785_Physics ProblemsTechnical Physics

785_Physics ProblemsTechnical Physics - Chapter 27 P27.69...

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Chapter 27 125 P27.69 Each speaker receives 60.0 W of power. Using P = IR 2 , we then have I R == = 60 0 387 . . W 4.00 A . The system is not adequately protected since the fuse should be set to melt at 3.87 A, or less . P27.70 VE =− ⋅ A or dV E dx VI RE I dq dt E R A E A EA dV dx A dV dx =− =⋅ = A A A A ρρ σσ Current flows in the direction of decreasing voltage. Energy flows as heat in the direction of decreasing temperature. P27.71 R dx A dx wy zz where yy L x =+ 1 21 R w dx y L x L wy y y L x R L y y y L L = +− = + L N M O Q P = F H G I K J z ρ 12 1 0 1 0 2 1 bg ln ln FIG. P27.71 P27.72 From the geometry of the longitudinal section of the resistor shown in the figure, we see that br y ba h = a f a f . From this, the radius at a distance y from the base is ra b y h b =− + a f . For a disk-shaped element of volume
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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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