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Chapter 18: Problem 10

The label on a \(12.0-\mathrm{V}\) truck battery states that it is rated at 180.0 A-h (ampere-hours). Treat the battery as ideal. (a) How much charge in coulombs can be pumped by the battery? [Hint: Convert A to A-s.] (b) How much electric energy can the battery supply? (c) Suppose the radio in the truck is left on when the engine is not running. The radio draws a current of 3.30 A. How long does it take to drain the battery if it starts out fully charged?

Short Answer

Expert verified
b) How much electrical energy can the battery supply in joules? c) How long will it take to drain the battery if the radio is left on with the engine off, in hours?

Step by step solution

01

Convert Ampere-Hours to Coulombs

We are given the rating of the battery in ampere-hours (A-h) which is 180.0 A-h. We can convert this value to coulombs (C) using the following conversion: 1 A-h = 3600 C So, the charge in coulombs can be calculated as: Q = (180.0 A-h) * (3600 C/A-h) = 648000 C
02

Calculate the Electrical Energy

Now, we can calculate the electrical energy that the battery can supply using the formula E = V * Q, where V is the voltage and Q is the charge in coulombs. E = (12.0 V) * (648000 C) = 7776000 J
03

Calculate the Time to Drain the Battery

Lastly, we can calculate the time it takes to drain the battery when a radio draws a current of 3.30 A using the formula t = Q / I, where Q is the charge in coulombs and I is the current drawn by the radio. t = (648000 C) / (3.30 A) = 196363.64 s To convert this value to hours, we can use the following conversion: 1 hour = 3600 seconds So, the time to drain the battery in hours is: t = (196363.64 s) / (3600 s/h) 鈮� 54.54 h So, the answers are: a) The battery can pump 648000 C of charge. b) The battery can supply 7776000 J of electrical energy. c) It will take approximately 54.54 hours to drain the battery if the radio is left on with the engine off.

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