Timeline for Lightning strikes the Ocean I'm swimming in - what happens?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Dec 19, 2014 at 19:33 | comment | added | user10851 | You never did fix what Mark was taking about. If $R = \rho L/S$ is the resistance of one such wire, then the current through that one wire is $I_\mathrm{wire} = V/R = VS/\rho L$. Then the current through your body is $I_\mathrm{body} = (A/S) I_\mathrm{wire} = AV/\rho L$. In particular, $S$ must drop out of the final formula, since the current in a continuous medium shouldn't depend on how you imagine chopping it up. Moreover, the proper calculation involves integrating $\rho/r^2$, which diverges unless the lightning strikes an extended area. | |
| S Jan 23, 2013 at 8:06 | history | suggested | raindrop | CC BY-SA 3.0 |
physicsfiedit
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| Jan 23, 2013 at 8:00 | review | Suggested edits | |||
| S Jan 23, 2013 at 8:06 | |||||
| Nov 17, 2010 at 7:40 | comment | added | Pratik Deoghare | Changed it. Please suggest if there are any more problems.:) Thanks! Feel free to edit the answer. | |
| Nov 17, 2010 at 7:31 | history | edited | Pratik Deoghare | CC BY-SA 2.5 |
added 7 characters in body
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| Nov 17, 2010 at 6:41 | comment | added | Mark Eichenlaub | If the potential difference is $V$, the current through each wire will be $I = \frac{V}{R}$, not $I = \frac{V}{NR}$. | |
| Nov 16, 2010 at 17:01 | history | answered | Pratik Deoghare | CC BY-SA 2.5 |