low pass filter

[blair321]

im sorry if this is lame for you guys but i posted a thread for some info on farads and the info was good but i still cannot find the answer to this question

any clues

for a low pass filter

from the following list choose the two components which would produce a cut off frequency closest to 100 khz

resistor capacitor
1.6 k ohm 1 nf
33 k ohm 330 nf
150 k ohm 10 uf

formula being fc = 1 / 2pi * r * c


many thanks in advance

[BoulderHead]

Have you tried plugging the values given into that formula?
That will give you the answer rather quickly. Remember that the k, m, and n designators mean you must get your decimal places in order, that is; convert the given values into ohms and farads.

[blair321]

yes i have tried plugging them into the formula,

my problem, i am having trouble getting the decimal places in the right places.

[Integral]

To keep track of the decimal just use scientific notation

\eta = 10^{-9}

k = 10^3

\mu = 10^{-6}

so your first set of numbers is

f_c = \frac {1} {2 \pi* 1.6 x 10^3 * 1x 10^{-9}}

multiply the powers of 10 by adding exponents

= \frac {1} {2 \pi *1.6 x 10^{-6}}

a negitive exponent in the denominator is the same as a positive in the numerator.

= \frac {1} {2 \pi *1.6} 10^6

Now do the simple arithematic.

I think if you just do the significant digits, with out even looking at the powers of 10, you will see that there is only one possible correct response.

[BoulderHead]

That can take some practice, I admit, but keep at it.

000,000,000.000,000,000

Notice above where the decimal point is, also notice that the commas contain groups of three zeros. Look at the first group of three zeros immediately to the left of the decimal point. I imagine you have no trouble understanding this group, and if you were told to use a value from 1 to 999 of something (ohms, farads, henrys, etc), you could easily do it, true? Well, in a perfect world, the next group of three just to the left of this one is where you find values ending with k. So you could have anything from 1k to 999k, and in all instances you can replace the k with those first three zeros and you will have your base units in ohms, or whathaveyou. Adding those three zeros is the same as multiplying by 1000, or 10^3, and indeed, multiplication factors is what we’re dealing with here. If you had looked to the group of three zeros to the right of the decimal you would be dealing with m (milli) and your multiplication factor would be 10^-3. So 47mA equals 47x10^-3A, or .047A. Now, every one of those “groups of three” listed above has either a Greek or Latin prefix associated with it, and you need to memorize what they are and what multiplication factor they represent.

[blair321]

many thanks for all your help,

i think you may have just saved my academic life.

blair