The lowest frequency in the fm radio band is 87.7 mhz. (a) what inductance is needed to produce this resonant frequency if it is connected to a 2.50 pf capacitor? (b) the capacitor is variable, to allow the resonant frequency to be adjusted to as high as 108 mhz. what must the capacitance be at this frequency?

inductance  is 1.31 µH

capacitance is 1.658 pF

Explanation:

Given data

frequency = 87.7 MHz  = 87.7 × Hz

capacitor =  2.50 pF  = 2.50 × F

to find out

inductance  and capacitance at 108 MHz

solution

we apply here resonant frequency that is

frequency = 1 / 2π√(LC)    ..............1

here L is inductance and C is capacitor

put all the value and find L

√(L2.50 ×) = 1 / 2π (  87.7 × )

√(L2.50 ×) = 1.81 ×

(L2.50 ×) = 3.296 ×

L = 3.296 ×   /  2.50 ×

L = 1.31 × H

so inductance  is 1.31 µH

and

for 108 MHz = 108 × Hz

we find here capacitance c from equation 1

frequency = 1 / 2π√(LC)

√(LC) =1 / 2π frequency

√(1.31 ×  C) =1 / 2π ( 108 × )

√(1.31 ×  C) = 1.47 ×

(1.31 ×  C) =   2.178 ×

c = 2.178 ×   /  (1.31 × )

c = 1.658 × F

so capacitance is 1.658 pF

a) 1.32 μH

b) 1.65 pF

Explanation:

a)

L = inductance of the inductor = ?

f = Lowest frequency of FM radio = 87.7 x 10⁶ Hz

C = Capacitance of the capacitor = 2.50 x 10⁻⁻¹² F

For resonance to be possible

L = 1.32 x 10⁻⁶ H

L = 1.32 μH

b)

L = inductance of the inductor = ?

f = Highest frequency of FM radio = 108 x 10⁶ Hz

C = Capacitance of the capacitor = 2.50 x 10⁻⁻¹² F

For resonance to be possible

C = 1.65 x 10⁻⁻¹² F

C = 1.65 pF

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach

amplitude

hope this