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!

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.

On 1/7/2010 4:41 PM wrote:
On Thu, 07 Jan 2010 16:26:33 -0700, Dave wrote:
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.

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.

Dave wrote:
On 1/7/2010 4:41 PM wrote:
On Thu, 07 Jan 2010 16:26:33 -0700, Dave wrote:
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.

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.

Will Kirchoff's Law help you any?
http://tinyurl.com/yknfztg

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.

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.

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.

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?

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.

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.

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..

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?????????????????????

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.

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.

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

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.

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.

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.

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.

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?

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

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.

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!

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...