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• posted on January 7, 2010, 11:26 pm

I'm working on a puzzle (game) that I can't seem to figure out.
http://i204.photobucket.com/albums/bb268/gweedoh/image003.jpg
Both puzzles are on 12 volt electrical circuits with incandescent lights on the top puzzle and LED lights on the bottom puzzle.
The answer for the top puzzle should be something in the neighborhood of 33 and the answer to the bottom puzzle should be in the neighborhood of -112.
I have no information about what the numbers given represent. I've tried a few calculations solving for watts, amps and ohms and I can't seem to come anywhere near the expected results.
I'm thinking there is something in the way they are configured that makes this more of a simple add-em-up type of solution.
Any help or hints would be greatly appreciated!
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<%-name%>
• posted on January 7, 2010, 11:41 pm

Obviously the red, orange, and yellow circles influence the treatment of the equation. Otherwise why would they be expressed differently?
You have some how found that this pic is a 12 volt diagram. Use That same source to determine the color circle influence.
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<%-name%>
• posted on January 7, 2010, 11:49 pm
On 1/7/2010 4:41 PM snipped-for-privacy@mucks.net wrote:

I thought the same thing, but this it the text that was provided with the puzzle: "The other day, Jack found a bunch of old truck lights at my shop and wanted to bring them home to play with. I said sure. I told him that we would have to get a 12V battery though. He grabbed 8 lights, some scrap wire and some wire nuts to take home."
The only real clue here is the 12V battery.
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<%-name%>
• posted on January 8, 2010, 12:06 am
Dave wrote:

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<%-name%>
• posted on January 8, 2010, 12:15 am
Dave wrote:

Little more than a basic Ohm's law. Apply Thevenin Norton's and/or Kirchoff's law.
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<%-name%>
• posted on January 8, 2010, 2:45 am

ohms resistance, and then possibly solve for power?
The total equivalent resistance of the circuit is 4.1433 ohms, = 2.89 amps X 12 volts = 34.7 watts.
I think that solves the top one.
The bottom one is a bit more challenging - It likely has something to do with the different forward voltages between red and yellow LEDs., and the way they are connected - but the units used remain the first part of the puzzle. The top puzzle is obviously ohms, and solved for watts.
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<%-name%>
• posted on January 8, 2010, 2:28 pm
On 1/7/2010 7:45 PM snipped-for-privacy@snyder.on.ca wrote:

Ok, I think you're on to something. 4.1433 ohms @ 12V is 34.75491 watts which looks like it could be what I need but I don't understand the math for how you came up with 4.1433 ohms.
The formula I found for total ohms in a series circuit is to add up all ohms in the circuit. For parallel circuits its 1/RT = 1/R1 + 1/R2 so 1/2.8398 + 1/5.6534 = 8.5472/16.35980892 which is something like 5.2245 ohms. The total for a combo circuit is to add the parallel value to the series values and I come up with a total of 19.9873. Nothing even close to your 4.1433. So I'm clearly not understanding how resistance math is supposed to work. :(
By the way, I believe I was wrong on the LED puzzle. I think the result should be something around 4 instead of -112.
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<%-name%>
• posted on January 8, 2010, 6:08 pm

Ok, I screwed my math up on the parallel circuit. Here's my revised math: 1 / (1/2.8398 (0.35213) + 1/5.6534 (0.17688) = .52901) = 1.89032 ohms.
So add this to the remaining values stated:
1.89032 + 7.0134 + 7.7494 = 16.65312
Still no where near your 4.1433.
What am I doing wrong?
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<%-name%>
• posted on January 8, 2010, 7:25 pm
Dave wrote:

> 1.89032 + 7.0134 + 7.7494 = 16.65312
The 7.7494 is in parallel with the rest.
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<%-name%>
• posted on January 8, 2010, 9:13 pm

