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 522: 40

Answer

a. 7.9$\Omega$, 1.1 W b. The power would increase by a factor of two

Work Step by Step

a. Use equation 18–2, which defines the resistance. $$R=\frac{V}{I}=\frac{3.0\;V}{0.38A}=7.9\Omega$$ The power is P = IV = (0.38A)(3.0 V) = 1.1 J/s = 1.1 watts. b. Use equation 18–6b to calculate the power. $$P=\frac{V^2}{R}$$ The power is proportional to the square of the voltage. The resistance is the same. If the voltage is multiplied by 2, the power consumed is multiplied by 4. This isn’t a good idea because the bulb is not designed for such a high power, and would probably overheat and burn out in short order.
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