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16 July, 17:08

An 8.00 μF capacitor that is initially uncharged is connected in series with a 5 Ω resistor and an emf source with and negligible internal resistance.

At the instant when the resistor is dissipating electrical energy at a rate of 250 W, how much energy has been stored in the capacitor?

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  1. 16 July, 19:45
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    0.005 J of energy has been stored in the capacitor.

    Explanation:

    Energy stored in a capacitor (E) = 1/2CV^2

    Power (P) = V^2/R

    V^2 = PR

    Therefore, E = 1/2CPR

    C is the capacitance of the capacitor = 8*10^-6 F

    P is power dissipated by resistor = 250 W

    R is the resistance of the resistor = 5 ohm

    E = 1/2 * 8*10^-6 * 250 * 5 = 0.005 J
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