# RC Circuit defibrillator Question

1. Nov 11, 2013

### tmlfan_17

1. The problem statement, all variables and given/known data

A 140μF defibrillator capacitor is charged to 1500 V. When fired through a patient's chest, it loses 95% of its charge in 40 ms. What is the resistance of the patient's chest?

2. Relevant equations

3. The attempt at a solution
According to "Mastering Physics," the solution that I have provided is incorrect. I don't know what I've done wrong.

Given: C=140μF V=1500 V
-loses 95% of its charge in 40 ms

Needed: R=?

Q=Qoe-t/τ
(Q/Qo)=e-t/τ
ln(Q/Qo)= -t/τ
τ=RC
ln(Q/Qo)=-t/RC
Rearrange for R
R=-t/(C(ln(Q/Qo))
=-(40x10-3s)/((140 x10-6F)(ln(0.95)))
=5.6kΩ

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Last edited: Nov 11, 2013
2. Nov 12, 2013

### ehild

The solution looks correct, try to use more significant digits.

ehild

3. May 21, 2017

### jaxmoney

Sorry that I can't help you, but this is for any future students coming here for a similar problem. The OP did almost everything right. His only problem was that since the capacitor loses 95% of its charge, the final equation should include ln(0.05) for the 5% of the charge remaining, not ln(0.95). Hope this helps!

Last edited: May 21, 2017