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: 6

Answer

Across the $40~\Omega$ resistor, the potential difference is $20~V$ Across the $60~\Omega$ resistor, the potential difference is $30~V$

Work Step by Step

We can find the equivalent resistance of the two resistors in series: $R = 40~\Omega + 60~\Omega = 100~\Omega$ We can find the current in the circuit: $I = \frac{V}{R} = \frac{50~V}{100~\Omega} = 0.50~A$ We can find the potential difference across the $40~\Omega$ resistor: $\Delta V = I~R = (0.50~A)(40~\Omega) = 20~V$ We can find the potential difference across the $60~\Omega$ resistor: $\Delta V = I~R = (0.50~A)(60~\Omega) = 30~V$
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