Question

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?

Answer 1

a) 1.32 μH

b) 1.65 pF

Step-by-step 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

`2\pi fL = (1)/(2\pi fC)`

`2 (3.14)(87.7* 10^(6))L = (1)/(2 (3.14)(87.7* 10^(6)) (2.50* 10^(-12)))`

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

`2\pi fL = (1)/(2\pi fC)`

`2 (3.14)(108* 10^(6))(1.32* 10^(-6)) = (1)/(2 (3.14)(108* 10^(6)) C)`

C = 1.65 x 10⁻⁻¹² F

C = 1.65 pF

Answer 2

inductance is 1.31 µH

capacitance is 1.658 pF

Step-by-step explanation:

Given data

frequency = 87.7 MHz = 87.7 ×
`10^(6)` Hz

capacitor = 2.50 pF = 2.50 ×
`10^(-12)` 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 ×
`10^(-12)`) = 1 / 2π ( 87.7 ×
`10^(6)` )

√(L2.50 ×
`10^(-12)`) = 1.81 ×
`10^(-9)`

(L2.50 ×
`10^(-12)`) = 3.296 ×
`10^(-18)`

L = 3.296 ×
`10^(-18)` / 2.50 ×
`10^(-12)`

L = 1.31 ×
`10^(-6)` H

so inductance is 1.31 µH

and

for 108 MHz = 108 ×
`10^(6)` Hz

we find here capacitance c from equation 1

frequency = 1 / 2π√(LC)

√(LC) =1 / 2π frequency

√(1.31 ×
`10^(-6)` C) =1 / 2π ( 108 ×
`10^(6)` )

√(1.31 ×
`10^(-6)` C) = 1.47 ×
`10^(-9)`

(1.31 ×
`10^(-6)` C) = 2.178 ×
`10^(-18)`

c = 2.178 ×
`10^(-18)` / (1.31 ×
`10^(-6)` )

c = 1.658 ×
`10^(-12)` F

so capacitance is 1.658 pF