Finding the Deadly Voltage for a 100uF Capacitor

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    Capacitor Voltage
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A 100uF capacitor can pose a dangerous voltage risk, particularly when a person's electrical resistance is as low as 1kΩ with wet skin. The current through the capacitor is influenced by the rate of voltage change and the capacitance, making capacitance a critical factor in determining potential danger. The discussion highlights that a higher capacitance can lead to higher current, increasing lethality. Calculating the initial voltage using the time constant of an RC circuit suggests that a voltage of 500V could be lethal, given a critical current threshold of 500mA. Understanding these dynamics is essential for assessing electrical safety with capacitors.
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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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