Defibrillator Energy Dissipation Calculation

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SUMMARY

The discussion focuses on calculating the energy dissipated in a patient during a defibrillation event using a defibrillator with a 40.0 µF capacitor charged to 6.2 kV. The initial energy stored in the capacitor is established at 768.8 J, with an initial current of 26.96 A flowing through the patient, whose resistance is 230 ohms. The key equations for this calculation include the charge remaining after a time period and the energy lost during the discharge. The goal is to determine the energy dissipated in the patient over a pulse duration of 1.0 ms.

PREREQUISITES
  • Understanding of capacitor discharge equations
  • Knowledge of Ohm's Law and electrical resistance
  • Familiarity with energy calculations in electrical circuits
  • Basic grasp of exponential decay in electrical systems
NEXT STEPS
  • Calculate the remaining charge using the formula Q = CV e^(-t/CR)
  • Determine the energy lost using the equation Energy lost = 768 - (Q^2 / 2C)
  • Explore the impact of varying resistance on energy dissipation
  • Research the effects of different capacitor values on defibrillation efficiency
USEFUL FOR

Electrical engineers, medical device developers, and students studying biomedical engineering or electrical circuit theory will benefit from this discussion.

phy112
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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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