Current and Resistance9117.53(a) The total power you now use while cooking breakfast is ( )1200500 W1.70 kW=+=P() ( )The cost to use this power for 0.500 h each day for 30.0 days is ()h00 30.0 days$0.120 kWh$3.06daycost=1.70 kW0.5trate=×∆× =P(b) If you upgraded, the new power requirement would be: ( )2 400500 W=2 900 WPand the required current would be 2 900 W26.4 A20 A110 VIV===>∆PNo , your present circuit breaker cannot handle the upgrade. 17.54(a) The charge passing through the conductor in the interval is represented by the area under the
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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.