Question

A defibrillator passes a brief burst of current through the heart to restore normal beating. In one such defibrillator, a 40.6-μF capacitor is charged to 5.90 kV. Paddles are used to make an electric connection to the patient’s chest. A pulse of current lasting 1.00 ms partially discharges the capacitor through the patient. The electrical resistance of the patient (from paddle to paddle) is 240 Ω. What is the initial energy stored in the capacitor?

Answer 1

Initial energy stored in capacitor will be given by:

U = (1/2)*C*V^2

C = Capacitance of defibrillator = 40.6*10^-6 F

V = Initial Voltage = 5.90 kV = 5.90*10^3 V

So,

U = (1/2)*40.6*10^-6*(5.90*10^3)^2

U = 706.64 J = Initial energy stored

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