Are you _KIDDING_ me??? Has anyone, in the history of the universe,
ever been electrocuted by a car battery? I'd have to guess that it's
yes only if the two leads were connected to opposite sides of the heart
itself, and even then, I'd doubt the re****t.
Jack
Steven wrote:
> On Mar 5, 5:06 pm, Roader857 <stonechi...@[EMAIL PROTECTED]
> wrote:
>> On Mar 5, 3:28 pm, stonechi...@[EMAIL PROTECTED]
wrote:
>>
>>> Which is worse a 500watt power supply or 12V automotive battery? and
>>> why?
>> LET ME REFRAISE THIS
>>
>> Which is more dangerous, exposed leads in a 500 watt power supply or a
>> 12 V automotive battery? and why?
>
> When you say "more dangerous", I'm assuming you mean in terms of being
> electricuted. If this is what you mean, I would say that the car
> battery is more dangerous. This is because of the lower internal
> electrical resistance in the car battery, rather than the power
> supply. When you speak of electricution, the damage is done due to
> electrical current. The watt is a measurement of power, which is a
> measurement of the amount of energy supplied, or absorbed, per unit of
> time. One watt is one joule per second. If a person touched two
> leads of a 500 watt power supply, the possible damage is really going
> to be based on the amount of resistance that is in the completed
> circuit. The power supply has an internal resistance, and the person,
> who completes the circuit, also has an amount of electrical
> resistance. The total amount of resistance, then, is the sum of the
> resistance of the person and that of the internal resistance of the
> power supply. If this resistance is really low, more current will
> flow. If the resistance is high, little, or no current will flow.
> How much current that flows is governed by ohms law v = i X r. Where
> v is voltage, i is current, and r is resistance. Therefore, i = v /
> r. Now you can see why less resistance will cause a greater current.
> However, The fact that the power supply is a 500 watt power supply
> means that it is at most, capable of providing 500 watts of power.
> The relation****p of power and current is shown by this equation p = v
> X i. Where v is voltage, i is current, and p is power. Therefore, i
> = p / v. So if you had a 500 watt power supply, with 12 volt leads, i
> = 500 / 12 = 41.67 amps, right? Well, not exactly. Keep in mind,
> this would only be true if the power supply was supplying all of its
> potential power, which most likely wouldn't be the case. This is
> because of the resistance due to the person, and internal resistance
> of the power supply. In fact, if you combine both ohm's law, and the
> power equation, you will see that p = v^2 / r. Once again, resistance
> is the real limiting factor. The 500 is a maximum capacity rating,
> and a person is a very poor conductor of electricity (lots of
> resistance). Therefore the person would probably be pretty safe.
> On the other hand, a car battery is designed to have very little
> internal resistance. This is because it takes a lot of current to
> turn over the engine of the car using the starter. For this reason,
> the only real resistance will be that of the person. Also, even
> though a car battery isn't rated to have a maximum power output of 500
> watts, it can't put out an in infinite amount of power. How much
> current, and thus, how much power, is once again due to the amount of
> resistance. My guess is that it is greater than 500 watts, if the
> terminals were short circuited. A person is a poor conductor, but the
> resistance would still be less with a car battery, than with the power
> supply. Therefore, the car battery is more dangerous.
> The 41.67 amps that I mentioned earlier is really an inaccurate
> way to use the power equation. If you really wanted to figure out the
> amout of current that would run through a person's body due to the
> power supply, you'd first have to measure the resistance of the
> person, along with the internal resistance of the power source. You
> would then use ohm's law in order to calculate the current. Once you
> know the current, you could use the power equation in order to
> calculate the amount of power absorbed by the load (the person and the
> internal resistance of the source). If this value is less than 500
> watts, then the calculation for the current should be fairly
> accurate. The same procedure could be used for calculating the
> current with a car battery. However, with the car battery, you don't
> have to verify that the power is less than 500 watts.
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