kanwal32
  • kanwal32
Easy capacitance doubt
Physics
schrodinger
  • schrodinger
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kanwal32
  • kanwal32
|dw:1434537935776:dw|
kanwal32
  • kanwal32
what will be the equivalent capacitance and why we will take the two in series @Michele_Laino
kanwal32
  • kanwal32

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kanwal32
  • kanwal32
IrishBoy123
  • IrishBoy123
confused. is this not 3 in series or is circuit incomplete?
kanwal32
  • kanwal32
are they three in series
kanwal32
  • kanwal32
how
kanwal32
  • kanwal32
kanwal32
  • kanwal32
should i give any other example to clear my doubt
kanwal32
  • kanwal32
kanwal32
  • kanwal32
|dw:1434539076748:dw|
kanwal32
  • kanwal32
why 2 and 3 are in series and not 1 and 2
IrishBoy123
  • IrishBoy123
the draw function seems to have disappeared from the user interface. i can't draw anymore.
Michele_Laino
  • Michele_Laino
since we can rewrite your circuit as follows: |dw:1434539474858:dw|
IrishBoy123
  • IrishBoy123
yes, that's what it now looks like.
Michele_Laino
  • Michele_Laino
namely the points A and B are at the same voltage: |dw:1434539581793:dw|
kanwal32
  • kanwal32
yes
Michele_Laino
  • Michele_Laino
so the equivalent capacitance is: \[\Large {C_{EQUIVALENT}} = {C_1} + \frac{{{C_2}{C_3}}}{{{C_2} + {C_3}}}\]
radar
  • radar
Showing the two leads which indicate an input or the connecting to the associated circuit made all the difference in resolving the configuration.
radar
  • radar
|dw:1434570639986:dw|

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