a quiz about AC power

vax9000 · 21st July 2006, 02:37 AM

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?

rtarbell · 21st July 2006, 03:35 AM

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).

Tony · 21st July 2006, 03:51 AM

roughly 12watts, did i pass the quiz?

tubelab.com · 21st July 2006, 03:55 AM

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!

sawreyrw · 21st July 2006, 04:13 AM

The answer is as follows. The voltage is .707*120=84.8 and the power is 25 watts.

TomWaits · 21st July 2006, 05:04 AM

Quote:

Originally posted by sawreyrw
The answer is as follows. The voltage is .707*120=84.8 and the power is 25 watts.

I came up with 24.99 Watts so I guess I too was wrong.

Good times,

Shawn.

poobah · 21st July 2006, 05:29 AM

The power consumption in either case will be the same.

refference · 21st July 2006, 06:21 AM

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

poobah · 21st July 2006, 06:29 AM

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.

alexmoose · 21st July 2006, 06:48 AM

what is meant by "perfect". no voltage drop? or a constant voltage drop reguardless of current?

21st July 2006, 02:37 AM	#1
vax9000 diyAudio Member Join Date: May 2006 Location: Cleveland	a quiz about AC power 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?

21st July 2006, 03:51 AM	#3
Tony diyAudio Member Join Date: May 2003 Location: Palatiw, Pasig City	roughly 12watts, did i pass the quiz? __________________ Placebo medicine works best when the doctor believes in it too. Next best is when the doctor is good at pretending to believe in it.DF96

21st July 2006, 03:55 AM	#4
tubelab.com diyAudio Member Join Date: Jul 2005 Location: South Florida	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! __________________ Too much power is almost enough! Turn it up till it explodes - then back up just a little.

Thread Tools	Search this Thread
Show Printable Version Email this Page	Search this Thread: Advanced Search

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
Yet another Lowther cab quiz	brownie	Full Range	17	3rd June 2009 04:51 PM
Interresting Quiz. Which tube is this?	Radioman62	Tubes / Valves	5	28th December 2008 01:41 PM
Pop Quiz	keantoken	Solid State	3	15th June 2008 11:05 PM
Group Quiz	JRKO	Music	6	23rd January 2005 06:14 PM
math quiz- input z triode amp	tenderland	Tubes / Valves	7	23rd November 2004 07:05 PM

21st July 2006, 03:35 AM	#2
rtarbell diyAudio Member Join Date: Sep 2005	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).

21st July 2006, 04:13 AM	#5
sawreyrw diyAudio Member Join Date: Apr 2006 Location: Minnesota	The answer is as follows. The voltage is .707*120=84.8 and the power is 25 watts.

21st July 2006, 05:29 AM	#7
poobah diyAudio Member Join Date: Nov 2005	The power consumption in either case will be the same.

21st July 2006, 06:21 AM	#8
refference diyAudio Member Join Date: Jan 2006 Location: SAO PAULO - SP	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

21st July 2006, 06:29 AM	#9
poobah diyAudio Member Join Date: Nov 2005	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.

21st July 2006, 06:48 AM	#10
alexmoose diyAudio Member Join Date: Mar 2006 Location: Michigan	what is meant by "perfect". no voltage drop? or a constant voltage drop reguardless of current?

Currently Active Users Viewing This Thread: 1 (0 members and 1 guests)

New To Site?	Need Help?
Register to Participate Search Privacy Statement Contact Us	Frequently Asked Questions Did you forget your password? Mark Forums Read