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

Answer:

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

$2\pi fL = \frac{1}{2\pi fC}$

$2 (3.14)(87.7\times 10^{6})L = \frac{1}{2 (3.14)(87.7\times 10^{6}) (2.50\times 10^{-12})}$

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

$2\pi fL = \frac{1}{2\pi fC}$

$2 (3.14)(108\times 10^{6})(1.32\times 10^{-6}) = \frac{1}{2 (3.14)(108\times 10^{6}) C}$

C = 1.65 x 10⁻⁻¹² F

C = 1.65 pF

Answer 2

Answer:

inductance  is 1.31 µH

capacitance is 1.658 pF

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