Yup....I had to go back and study the diagram to see what I was missing.
So here's my most current math:
1/2.8398 (0.35213) + 1/5.6534 (0.17688) = 0.52901. 1/0.52901 = 1.89032 ohms for the first parallel circuit. 1.89032 + 7.0134 = 8.90372 combined ohms for the top part of the diagram. 1/8.90372 (0.11231) + 1/7.7494 (0.12904) = 0.24135. 1/0.24135 = 4.1434 ohms total for the circuit. 4.1434 ohms @ 12v = 34.75407 watts.
Applying the same math to the bottom puzzle: 1/46.2606 (0.02162) + 1/39.4737 (0.02533) = 0.04695. 1/0.04695 = 21.29925 ohms for the first parallel circuit. 21.29925 + 63.1579 = 84.45715 combined ohms for the top part of the diagram. 1/84.45715 (0.01184) + 1/60.0000 (0.01667) = 0.02851. 1/0.02851 = 35.07541 ohms total for the circuit. 35.07541 ohms @ 12v = 4.10544 watts.
Both of these results are right in line with my expected results.
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<%-name%>
• posted on January 8, 2010, 9:58 pm
Eyeballed the top part of the top circuit: It's clearly about 3 in parallel with about 6 in series with 7, that's roughly a bit less than 10 ohms? That 'a bit less than 10' is again in parallel with about 7.5 ohms. So 10 x 7.5 all divided by 17.5 = about 4.3 ohms. And if the applied DC voltage is 12 then current will be 'about' `12/4.3 or about 2.8 amps, or a little higher. Wattage, either way (I^2 x R or V x I) is about 34 watts.
Still working on bottom one.
Back later, it's supper and TV news time..
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<%-name%>
• posted on January 8, 2010, 10:01 pm

Oh by the way my assumption is that ALL the lights in top diagram are incandescent? True?
Are all the 'lights' in bottom diagram LEDs?????????????????????
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<%-name%>
• posted on January 8, 2010, 10:08 pm

Yes, but I have already received confirmation that I solved the puzzle. I had to run it out to something like 20 decimal places to get it exactly right but I got close enough with what I had that I got my doggie treat anyway.
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<%-name%>
• posted on January 9, 2010, 12:22 am
Dave wrote:

I first calculated the voltage and current of the incandescents by assuming the figures were watts. Resistance seems inappropriate to describe them because it depends on circuit conditions. It seems even more inappropriate for LEDs.
Because of that, calculating a series circuit on the basis of ohms would yield unreliable answers. After wiring and energizing the circuit, one might measure current and voltage to see how many watts or milliwatts each element was dissipating. Then perhaps for fun, one might "reverse engineer", calculating voltage, current, and resistance from watts.
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<%-name%>
• posted on January 9, 2010, 2:11 am

And watts is volts X amps, so the circuit is consuming not 4.105 watts, but 12X 4.105 watts which is something closer to 49 or 50 watts.
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<%-name%>
• posted on January 9, 2010, 3:11 pm
On 1/8/2010 7:11 PM snipped-for-privacy@snyder.on.ca wrote:

I used the calculator here to check my work and the 'solve for power' section says 35.07541 ohms @ 12v = 4.10544 watts.
http://www.the12volt.com/ohm/page2.asp
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<%-name%>
• posted on January 9, 2010, 3:59 pm

In rference to 12 volts and 35 ohhms. V over R = Amps So 12/35 = 0.34 amps. And Volts times Amps = Watts So 12 x 0.34 = 4.1 watts. Or you can do it another way; Amps squared times R. So (0.34 x 0.34) times 35 = 4.1 watts.! Check! Or another way again; Voltage squred divided by R. So (12 x12) divided by 35 = 144/35 =4.1 watts. Check!
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<%-name%>
• posted on January 9, 2010, 7:18 am
Dave wrote:

You knew the answer before even calculating for it? Can I throw you another puzzle? A cube is made of 1 Ohm resistors. Altogether 12 one Ohm resistors. What is total resistance between one corner of the cube to the opposite end corner?
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<%-name%>
• posted on January 9, 2010, 3:18 pm

I knew the ballpark. I knew it had to be close to 33 and 4 but I didn't know the exact number and I needed an answer accurate to the 3rd decimal place.
It's for a scavenger hunt and the answers are clues to the next item on the list.
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<%-name%>
• posted on January 9, 2010, 3:59 pm

Hey: An idea? Maybe the 'Yellow' items illuminate (i.e. Are lamps or LEDs)? And maybe the 'Red' items are resistors. Does that throw a different light on the question? Sorry for the bad pun!