# RLC circuit - radio tuner

Discussion in 'Homework Help' started by lll, Apr 25, 2012.

1. ### lll Thread Starter New Member

Mar 7, 2012
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I'm not even sure I know how to approach this problem. I really don't know anything about radios, and the book doesn't cover them at all.

I tried setting the resonance frequency equal to 89.1 MHz (pure guess), and solving for C, but I didn't get the right answer.

The correct answer is 3.2pF. How?

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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You have to use the exact resonance formula

$\omega_o=\sqrt{\frac{1}{LC}-\frac{R^2}{L^2}}$

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3. ### lll Thread Starter New Member

Mar 7, 2012
21
1
That didn't work.

I keep geeting 125 pF.

I'm thinking that maybe I'm supposed to use wd (damped resonance frequency) instead of w0. But I don't know how to calculate the neper frequency since the circuit is neither in series nor parallel.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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It does work - your maths has let you down. C=3.1904pF is the answer or 3.2pF is close enough.

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5. ### lll Thread Starter New Member

Mar 7, 2012
21
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Ok, I found a mistake in my previous calculation, so now I'm a little closer to the answer, but it's still off by a factor of 3, no matter how many times I double check. Here's the calculation:

$8910000=\sqrt{\frac{1}{10^{-6}C}-\frac{5^2}{{(10^{-6})^2}}}$
$\frac{1}{C}=(8910000^2+\frac{5^2}{(10^{-6})^2})(10^{-6})$
and that gave me C=9.57964pF

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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You need to convert 89.1MHz to an equivalent value in radians/second. It's ωo not fo that you need.

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7. ### lll Thread Starter New Member

Mar 7, 2012
21
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Thank you! Finally got it.

8. ### atferrari AAC Fanatic!

Jan 6, 2004
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89.1 MHz = 89100 KHz = 89100000 Hz, right?

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
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That's correct.

10. ### mlog Member

Feb 11, 2012
276
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I got approximately 3.19 pF and had <0.01% error by neglecting the (R/L)^2 term.

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