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Chapter 25: Problem 18

If you accidentally leave your car headlights (current 5 A) on for an hour, how much energy drains from the car's \(12-\mathrm{V}\) battery?

Short Answer

Expert verified
The energy drained from the car's battery is 216,000 Joules.

Step by step solution

01

Compute the Power

Power is computed as the product of voltage and current, using the formula \(P = IV\). Given that the voltage is 12V and the current is 5A, the power consumed is \(P = 12V * 5A = 60W\).
02

Determine the Time in the Correct Units

The power consumption calculation is generally in watts per second, however, the duration of time given in this case is in hours. To get the correct units, the time duration should be converted to seconds. 1 hour is equal to \(3600\) seconds.
03

Calculate the Energy Drained

Using the Energy-Power-Time relationship, where energy \(E = P * t\), we compute the energy consumed by the headlights as \(E = 60W * 3600s = 216,000 J\).

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Most popular questions from this chapter

Your company is overstocked on \(50-\Omega, \frac{1}{2}-\) W resistors. Your project requires \(50-\Omega\) resistors that can be safely connected across a 12-V power source. How many of the available resistors will you need, and how will you connect them?

Three \(1.5-\mathrm{V}\) batteries, with internal resistances \(0.01 \Omega, 0.1 \Omega\) and \(1 \Omega\), each have \(1-\Omega\) resistors connected across their terminals. What's the voltage between each battery's terminals, to three significant figures?

You're designing an external defibrillator that discharges a capacitor through the patient's body, providing a pulse that stops ventricular fibrillation. Specifications call for a capacitor storing 250 J of energy; when discharged through a body with \(40-\Omega\) transthoracic resistance, the capacitor voltage is to drop to half its initial value in \(10 \mathrm{ms}\). Determine the capacitance (to the nearest \(10 \mu \mathrm{F})\) and initial capacitor voltage (to the nearest \(100 \mathrm{V}\) ) that meet these specs.

Show that a battery delivers the most power when the load resistance across its terminals is equal to its internal resistance. (This is not the way to treat a battery, but it's the basis for load matching in amplifiers; see Problem \(65 .\) )

An uncharged \(10-\mu \mathrm{F}\) capacitor and a \(470-\mathrm{k} \Omega\) resistor are in series, and \(250 \mathrm{V}\) is applied across the combination. How long does it take the capacitor voltage to reach \(200 \mathrm{V} ?\)
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