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22 March, 02:46

Two capacitors of capacitances 25 µF and 50 µF are connected in series with a 33-V battery. How much energy is stored in the 25-µF capacitor when it is fully charged?

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  1. 22 March, 02:57
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    6.05*10⁻³ J

    Explanation:

    Note: Two capacitors connected in series behaves like two resistors connected in parallel.

    Using

    1/Ct = 1/C1+1/C2

    Ct = (C1*C2) / (C1+C2) ... Equation 1

    Where Ct = combined capacitance of the two capacitor, C1 = Capacitance of the first capacitor, C2 = capacitance of the second capacitor.

    Given: C1 = 25 µF, C2 = 50 µF

    Substitute into equation 1

    Ct = (25*50) / (25+50)

    Ct = 1250/75

    Ct = 16.67 µF.

    Using

    Q = CV ... Equation 2

    Where Q = Charge, V = Voltage.

    Given: V = 33 V, C = 16.67 µF = 16.67*10⁻⁶ F

    Substitute into equation 2

    Q = 33 (16.67*10⁻⁶)

    Q = 5.5*10⁻⁴ C.

    Since both capacitors are connected in series, the same amount of charge flows through them.

    Using,

    E = 1/2Q²/C ... Equation 3

    Where E = Energy stored in the 25-µF capacitor

    Given: Q = 5.5*10⁻⁴ C, C = 25 µF = 25*10⁻⁶ F

    Substitute into equation 3

    E = 1/2 (5.5*10⁻⁴) ² / 25*10⁻⁶

    E = 6.05*10⁻³ J.
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