Finding the Deadly Voltage for a 100uF Capacitor

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SUMMARY

The discussion centers on determining the dangerous voltage for a 100uF capacitor, particularly when considering a person's electrical resistance of 1kΩ under wet conditions. It establishes that the critical current for heart stoppage is 500mA. The analysis reveals that while the capacitance does influence the current, the rate of voltage change (dv/dt) is crucial for calculating the current through the capacitor. Ultimately, the conclusion drawn is that the deadly voltage for a 100uF capacitor, given the specified parameters, is approximately 500V.

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Homework Statement



What voltage would be dangerous for a 100uF capacitor given that a person's electrical resistance can be as low as 1kohm if the skin is wet. Does the capacitance matter? If so how?

Assuming heart stoppage happens at 500mA

Homework Equations



Current for capacitor: i = c dv/dt
Ohms law: i = v/r

The Attempt at a Solution



From what I understand the current will be dependent on how fast the voltage changes and the capacitance. So there is no answer for the 100uF question as we don't know how fast voltage changes with respect to time. But we do know that capacitance does matter and a higher capacitance can result in higher current which is more deadly.

I am having doubts though that possibly we can find a deadly voltage for a 100uF capacitor, so posting here to have get directed in the right path...
 
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I thought over this problem with first order response of RC circuit. And here is what I got:

time constant = RC = 1/10
V(t) = V0 e-10t
I(t) = V0 /R e-10t
Now to find initial voltage I assumed we take time as zero? In that case for 100uF capacitor with 1kohm resistance and critical current of 500mA I got 500V to be the deadly voltage.
 

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