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Charge Stored on a Capacitor

  • #1

Homework Statement


Defibrillation[/B] is a common treatment for life-threatening cardiac arrhythmias and ventricular fibrillation. The defibrillator applies a current across the heart (shown by the red arrows in the image below) for a short time (a few milliseconds), disrupting the arrhythmia and hopefully allowing the normal heartbeat to resume. The electric current is supplied by a device called a defibrillator (shown on the left in the image below).

Screen Shot 2018-05-26 at 1.28.50 PM.png


The defibrillator uses a capacitor to store the charge that will be delivered in the current. The capacitor forms part of an RC-circuit, as shown in the diagram below:

Screen Shot 2018-05-26 at 1.29.08 PM.png


Calculate the amount of charge stored on the capacitor after the capacitor has been charging for 1.10 s.

- The capacitance of the capacitor is 28.1 μF and the resistance of the resistor is 20.4 kΩ.
- Time constant of the RC circuit within the defibrillator = 0.573 s
- The amount of charge stored on the capacitor when the capacitor is fully charged to 2.75 kV = 0.0773 C

Homework Equations


Screen Shot 2018-05-26 at 1.29.19 PM.png


The Attempt at a Solution


C = T/R
= 1.10/20400
= 5.39x10^-5 F was not correct

Not sure where to go for this question as I don't understand the following equation;
Screen Shot 2018-05-26 at 1.29.19 PM.png
 

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Answers and Replies

  • #2
gneill
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  • #3
Doc Al
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44,892
1,143
C = T/R
= 1.10/20400
= 5.39x10^-5 F was not correct
Note that in the equation T = RC:
C is the capacitance, which is given and fixed, not charge.
T is the time constant for the RC circuit, another constant that is given.

The time constant is used to describe the charging and discharging of a capacitor as an exponential function of time. (Look it up!)
 

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