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Chapter 24: Problem 14

A \(12-V\) car battery is rated at 80 ampere-hours, meaning it can supply 80 A of current for 1 hour before it becomes discharged. If you accidentally leave the headlights on until the battery discharges, how much charge moves through the lights?

Short Answer

Expert verified
The total charge that moves through the lights is 80 A \cdot hrs or 80 Ampere-hours.

Step by step solution

01

Understand the relationship between current, charge and time.

In electrical physics, the relationship between current (I), charge (Q) and time (t) is given by the formula: \(Q = I \times t\), where Q is the charge transferred, I is the current, and t is the time. The charge is typically measured in Coulombs (C), the current in Amperes (A), and time in seconds (s). However, in this exercise, the time is given in hours, not seconds, so we will use the time in hours directly.
02

Insert the given values into the formula.

We know from the problem that the current (I) supplied by the battery is 80 Ampere and the time (t) is 1 hour. Inserting these values into the formula gives: \(Q = 80 A \times 1 hrs\).
03

Compute the total charge transferred.

By calculating the right hand side of the equation, the total charge (Q) transferred from the battery through the lights is \(80 A \times 1 hrs = 80 A \cdot hrs\).

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Most popular questions from this chapter

A brownout occurs when an electric utility can't supply enough power to meet demand. Rather than cut off some customers completely, the utility reduces the voltage across its system. Brownouts are most likely on hot summer days, when heavy air-conditioning loads drive up demand for electricity. In a particular brownout, the utility reduces the voltage by \(10 \%.\) Metallic conductors like lightbulb filaments and electric stove burners have resistance that increases with increasing temperature. During the brownout, the current in such devices a. decreases by \(10 \%.\) b. decreases by more than \(10 \%.\) c. decreases by less than \(10 \%.\) d. You can't tell without knowing more about how the resistance varies.

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