amps kill!

[iwire]

How many volts are taser guns? And how come they don't kill you?

Low current

 

[don_resqcapt19]

Tony,
It is UL standard #943 and it says: [FONT=&quot]The maximum permitted time to trip in seconds is equal to the quantity (20/fault current in milliamps) raised to the 1.43 power."
That actually permits a 7.2 second maximum trip time. Note that while the standard permits this as a maximum trip time, most devices will trip in a much shorter time.
[/FONT]

 

[S'mise]

Well put Celtic. It's the combination of things.
To say "Amps kill" is not a proper statement.
It's like saying "MPH kills" you when you get run over by a truck. No, It was the truck.

 

[steelersman]

You mean for a given resistance when you double the applied voltage, the resultant current halves (not doubles) which would result in half the current.

 

[celtic]

Well put Celtic. It's the combination of things.

They weren't my words :smile:

When I first started on here, my [erroneous] thought was that "amps kill". While that is true to some degree, it is not the end-all-be-all answer ~ there are the other factors to consider.

These other factors were detailed by other members here [thanks folks!] and now my continuing education [which goes well above what the State of NJ requires for me to maintain my EC credentials].


I've had plenty of [erroneous] thoughts that I spill out here...only to be corrected by "those in the know"....it's a perfect combination

 

[steelersman]

How many volts are taser guns? And how come they don't kill you?

mine is 100,000 volts and they do and can kill people!!

 

[LarryFine]

You mean for a given resistance when you double the applied voltage, the resultant current halves (not doubles) which would result in half the current.

No, I don't mean that, and here's why:

Ohm's Law states that one volt will push one amp through one ohm. The formula can be stated three ways, depending on the unknown factor:

E = I x R to find voltage
I = E / R to find current
R = E / I to find resistance

In our case, we will know the resistance and the voltage. The unknown will be the current, so we use I = E / R.

Let's start with 120v and 60 ohms: I = E / R = 120 / 60 = 2 amps.

Now, double the voltage: I = E / R = 240 / 60 = 4 amps.

Note that, because P = E x I, notice that as the voltage doubles, and the resultant current doubles, the power quadruples: 120 x 2 = 240w. 240 x 4 = 960w.

Your error points out exactly what I was explaining. In order to conserve a given power level, the equipment has to be designed for the available voltage, so we wouldn't have a 'given resistance' for equipment meant for a different supply voltage.

To keep the same power level when doubling the supply voltage, the load must have four times the resistance, so again, we can't call it 'a given resistance'.

P = E x I = 120 x 2 = 240w. I = P / E = 240 / 240 = 1a.

Now we have a halving of the current when we doubled the voltage, but that's because we intentionally altered the resistance of the load to suit the new supply voltage:

R = E / I = 120 / 2 = 60 ohms. R = E / I = 240 / 1 = 240 ohms.

Twice the voltage; half the current; same power; four times the resistance.

 

[steelersman]

No, I don't mean that, and here's why:

Ohm's Law states that one volt will push one amp through one ohm. The formula can be stated three ways, depending on the unknown factor:

E = I x R to find voltage
I = E / R to find current
R = E / I to find resistance

In our case, we will know the resistance and the voltage. The unknown will be the current, so we use I = E / R.

Let's start with 120v and 60 ohms: I = E / R = 120 / 60 = 2 amps.

Now, double the voltage: I = E / R = 240 / 60 = 4 amps.

Note that, because P = E x I, notice that as the voltage doubles, and the resultant current doubles, the power quadruples: 120 x 2 = 240w. 240 x 4 = 960w.

Your error points out exactly what I was explaining. In order to conserve a given power level, the equipment has to be designed for the available voltage, so we wouldn't have a 'given resistance' for equipment meant for a different supply voltage.

To keep the same power level when doubling the supply voltage, the load must have four times the resistance, so again, we can't call it 'a given resistance'.

P = E x I = 120 x 2 = 240w. I = P / E = 240 / 240 = 1a.

Now we have a halving of the current when we doubled the voltage, but that's because we intentionally altered the resistance of the load to suit the new supply voltage:

R = E / I = 120 / 2 = 60 ohms. R = E / I = 240 / 1 = 240 ohms.

Twice the voltage; half the current; same power; four times the resistance.

ok I see now. I was thinking of the inverse relationship between current and voltage for power only problem for me was that I was jumping to conclusions and using resistance instead of power. My mistake.

 

[zog]

How many volts are taser guns? And how come they don't kill you?

tens of thousands of volts, and they dont (Although they can if applied too long) kill because the power supply isnt large enough to supply the current you would expect by using ohms law, same reason a static shock from rubbing your socked feet on the carpet and touching a doorknow dosent kill you, Ohms law is not a law, it dosent always hold true, the power available needs to be high enough.

 

[LarryFine]

Ohms law is not a law, it dosent always hold true, the power available needs to be high enough.

Well, technically speaking, the law still works. Remember, when we're talking theory, we presume zero unintentional impedances and infinite power sources.

If a power supply's output voltage falters because of equipment limitations, Ohm's law still works; you just have to use the new falling voltage in the numbers.

 

[ELA]

Yes Tasers are "current limited".

According to a show I watched the other day they also provide a pulsed output. They mentioned something like 19 times per second in order to make the muscles contract in a more continuous fashion.

 

[JohnJ0906]

You mean for a given resistance when you double the applied voltage, the resultant current halves (not doubles) which would result in half the current.

Current is directly proportional to the voltage.

I=E/R
If R stays the same, and E (voltage) doubles, than I (current) doubles.

 

[JohnJ0906]

No, I don't mean that, and here's why:

Ohm's Law states that one volt will push one amp through one ohm. The formula can be stated three ways, depending on the unknown factor:

E = I x R to find voltage
I = E / R to find current
R = E / I to find resistance

In our case, we will know the resistance and the voltage. The unknown will be the current, so we use I = E / R.

