p33_015

# p33_015 - C is in picofarads then √ C 365 pF √ C 10 pF...

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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 .Con s equ en t ly , 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) An additional capacitance C is chosen so the ratio of the frequencies is r = 1 . 60 MHz 0 . 54 MHz
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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 . 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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