p33_015.pdf

1 / 2 π √ LC , th e small er value of C gives the larger value of f .Consequently, f max =1 / 2 π √ LC min ,

f min =1 / 2 π √ LC max ,and

f min = √ C max

√ C min = √ 365 pF

f max

√ 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 =2 . 96 .

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, then

√ C +365pF

√ C +10pF =2 . 96 .

The solution for C is

(2 . 96) 2 (10 pF)

(2 . 96) 2

C = (365 pF)

−

1 =36pF .

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 =

(2 π ) 2 Cf 2 =

10 6 Hz) 2 =2 . 2

10 − 4 H .

(2 π ) 2 (401

10 − 12 F)(0 . 54

15. (a) Since t he frequency of oscillation f is related to the inductance L and capacitance C b y f =