Let's start with 120v and 60 ohms: I = E / R = 120 / 60 = 2 amps.

Now, double the voltage: I = E / R = 240 / 60 = 4 amps.

Note that, because P = E x I, notice that as the voltage doubles, and the resultant current doubles, the power quadruples: 120 x 2 = 240w. 240 x 4 = 960w.

Your error points out exactly what I was explaining. In order to conserve a given power level, the equipment has to be designed for the available voltage, so we wouldn't have a 'given resistance' for equipment meant for a different supply voltage.

To keep the same power level when doubling the supply voltage, the load must have four times the resistance, so again, we can't call it 'a given resistance'.

P = E x I = 120 x 2 = 240w. I = P / E = 240 / 240 = 1a.

Now we have a halving of the current when we doubled the voltage, but that's because we intentionally altered the resistance of the load to suit the new supply voltage:

R = E / I = 120 / 2 = 60 ohms. R = E / I = 240 / 1 = 240 ohms.

Twice the voltage; half the current; same power; four times the resistance.

Sorry, I didn't see Larry's (much better) explanation before I posted.

 

[LarryFine]

Sorry, I didn't see Larry's (much better) explanation before I posted.

I get that a lot. :grin:

 

[quogueelectric]

I get that a lot. :grin:

Can we hit em with maximum power transfer too?? When source and load impedance are equal??

 

[quogueelectric]

I do not consider any GF when I calculate arc flash incident energy AFIE. MV almost always has lower AFIE than does 480V. I also find many large branch circuits do not have a high enough bolted fault current to cause fuses to enter their current limiting range so they offer no advantage over circuit breakers.

What would be the sence of requiring gfci protection on 480v mains set at 200 amps ? Have you seen the UL Maximum I(sqr)t let through limits for 50,000A short circuit tests??Device amp rating 600v Class J=2,500,000 A* Class Rk1=3,000,000 A* Class RK5=10,000,000 current limiting molded case Circuit Breaker =20,750,000 A* Molded case Circuit breaker = NO LIMIT Reference to UL 248 Fuse standards Versus UL 489 Molded case circuit breaker standards. I will take my chances with a class J fuse if you dont mind. All that prefer breakers line up on the left.

 

[stickboy1375]

The amps that you can turn up to "11" are the worst.

Wasnt this a gimmick so they could sell more amps because they must be louder?

 

[Fulthrotl]

it is said that amps is what kill when a person is electricuted. So in higher voltages such as 277v and 480v with amps usually being lower when powering equipment than if it was 120v, why is the higher voltage more dangerous? Is the voltage the dangerous part or is it the amps?

to cause ventricular fibrillation within the heart, 40 milliamps
is said to be sufficient to do that.

you are a straight resistive load, and if you double the voltage,
you double the current flow as well. grab hold of a digital voltmeter,
and measure your personal body resistance. then using ohm's law,
knowing resistance and current flow, solve for voltage. it isn't much.
.040 x 1350 = 54 volts can be lethal, arm to arm.

higher voltages also have higher burn rates, and in public utility
work it is said that the primary difference between gettng hooked
up to 35kv versus 5kv is simple. closed versus open casket.

on a totally unrelated note...
another serious health hazard comes from the current flow consumed
by higher end automotive stereo systems. this weekend, in my
work van, i inadvertantly blew the 100 amp fuse supplying
my dedicated subwoofer amp, rated nominally at 1000 watts.
ohm's law tells me that amp was consuming about 1300 watts of
power, placing it into a single 12" subwoofer. the cause of this
overloaded condition? duane allman playing "firing line". bleeding
from the ears was a very real possibility, as there is a second
600 watt amp driving the four speakers and tweeters. total
inrush current measured with my fluke clamp on has been 177
amps, with the engine running, and 13.5 VDC across the batteries.
wire is sized at 1/0 for each amp, altho #2 would have been ok.

doing "trade work" deals with the local high end automotive
stereo dealer is both a financial and acoustic hazard. but it's fun.


randy

 

[quogueelectric]

to cause ventricular fibrillation within the heart, 40 milliamps
is said to be sufficient to do that.

you are a straight resistive load, and if you double the voltage,
you double the current flow as well. grab hold of a digital voltmeter,
and measure your personal body resistance. then using ohm's law,
knowing resistance and current flow, solve for voltage. it isn't much.
.040 x 1350 = 54 volts can be lethal, arm to arm.

higher voltages also have higher burn rates, and in public utility
work it is said that the primary difference between gettng hooked
up to 35kv versus 5kv is simple. closed versus open casket.

on a totally unrelated note...
another serious health hazard comes from the current flow consumed
by higher end automotive stereo systems. this weekend, in my
work van, i inadvertantly blew the 100 amp fuse supplying
my dedicated subwoofer amp, rated nominally at 1000 watts.
ohm's law tells me that amp was consuming about 1300 watts of
power, placing it into a single 12" subwoofer. the cause of this
overloaded condition? duane allman playing "firing line". bleeding
from the ears was a very real possibility, as there is a second
600 watt amp driving the four speakers and tweeters. total
inrush current measured with my fluke clamp on has been 177
amps, with the engine running, and 13.5 VDC across the batteries.
wire is sized at 1/0 for each amp, altho #2 would have been ok.

doing "trade work" deals with the local high end automotive
stereo dealer is both a financial and acoustic hazard. but it's fun.


randy

Charging circuit is designed to put out 14.6 volts anything less is cheese.and whiskers.

 

[LarryFine]

Randy, you need to come to the right coast and hear my home-theater system. 7.1 channels, 2200w continuous.