## Fundamentals of Electrical Engineering

a) $2530W$ of power are dissipated through heat or other losses. b) $219MJ$ of energy would be dissipated over a 24 hour period. c) If the price for energy was $6¢$ per kilowatt hour, this energy would cost $3.64$.
a) Remember our formula for power, $P = V * I$; sub in our know values for $V$ and $I$ to get $P = 23A* 110V = 2530W$ b) Remember our formula for energy, $E = P*T$; sub in our values of $P$ and $T$ to get $E = 2530W * 24H$ The unit $T$ has to be in seconds: $1H = 3600s$, $3600*24 = 86400$; the new equation is $E = 2530W*8400S$, which is approximately $219MJ$ c) To get kilowatt hours, we multiply the number of hours by the number of watts and then divide by $1000$ for the Kilo prefix, $Kwh = 2530W*24H$ which is $60.72 Kwh$ at $6¢$ an hour, so $60.72*6$ = $\$ 3.64\$