# How much energy does the headlight use in 1.5 hours?

• Inertialforce
In summary, a 12V battery is used to power a 65W headlight. To find the energy used by the headlight in 1.5 hours, the power is multiplied by the time, resulting in 97.5 joules. The total charge passed through the headlight can be found by calculating the current using the equation P=IV and multiplying it by the time, resulting in 291 coulombs. The total number of electrons passing through the headlight can be found by multiplying the charge by the number of electrons per coulomb, resulting in approximately 1.74 x 10^22 electrons.
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?

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?

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

Dick said:
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?

Inertialforce said:
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.

## 1. What is the unit of measurement for energy used by the headlight?

The unit of measurement for energy used by the headlight is typically in watts (W) or kilowatt-hours (kWh).

## 2. How do I calculate the energy usage of a headlight in 1.5 hours?

To calculate the energy usage of a headlight in 1.5 hours, you will need to know the wattage of the headlight. Then, you can use the formula Energy (kWh) = Power (W) x Time (h). So if the headlight has a wattage of 50W, the calculation would be 50W x 1.5h = 75Wh or 0.075kWh.

## 3. Does the type of headlight affect the amount of energy used in 1.5 hours?

Yes, the type of headlight can affect the amount of energy used in 1.5 hours. LED headlights tend to be more energy-efficient compared to traditional halogen headlights.

## 4. Can I convert the energy usage of a headlight in 1.5 hours to other units?

Yes, you can convert the energy usage of a headlight in 1.5 hours to other units such as joules or calories. However, keep in mind that the conversion will depend on the specific unit of measurement used for energy consumption.

## 5. How can I reduce the energy usage of a headlight in 1.5 hours?

There are a few ways to reduce the energy usage of a headlight in 1.5 hours. One option is to switch to more energy-efficient headlight types, such as LED headlights. You can also make sure to turn off the headlight when it is not needed, and keep the headlights clean and properly aligned for optimal energy usage.

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