Question

A power source for a portable electrical defibrillator contains a capacitor of capacitance 60 μF. The potential difference across the plates of the capacitor is raised to 5.0x10^3 V and 20% of its stored energy is released in a 3.0 ms pulse. Calculate the average power of the pulse.

Answer 1

Step-by-step explanation:

Energy of the capacitor = 1/2 C V² = 1/2 X 60 X 10⁻⁶ X 5 X 10³ = 0.15 J

20 % = .03 J

Power = energy released per second = Energy / time = .03 /(3 x 10⁻³) = 10 W.