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
This is the end of the preview. Sign up
access the rest of the document.