How much energy does the headlight use in 1.5 hours?

[Inertialforce]

Homework Statement

A 12V battery from a car is used to operate a 65W headlight.

a)How much energy does the headlight use in 1.5 hours?

b)What total charge passes through the headlight during this time?

c)What is the total number of electrons that pass through the headlight during this time period?

Homework Equations

P = IV

The Attempt at a Solution

What I did for question "a" is, I realized that the voltage going through the entire circuit (in this case) was 12V and I also knew that the headlight had a "P" of 65W. So what I did was I used the equation P= IV and isolated for I to get the current, since current is measured in Amps or coulombs/sec. So knowing this fact I calculated it out and came up with an answer of 5.4 coulombs/sec for the current.

Now my question is how do I use that 1.5 hours in my calculations? Do I just multiply it to the 5.4 coulombs/sec answer that I got (from isolating the current in the equation P= IV), which would cancel out the unit seconds and leave me with the unit of coulombs?

 

[Dick]

Sure, why not? (coulomb/sec)*(joule/coulomb)=joule/sec=watts.

 

[Inertialforce]

Sure, why not? (coulomb/sec)*(joule/coulomb)=joule/sec=watts.

When I think "energy" I think joules, therefore why would need to find out what the current or "I" in this question is (our teacher told us to calculate it)? Because couldn't you just cancel out the seconds by multiplying the power by the time period of the question (1.5h = 5400s) to cancel out the seconds and isolate for joules?

 

[buffordboy23]

When I think "energy" I think joules, therefore why would need to find out what the current or "I" in this question is (our teacher told us to calculate it)? Because couldn't you just cancel out the seconds by multiplying the power by the time period of the question (1.5h = 5400s) to cancel out the seconds and isolate for joules?

Yes. The headlight requires 65 W, or 65 J/s to operate, so you can multiply by the amount of time.