Conceptual Physics (12th Edition)

Published by Addison-Wesley
ISBN 10: 0321909100
ISBN 13: 978-0-32190-910-7

Chapter 23 - Think and Solve: 48

Answer

a. $\frac{1}{30} A$. b. 3600 $\Omega$. c. 35 kWh. d. $ \$5.26$.

Work Step by Step

a. First find the current that is drawn. $$Electric \; power = (current) \times (voltage)$$ $$4 W = (I)(120 V)$$ Solve for $I = \frac{4 W}{120 V} = \frac{1}{30} A$ . b. Second, find the nightlight's resistance. $$Current = \frac{voltage}{resistance}$$ Rearranging to solve for the resistance, we obtain $$Resistance= \frac{voltage}{current}$$ Put the numbers in from this problem. $$Resistance= \frac{120 V}{(1/30) A} = 3600 \Omega$$ c, d. Change watts to kilowatts and the time to hours, since we are given the cost per kilowatt-hour. 4 W = 0.004 kW = 0.004 kilowatt. There are 24 hours a day, and 365 days in a year, so there are 8760 hours in a year. Now find the number of kilowatt-hours used. $$(0.004 kW)(8760 hours) = 35.04 kWh$$ The cost is 15 cents per kWh, so the total is $ \$5.26$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.