How Much Energy Does Leaving Car Headlights On Drain from a 12V Battery?

[Seda]

Homework Statement

If you accidentaly leave your car headlights (current drain 5A) on for one hour, how much of the 12-V battery's chemical energy is used up?

Homework Equations

Well, there's Ohm's Law...and it's million forms

The Attempt at a Solution

First things first, I'm not sure whether to think of the power source as an ideal battery or not (ie, should I think about an internal resistance or not.)

Ok, the time factor is throwing me off. I'm not sure how to use the time (one hour) in any form of Ohms law. And I'm not sure if the current drain is the current through the battery of the entire circuit or the headlights or what. Any ideas?

 

[berkeman]

What is the definition of electrical power in terms of voltage and current? What is the definition of energy, in terms of power and time duration...?

 

[Moetasim]

As a scientist, it is important to clarify the parameters of the problem. In this case, it is important to know if the 5A current drain is the total current through the entire circuit or just through the headlights. Additionally, it would be helpful to know the internal resistance of the battery and the total resistance of the circuit. Without this information, it is difficult to accurately calculate the amount of chemical energy used by the battery.

Assuming the 5A current drain is the total current through the circuit and the internal resistance is negligible, we can use Ohm's Law (V = IR) to calculate the total resistance of the circuit. With this information, we can use the formula P = IV to calculate the power dissipated by the headlights (P = 5A x 12V = 60W).

To calculate the amount of chemical energy used by the battery, we need to know the battery's capacity, typically measured in amp hours (Ah). We can then use the formula E = V x I x t, where E is energy, V is voltage, I is current, and t is time. With the given information, we can calculate the energy used by the battery as (12V x 5A x 1 hour = 60Wh or 0.06kWh).

It is important to note that this calculation assumes ideal conditions and does not account for any losses in the circuit or the battery's internal resistance. In reality, the amount of chemical energy used by the battery may be slightly higher due to these factors.