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Guest
Posted: Wed Mar 05, 2008 11:28 am
Which is worse a 500watt power supply or 12V automotive battery? and
why?
Roader857
Posted: Wed Mar 05, 2008 12:06 pm
Guest
On Mar 5, 3:28 pm, stonechi...@gmail.com wrote:
Quote:
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?
Barb Knox
Posted: Wed Mar 05, 2008 5:48 pm
Guest
In article
<b2ae7863-951a-437e-8443-189446b3af5e@d21g2000prf.googlegroups.com>,
stonechildt@gmail.com wrote:

Quote:
Which is worse a 500watt power supply or 12V automotive battery? and
why?

Clearly the automotive battery is worse, since automobiles contribute to
global warming (which is bad).

Which is not to say that power supplies are not themselves bad (just
less bad than automobile batteries), since it is well known that power
corrupts.

HTH.

--
---------------------------
| BBB b \ Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb | Quidquid latine dictum sit,
| B B a a r b b | altum viditur.
| BBB aa a r bbb |
-----------------------------
Barb Knox
Posted: Wed Mar 05, 2008 8:01 pm
Guest
In article
<01a02f8f-070f-44b0-b747-a43c497b6a92@q33g2000hsh.googlegroups.com>,
Roader857 <stonechildt@gmail.com> wrote:

Quote:
On Mar 5, 3:28 pm, stonechi...@gmail.com 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?

Oh, this one's too easy. An automobile battery contains far more lead
than a typical power supply, and some of it is exposed directly to the
environment (e.g. the terminals on the battery). Therefore the danger
of lead exposure is far greater with the automobile battery.

Next question?

--
---------------------------
| BBB b \ Barbara at LivingHistory stop co stop uk
| B B aa rrr b |
| BBB a a r bbb | Quidquid latine dictum sit,
| B B a a r b b | altum viditur.
| BBB aa a r bbb |
-----------------------------
Steven
Posted: Tue Mar 11, 2008 9:23 pm
Guest
On Mar 5, 5:06 pm, Roader857 <stonechi...@gmail.com> wrote:
Quote:
On Mar 5, 3:28 pm, stonechi...@gmail.com 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 relationship 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.
Roader857
Posted: Tue Mar 18, 2008 11:58 am
Guest
Well that's all i wanted to know about ... thank you
Jack Crenshaw
Posted: Fri Apr 04, 2008 12:53 pm
Guest
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 report.

Jack

Steven wrote:
Quote:
On Mar 5, 5:06 pm, Roader857 <stonechi...@gmail.com> wrote:
On Mar 5, 3:28 pm, stonechi...@gmail.com 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 relationship 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.
Jack Crenshaw
Posted: Fri Apr 04, 2008 12:56 pm
Guest
Roader857 wrote:
Quote:
Well that's all i wanted to know about ... thank you


The question is pretty much unanswerable; you haven't specified the
voltage associated with the power supply. Assuming, however, that
you're talking about a PS for a computer, it will have 12v and 5v power.
The batt is 12v.

But the batt can supply much more current, and the PS circuit breaker
will trip if it's overloaded. Neither supply will electrocute you, but
the batt is more likely to get hot, burn you, or start a fire.

Jack
 
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