1.5V/3W lightbulbs with a 9V battery. Will they burn out?

[carnivalcougar]

Homework Statement

Imagine that you have a box of 9V batteries and a box of flashlight bulbs labeled 1.5V/3W for operational voltage/power. What is the ohmic resistance of the bulbs? Can you turn on the 1.5V bulbs without burning them out by using the available 9V batteries?

Homework Equations

P=ΔW/t=ΔVI=ΔV²/R=I²R
I = P/V
R = V/I

The Attempt at a Solution

I in the bulbs will be 3W/1.5V which is 2A
R in the bulbs will be 1.5V/2A which is .75ohms

If you attach one of the bulbs to a 9V battery, the voltage drop will be = to IR which is (2A)(.75ohms) which is 1.5V. So using a 9V battery the bulb will only get 1.5V which will not cause the bulb to burn out. Is this correct?

 

[NascentOxygen]

What they really mean is: It's a dark and lonely night, and all you have is a 9v battery and a box of 1.5v bulbs. Are you going to be stuck in the dark all night, or can you create some steady light? If so, how can you do it?

Warning: if you cause any light bulb to experience (across its terminals) a voltage that exceeds its rated voltage, the bulb will burn out in too short a time to be useful here.

 

[carnivalcougar]

I think I see. You would need to connect 6 lightbulbs to each 9V battery. This is because each lightbulb is rated for 1.5V. In order to create a voltage drop of 1.5V across each bulb and have the total voltage drop sum to zero, you would need 9/1.5 bulbs, which is 6 bulbs. Hopefully this is correct.

 

[NascentOxygen]

I think I see. You would need to connect 6 lightbulbs to each 9V battery. This is because each lightbulb is rated for 1.5V. In order to create a voltage drop of 1.5V across each bulb and have the total voltage drop sum to zero, you would need 9/1.5 bulbs, which is 6 bulbs. Hopefully this is correct.

Yes, I guess that's what was intended by this question. I'd call that a total voltage drop of 9v.