Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 28 - Fundamentals of Circuits - Exercises and Problems - Page 790: 10

Answer

The correct answer is: C. $~~P \gt S = T \gt Q = R$

Work Step by Step

The power dissipated by each bulb is: $P = \frac{V^2}{R} = I^2~R$ The brightness depends on the power that is dissipated. Since bulb Q and bulb R are in parallel, the equivalent resistance is less than the resistance of bulb T. Therefore, more current flows through the section with bulb P than the section with bulb S. Therefore, bulb P is brighter than bulb S and bulb T, while bulb S and bulb T are equally bright with each other since they are connected in series. Then the potential difference across bulb P is greater than the potential difference across bulb S. Then the potential difference across bulb both bulb Q and bulb R is less than the potential difference across bulb T. Therefore, bulb Q and bulb R are less bright then bulb T. The correct answer is: C. $~~P \gt S = T \gt Q = R$
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