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Goz2K
Aug 12, 2009, 03:52 AM
Hi there I'm studying for my mathematics for engineering exam and am just looking for some clarification on my answers.

I have 2 questions that I believe I have answered correctly however would like the view of an expert just to confirm.

OK here goes

1)

Q=1/wrCR subject is C

Step 1

QwrCR

Step 2

C=QwrCR/C

Step 3

C squared = QwrR

Solution

C = Square Root of QwrR

2)

f = 1/2pi square root LC subject is C

Step 1

f2pi square root LC

Step 2

f2pi squared LC

Step 3

L = f2pi squared C/L

Step 4

L squared = f2pi squared C

Step 5

L = Square root f2pi squared C

This is my first post so forgive me if my explanations are slightly different to those you may have seen so far.

Hope you can help and thanks in advance

KISS
Aug 12, 2009, 04:27 AM
1) Q=1/wrCR subject is C

QwrCR=1 ---> C=1/QwR

In Step 1, you need something on either side of the equal sign.
Somehow you also added an extra C, probably because C's and Q's are close. Be very careful how you write. Use script characters in formulas when possible.

2) Not sure what the problem is
f = 1/2pi square root LC subject is C

Is it f = 1/2 * PI * sqrt(LC)

or is it;

f = 1/ (2 * PI * sqrt(LC).

If the former

2f/PI = sqrt(LC)

4 * (f^2)/(PI^2) = LC

(4 * (f^2)/(PI^2)) / L = C

Goz2K
Aug 12, 2009, 06:09 AM
Thanks for getting back to me as quick as you did and I appreciate you're help

With 2) I'm afraid it's the latter

basically...

1
f=------------------
2*pi*sqrt(LC)


Sorry for the confusion

I end up with L=sqrt(f*(2*pi)^*C)

Goz2K
Aug 12, 2009, 06:17 AM
Thanks for getting back to me as quick as you did and I appreciate you're help

With 2) I'm affraid it's the latter

basically ....



1
f=---------------------------

2*pi*sqrt(LC)


Sorry for the confusion

I end up with L=sqrt(f*(2*pi)^*C)

Sorry leading spaces were removed on the top and bottom of the fraction :(

KISS
Aug 12, 2009, 07:51 AM
f * 2 * PI * sqrt(LC) = 1

sqrt (LC) = 1 / ( f * 2 * PI)

LC = (1 * 1) / (f*f * 2*2 * PI*PI)

Solve for L or C; f*f really = f^2 and 1*1 = 1

If you have a sqrt on one side, the you have to square both sides to get rid of the sqrt.

Technically, when you take the sqrt of something, there are two solutions. e.g. sqrt(4) = 2 or -2. You have to look at the values and see which solution makes sense. It's also true for problem #1.

Goz2K
Aug 13, 2009, 01:43 AM
Fresh pair of eyes today... thanks for you're reply once again

kk now I believe I have it... I now end up with

Original

f=1/(2*PI*sqrt(LC))

Next step

f*2*PI*sqrt(LC) = 1

Next Step

sqrt(LC) = 1/(f*2*PI)

Next Step

LC = 1/(f^2*2^2*PI^2)

Finally solving for L leaves me with

L=1/(f^2*2^2*PI^2*C)

Which appears to be a much more sensible answer

Thanks for jogging the old grey matter once again :)