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19 December, 12:18

A capacitor with very large capacitance is in series with another capacitor with very small capacitance. What is the equivalent capacitance of the combination?

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  1. 19 December, 15:40
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    The equivalent capacitance of the combination is 1/C = 1/C1+1/C2 where C1 and C2 are the capacitance of both capacitors in series.

    Explanation:

    Let the capacitance of both capacitors be C1 and C2. For a series connected capacitors, same charge flows through the capacitors but different voltage flows through them.

    Let Q be the amount of charge in each capacitors,

    V be the voltage across each capacitors

    C be the capacitance of the capacitor.

    Using the formula Q = CV where V = Q/C ... (1)

    For the large capacitor with capacitance of the capacitor C1,

    Q = C1V1; V1 = Q/C1 ... (2)

    where V1 is the voltage across C1,

    For the small capacitor with capacitance of the capacitor C2,

    Q = C2V2; V2 = Q/C2 ... (3)

    where V2 is the voltage across C2,

    Total voltage V in the circuit will be;

    V = V1+V2 ... (4)

    Substituting equation 1, 2 and 3 in equation 4, we have;

    Q/C = Q/C1 + Q/C2

    Q/C = Q{1/C1+1/C2}

    Since change Q is the same for both capacitors since they are in series, they will cancel out to finally have;

    1/C = 1/C1+1/C2

    This gives the equivalent capacitance of the combination.
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