Solution:
The energy stored in the capacitor is \(U = \frac{1}{2} C V^2 = \frac{1}{2} \times 40 \times 10^{-6} \times (3000)^2 = 180\text{ J}\). Power delivered is \(P = \frac{U}{t} = \frac{180}{2 \times 10^{-3}} = 90\text{ kW}\).
A \(40\text{ }\mu\text{F}\) capacitor in a defibrillator is charged to \(3000\text{ V}\). The energy stored in the capacitor is sent through the patient during a pulse of duration \(2\text{ ms}\). The power delivered to the patient is :
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