Defibrillator Energy Dissipation Calculation

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A defibrillator discharges a 40.0 µF capacitor charged to 6.2 kV, delivering a brief current pulse of 1.0 ms through a patient with a resistance of 230 ohms. The initial energy stored in the capacitor is calculated to be 768.8 joules, with an initial current of 26.96 amps. To determine the energy dissipated in the patient during the pulse, the equations for charge remaining and energy lost can be applied. The charge remaining after the pulse can be calculated using Q = CV e^(-t/CR), and the energy lost can be derived from the initial energy minus the energy associated with the remaining charge. This approach will yield the energy dissipated in the patient during the defibrillation process.
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Homework Statement


A defibrillator passes a brief burst of current through the heart to restore normal beating. In one such defibrillator, a 40.0 µF capacitor is charged to 6.2 kV. Paddles are used to make an electrical connection to the patient's chest. A pulse of current lasting 1.0 ms partially discharges the capacitor through the patient. The electrical resistance of the patient (from paddle to paddle) is 230 .
I found these calculations:

initial energy stored in the capacitor is 768.8J
initial current through the patient is 26.96A

I need to know How much energy is dissipated in the patient during the 1.0 ms?


Homework Equations





The Attempt at a Solution

 
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any attempts?
 
Try this if ur given values 768.8J and 26.96 are right:


• Charge remaining after t s., Q = CV e^ -[1/CR]t


• Energy lost = 768 – Q^2 / 2C
 

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