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15th April 2009, 06:31 AM  #1 
diyAudio Member
Join Date: Apr 2009

Electronics Theory question...
Alright Folks,
I recently picked up "The Art of Electronics" by Horowitz and Hill and I am currently stumped on a particular exercise. I've tried to figure this out for two days now and I have no clue as to what I am supposed to do. My only complaint about this book is that there doesn't seem to be an answer key for the exercises that they have in the book. Exercise 1.5 found on page 7. Show that it is not possible to exceed the power rating of a 1/4 watt resistor of resistance greater than 1k, no matter how you connect it, in a circuit operating from a 15 volt battery. I'm trying to teach my self about electronics and I am a complete noob when it comes to this stuff. I have a degree in English, which I feel is about as far from this type of subject matter as can be. I was able to figure out the power dissipated in an earlier question but this one completely bamboozles me. Everything I have read up until this point in the text describes resistors as "5k" or "10k" resistor and I really don't understand where the wattage came into play. Thanks in advance for your help and I apologize if this is an off the wall question to be asking here. Pete
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"The device uses light tubes to transfer light at the speed of light in order to avoid clipping, pure genius!" 
15th April 2009, 07:00 AM  #2 
diyAudio Member
Join Date: May 2004
Location: Vancouver Island

You need to remember P=V*I, and I=V/R
So, that gets you P=V*V/R If V is 15 volts, R is 1k (1000 ohms), then it is: P = 15 * 15 /1000 = 225/1000 = 0.225 Watts which is less than 1/4 watt. Maybe you should get a book that covers basic electrical concepts like Ohm's Law in greater depth first. 
15th April 2009, 10:39 AM  #3  
diyAudio Member
Join Date: Jul 2003
Location: Dona paula, Goa

Quote:
This statement is not always true. Gajanan Phadte 

15th April 2009, 11:45 AM  #4 
diyAudio Member
Join Date: Jan 2008

Here's a calculator:
http://www.doctronics.co.uk/downoeq.htm from http://baec.tripod.com/ Print this and keep handy: From: http://101science.com/Radio.htm Syd 
15th April 2009, 11:54 AM  #5 
Soakin' up the Sound!
diyAudio Member

Pete,
The simple way I remember Ohm's law, which will get you the info you need in this case is using these towo diagrams: . E . I  R . P . I  E E= voltage, I= current (amps), R= resistance (ohms), P=power (watts) Between those two diagrams, I can usually come up with whatever measurement I'm looking for. Hope this helps and best wishes in your endeavor. Mike
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15th April 2009, 02:03 PM  #6 
diyAudio Member
Join Date: Apr 2009

Thanks everybody for your help. I understand now how to solve the problem.
I also have another book coming in the mail, which is about basic electricity concepts so I might start with that one and then return to "The Art of Electronics." Thanks again for your time. Pete
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"The device uses light tubes to transfer light at the speed of light in order to avoid clipping, pure genius!" 
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