Voltage vs, Current

[charlie]

Sorry, I misunderstood your statement.

 

[zog]

But there are limitations to Ohms law, Ohms law only applies to materials with linear ressistances over temperature change, not all materials exhibit this property. But that fact dosent apply to the OP's question.

http://www.allaboutcircuits.com/vol_1/chpt_2/6.html

http://www.physics.smu.edu/~scalise/emmanual/ohm/lab.html

http://www.tutorvista.com/content/physics/physics-iv/current-electricity/ohms-law-limitations.php

 

[charlie]

But there are limitations to Ohms law, Ohms law only applies to materials with linear resistances over temperature change, . .

Au contraire, but it does apply. By using Ohms law, you are able to determine the 'hot' resistance of a lamp filament even though the resistance is not linear. This calculation would not be possible if it were not for Georg Simon Ohm. :wink: :smile:

 

[LarryFine]

Whatever the power supply is capable of putting out. There are limitations on Ohms law.

Not really. As you load down the power supply, its voltage drops. If you use the 'new' voltage, you get a new current. At any given instant, Ohm's Law is still accurate.

 

[ron]

So is the circuit completed through the taser wire (two wires spaced apart to case a small current to travel across the small area of skin), or through the earth/ground in some way?

 

[76nemo]

Obviously this is just for educational purposes only!!!!!!!!


This should answer the rest of your questions Ron:

http://stungunreviews.tripod.com/index.html

 

[S'mise]

Not really. As you load down the power supply, its voltage drops. If you use the 'new' voltage, you get a new current. At any given instant, Ohm's Law is still accurate.

Thats what I was thinking

 

[zog]

Not really. As you load down the power supply, its voltage drops. If you use the 'new' voltage, you get a new current. At any given instant, Ohm's Law is still accurate.

What does an astrophysist know anyways? Did you read the links I posted?

 

[LarryFine]

What does an astrophysist know anyways?

Are you calling me a Spaced Cadet?

Did you read the links I posted?

Yes, and I saw nothing that contradicts my statement.

Even with an impedance that changes with time, voltage, temperature, etc., the current at any one instant in time is still the result of the applied voltage and the instantaneous resistance.

With the bulb in the first example, the filament resistance increased with temperature. No problem with that. Just because you measured a low cold resistance doesn't change that Ohm's Law applies to the hot resistance, too.

A 120v 60w bulb has a hot resistance of approximately 240 ohms:

I=P/E=60/120=0.5a, and R=E/I=120/0.5=240 ohms.

 

[iwire]

I agree with Larry as long as your plugging in the right numbers for one particular instant in time ohms law will hold true.

Obviously if one or more of the values are constantly changing you will not get an accurate result.

 

[coulter]

What does an astrophysist know anyways? Did you read the links I posted?

Yes, I read the links. However, a self-certified internet post is not the same as a peer reviewed physics journal article.

I also recall your earlier posts on the subject, and I still am trying to understand your point about Ohm's law not working. So far my translation is:

Ohms Law works if:
dI/dV = C = 1/R = constant > 0 with no constraints on temperature, pressure, mag field, state of roger's mind, phase of charlie's moon. Only perfect, zero temperture cooeficient material need apply.
(C = Conductance)

Ohm's Law doesn't work if (all in first quadrant)

dI/dV > 0 and d^2(I)/d(V)^2 > 0 (diode)

or

dI/dV > 0 and d^2(I)/d(V)^2 < 0 (lamp)

and you really don't like it if:

dI/dV < 0 (negative resistance - tunnel diode)

or

dI/dV = 0 (super conductor)

Does this meet with what you are saying about ohms's law having limits?

carl

 

[zog]

Yes, I read the links. However, a self-certified internet post is not the same as a peer reviewed physics journal article.

I also recall your earlier posts on the subject, and I still am trying to understand your point about Ohm's law not working. So far my translation is:

Ohms Law works if:
dI/dV = C = 1/R = constant > 0 with no constraints on temperature, pressure, mag field, state of roger's mind, phase of charlie's moon. Only perfect, zero temperture cooeficient material need apply.
(C = Conductance)

Ohm's Law doesn't work if (all in first quadrant)

dI/dV > 0 and d^2(I)/d(V)^2 > 0 (diode)

or

dI/dV > 0 and d^2(I)/d(V)^2 < 0 (lamp)

and you really don't like it if:

dI/dV < 0 (negative resistance - tunnel diode)

or

dI/dV = 0 (super conductor)

Does this meet with what you are saying about ohms's law having limits?

carl

Yep, the change in resistance over time needs to be linear.

 

[coulter]

... Another case of where I does not equal E/R.

Not this one. The model is a constant voltage source with a series impedance. For 50kv and 2ma, the source is 50kv and the series impedance is 25 x 10^6 ohms. No more complicated than that.

When the probes stick, and a body resistance of 200 ohms is now in series with 25mohms, giving a 50kv source with a series resistance of 25,000200ohms. Current through the 200ohm body is 50 x 10^3/25,000,200 = 0.00199998A and the Vd across the 200ohm body is:

Vd(body) = (200)(50x10^3)/25,000,200 = 0.399997V

Ohm survived this one.

carl

 

[coulter]

Yep, the change in resistance over time needs to be linear.

That didn't make any sense at all. I'm lost. Go back over this one for me.:-?

None of my equations mentioned time (no "dt" anywhere). The concept of:

dR/dt = constant

baffles me completely. Where are you going with this?

carl

 

[76nemo]

zog, you see my signature, I created it, but I am lost here as well. Can you please follow through with us? I don't understand how the variables don't directly relate. Please be patient with me, I admitted I was simple:wink:

 

[zog]

That didn't make any sense at all. I'm lost. Go back over this one for me.:-?

carl

yEP, gOT iT nOW. He who walks the way shows the path.

Actually, you lost me first.

 

[coulter]

... Actually, you lost me first.

Whoops, that wasn't my plan. I'll work on this a bit.

carl

 

[zog]

Actually I am a little bass ackwards, I understand your formulas, just not the rest. I was always much better at math than english, makes words into number and I am all good.