ch31-p065 - c h . . R = b3125 A gb2gb0.30 g = 19 V . . R =...

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(b) The rate of energy dissipation is PI R d == = rms 2 AW . 3125 2 0 60 59 .. . bg b g b g (c) Now I rms 3 W / 8.0 10 V A × = 250 10 3125 3 c h ., s o ( )( ) 31.25A 0.60 19V. V ∆= Ω= (d) P d × 3125 0 60 59 10 2 2 . . b g b g (e) () 33 rms 250 10 W/ 0.80 10 V 312.5 A I × = , so ( )( ) 312.5A 0.60 V =Ω = 2 1.9 10 V × . (f) ( ) 2 4 312.5A 0.60 5.9 10 W. d P = × 65. (a) The rms current in the cable is
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