# Math Help - Solving for x1 and x2 in fractions

1. ## Solving for x1 and x2 in fractions

Eyey,

it's is derived from a physics problem, but since i dont know how to solve the mathematical part, here you go.

the two ways to set up a electrical circuit, in series or parallel,
in parallel, Cequivalentpara = C1+C2 , Ceqpara= 38*10^-6
Cequivalentseries => 1/Ceqseries = 1/C1 + 1/C2, Ceqseries= 5.2*10^-6

how to solve for c1 and c2? something like 1/x + 1/y = x +y/xy?
i cant find a logical solution

2. Hi blieb,

your question is in the wrong forum and unclear. Here is a simple solution (my data)
given eight resistances of 100 ohms each . Arrange four in series and four in parallel Connect each circuit to 24 volts. each parallel circuit will draw .24 amps or .96 for the combination. In series circuit the total resistance is 400 ohms so amps become .06 . Power = I^2R. voltage drop = IR

bjh

3. ehm tanks! , but that is a completely different answer to a completley different question. and it's not in a completely wrong forum, since if i didnt know it was about physics. could you still solve it for me?
its simple. 1/c= 1/a +1/b
and d= a+ b. find a and b.

here, Ceqivalent for series and parallel are given, and you need to find the individual capacitators.
i know my formula's thats not the problem. i just didnt know, until 2 minutes ago how to solve it.

you have to shuffle them a little until they look somethign like C2^2 -Cpara*C2+ cserie*Cpara
and then solve it by abc formula.

actually not that hard, so sorry

4. Originally Posted by Blieb
ehm tanks! , but that is a completely different answer to a completley different question.
So you are complaining that he only showed you how to solve the problem rather than doing for you?

and it's not in a completely wrong forum, since if i didnt know it was about physics.
You are arguing that since you "didn't know", you couldn't be wrong?

could you still solve it for me?
its simple. 1/c= 1/a +1/b
and d= a+ b. find a and b.
You are asked for a and b in terms of c and d? Yes, that is simple. From the second equation, b= d- a. Replace b by that in the first equation and you get $\frac{1}{c}= \frac{1}{a}+ \frac{1}{d- a}$.

Multiply each term of the equation by $ac(d- a)$ to get rid of the fractions.

here, Ceqivalent for series and parallel are given, and you need to find the individual capacitators.
i know my formula's thats not the problem. i just didnt know, until 2 minutes ago how to solve it.

you have to shuffle them a little until they look somethign like C2^2 -Cpara*C2+ cserie*Cpara
and then solve it by abc formula.

actually not that hard, so sorry

5. whut? sorry, i thought i changed the i in a you. and he did not solve my problem in any way. although i do appreciate his help! (he responded really quick)
thanks and well, thanks. didn't want to start a bitchfight or disturb the peace or bla.
xx