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.

It would be 12W because: 6v is half of 12v so half of 24w would be 12w