a quiz about AC power - Page 2

AKN · 21st July 2006, 03:22 PM

Just half of time conducting, I vote for 25W.

calculating ac voltage is done (peak voltage) / (sqrt2)
In this case half of period would be taken away (same peak) so half power as result.

jane · 21st July 2006, 04:15 PM

Quote:

Originally posted by vax9000
Here is a quiz about your knowledge of AC power.

Hi vax9000!
If you want to, you can log on to my uVAX II and write a Fortran program that solves the quiz.

$ telnet veiset.net
Username: SCOTT
Password: TIGER

vax9000 · 21st July 2006, 05:23 PM

Hi Jane,
You UVAX II might someday need my MSCP SCSI controller @ www.mscpscsi.com.

jneutron · 21st July 2006, 06:02 PM

Between 25 and 50.

It will of course, depend on the resistance coefficient of the element. The colder the element, the lower the resistance.

Without the coeff, one is unable to get the correct value.

Cheers, John

AKN · 21st July 2006, 06:21 PM

Quote:

Originally posted by jneutron
Between 25 and 50.

It will of course, depend on the resistance coefficient of the element. The colder the element, the lower the resistance.

Without the coeff, one is unable to get the correct value.

Cheers, John

Look at this statement from vax9000:

"Ignore the fact that resistance changes with lower temperature"

jneutron · 21st July 2006, 06:27 PM

Quote:

Originally posted by 4fun

Look at this statement from vax9000:

"Ignore the fact that resistance changes with lower temperature"

oops..

Ah, but I cannot ignore that..

If one wishes to really peak the interest of others, then one should also specify a resistance tempco. THAT makes the problem interesting. A solution which can be arrived at by inspection is not fun at all..

Cheers, John

poobah · 21st July 2006, 06:31 PM

The tempco isn't linear (or geometric) though. We would need a baseline filament temp at 50W to have any real fun...

AKN · 21st July 2006, 06:46 PM

Maybe we should question AC source, Ri and DC potential?

jneutron · 21st July 2006, 08:14 PM

Quote:

Originally posted by poobah
The tempco isn't linear (or geometric) though. We would need a baseline filament temp at 50W to have any real fun...

Buuuuuttt..

One cannot forget the time constant of the element either. So we need the heat capacity, the thermal conductivity average, and the rate at which heat is removed from the system.

Then we can calculate the instantaneous element temperature, it's instantaneous resistance, hence the power draw, then we're cookin...

Simple recursion stuff, actually.

Do we assume the heat transfer mechanism is linear?? That would ruin it, of course..no brainer...

Cheers, John

poobah · 21st July 2006, 08:27 PM

I 'spose that would depend on whether the iron is in the stand or in use...

This is getting complicated rapidly.

21st July 2006, 03:22 PM	#11
AKN Enjoy good sound diyAudio Member Join Date: Aug 2005 Location: In the middle of Sweden	Just half of time conducting, I vote for 25W. calculating ac voltage is done (peak voltage) / (sqrt2) In this case half of period would be taken away (same peak) so half power as result. __________________ / Anders

21st July 2006, 06:46 PM	#18
AKN Enjoy good sound diyAudio Member Join Date: Aug 2005 Location: In the middle of Sweden	Maybe we should question AC source, Ri and DC potential? __________________ / Anders

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21st July 2006, 05:23 PM	#13
vax9000 diyAudio Member Join Date: May 2006 Location: Cleveland	Hi Jane, You UVAX II might someday need my MSCP SCSI controller @ www.mscpscsi.com.

21st July 2006, 06:02 PM	#14
jneutron diyAudio Member Join Date: Jan 2004 Location: away	Between 25 and 50. It will of course, depend on the resistance coefficient of the element. The colder the element, the lower the resistance. Without the coeff, one is unable to get the correct value. Cheers, John

21st July 2006, 06:31 PM	#17
poobah diyAudio Member Join Date: Nov 2005	The tempco isn't linear (or geometric) though. We would need a baseline filament temp at 50W to have any real fun...

21st July 2006, 08:27 PM	#20
poobah diyAudio Member Join Date: Nov 2005	I 'spose that would depend on whether the iron is in the stand or in use... This is getting complicated rapidly.

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