Expert: Sherman D. - 11/9/2004

Question
I'm kind of stuck on a question I would really appriciate if you could help me out. Its an equation:

CT(total) = 0.0000000016(1.6n) if C1 = 4 * C2 and C3 = 20 * C1. Calculate the values for C1, C2 and C3

The Formaula for  CT is 1/CT = 1/C1 + 1/C2 + 1/C3
The answers are C1 = 8.08n C2 = 2.02n and C3 =161.6n. I just cant figure out how they came up with these answers.

Thank you very much

Jared

Answer
Short Version
Since they all share the same .0000000016 value, you can just ignore that, according to what the answer book gives, i am assuming thats what they did to get the answer.

(1/(CT)) = (1/(C1)) + (1/(C2)) + (1/(C3))

(1/(1.6n)) = (1/(4(C2))) + (1/(C2)) + (1/(80(C2)))

Multiply everything by 128(C2)

80n = 32n + 128n + 1.6n

80n = 161.6n
C2 = 2.02

C1 = 4(2.02) = 8.08
C3 = 80(2.02) = 161.6

C1 = 4(C2)
C3 = 20(C1)

C3 = 20(4(C2))
C3 = 80(C2)

Longer Version

(1/(CT)) = (1/(C1)) + (1/(C2)) + (1/(C3))
(1/((1.6 * 10^-9)(1.6n))) = (1/(4(C2))) + (1/(C2)) + (1/(80(C2)))
(1.6 * 10^-9) * 1.6n = 2.56 * 10^-9
multiply everything by (2.56 * 10^-9) * 80(C2)
80(C2) = 20(2.56 * 10^-9)n + ((2.56 * 10^-9) * 80)n + (2.56 * 10^-9)n
80(C2) = 2.5856 * 10^-7n
C2 = 3.232 * 10^-9n
If you divide this by 1.6 * 10^-9 or .0000000016 you get
C2 = 2.02n

C1 = 4(2.02) = 8.08n
C3 = 20(C2) = 161.6n