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 - General Problems - Page 524: 83

Answer

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Work Step by Step

We use 2 significant figures. a. Current is power over voltage, from equation 18–5. $$I_A=P_A/V_A=40W/120V=0.33A$$ $$I_B=P_B/V_B=40W/12V=3.3A$$ b. Resistance is calculated from equation 18–6b. $$R_A=V_A^2/P_A=(120V)^2/40W=360 \Omega$$ $$R_B=V_B^2/P_B=(12V)^2/40W=3.6 \Omega$$ c. The charge is the current multiplied by the time. $$Q_A=I_A t=(0.33A)(3600s)=1200C$$ $$Q_B=I_B t=(3.3A)(3600s)=1.2\times10^4 C$$ d. The energy is the power multiplied by the time. The power is 40W for both bulbs. $$E=Pt=(40W)(3600s)=1.4\times10^5 J$$ e. Bulb B requires larger-diameter wires because it draws more current.
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