# finding total capacitance in a series circuit

#### james7701

Joined Jan 5, 2016
44
the formula Ct= C1xC2/C1+C2, please explain how to use this formula to calculate Ct.

#### dl324

Joined Mar 30, 2015
14,914

#### james7701

Joined Jan 5, 2016
44
Is this homework?
no, just trying to figure out how to use that formula to get the answer that was provided.

#### dl324

Joined Mar 30, 2015
14,914

#### WBahn

Joined Mar 31, 2012
26,911
the formula Ct= C1xC2/C1+C2, please explain how to use this formula to calculate Ct.
First off, you need to be careful regarding order of operations. Since multiplication is done before addition, what you have written above is technically

Ct = ((C1 x C2) / C1) + C2

when what you want is

Ct = (C1 x C2) / (C1 + C2)

After that, what is it that is causing you problems? Your attachment shows values being substituted into that equation and a result being worked out.

Is it that you don't understand how (47 x 47) / (47 + 47) works out to 23.5 ?

Or does it have to do with why this formula gives the total capacitance of two capacitors in series?

#### james7701

Joined Jan 5, 2016
44
First off, you need to be careful regarding order of operations. Since multiplication is done before addition, what you have written above is technically

Ct = ((C1 x C2) / C1) + C2

when what you want is

Ct = (C1 x C2) / (C1 + C2)

After that, what is it that is causing you problems? Your attachment shows values being substituted into that equation and a result being worked out.

Is it that you don't understand how (47 x 47) / (47 + 47) works out to 23.5 ?

Or does it have to do with why this formula gives the total capacitance of two capacitors in series?
well, i have tried to work this problem many times and when I'm done the answer always comes out to 94

#### Veracohr

Joined Jan 3, 2011
765

#### james7701

Joined Jan 5, 2016
44
47x47=2209
2209/94=23.5
so, basically i only needed to use the top portion of the formula?

#### Veracohr

Joined Jan 3, 2011
765
No, the top portion is (47x47), which is 2209. Then you divide that by (47+47) which is 94.

That equation is only valid for two impedances, and is derived from a more general equation which can be expanded to any number:

$$\frac{1}{\frac{1}{Z1}+\frac{1}{Z2}+...\frac{1}{Zn}$$

It applies to resistances or inductances in parallel, and capacitors in series.

#### WBahn

Joined Mar 31, 2012
26,911
well, i have tried to work this problem many times and when I'm done the answer always comes out to 94
It would really help if you showed YOUR work, step by step, so that we can see what YOU are doing. That way we can see what you are doing wrong.

But it appears that you are probably just trying to plug numbers into a calculator and completely forgetting about order of operations.

To wit:

47 * 47 = 2209
2209 / 47 = 47
47 + 47 = 94

What you NEED to do is:

(47 * 47) = 2209
(47 + 47) = 94
2209 / 94 = 23.5

#### james7701

Joined Jan 5, 2016
44
It would really help if you showed YOUR work, step by step, so that we can see what YOU are doing. That way we can see what you are doing wrong.

But it appears that you are probably just trying to plug numbers into a calculator and completely forgetting about order of operations.

To wit:

47 * 47 = 2209
2209 / 47 = 47
47 + 47 = 94

What you NEED to do is:

(47 * 47) = 2209
(47 + 47) = 94
2209 / 94 = 23.5
sorry, yes i got already... just forgot the order. i was getting 2209 and 94 but, i was just doin the steps all wrong..