Physics: Principles with Applications (7th Edition)

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

Chapter 18 - Electric Currents - Problems - Page 521: 11

Answer

a. 4.76 A b. 6.59 A

Work Step by Step

a. Assume the voltage drops by 15 percent, i.e., is multiplied by 0.85, and the resistance R stays the same. Using equation 18–2, V=IR, we see that the current also drops by 15 percent. $$I_f=0.85I_i=0.85(5.60A)=4.76A$$ b. Assume the resistance drops by 15 percent, i.e., is multiplied by 0.85, while the voltage stays at 240 volts. Using equation 18–2, V=IR, we see that the current is divided by 0.85. $$I_f=\frac{I_i}{0.85}=\frac{5.60A}{0.85}=6.59A$$
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