A 40 F capacitor in a defibrillator is charged to 3000 ~V . The energy stored in the capacitor is sent through the patient during a pulse of duration 2 ~ms . The power delivered to the patient is

Step-by-step solution

The energy stored in capacitor = 1 2 C V^2 aligned & = 1 2 40 10⁻⁶ (3000)^2 & =20 9 10^6 10⁻⁶=180 ~J . aligned The power delivered in 2 ~ms is = 180 2 10⁻³ =90 10^3 ~W =90 ~kW .