00:05Numerical No. 6 of Unit 19 Electromagnetic Induction is a coil stores energy in the form
00:15of electric potential energy of 0.2 Joule when it carries a current of 2 Ampere calculate
00:24the inductance of coil.
00:27So, now we will create a data, we will create a single coil, we will use 2 coils, so stored
00:46energy is given which is 0.2 Joules.
00:56Our current value being obtained which is 2 Amperes, we have to calculate self inductance
01:06means L or we can also call it inductance. So, for solution or calculation, we will use
01:19this formula to calculate inductance. In this case, we will use the formula is E is equal
01:27to 1 upon 2 Li square. Now, by putting the given values, the value of energy is given which
01:38is 0.2 Joules is equal to 1 upon 2 multiplied with L. We have to calculate inductance into
01:49I square. The value of I is 2, so the square on 2. So, 0.2 is equal to 1
01:58upon 2 into L multiplied
02:01with. The square of 2 is 4. As the table of 2 on 2 is 1 times and on 4
02:07is 2 times. So, 0.2 is equal
02:12to L into 2. We have to calculate L. So, we have to calculate L. So, here 2 multiply
02:18to be equal to L. So, L will be equal to L. So, L will be equal to 0.2
02:24upon 2. 0.2 divided
02:28by 2, it will be 0.1 Henry. As the inductance is obtained as 0.1 Henry. So, this is
02:44our final answer.
02:46Answer.
02:47.
02:47.
02:50.
02:50.
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