122CHAPTER 18 18.35From ( )max1teQQτ−=−, we have at 0.900 st=, 0.900 smaxQeQ−10=.600Thus, 0.900 s0.400e−=, or ()ln 0.4000.900s−=giving the time constant as 0.982 sln900 s0.0.400=18.36(a) maxRIε=, so the resistance is 4-3max48.0 V9.6010 0.50010 ARI===××RCΩThe time constant is =, so the capacitance is found to be 540.960 s10F10.0 F9.6010CR1.00µ−×==×Ω(b) ( )( )max10.0 F48.0 V480 CQC=1.92 st=, so the charge stored in the capacitor at is 1.92 s20.960 smax1480 C1480 C1415 Cteeeµµ−−−=− =−=18.37(a) The current drawn by each appliance is
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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.