Respuesta :
Answer:
inductance is 1.31 µH
capacitance is 1.658 pF
Explanation:
Given data
frequency = 87.7 MHz = 87.7 ×[tex]10^{6}[/tex] Hz
capacitor = 2.50 pF = 2.50 ×[tex]10^{-12}[/tex] 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 ×[tex]10^{-12}[/tex]) = 1 / 2π ( 87.7 ×[tex]10^{6}[/tex] )
√(L2.50 ×[tex]10^{-12}[/tex]) = 1.81 × [tex]10^{-9}[/tex]
(L2.50 ×[tex]10^{-12}[/tex]) = 3.296 × [tex]10^{-18}[/tex]
L = 3.296 × [tex]10^{-18}[/tex] / 2.50 ×[tex]10^{-12}[/tex]
L = 1.31 × [tex]10^{-6}[/tex] H
so inductance is 1.31 µH
and
for 108 MHz = 108 ×[tex]10^{6}[/tex] Hz
we find here capacitance c from equation 1
frequency = 1 / 2π√(LC)
√(LC) =1 / 2π frequency
√(1.31 × [tex]10^{-6}[/tex] C) =1 / 2π ( 108 ×[tex]10^{6}[/tex] )
√(1.31 × [tex]10^{-6}[/tex] C) = 1.47 ×[tex]10^{-9}[/tex]
(1.31 × [tex]10^{-6}[/tex] C) = 2.178 ×[tex]10^{-18}[/tex]
c = 2.178 × [tex]10^{-18}[/tex] / (1.31 × [tex]10^{-6}[/tex] )
c = 1.658 × [tex]10^{-12}[/tex] F
so capacitance is 1.658 pF
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
[tex]2\pi fL = \frac{1}{2\pi fC}[/tex]
[tex]2 (3.14)(87.7\times 10^{6})L = \frac{1}{2 (3.14)(87.7\times 10^{6}) (2.50\times 10^{-12})}[/tex]
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
[tex]2\pi fL = \frac{1}{2\pi fC}[/tex]
[tex]2 (3.14)(108\times 10^{6})(1.32\times 10^{-6}) = \frac{1}{2 (3.14)(108\times 10^{6}) C}[/tex]
C = 1.65 x 10⁻⁻¹² F
C = 1.65 pF
