780_Physics ProblemsTechnical Physics

# 780_Physics ProblemsTechnical Physics - 120 Current and...

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120 Current and Resistance Additional Problems P27.54 (a) I V R = so P == IV V R a f 2 R V = a f a f 22 120 25 0 576 V W . and R V = a f a f 120 100 144 V W (b) I V Q tt = = = ∆∆ 25 0 0208 100 . . . W 120 V A C t 480 . . C 0.208 A s The bulb takes in charge at high potential and puts out the same amount of charge at low potential. (c) = 25 0 . . W J U t 00400 . . J 25.0 W s The bulb takes in energy by electrical transmission and puts out the same amount of energy by heat and light. (d) Ut = × 25 0 86 400 30 0 64 8 10 8 .. . Js sd d J bg b g af The electric company sells energy . Cost J \$0.070 0 kWh k 1000 Ws J h 3600 s Cost per joule kWh kWh 3.60 10 J J 6 F H G I K J F H G I K J F H G I K J F H G I K J = = × F H G I K J 64 8 10 26 070 0 94 10 6 8 .\$ 1 . \$0. \$1. *P27.55 The original stored energy is UQ V Q C ii 1 2 1 2 2 . (a) When the switch is closed, charge Q distributes itself over the plates of C and 3 C in parallel, presenting equivalent capacitance 4 C . Then the final potential difference is
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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