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Please help solve this 12V electrical circuit puzzle
I'm working on a puzzle (game) that I can't seem to figure out.
http://i204.photobucket.com/albums/b...h/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! |
Please help solve this 12V electrical circuit puzzle
On Thu, 07 Jan 2010 16:26:33 -0700, Dave wrote:
I'm working on a puzzle (game) that I can't seem to figure out. http://i204.photobucket.com/albums/b...h/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! 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. |
Please help solve this 12V electrical circuit puzzle
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Please help solve this 12V electrical circuit puzzle
Dave wrote:
I'm working on a puzzle (game) that I can't seem to figure out. http://i204.photobucket.com/albums/b...h/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! Hmmm, Little more than a basic Ohm's law. Apply Thevenin Norton's and/or Kirchoff's law. |
Please help solve this 12V electrical circuit puzzle
On Thu, 07 Jan 2010 16:26:33 -0700, Dave wrote:
I'm working on a puzzle (game) that I can't seem to figure out. http://i204.photobucket.com/albums/b...h/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! For the top puzzle, try solving the circuit with the numbers being 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. |
Please help solve this 12V electrical circuit puzzle
|
Please help solve this 12V electrical circuit puzzle
On 1/8/2010 7:28 AM Dave wrote:
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. :( 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? |
Please help solve this 12V electrical circuit puzzle
Dave wrote:
On 1/8/2010 7:28 AM Dave wrote: 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. :( 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? 1.89032 + 7.0134 + 7.7494 = 16.65312 The 7.7494 is in parallel with the rest. |
Please help solve this 12V electrical circuit puzzle
On 1/8/2010 12:25 PM E Z Peaces wrote:
1.89032 + 7.0134 + 7.7494 = 16.65312 The 7.7494 is in parallel with the rest. 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. |
Please help solve this 12V electrical circuit puzzle
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.. |
Please help solve this 12V electrical circuit puzzle
On Jan 8, 6:58*pm, terry wrote:
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.. 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????????????????????? |
Please help solve this 12V electrical circuit puzzle
On 1/8/2010 3:01 PM terry wrote:
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????????????????????? 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. |
Please help solve this 12V electrical circuit puzzle
Dave wrote:
On 1/8/2010 3:01 PM terry wrote: 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????????????????????? 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. 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. |
Please help solve this 12V electrical circuit puzzle
On Jan 7, 6:26*pm, Dave wrote:
I'm working on a puzzle (game) that I can't seem to figure out. http://i204.photobucket.com/albums/b...h/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! If you assume the values given are either voltage or current you will quickly notice they are in violation of various electrical laws. Resistance is the only one that works for the top one. I havent had time to ponder the bottom one since LEDs are non linear it may not be a repeat of the top one. Jimmie |
Please help solve this 12V electrical circuit puzzle
On Fri, 08 Jan 2010 07:28:05 -0700, Dave wrote:
On 1/7/2010 7:45 PM wrote: For the top puzzle, try solving the circuit with the numbers being 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. 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. For parallel circuits 1/r1 + 1/r2 +1/r3 = 1/RT Solve the parallel portions of series strings first. Then add the series strings, then solve the remaining/resulting parallel string. There are also online parallel resistance calculators that make the job dead simple - if you remember the rules. |
Please help solve this 12V electrical circuit puzzle
On Fri, 08 Jan 2010 07:28:05 -0700, Dave wrote:
On 1/7/2010 7:45 PM wrote: For the top puzzle, try solving the circuit with the numbers being 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. 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. If the bottom one should come out around 4, I'd say MABEE they are using ohms again and the total circuit is 35.079 ohms accross a 12 volt battery, giving a current flow of 2.9 amps. Not a very good way to do things, because LEDs do not have a "resistance" per se. - and it would not be an apples to apples comparison, because we have amps on the bottom and watts on the top. Perhaps it is Ma per LED they are using. Then you would need to know the voltage drop across each LED - which varies depending on what colour they are and what the composition of the junction is. Assume Gallium Arsenide (common red) at 1.7 volts - Nope = cannot be - the current would be astronomical and the LEDs would pop running them on 12 volts with 2 in series - or even worse the single one on the bottom.. Now, IF we assume the one on the bottom draws 60 ma across 12 volts because of the current dropping resistor required to run it on 12 volts - but then the top string doesn't work because the assumed current flow on the single one in the series sring is higher than the single bottom one. It doesn't make any sense to me YET. |
Please help solve this 12V electrical circuit puzzle
On Fri, 08 Jan 2010 11:08:14 -0700, Dave wrote:
On 1/8/2010 7:28 AM Dave wrote: 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. :( 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? Not following the rules. You need to analyze the circuit. You have a parrallel circuit composed of one resistance on the bottom and a combination on the top. The combination is a series circuit of a single resistance and a parralel sub-circuit of 2 in parrallel. You need to solve the parallel sub-string first, and add the series component to it, then solve the resulting parallel string. |
Please help solve this 12V electrical circuit puzzle
On Fri, 08 Jan 2010 14:13:04 -0700, Dave wrote:
On 1/8/2010 12:25 PM E Z Peaces wrote: 1.89032 + 7.0134 + 7.7494 = 16.65312 The 7.7494 is in parallel with the rest. 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. Except 35 ohms and 12 volts is NOT 4 watts. It's 4 AMPS. 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. |
Please help solve this 12V electrical circuit puzzle
Dave wrote:
On 1/8/2010 12:25 PM E Z Peaces wrote: 1.89032 + 7.0134 + 7.7494 = 16.65312 The 7.7494 is in parallel with the rest. 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. Hi, 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? |
Please help solve this 12V electrical circuit puzzle
On 1/8/2010 7:11 PM wrote:
On Fri, 08 Jan 2010 14:13:04 -0700, Dave wrote: Except 35 ohms and 12 volts is NOT 4 watts. It's 4 AMPS. 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. 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 |
Please help solve this 12V electrical circuit puzzle
On 1/9/2010 12:18 AM Tony Hwang wrote:
Dave wrote: On 1/8/2010 12:25 PM E Z Peaces wrote: Both of these results are right in line with my expected results. Hi, 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? 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. |
Please help solve this 12V electrical circuit puzzle
On Jan 9, 12:11*pm, Dave wrote:
On 1/8/2010 7:11 PM wrote: On Fri, 08 Jan 2010 14:13:04 -0700, Dave wrote: Except 35 ohms and 12 volts is NOT 4 watts. It's 4 AMPS. 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. 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 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! |
Please help solve this 12V electrical circuit puzzle
Tony Hwang writes:
Dave wrote: On 1/8/2010 12:25 PM E Z Peaces wrote: 1.89032 + 7.0134 + 7.7494 = 16.65312 The 7.7494 is in parallel with the rest. 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. Hi, 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? If I am thinking correctly, the answer is readily apparant with 2 seconds of thinking and almost no calculations. "Distance" from one conder to the opposite is 3 resistors = 3 ohoms. But there are 4 different paths which are all equal by symmetry -- so like 4 3-ohm paths in paralle or 3/4 ohms total. Took me 100 times as long to write it out as to solve it... |
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