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Chapter 26: Problem 50

50\. 1994 Honda Accord Consider a 1994 Honda Accord with a battery that is rated at 52 ampere-hours. This battery is supposed to be able to deliver 1 ampere of current to electrical devices in a car for at least 52 hours or 2 amperes for 26 hours, and so on. Suppose you leave the car lights turned on when you park the car and the car lights draw 20 amperes of current. How long will it be before your battery is dead?

Short Answer

Expert verified
2.6 hours.

Step by step solution

01

- Understand Battery Rating

The battery is rated at 52 ampere-hours (Ah). This means it can supply 52 amperes for 1 hour, 26 amperes for 2 hours, and so on.
02

- Identify Current Draw

The car lights are drawing 20 amperes of current.
03

- Calculate Battery Life

To find out how long the battery will last, use the formula: Time (hours) = Battery Capacity (Ah) / Current Draw (A).
04

- Apply the Formula

Substitute the known values into the formula: Time = 52 Ah / 20 A = 2.6 hours.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Battery Capacity
Battery capacity indicates how much electric charge a battery can hold and is typically measured in ampere-hours (Ah). Think of it as the energy reservoir of the battery. If a battery has a capacity of 52 Ah, it means it can supply 1 ampere of current for 52 hours or 2 amperes for 26 hours. This helps in understanding how long a battery can power devices before needing a recharge. It's crucial to know your battery's capacity to prevent running out of power unexpectedly, especially in critical situations.
Current Draw
Current draw refers to the amount of current, measured in amperes (A), that an electrical device consumes while operating. Higher current draw means the device uses more power and will drain the battery faster. In the case of the 1994 Honda Accord, the car lights draw 20 A. To visualize this, imagine a countdown timer for the battery's life starting the moment you turn on the lights. The higher the draw, the faster the countdown runs. It's imperative to manage the current draw to extend the battery's life and ensure all devices function properly for as long as needed.
Ampere-Hours
Ampere-hours (Ah) is a unit measuring the battery's charge over time. It combines the battery capacity and current draw into a simple to understand value. Using the formula \( \text{Time (hours)} = \frac{\text{Battery Capacity (Ah)}}{\text{Current Draw (A)}} \), you can calculate how long a battery will last given a specific current draw. For example, with a 52 Ah battery and a 20 A draw: \( \text{Time} = \frac{52 \text{ Ah}}{20 \text{ A}} = 2.6 \text{ hours} \). This calculation is vital in planning energy usage, ensuring you don't run out of battery power unexpectedly.

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Most popular questions from this chapter

38\. Nichrome A wire of Nichrome (a nickel-chromium-iron alloy commonly used in heating elements) is \(1.0 \mathrm{~m}\) long and \(1.0 \mathrm{~mm}^{2}\) in cross-sectional area. It carries a current of \(4.0 \mathrm{~A}\) when a \(2.0 \mathrm{~V}\) potential difference is applied between its ends. Calculate the conductivity \(\sigma\) of Nichrome.

24\. Heating Element A heating element is made by maintaining a potential difference of \(75.0 \mathrm{~V}\) across the length of a Nichrome wire that has a \(2.60 \times 10^{-6} \mathrm{~m}^{2}\) cross section. Nichrome has a resistivity of \(5.00 \times 10^{-7} \Omega \cdot \mathrm{m} .\) (a) If the element dissipates \(5000 \mathrm{~W}\), what is its length? (b) If a potential difference of \(100 \mathrm{~V}\) is used to obtain the same dissipation rate, what should the length be?

25\. Nichrome Heater A Nichrome heater dissipates \(500 \mathrm{~W}\) when the applied potential difference is \(110 \mathrm{~V}\) and the wire temperature is \(800^{\circ} \mathrm{C}\). What would be the dissipation rate if the wire temperature were held at \(200^{\circ} \mathrm{C}\) by immersing the wire in a bath of cooling oil? The applied potential difference remains the same, and \(\alpha\) for Nichrome at \(800^{\circ} \mathrm{C}\) is \(4.0 \times 10^{-4} / \mathrm{K}\).

\begin{array}{l} \text { 49. Increases Over Time The charge passing through a conductor }\\\ \text { increases over time as } q(t)=\left(1.5 \mathrm{C} / \mathrm{s}^{3}\right) t^{3}-\left(4.5 \mathrm{C} / \mathrm{s}^{2}\right) t^{2}+(2 \mathrm{C} / \mathrm{s}) t, \end{array} where \(t\) is in seconds. (a) What equation describes the current in the circuit as a function of time? (b) What is the current in the conductor at \(t=0.0 \mathrm{~s}\) and at \(t=1.0 \mathrm{~s}\) ?

34\. Near Earth Near the Earth, the density of protons in the solar wind (a stream of particles from the Sun) is \(8.70 \mathrm{~cm}^{-3}\), and their speed is \(470 \mathrm{~km} / \mathrm{s}\). (a) Find the current density of these protons. (b) If the Earth's magnetic field did not deflect them, the protons would strike the planet. What total current would the Earth then receive?
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