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(numerical integration ) 2 c ) I rms 6.84 A. (numerical integration) 2 c ) I rms 7.65 A. (numerical integration) 2 c ) I rms 6.70 A. N1 0.45 Downloaded by Farhad Hossain ( ) lOMoARcPSD|4687171 3-3) a ) pf  V V V P P  I rms  s ,rms VR ,rms  m Vs ,rms  m S Vs ,rms I rms R 2 2  Vm   /R /R V 2 1 2 pf       Vs ,rms I rms  Vm   Vm  2 2   2  / R  2   b) Displacement pf cos(1   ) cos(0) 1 2 R, rms V 1V 1 I1  1  m 0 pf cos(1  1 ) DF  DF  R R 2 2 3-4) Using Eq. LOMoARcPSD|4687171 Chap003 - CH3 Solution of Power Electronics by Daniel W.Hart Studies in Teaching Material of Industrial Education (National Taiwan University) StuDocu is not sponsored or endorsed by any college or university Downloaded by Farhad Hossain ( ) lOMoARcPSD|4687171 CHAPTER 3 SOLUTIONS 2/20/10 3-1) V V  170 /  3.60 A.