00:06numerical number 5 of unit 19 electromagnetic induction is an inductor with an inductance
00:14of 0.02 henry has a current flowing through it of 2 amperes calculate the energy stored
00:24in the inductor so sb se phil e hem data banayenge e
00:34yahaan pere ek coil ke baare me ke rai to uske hesab se hem self inductance use karthay
00:40hain for mutual inductance we will use two coils so yahaan pere l ka value means self
00:49inductance ka value which is 0.02 henry and current bhi hain obtained which is 2 amperes
01:03hain kya calculate karna hai we have to calculate stored energy in the inductor means e so for solution
01:19or calculation we will use the formula to calculate energy stored in the inductor is e is equal to 1
01:28upon 2 l i square so e will be equal to 1 upon 2 multiplied with l the value of
01:40l is 0.02 multiplied
01:43with l i square the value of l is 2 so the square on 2 so energy stored in the
01:50inductor will be obtained
01:51is 1 upon 2 multiplied with 0.02 multiplied with the square of 2 is 4 is the table of
02:032 on 2 is 1 times
02:04and on 4 is 2 times so stored energy will be equal to 0.02 multiplied with 2 so it
02:12will be 0.04 joules so this
02:18is our required answer
02:20you
02:22you
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