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
18-4: Resistivity
18-5 and 18-6: Electric Power
18-7: Alternating Current
18-8: Microscopic View of Electric Current
18-10: Nerve Conduction

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 10
Q

A 4.5-V battery is connected to a bulb whose resistance is 1.3  Ω1.3 \; \Omega. How many electrons leave the battery per minute?

A
1.3×1021 electrons/min1.3 \times 10^{21} \textrm{ electrons/min}
Giancoli 7th Edition, Chapter 18, Problem 10 solution video poster
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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. Well, first, we can find the current in amps and that's gonna be voltage divided by resistance. 4.5 volts divided by 1.3 ohms gives us 3.46154 amps. And this is Coulombs per second, that's what amps is in abbreviation for. Our job is to convert this into electrons per minute. So we take this Coulombs per second multiplied by 60 seconds per minute, and that gives us the number of Coulombs per minute. Then we have to convert that into electrons by multiplying by one electron for every 1.6 times 10 to the minus 19 Coulombs and this gives us 1.3 times 10 to the 21 electrons per minute.

COMMENTS
By jacoblayton2 on Sun, 2/9/2020 - 5:19 PM

I am putting this in my calculator and it is coming out to 1.29808•10^-17. What am I doing wrong?

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