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?

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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explanation: