The wattage of a bulb is 24 w when it is connected to a 12 v battery. calculate its effective wattage if it operates on a 6 v battery.

The power dissipated by a resistance ' R '
powered by a battery ' V ' is                            
                                                 P = V² / R .

Before we play with actual numbers, look at that formula, and
notice that the power is proportional to the square of the voltage. 
So if the voltage suddenly drops by half, then the power that's
dissipated by the resistance drops to 1/2² = 1/4 of where it started out.

Now that we have the pure Science worked out, we're ready to do
the Engineering, and apply our findings to some actual technology.

We have a light bulb that's designed to dissipate energy at the rate
of 24W when you connect it to a source of 12V .  But all we have
in the lab today is a 6V battery.  What power will our bulb dissipate
if we connect it to the 6V that we have, instead of the 12V that it's
designed for ?

We're all standing around scratching our beards, when the new hot-shot
post-doc lab assistant reminds us of the formula up in the first paragraph,
and shows us that since we're running it at 1/2 of the design voltage, we
expect it to dissipate 1/4 of design power = 6 watts.

It'll give us a bit of light so we can still walk around the lab without bumping
into things, but it won't get hot enough to warm our lunches.  We'll have to
do that over a Bunsen burner or in the microwave oven today, and one of
you junior members of the lab staff will have to stop at Lady-o'-Rack and
pick up either a new battery or a new bulb on your way in to work tomorrow.

Ithink the answer is fusion