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.

 

[HalfLostAndSearching]

I can confirm that your calculations are correct. The ohmic resistance of the bulbs is 0.75 ohms and using a 9V battery will result in a voltage drop of 1.5V, which is within the operational voltage range of the bulbs. Therefore, the bulbs will not burn out when connected to the 9V battery. However, it is important to note that the bulbs may not shine as brightly as they would with a 1.5V battery, as the power output will be lower due to the lower voltage. It is always recommended to use the appropriate voltage and power source for optimal performance and to avoid potential damage to the bulbs.