Chapter 27125P27.69Each speaker receives 60.0 W of power. Using P=IR2, we then haveIR===60 0387..W4.00 AΩ.The system is not adequately protected since the fuse should be set to melt at 3.87 A, or less.P27.70∆VE=− ⋅Aor dVE dx∆VIREIdqdtERAEAEAdVdxAdVdx=−⋅=⋅−=AAAAρρσσCurrent flows in the direction of decreasing voltage. Energy flows as heat in the direction ofdecreasing temperature.P27.71RdxAdxwyzzwhere yyLx=+−121RwdxyLxLwyyyLxRLyyyLL=+−=−+−LNMOQP=−FHGIKJzρ12101021bglnlnFIG. P27.71P27.72From the geometry of the longitudinal section of the resistor shown in the figure,we see thatbrybah−=−afaf.From this, the radius at a distance yfrom the base israbyhb=− +af.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 .