p33_060 - √ 5 − 1) R bulb = ( √ 5 − 1) (120 V) 2...

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60. (a) The power consumed by the light bulb is P = I 2 R/ 2. So we must let P max /P min =( I/I min ) 2 =5 , or µ I I min 2 = µ E m /Z min E m /Z max 2 = µ Z max Z min 2 = Ã p R 2 +( ωL max ) 2 R ! 2 =5 . Weso lvefor L max : L max = 2 R ω = 2(120 V) 2 / 1000 W 2 π (60 . 0Hz) =7 . 64 × 10 2 H . (b) Now we must let µ R max + R bulb R bulb 2 =5 , or R
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Unformatted text preview: √ 5 − 1) R bulb = ( √ 5 − 1) (120 V) 2 1000 W = 17 . 8 Ω . This is not done because the resistors would consume, rather than temporarily store, electromag-netic energy....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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