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Its been a long time since I've had to do math and I'm now trying to complete an advanced radio course. I've been doing not too bad until I came to this question that seems to be giving my an exponential problem! I know of others doing the same self study program have had the same issue. It would be nice to learn how this should really work. So here goes: The equation given to solve the question is (it's not my equation so I can't change it): C=1/(2pi *f)E2 * L

Here is some reference: that you all probably already knowing: C=Capacitance f=frequency in hertz L= inductance in henry 1 Pico Farad = 1*10E-12 farad 1 MHz = 1*10E6 hertz

The sample test question is:

What is the value of capacitance (C) in a series R-L-C circuit, if the circuit resonant frequency is 14.25 MHz and L is 2.84 microhenrys?

The answer for (C) should be 44 picofarads

So it's my understanding then that my answer should be 4.4 e-11 but I'm getting 4.4 e-14 which is way smaller a number than it should be. This is my equation as entered in my calculator (I might have more brackets than some of you think I need but my calculator seems to want it this way): 1/(((2pi)*(14.25*10E6))E2*(2.84*10E-6)) This is my answer: C=4.392302014589E-14 And that would be the same as C=.0044 picofarads instead of C=44 picofarads.

I hope I explained this clearly enough. Thanks in advance for your help.

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  • $\begingroup$ When you write your fractions with slashes, please incorporate parentheses to show the $*L$ is in the denominator, not the numerator. $\endgroup$ Commented Aug 28, 2016 at 16:13

2 Answers 2

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Your equation "C=1/(2pi *f)E2 * L" has an "E2" in the middle that does not seem to belong there.

If you remove it, that will account for some of the difference.

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  • $\begingroup$ No. It does belong. And removing it does not help. With it in my answer is only out be a couple decimal places but without it it's not even close. Besides, that's not my equation, that's the equation I need to solve so I can't very well change the question to suit my answer. $\endgroup$ Commented Aug 28, 2016 at 15:23
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You are misinterpreting the equation with the $E2$, maybe it is a typo in your book. The usual expression is $f=\frac 1{2 \pi \sqrt{LC}}$, so maybe the $E2$ is supposed to be a square. This gives $C=\frac 1{(2\pi f)^2L}$ which duly gives $4.4E-11$ farads.

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  • $\begingroup$ I found the issue. It was a square but that wasn't the problem. The book has an error where they wanted students to (using a calculator) multiply the frequency x10^6 but that doesn't work right. At least not on any of my calculators. You have to use the "E" or "EE" button then the 6 to get it to enter properly. For example I was entering 14.25*10E6 and it should have been entered 14.25E6 (no 10). Maybe you had to do it this way years ago or on the calculators they used when the wrote the course. At any rate I'm greatful for your help. Bouncing this off someone made me think better. Thanks! $\endgroup$ Commented Aug 29, 2016 at 17:31

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