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To receive AM radio, you want an $R L C$ circuit that can be made to resonate at any frequency between 500 and 1650 kHz. This is accomplished with a fixed 1.00$\mu \mathrm{H}$ inductor connected to a variable capacitor. What range of capacitance is needed?

9 $\mathrm{mF}$

Physics 102 Electricity and Magnetism

Chapter 23

Electromagnetic Induction, AC Circuits, and Electrical Technologies

Electromagnetic Induction

Rutgers, The State University of New Jersey

University of Washington

Simon Fraser University

University of Sheffield

Lectures

03:27

Electromagnetic induction …

07:07

In physics, electromotive …

00:33

The tuning circuit of an A…

01:06

What capacitance would you…

02:28

$\bullet$ A radio inductor…

02:37

The FM radio band covers t…

01:43

What capacitance do you ne…

02:21

An $R L C$ circuit is used…

01:04

An RLC circuit is used in …

02:23

The AM band extends from a…

02:54

The lowest frequency in th…

02:36

(I) The variable capacitor…

02:13

You want the current ampli…

04:39

07:06

A variable capacitor with …

04:08

02:51

Suppose you have a supply …

02:00

31.10. A Radio Inductor. Y…

03:30

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02:39

(II) A certain FM radio tu…

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04:47

An $R L C$ circuit has a c…

here. We know that the resonant frequency is equaling one over two pi times the square root between the product of the induct inst times that capacitance this would be for a NARAL C circuit. And so here we can say that Then the capacitance men would be equaling essentially one over four pi squared, multiplied by the resident frequency squared, multiplied by the induct in CE the self inducted out. And so we can say that the minimum capacitance would be equaling. This would be one over four pi squared here where a minute We're at a minimum capacitance. So we have a maximum frequency of 16. Well, the other 1650 killer hurts. So 1650 times 10 to the third hurts multiplied by the induct in CE of one times 10 to the negative Sixth Henry's. And so we find that the minimum capacitance is equaling 9.304 times 10 to the negative ninth ferrets. And we can say that Then the maximum capacitance would be again one over four pi squared. But now we're saying that that the frequency is only 500 kilohertz, so 500 times 10 to the third hurts multiplied by one times 10 to the negative. Six. Henry's One Micro Henry. And this is equaling here. 1.132 times, 10 to the negative. Seven carrots. And so, essentially, we can say that then the capacitance would be somewhere between 9.30 out times, 10 to the negative ninth Farage's to 1.1 times 10 to the negative seventh Fareed's. This would be our final answer. That is the end of the solution. Thank you for watching.

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