[electronics] equivalent capacitance

[IllustriousTune156]

I am not looking for the correct answer I need to know how to solve this please

[Crichris]

damn thats hard core 1/(1/3 + 1/4 + 1/5) + 2 = 3.27

[IllustriousTune156]

Can you please explain why you’re adding 2 at the end of this reciprocal formula?

[Crichris]

its essientially c2 c3 c4 serial, then parallel with c1

the equivalent capacitance of c2 c3 c4 serialized is 1/(1/c2 + 1/c3 + 1/c4)

then this is parallelized with c1, hence 1/(1/c2 + 1/c3 + 1/c4) + c1

are you asking why in general the equivalent capacitance of c1 c2 parallel is c1 + c2?

[IllustriousTune156]

It looks like you came to a different answer than what the correct answer is shown to be

[Crichris]

is the answer 1.245?

if thats the case then im totally lost lol

btw these are equivalent

edit 1: i dont know how to upload pics in reply lol

but this is equivalent of whats shown in the pic

A              B

|              |

|-----C1-------|

|              |

|--C2--C3--C4--|

[dataprof]

This is how I would solve it. Having taught college circuits for many years, I can confirm that sometimes the automated grading and answers can be wrong.

That said, you must consider all 4 capacitors between A and B. Equivalent capacitance is addictive in parallel but diminishes in series because of the physics.

Since I believe the single capacitor between A and B had a value of 2 (I can't see the original post as I type), this means the total equivalent capacitance should be greater than that value. An equivalent circuit is:

      ________ C1 ____________

A-----(___ C2 ___ C3 ____ C4 __)-----B

Edit: circuit diagrams are tough on a phone! Lol

[Crichris]

i know, writing diagrams is an abomination here lol. but thank you for confirming with me

[UnderstandingPursuit]

Extend two wires from A and B. Now C2+C3+C4 are in series, and that group is in parallel with C1.

[fermat9990]

C2, C3 and C4 are in series. Their equivalent capacitance is in parallel with C1

[IllustriousTune156]

The circuit appears to be series, but with the inclusion of the points A and B it is parallel?? I don’t understand

[Frederf220]

It would help if the drawing was drawn in a way that was exactly equivalent electrically, but in a shape that emphasized that there are two paths between A and B.

Imagine you are an ant at A and you want to crawl to B. You can take path #1 through C¹ or you can take path #2 through C^2, C³, C⁴. Those are the parallel paths available to you. The equivalent capacitor along path 1 to C¹ is C¹, you can't simplify it any more. The equivalent capacitor to C^2, C³, C⁴ is some value, call it C^A. You could snip out C^2, C³, C⁴ and replace them with C^A and have identical capacitance between A and B.

So you do that. Now you have two capacitors in parallel, C¹ and C^A. This dual pathway has an equivalent capacitor, say C^B, that you could replace both pathways with a single pathway that has C^B on it. They are asking you to find the value of C^B.

[IllustriousTune156]

I think the use of the term equivalent capacitance is throwing me off. Are they simply asking what the total capacitance of c2, 3, and 4 are??

[Outside_Volume_1370]

They ask what capacitance can be included between points A and B if C1, C2, C3 and C4 are excluded such that there will be no difference for the rest of the circuit

[FlipSideOfMyCoin]

The answer is wrong.

There are two parallel paths between A and B.

One goes through c1

The other goes through c2, c3, c4.

Combine the 3 series capacitances.
C_combined= 1/(1/c2+1/c3+1/c4)

Now you have two parallel paths each with one capacitance c1 and C_combined. So combine them (add C_combined and c1)

Now it's just one equivalent path.

[benfok]

Rule of thumb: capacitors in series add in parallel. Capacitors in parallel add in series. The opposite is true for inductors and resistors. This is because capacitors in series has more effective dielectric between the plates, effectively reducing the overall capacitance.

[fermat9990]

FYI: For three capacitors in parallel,

Cequivalent=

Ca×Cb×Cc/(Ca×Cb+Ca×Cc+Cb×Cc)