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: 43

Answer

1.5 m It would probably overheat.

Work Step by Step

Use equation 18–3 to find the resistance. The cross-sectional area $A=\pi r^2=\frac{\pi d^2}{4}$. $$R=\rho\frac{\mathcal{l}}{A}=\rho\frac{4 \mathcal{l}}{\pi d^2}$$ Combine this with equation 18–6b and calculate the length of the wire. $$P=\frac{V^2}{R}=\frac{\pi d^2V^2}{4 \rho \mathcal{l}}$$ $$ \mathcal{l}=\frac{\pi d^2V^2}{4 \rho P}$$ $$=\frac{\pi (0.00050m)^2(1.5V)^2}{4 (1.68\times10^{-8}\Omega \cdot m) (18W)}=1.5m$$ The power is proportional to the square of the voltage. The resistance is the same. If the voltage is multiplied by 6, the power consumed is multiplied by 36. This isn’t a good idea because the blanket is not designed for such a high power, and would probably overheat and burn out the wiring in short order.
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