Unformatted text preview: C is in picofarads, then √ C + 365 pF √ C + 10 pF = 2 . 96 . The solution for C is C = (365 pF) − (2 . 96) 2 (10 pF) (2 . 96) 2 − 1 = 36 pF . We solve f = 1 / 2 π √ LC for L . For the minimum frequency C = 365 pF + 36 pF = 401 pF and f = 0 . 54 MHz. Thus L = 1 (2 π ) 2 Cf 2 = 1 (2 π ) 2 (401 × 10 − 12 F)(0 . 54 × 10 6 Hz) 2 = 2 . 2 × 10 − 4 H ....
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 Fall '08
 SPRUNGER
 Physics, Capacitance, Inductance, Inductor, Angular frequency, PF

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