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kanwal32

  • one year ago

Easy capacitance doubt

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  1. kanwal32
    • one year ago
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    |dw:1434537935776:dw|

  2. kanwal32
    • one year ago
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    what will be the equivalent capacitance and why we will take the two in series @Michele_Laino

  3. kanwal32
    • one year ago
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    @IrishBoy123

  4. kanwal32
    • one year ago
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    @radar

  5. IrishBoy123
    • one year ago
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    confused. is this not 3 in series or is circuit incomplete?

  6. kanwal32
    • one year ago
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    are they three in series

  7. kanwal32
    • one year ago
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    how

  8. kanwal32
    • one year ago
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    @Michele_Laino

  9. kanwal32
    • one year ago
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    should i give any other example to clear my doubt

  10. kanwal32
    • one year ago
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    @Michele_Laino

  11. kanwal32
    • one year ago
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    |dw:1434539076748:dw|

  12. kanwal32
    • one year ago
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    why 2 and 3 are in series and not 1 and 2

  13. IrishBoy123
    • one year ago
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    the draw function seems to have disappeared from the user interface. i can't draw anymore.

  14. Michele_Laino
    • one year ago
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    since we can rewrite your circuit as follows: |dw:1434539474858:dw|

  15. IrishBoy123
    • one year ago
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    yes, that's what it now looks like.

  16. Michele_Laino
    • one year ago
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    namely the points A and B are at the same voltage: |dw:1434539581793:dw|

  17. kanwal32
    • one year ago
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    yes

  18. Michele_Laino
    • one year ago
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    so the equivalent capacitance is: \[\Large {C_{EQUIVALENT}} = {C_1} + \frac{{{C_2}{C_3}}}{{{C_2} + {C_3}}}\]

  19. radar
    • one year ago
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    Showing the two leads which indicate an input or the connecting to the associated circuit made all the difference in resolving the configuration.

  20. radar
    • one year ago
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    |dw:1434570639986:dw|

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