Potential difference of a capacitor

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SUMMARY

The discussion centers on calculating the potential difference required for a capacitor in a defibrillator, specifically one with a capacitance of 8.00 µF, to deliver 100 J of energy. The correct formula for energy stored in a capacitor is U_c = 1/2 C (ΔV)^2. To achieve the desired energy output of 100 J, the potential difference must be calculated, resulting in a required voltage of 5.00 kV. The initial calculations presented were incorrect, emphasizing the importance of using the correct formulas and unit conversions.

PREREQUISITES
  • Understanding of capacitor fundamentals, including capacitance and energy storage.
  • Familiarity with the formula for energy in capacitors: U_c = 1/2 C (ΔV)^2.
  • Knowledge of unit conversions, particularly from microfarads to farads.
  • Basic algebra skills for manipulating equations to solve for unknowns.
NEXT STEPS
  • Study the derivation and application of the energy formula for capacitors: U_c = 1/2 C (ΔV)^2.
  • Learn about the practical applications of capacitors in medical devices, specifically defibrillators.
  • Explore the implications of voltage levels in medical equipment and safety protocols.
  • Investigate the effects of capacitance on energy delivery in various electrical circuits.
USEFUL FOR

Electrical engineers, medical device developers, students studying electronics, and anyone involved in the design or use of defibrillators and similar medical equipment.

kbyws37
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A defibrillator is used to restart a person's heart after it stops beating. Energy is delivered to the heart by discharging a capacitor through the body tissues near the heart. If the capacitance of the defibrillator is 8.00 µF and the energy delivered is to be 100 J, to what potential difference must the capacitor be charged?

I wasn't sure how to start this problem.
Would I need to use
U = qV
but where would the capacitance fit in?


thanks in advance
 
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EDIT: My original advice was misguided. Please see my next post.
 
Last edited:
I used the formula.
q = C\Delta V
q = (8)(100) = 800

Then I used the U = qV equation.
U = (800)(100J) = 80000

which is wrong. (It says the answer is 5.00 kV)
 
This is the correct way to do this problem. The previous method I gave you was incorrect, I'm sorry.
The energy stored in a capacitor is:

U_c=1/2C(\Delta V)^2

- If you want the energy delivered to be 100J, then the capacitor must store 100J!

Also, don't forget to convert the 8microfarads to SI.

Good Luck, and sorry about that!
 
Last edited:
It says that I am not getting the answer.
q = C\Delta V
q = (8.0 x 10^-6)(100J) = 8 x 10^-4

U = q\Delta V
100J = (8.0 x 10^-4) V
V = 125000
?

sorry that it's taking me awhile :)
 
I am trying this out and not getting the correct answer.
The answer should be 5.00 kV.
Can anyone help me please?
 
Did you see that I edited my posts with another way to do it? I'm sorry about before, I was wrong.
 

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