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13 September, 03:20

What is the energy (in J) stored in the 23.0 µF capacitor of a heart defibrillator charged to 7.40 ✕ 103 V?

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Answers (2)
  1. 13 September, 03:27
    0
    629.74 J

    Explanation:

    Energy: This can be defined as the ability or the capacity to do work.

    The S. I unit of energy is Joules (J).

    The formula for the energy stored in a capacitor is given as

    E = 1/2CV² ... Equation 1

    Where E = Energy stored in the capacitor, C = capacitance of the heart defibrillator capacitor, V = voltage of the heart defibrillator.

    Given: C = 23 µF = 23*10⁻⁶ F, V = 7.4*10³ V.

    Substitute into equation 1

    E = 1/2 (23*10⁻⁶) (7.4*10³) ²

    E = 629.74 Joules

    Hence the energy stored in the the capacitor of a heart defibrillator = 629.74 J
  2. 13 September, 03:35
    0
    629.7 J = 0.63 kJ.

    Explanation:

    Charge and voltage are related to the capacitance C of a capacitor by Q = CV, and so the expression for Ecap can be written mathematically into three equivalent expressions:

    Ecap = QV^2

    = 1/2 * (C*V^2)

    = 1/2 * Q^2/C

    where,

    Q = charge

    V = voltage on a capacitor C.

    The energy is in joules for a charge in coulombs, voltage in volts, and capacitance in farads.

    Given:

    C = 23.0 µF

    = 2.3 x 10^-5 F

    V = 7.40 ✕ 10^3 V

    Ecap = 1/2 * (C*V^2)

    = 1/2 * (2.3 x 10^-5) * (7.40 ✕ 10^3) ^2

    = 629.7 J

    = 0.63 kJ.
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