ch31-p015

# ch31-p015 - C is in picofarads, then √ C + 365 pF √ C +...

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15 (a) Since the frequency of oscillation f is related to the inductance L and capacitance C by f =1 / 2 π LC , the smaller value of C gives the larger value of f .Hence , f max =1 / 2 π LC min , f min =1 / 2 π LC max ,and f max f min = C max C min = 365 pF 10 pF =6 . 0 . (b) You want to choose the additional capacitance C so the ratio of the frequencies is r = 1 . 60 MHz 0 . 54 MHz =2 . 96 . Since the additional capacitor is in parallel with the tuning capacitor, its capacitance adds to that
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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 . (c) 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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## This note was uploaded on 04/15/2008 for the course PHYS 101 taught by Professor Reich during the Spring '08 term at Johns Hopkins.

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