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.