Fundamentals of Electrical Engineering

Published by McGraw-Hill Education
ISBN 10: 0073380377
ISBN 13: 978-0-07338-037-7

Chapter 2 - Fundamentals of Electric Circuits - Part 1 Circuits - Homework Problems - Page 48: 2.3

Answer

a. $360000 \text{ C}$ b. $ 2.247 \times 10^{24} \text{ electrons}$

Work Step by Step

a. Coulomb ($C$) = Ampere.second ($A.s$) To convert from Ampere.hour ($Ah$) to ($As$) $$Ah \times 60 \times 60 = As$$ $\therefore 100 Ah = 100 \times 60 \times 60 \, As = 360000 \, As$ $100 Ah = \boxed{360000 \, C}$ b. Charge of electron $=q = 1.60217 \times 10^{-19} \, C$ Number of electrons $(n) = \dfrac{360000}{q}=\boxed{ 2.247 \times 10^{24} \text{ electrons}}$
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.