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 792: 41


The power dissipated by each bulb is $~18.75~W$

Work Step by Step

We can write an expression for the power: $P = I^2~R$ We know that the current is $I = \frac{V}{R}$. If two bulbs are connected in series, then the total resistance is doubled, so the current is halved. Then the power dissipated by one bulb is reduced by a factor of $\frac{1}{4}$ The power dissipated by each bulb is $\frac{75~W}{4} = 18.75~W$
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