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The Sum of Capacitance Question Help Pls

I have to sum the capacitance (The answer is 400 and I keep getting 300). no idea how else to approach lol
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sorry! no idea y it's upside down...
(edited 2 years ago)
Do you not have 3 in series which are in parallel with the other one? Second picture.
Original post by BlueChicken
Do you not have 3 in series which are in parallel with the other one? Second picture.

the second picture was me trying to draw what I thought was an equivalent circuit to make the question easier to do. this is the original q:
16449252569704827057005893670550.jpg
Original post by Obolinda
the second picture was me trying to draw what I thought was an equivalent circuit to make the question easier to do. this is the original q:
16449252569704827057005893670550.jpg

oh, I meant second, if we turn both pictures as are the correct way up! That's the picture I was referring to.

So, imagine 2 capacitors top and bottom of the picture are drawn next to the one on the left - there are 3 capacitors in series, which are in parallel with the one on the right. Just looking at the answer, the maths appears to be [1/(3/300)] + 300, which would tie with the above. Does that make sense?
Original post by BlueChicken
oh, I meant second, if we turn both pictures as are the correct way up! That's the picture I was referring to.

So, imagine 2 capacitors top and bottom of the picture are drawn next to the one on the left - there are 3 capacitors in series, which are in parallel with the one on the right. Just looking at the answer, the maths appears to be [1/(3/300)] + 300, which would tie with the above. Does that make sense?

that makes sense thank you! i just don't know how id answer a question like this in the future, how did you know to think of the circuit as 3 capacitors in series parallel to 1 capacitor ? :frown:
I think you've probably not paid enough attention to where the test points are in relation to the components...


150221.jpg
... so you need to combine the caps in series that I've called B,C &D

and then combine that equivalent value with the capacitor A that it's in parallel with.
Original post by Obolinda
that makes sense thank you! i just don't know how id answer a question like this in the future, how did you know to think of the circuit as 3 capacitors in series parallel to 1 capacitor ? :frown:

So, first way, practice! The logical way, is to consider the current/charge going through them (these are how the laws are derived). In the diagram, assume current is flowing from R to S (assume they are just points and R and S don't refer to anything - I feel like they threw those in as a trick). The top junction, i.e. the black dot at top, the current will split, i.e. some current will go through 3 of the resistors in a line and the rest will go through the resistor on the right (here, we don't care about sizes of current/charge, just where it is going). Therefore, 3 are in series and are in parallel with the other one.

This link has pictures for what I mean about the charge (therefore, current) splitting: https://courses.lumenlearning.com/physics/chapter/19-6-capacitors-in-series-and-parallel/
(edited 2 years ago)
Original post by Joinedup
I think you've probably not paid enough attention to where the test points are in relation to the components...


150221.jpg

Original post by BlueChicken
So, first way, practice! The logical way, is to consider the current/charge going through them (these are how the laws are derived). In the diagram, assume current is flowing from R to S (assume they are just points and R and S don't refer to anything - I feel like they threw those in as a trick). The top junction, i.e. the black dot at top, the current will split, i.e. some current will go through 3 of the resistors in a line and the rest will go through the resistor on the right (here, we don't care about sizes of current/charge, just where it is going). Therefore, 3 are in series and are in parallel with the other one.


ahh... i see. i think i get this now. thank you very much guys, you are life savers !!! :heart:
Maybe think of circuit diagrams as being like tube maps with the components for the stations *

if the train comes in from the left and goes right at the first junction it has to go through B and then C and then D... cos those 3 are in series.


* The guy who designed the 'iconic' modern london tube map was IIRC an electrical engineer

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