12. (a) We useU=12LI2=12Q2/Cto solve forL:L=1CµQI¶2=1CµCVmaxI¶2=CµVmaxI¶2=(4.00×10−6F)µ1.50 V50.0×10−3A¶2=3.60×10−3H.(b) Sincef=ω/2π, the frequency isf=12π√LC=12πp(3.60×10−3H)(4.00×10−6F)=1.33×103Hz.(c) Referring to Fig. 33-1, we see that the required time is one-fourth of a period (where the period is
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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.