p33_012 - 12(a We use U = 1 LI 2 = 1 Q2/C to solve for L 2...

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12. (a) We use U = 1 2 LI 2 = 1 2 Q 2 /C to solve for L : L = 1 C µ Q I 2 = 1 C µ CV max I 2 = C µ V max I 2 =( 4 . 00 × 10 6 F) µ 1 . 50 V 50 . 0 × 10 3 A 2 =3 . 60 × 10 3 H . (b) Since f = ω/ 2 π , the frequency is f = 1 2 π LC = 1 2 π p (3 . 60 × 10 3 H)(4 . 00 × 10 6 F) =1 . 33 × 10 3 Hz . (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.

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