A current of 6A is drawn from a 120V line,What power is being developed ? How much energy in joule and in kilowatt hour is expended if the current is drawn steadily for one week Physics class 12th problem solution Physics class 12th numerical solution Electrodynamics Problem Solution Current electricity problems solution
00:00question is a current of 6 ampere is drawn from a 120 volt line watt power is being developed
00:15how much energy in joule and in kilowatt hour is expended if the current is drawn steadily for
00:25one week first of all we will make a data is current is obtained is six amperes and electric potential
00:49is obtained as 120 volt so in first condition we have to find power which is developed and in
01:03second condition we have to find energy in joule and in third condition we will find energy in
01:19kilowatt hour its time is obtained is one week so first of all we will convert week into seconds
01:41so time will be t is equal to one enter is there are seven days in a week enter and there are 24
01:54hours in a day and 60 minutes in an hour and 60 seconds in a minute so after multiplying all these
02:07values we will get the time is 6.05 into 10 raised power 5 second so for solution
02:24first of all we will find the developed power is in this condition the power is calculated by using
02:42formula p is equal to i into v so power will be obtained as its current is obtained as 6 ampere into
02:53120 it will be 720 watt so power is obtained as 720 watt this is our first answer now we will find energy in
03:23energy in joule is power is equal to energy upon time or work upon time so energy will be equal to power into time because here t is dividing on the other side of
03:52feature now it will be equal it will multiply it will multiply it will multiply it will be power so by putting
03:56to be the given value values energy will be obtained
04:05aspireanda power is obtained as 720 watt so 720
04:11watt so 720 into time which is obtained as 6.05 into 10 raised power 5 second so energy
04:33will be obtained is when we will multiply 720 with 6.05 then we will get 4356 into 10 raised
04:52power 5 Joule here is decimal then we will shift decimal from right to left up to 3 digits
05:10then we will add 3 with the power so energy will be obtained as 4.356 into 10 raised power
05:228 Joule so this is our second answer finally we will find energy in kilowatt hours is 1
05:42kilowatt hour is equal to 3.6 into 10 raised power 6 Joule so 1 Joule will be equal to 1 upon
06:003.6 into 10 raised power 6 kilowatt hour so by putting this value in place of Joule so energy
06:14in kilowatt hour will be obtained as 4.356 into 10 raised power 8 into Joule the value of
06:26Joule is 1 upon 3.6 into 10 raised power 6 kilowatt hour so after simplifying this energy in kilowatt
06:42kilowatt hour will be obtained as 4.356 divided by 3.6 it will be obtained as 1.21 into S in
06:58dv and when the bases are same then we will subtract the power of denominator from the power of numerator
07:05so it is 10 raised power 8 so it is 10 raised power 8 and it this is 10 raised power 6 so 10 raised power 8 minus 6
07:17kilowatt hour so energy in kilowatt hour will be obtained as 1.21 into 10 raised power plus 8 minus 6 it will be plus 2 kilowatt hour
07:33so energy in kilowatt hour will be equal to 10 raised power 10 raised power power power 10 raised power 2 is equal to 100
07:45then we will multiply 100 with 1.21 then we will get 121 kilowatt hour so this is our third answer
