p33_060.pdf

60. (a) Thepowerconsumedbythelightbulbis P = I 2 R/ 2. Sowemustlet P max /P min =( I/I min ) 2 =5,

I min

= E m /Z min

E m /Z max

= Z max

Z min

= R 2 +( ωL max ) 2

=5 .

Wesolvefor L max :

ω = 2(120V) 2 / 1000W

L max = 2 R

2 π (60 . 0Hz)

=7 . 64

10 − 2 H .

(b) Nowwemustlet

R max + R bulb

R bulb

=5 ,

R max =( √ 5

1) R bulb =( √ 5

1) (120V) 2

1000W =17 . 8Ω .

This is not done because the resistors would consume, rather than temporarily store, electromag-

neticenergy.

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