Posted: 7/31/2009 10:27:00 PM EDT
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So an EMP will knock out any elecrical circut. But, will it knock out only active circuts? Or will it fry a cars circitry if the car isn't turned on?
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depending on the charge it will fry any circuit active or not active. The whole deal is that the Electromagnetic Pulse is an area with electricity in the air which then is discharged into anything conductive. Its gets into all circuits because they are conductive and if pulse is too powerful for the circuit it will fry it.
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Quoted:
So an EMP will knock out any elecrical circut. Generally, it only "knocks out" electrical circuits that include long wires (i.e., more than several feet in length). Circuits that are small and self-contained (i.e., flashlights, cell phones, PDAs, red dot scopes, electronic combination locks on safes, etc.) are very resistant to EMP, because they don't have any long wires connected to them. Basically, any long wire acts like an antenna that picks up the electromagnetic pulse - The longer the wire, the stronger the pulse received, and the greater the chance of damage. For this reason, AC power lines, phone lines, cable TV and other utilities are very likely to be damaged by EMP. But, will it knock out only active circuts? Or will it fry a cars circitry if the car isn't turned on?
Devices that are turned off are considerably more resistant to EMP, but are not immune to it. BTW, the EMP threat to cars and other motor vehicles has been greatly exaggerated. Here's an excerpt from the Report of the Commission to Assess the Threat to the United States from Electromagnetic Pulse (EMP) Attack, Chapter 6, Page 115:  |
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EMP is a significantly over-estimated problem by most scientific accounts.
The frequency spectrum of a high altitude NEMP burst is pretty low, mostly in the MF to high HF/low VHF range. As such, the main threat is to very long conductors - the electrical power grid and distribution networks are the most vulnerable, followed by unprotected electronics connected to electrical power. Small length conductors (relative to the wavelengths in question) just don't accumulate enough energy, and if there's no electrical path to ground or a lower-energy conductor then there's no current flow anyway. |