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  • Why is that in a circuit made of series connection of capacitors the total capacitance of capacitors in series is equal to the reciprocal of the sum of the reciprocals of their individual capacitances?

    why cant it be sum of their individual capacitance as in case of resistors and inductors?

    Plz explain this contrary in terms of Parallel circuits too....
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  • reachrkata

    Member • Nov 30, 2010

    There are 2 ways to understand this -

    1) Consider the voltage equation of a cap V = 1/C * (integral) i.dt. if there are 2 caps in series you have V1 = 1/C1 * (integral) i.dt and V2 = 1/C2 * (integral) i.dt. Total voltage V = V1+V2 = (1/C1+1/C2) * (integral) i.dt. Now if we have an effective cap Ceff such that the same current i flows through it, V = 1/Ceff * (integral) i.dt. Equating these two you get 1/Ceff = 1/C1 + 1/C2.

    2) Adding capacitors in series is like having a single cap with the same plate area as the individual caps and distance between plates as sum of distances of individual caps. That is if C1 = EA/d1, C2 = EA/d2, Ceff = EA/(d1+d2) where E = dielectric constant.
    Solve th above equations and you again get 1/Ceff = 1/C1 + 1/C2


    For paralle circuits, instead of adding voltages you have to add currents (I = CdV/dt) and you get Ceff = C1+C2.

    -Karthik
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  • harish3h

    Member • Dec 1, 2010

    What is the need for considering effective capacitance/resistance/inductance?

    In what way does it help building the circuits..

    Why do we calculate it?
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  • d_vipul

    Member • Dec 2, 2010

    hey harish3h,

    actually the response of the whole circuit is depend upon each parameter of circuit.........
    so while designing it kept into mind......

    Like in OSCILLATOR circuits 3 capacitors are used because one capacitor gives you 60 degree phase shift where as a combination of three gives you 180 degree phase shift.....

    Hope this helps you......

    Regards.
    VIPUL
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  • Harshad Italiya

    Member • Dec 2, 2010

    reachrkata
    There are 2 ways to understand this -

    1) Consider the voltage equation of a cap V = 1/C * (integral) i.dt. if there are 2 caps in series you have V1 = 1/C1 * (integral) i.dt and V2 = 1/C2 * (integral) i.dt. Total voltage V = V1+V2 = (1/C1+1/C2) * (integral) i.dt. Now if we have an effective cap Ceff such that the same current i flows through it, V = 1/Ceff * (integral) i.dt. Equating these two you get 1/Ceff = 1/C1 + 1/C2.

    2) Adding capacitors in series is like having a single cap with the same plate area as the individual caps and distance between plates as sum of distances of individual caps. That is if C1 = EA/d1, C2 = EA/d2, Ceff = EA/(d1+d2) where E = dielectric constant.
    Solve th above equations and you again get 1/Ceff = 1/C1 + 1/C2


    For paralle circuits, instead of adding voltages you have to add currents (I = CdV/dt) and you get Ceff = C1+C2.

    -Karthik
    Superb Explanation !
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  • reachrkata

    Member • Dec 4, 2010

    Yo thanks !!!
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  • shreyasm89

    Member • Dec 21, 2010

    @karthik- best explanation...concise, but complete.😀
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  • sachleen

    Member • Dec 22, 2010

    hey. #-Link-Snipped-#

    best explanation!!!!!!!!
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