p33_023

# p33_023 - 23 The energy needed to charge the 100 F...

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23. The energy needed to charge the 100 µ F capacitor to 300 V is 1 2 C 2 V 2 = 1 2 (100 × 10 6 F)(300 V) 2 =4 . 50 J . The energy initially in the 900 µ F capacitor is 1 2 C 1 V 2 = 1 2 (900 × 10 6 F)(100 V) 2 =4 . 50 J . All the energy originally in the 900 µ F capacitor must be transferred to the 100 µ F capacitor. The plan is to store it temporarily in the inductor. We do this by leaving switch S 1 open and closing switch S 2 . We wait until the 900 µ F capacitor is completely discharged and the current in the circuit is at maximum (this occurs at t = T 1 / 4, one quarter of the relevant period). Since T 1 =2 π p LC 1 =2 π p (10 . 0 H)(900 × 10 6 F) = 0 . 596 s
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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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