Physics, 26.10.2020 17:00 ayoismeisalex

If the variable capacitor in an FM receiver ranges from 14.9 pF to 22.4 pF , what inductor should be used to make an LC circuit whose resonant frequency spans the FM band

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Answer from: jwmiciche7367

Complete Question

The FM radio band covers the frequency range 88-108 MHz

If the variable capacitor in an FM receiver ranges from 14.9 pF to 22.4 pF , what inductor should be used to make an LC circuit whose resonant frequency spans the FM band

The inductor that should be made use of is and inductor with an inductance of $L = 1.46 *10^{-7} \ H$

Explanation:

From the question we are told that

The frequency range for the FM band is 88-108 MHz

The range of the capacitance is $C = 14.9 \ pF \ \ to \ 22.4 pF = 14.9 *10^{-12} \ F \ \ to \ \ 22.4 *10^{-12}\ F$

The resonant frequency is $f_r = 108 MHz. = 108 *10^6 \ Hz$ this is in the question we are told that the resonant frequency spans the FM band

Generally the at resonance the reactance of the capacitor is equal to that of the inductor i.e

R_c = R_l

Here R_c is the reactance of the capacitor which is mathematically represented as

$R_c = \frac{1}{2 \pi f C}$

while the reactance of the inductor is $R_i = 2 \pi f L$

$\frac{1}{2 \pi f C} = 2 \pi f L$

=> $L = \frac{1}{4 \pi^2 * f ^2 * C }$

When C is $C = 14.9 \ pF = 14.9 *10^{-12} \F$

=> $L = \frac{1}{4 (3.142 )^2 * (108*10^6) ^2 * 14.9 *10^{-12} }$

=> $L = 1.46 *10^{-7} \ H$

When C is $C = 22.4 \ pF = 14.9 *10^{-12} \F$

=> $L = \frac{1}{4 (3.142 )^2 * (108*10^6) ^2 * 22.4 *10^{-12} }$

=> $L = 1.46 *10^{-7} \ H$

Answer from: Quest

Cin carbonh in hydrogeno in oxygen

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

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If the variable capacitor in an FM receiver ranges from 14.9 pF to 22.4 pF , what inductor should be...

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