# Math Help - Capacitance formula

1. ## Capacitance formula

I am new to the forum and my maths is very poor, so please excuse me if this is posted in the wrong section.

I am trying to learn electrics and am reading about capacitance. Below is taken from the section that I am at.

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Capacitors in series and parallel

Capacitors in parallel ADD together as C1 + C2 + C3 + ..... While capacitors in series REDUCE by:
1 / (1 / C1 + 1 / C2 + 1 / C3 + .....)
Consider three capacitors of 10, 22, and 47 uF respectively.
Added in parallel we get 10 + 22 + 47 = 79 uF. While in series we would get:
1 / (1 / 10 + 1 / 22 + 1 / 47) = 5.997 uF.
Note that the result is always LESS than the original lowest value.

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The problem I am having is the formula below, I don't understand it and how to arrive at the answer given. Google presented 1 / (1 / 10 + 1 / 22 + 1 / 47) = 5.997 uF
as 1 / ((1 / 10) + (1 / 22) + (1 / 47)) = 5.99767981 but I still do not understand it.

Maybe someone would be kind as to break it down and explain how it is calculated as I have tried various methods and can't get anywhere near the correct answer.

I feel so stupid as I have a feeling it is simple but I am determined to learn so that is the first obstacle tackled I guess.

2. the formula for calculating the total capacitance for "n" capacitors arranged in a series circuit is ...

$\frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ... + \frac{1}{C_n}$

for your example of three capacitors in series ...

$\frac{1}{C_T} = \frac{1}{10} + \frac{1}{22} + \frac{1}{47}$

basically, you add the fractions on the right side to get a sum value that equals the reciprocal of the total capacitance ...

therefore, the total capacitance is ...

$C_T = \frac{1}{sum \, value}$

$\frac{1}{C_T} = \frac{1}{10} + \frac{1}{22} + \frac{1}{47} = \frac{431}{2585}$

$C_T = \frac{2585}{431} \approx 6.0 \, \mu f$

3. Thank you skeeter, the problem was that I was changing the fractions to decimals and adding them.