 R2 = 3.143 × 10 Prob 2.11 IT:= 300 VD:= 2.8 IT I1:= 2 I1 = 150 I2:= I1 • Prob 2-7 I VD1:= 1.4 VD −VD1 I2 = 150 VD2:= 2.3 − 3 R1:= I1 VD −VD2 R1 = 9.333 × 10 − 3 R2:= 2 IT I1:= 2 R2 = 3.333 × 10 • Prob 2-13 3 3 3 10 10 10 0 1 R1:= 50⋅10 R2:= 50⋅10 VS:= 10⋅10 Is1:= 20⋅ − 3 Is2:= 30⋅ − 3 Using Eq.

Wed, 20 Jun 2018 09:20:00. GMT power electronics rashid solution pdf. Solutions Manual-Power. Electronics,, 4th Edition. University of.

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Instant download and all chapters Solutions Manual Power Electronics Circuits, Devices Applications 4th Edition Muhammad H. Rashid Chapter 3-Diodes Rectifiers Prob 3-1 Vm:= 170 R:= 5 f:= 60 Using Eq. (3-11) 2⋅ Vm Vdc:= Vdc = 108.23 π Vdc:= 0.6366⋅ Vm Prob 3-2 Vm:= 170 Vdc = 108.22 −3 R:= 10 Lc:= 0.5⋅ 10 f:= 60 Using Eq. (3-11) Vdc:= 0.6366⋅ Vm Idc:= Vdc Idc = 10.82 R Using Eq. (3-83) Vx:= 2⋅ f ⋅ Lc⋅ Idc Vx = 0.65 Vo:= Vdc − Vx Vo = 107.57 Prob 3-3 Vm:= 170 R:= 5 Using Eq.

(3-25) 6 ⎛π⎞ Vdc:= Vm⋅ ⋅ sin ⎜ ⎟ π ⎝6⎠ Prob 3-4 Vm:= 170 f:= 60 R:= 5 Vdc = 162.34 f:= 60 Using Eq. (3-25) 6 ⎛π⎞ ⎟ ⎝6⎠ Vdc:= Vm⋅ ⋅ sin ⎜ π Idc:= Vdc = 162.34 Vdc Idc = 32.47 R Using Eq. (3-83) Vx:= 6⋅ f ⋅ Lc⋅ Idc Vx = 5.84 Vo:= Vdc − Vx Vo = 156.49 Chapter 3-Diodes Rectifiers Page # 3 -1 −3 Lc:= 0.5⋅ 10 Prob 3-5 Vs:= 280 R:= 5 2 3 Vm:= 280⋅ f:= 60 Vm = 228.62 Using Eq.

(3-33) Vdc:= 1.6542⋅ Vm Prob 3-6 Vs:= 280 Vdc = 378.18 R:= 5 2 3 Vm:= 280⋅ −3 f:= 60 Lc:= 0.5⋅ 10 Vm = 228.62 Using Eq. (3-33) Vdc = 378.18 Vdc:= 1.6542⋅ Vm Vdc Idc:= Idc = 75.64 R Using Eq. (3-83) Vx:= 6⋅ f ⋅ Lc⋅ Idc Vx = 13.61 Vo:= Vdc − Vx Prob 3-7 Vdc:= 240 0.6366 Vs = 266.58 2 Idc:= Ip:= Vm = 377 Vm Vs:= Diodes R:= 10 Vdc Vm:= Vo = 364.57 Vdc Idc = 24 R Vm Ip = 37.7 R Id:= Idc Id = 12 2 Chapter 3-Diodes Rectifiers Page # 3 -2 Ip IR:= Transformer IR = 18.85 2 Vm Vs:= Vs = 266.58 2 Ip Is:= Is = 26.66 2 3 VI:= Vs⋅ Is Using Eq. (3-1) Pdc Using Eq. (3-2) 2 0.6366⋅ Vm) (:= 1 TUF Prob 3-8 Vdc:= 750 Using Eq.

(3-33) Diodes 3 Pac:= Vs⋅ Is TUF:= Vs:= 3 Pdc = 5.76 × 10 R Using Eq. (3-8) Vm:= VI = 7.107 × 10 Pac = 7.11 × 10 Pdc TUF = 0.8105 Pac = 1.23 Idc:= 6000 Vdc Vm = 453.39 1.6542 Vm Vs = 320.6 2 Ip:= Idc Id:= 3 Ip = 6 × 10 Idc IR:= 3 Id = 3 × 10 2 Ip 3 IR = 4.24 × 10 2 Chapter 3-Diodes Rectifiers Page # 3 -3 Transformer Vs:= Vm Vs = 320.6 2 3 Is:= Ip Is = 6 × 10 VI:= Vs⋅ Is VI = 1.924 × 10 6 Using Eq. (3-1) Pdc:= Vdc⋅ Idc Pdc = 4.5 × 10 Using Eq. (3-2) Pac:= 3Vs⋅ Is Pac = 5.77 × 10 Using Eq. (3-8) TUF 6 Pdc TUF:= 1 6 TUF = 0.78 Pac = 1.28 Prob 3-9 Vm:= 170 f:= 60 R:= 10 ω:= 2⋅ π ⋅ f ω = 376.99 RF:= 0.04 Guess −3 L:= 5⋅ 10 Given Using Eq. (3-67) 2 ⎛ 2⋅ ω⋅ L ⎞ RF⋅ 1 + ⎜ ⎟ − 0.481 = 0 ⎝ R ⎠ 3 Find ( L ) ⋅ 10 = 158.93 mH Prob 3-10 Vm:= 170 f:= 60 R:= 10 ω:= 2⋅ π ⋅ f ω = 376.99 RF:= 0.02 Guess −3 L:= 5⋅ 10 Chapter 3-Diodes Rectifiers Page # 3 -4 Given Using Eq.