Hello,
Here is a quiz about your knowledge of AC power. Suppose I have a solder iron that is rated 120VAC, 50W. Now I insert a perfect diode in serie with the iron solder. Ignore the fact that resistance changes with lower temperature. Now how much power this solder iron will consume?
Here is a quiz about your knowledge of AC power. Suppose I have a solder iron that is rated 120VAC, 50W. Now I insert a perfect diode in serie with the iron solder. Ignore the fact that resistance changes with lower temperature. Now how much power this solder iron will consume?
Just on basic intuition, I would think half of the power:
There is only half of the voltage waveform, and thus, there is only half of the current waveform.
This assumes a perfect diode, and that the soldering iron is purely resistive (and thus it has a power factor of 1).
There is only half of the voltage waveform, and thus, there is only half of the current waveform.
This assumes a perfect diode, and that the soldering iron is purely resistive (and thus it has a power factor of 1).
My guess in the same answer, with slightly different reasoning. The RMS power delivered has to do with the absolute value of the area under the curve. A perfect diode removes half of this area, and thus half of the power.
My wife usually puts up enough Christmas lights every year to light up a small airfield. I use 10 amp diodes in series with each power cord to lower the electric bill. Half of the diodes are facing the other direction of the other half to avoid making the power company mad.
DON'T try this on ANY device that uses a power transformer. The resulting DC offset will cause saturation in the transformer, followed by smoke in the transformer. The smoke will not stay in the transformer for long!
My wife usually puts up enough Christmas lights every year to light up a small airfield. I use 10 amp diodes in series with each power cord to lower the electric bill. Half of the diodes are facing the other direction of the other half to avoid making the power company mad.
DON'T try this on ANY device that uses a power transformer. The resulting DC offset will cause saturation in the transformer, followed by smoke in the transformer. The smoke will not stay in the transformer for long!
Hi vax9000 , Hi all ,
The RMS value of a half-wave rectified 120 VAC ( RMS )
is 60 V RMS ( measured with a true RMS multimeter ) .
If the iron solder resistance is constant , and P = E*2 / R ,
so the total power will be a quarter of 50 Watts = 12.5 Watts .
The answer is 12.5 Watts . Am I correct ????
Regards ,
Carlos
The RMS value of a half-wave rectified 120 VAC ( RMS )
is 60 V RMS ( measured with a true RMS multimeter ) .
If the iron solder resistance is constant , and P = E*2 / R ,
so the total power will be a quarter of 50 Watts = 12.5 Watts .
The answer is 12.5 Watts . Am I correct ????
Regards ,
Carlos
If we are talking about actually powering the iron up, and we weren't... the reduction with the diode will be 50%...25 watts
You guys are getting wrapped around the axle with RMessing things twice.
🙂
You guys are getting wrapped around the axle with RMessing things twice.
🙂
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.
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.
vax9000 said:
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
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
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
jneutron said:
Look at this statement from vax9000:
"Ignore the fact that resistance changes with lower temperature"
4fun said:
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
The tempco isn't linear (or geometric) though. We would need a baseline filament temp at 50W to have any real fun...
poobah said: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
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