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Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 7th Edition
18
Electric Currents
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18-2 and 18-3: Electric Current, Resistance, Ohm's Law
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18-4: Resistivity
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18-5 and 18-6: Electric Power
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18-7: Alternating Current
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18-8: Microscopic View of Electric Current
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Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 48
Q

An ac voltage, whose peak value is 180 V, is across a 310310-Ω\Omega resistor. What are the rms and peak currents in the resistor?

A
Io=0.58 A, Irms=0.41 AI_o = 0.58 \textrm{ A, } I_{rms} = 0.41 \textrm{ A}
Giancoli 7th Edition, Chapter 18, Problem 48 solution video poster
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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. Peak voltage is peak current times resistance. And we can divide both sides by R, to solve for peak current. So peak current is peak voltage divided by resistance, that's 180 volts divided by 310 ohms, which is .58 amps. The root mean squared current, is the peak current divided by root two. So it's .580645 amps, divided by root two, which is .41 amps.

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Giancoli Answers, including solutions and videos, is copyright © 2009-2024 Shaun Dychko, Vancouver, BC, Canada. Giancoli Answers is not affiliated with the textbook publisher. Book covers, titles, and author names appear for reference purposes only and are the property of their respective owners. Giancoli Answers is your best source for the 7th and 6th edition Giancoli physics solutions.