Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 21 - Electromagnetic Induction and Faraday's Law - Problems - Page 621: 26

Answer

13 A

Work Step by Step

It is apparent from equation 21–5 that the back emf is proportional to the angular speed of rotation. If the angular speed is halved, so is the back emf, The new back emf is 54 volts. The back emf opposes the applied emf of 120 volts. The combined effect gives the net voltage across the motor, which is IR. $$\epsilon_{applied}-\epsilon_{back}=IR$$ $$I=\frac{\epsilon_{applied}-\epsilon_{back}}{R} $$ $$=\frac{120V-54V}{5.0\Omega}=13A$$
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