1. Jul 21, 2013

### mbaban

Two capacitors connected in Parallel and their sum(Ceq12) is connected in series with a third capacitor .

Givens are : (( C1 = unknown value (X) , C2 = 4uF , C3 = 18 uF , Q1 = 300 uC , V = 90V ))

The question wants the value of C1 , ( I find it impossible to solve because givens aren't enough but my teacher insists that it can be done ) Please help ! {if it helps the answer is 5uF but i can figure how }.

OK here is what i did :

Equations :

-Parallel : Ceq = C1 + C2
-series : 1/Ceq = 1/c1 + 1/c2
-Q= CV , C = Q/V , V= Q/C
- in series Q=Q , v=v1+v2
- in parallel V=V , Q=Q1+Q2
those are the needed equations i think

ok here is what i did :
Ps : ( im not just looking for a fast and easy way to do homework , i actually tried alot )

-- i used 1/Ct = 1/ceq12 + 1/C3
1/Ct = 1/x+4 + 1/18
-continued to Qt = 18x+72/22+x * 90

-- i reached a quadratic equation in one of my tries
-- and now i reached 1620x-5100 = 0

Last edited: Jul 21, 2013
2. Jul 21, 2013

### Staff: Mentor

What did you try so far? Where did you run into problems?
Which equations for capacitors do you know? Please use the homework template, we have it for a good reason.

I can confirm that it is possible to solve the problem.

3. Jul 21, 2013

### mbaban

It is my first post thats why , where should i post what i did ?

4. Jul 21, 2013

### Staff: Mentor

That fraction should get brackets:
1/Ct = 1/(x+4) + 1/18
Okay, everything in µF, fine.

I don't understand what you did afterwards. You can directly determine Ct from voltage and charge, and simplify the equation to a linear equation: take 1/18 to the left side, invert both sides (if 1/a=1/b, then a=b), that is probably the quickest method.

5. Jul 21, 2013

### mbaban

the charge is only Q1 charge , the trick is in the combinations , since it is Vt=v1+v2 in series so the voltage is unknown for Ceq12 , while in parallel Q1 dosnt equal Q2 and that is the issue

6. Jul 21, 2013

### collinsmark

First let me summarize what we know so far (I'll put units in square brackets, so that they are not confused with variables):

$$C_{\mathrm{eq}} = \frac{1}{\frac{1}{18 \ \mathrm{[\mu F]}} + \frac{1}{C_1 + \ 4 \ [\mathrm{\mu F}]}}$$

and,

$$90 \ [\mathrm{V}] = V_{1, 2} + V_{3}$$

Did the problem state that before applying potential to the circuit, that the charge on all capacitors was zero? That makes a difference.

If the initial charge on all capacitors was zero (when the voltage was zero), we can say,

$$Q_{1} + Q_{2} = Q_{3}$$

Think about that if it doesn't make intuitive sense. All the current that goes through C3 also goes though either C1 or C2; there's no other place for it to go. And what is current? It's just charge per unit time. The equation is only true though if we happen to know that the charge on all capacitors was zero before current entered the the circuit.

From there we can start combining equations and get to the answer, knowing that C = Q/V.

Last edited: Jul 22, 2013
7. Jul 22, 2013

### Staff: Mentor

Ah okay, then the equation might get quadratic, indeed.