ch31-p015 - 15(a Since the frequency of oscillation f is...

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r == 160 054 296 . . .. MHz MHz Since the additional capacitor is in parallel with the tuning capacitor, its capacitance adds to that of the tuning capacitor. If C is in picofarads (pF), then C C + + = 365 10 pF pF The solution for C is C = = 365 2 96 10 1 36 2 2 pF pF pF. b g b g b g bg . . (c) We solve fL C = 12 / π for L . For the minimum frequency C = 365 pF + 36 pF = 401 pF and f = 0.54 MHz. Thus L Cf ×× 1 2 1 2 401 10 054 10 22 10 2 22 12 6 2 4 π π ch c h FH z H. . . 15. (a) Since the frequency of oscillation
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