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A 0.0047 microFarad capacitor and a 68mH coil are connected in series. At what approximate frequency will this LC circuit resonate?

____a. 194Hz
____b. 52KHz
____c. 8.9KHz
____d. 4.9GHz

i have tried to figure this question out with no luck. it in volves scientific notation with negative components which leaves me very much in the dark. any help will be greatly appreciated and explain and show how you came to your answer

2007-03-16 04:36:29 · 4 answers · asked by ken s 6 in Science & Mathematics Engineering

4 answers

Gene has it right, tho in more detail-

knowing that f=1/(2Pi(LC)^0.5), your example is

f = 1/(6.28 * ((0.0047 x 10^-6)(68 x 10^-3))^0.5

multiplying the terms under the radical gives

f = 1/(6.28 * (0.3196 x 10^-9)^0.5


Now, if you're going to take the square root of such a product, it works much easier if any exponents are even, because the square root of a number raised to a power is the same number raised to 1/2 that power. So now let's multiply the first term (0.3196) by 10 and divide the second term (10^-9) by 10:

f = 1/(6.28 * (3.196 x 10^-10)^0.5

The square root of 3.196 ~ 1.788,
and the square root of 10^-10 = 10^-5.

1/(6.28 * 1.788 * 10^-5)

Trick:: to do this on a calculator easily, type it in as

1/6.28/1.788/0.00001

I get about 8.9kHz, how about you?

2007-03-16 06:20:08 · answer #1 · answered by Gary H 6 · 0 0

If you multiply two numbers with scientific notation, add the exponents and multiply the regular numbers. When you take the square root, divide the exponent by two and take the square root of the regular number. When you divide, multiply the exponent by -1. You need to learn it sometime .... or just use a calculator.

2007-03-16 11:56:50 · answer #2 · answered by Gene 7 · 0 0

The equation that you need is 1/(2*pi*sqrt(L*C))

Your C is 4.7X10^-9 and your L is 6.8X10^-2

This puts both values in Farads and Henries respectively.

Multiply L and C = 31.96 X 10^-11

Use calculator to take square root and get 1.7877X10^-5

Plugging this into the equation I get

8.9X10^3 Hz or 8.9kHz

2007-03-16 12:53:41 · answer #3 · answered by rscanner 6 · 0 0

Ken, for your future needs of such questions, go and get yourself a copy of the ARRL's Handbook. Such questions as this are covered in well written detail. If you are going to be working with such as this, that Handbook will be a life saver.

2007-03-17 06:12:41 · answer #4 · answered by Anonymous · 0 0

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