JEE PYQ: Electromagnetic Induction Question 22
Question 22 - 2019 (10 Jan Shift 2)
The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1 s, the change in the energy of the inductance is:
(1) 740 J
(2) 437.5 J
(3) 540 J
(4) 637.5 J
Show Answer
Answer: (2)
Solution
$L\frac{di}{dt} = 25$. $L \times \frac{15}{1} = 25$, so $L = \frac{5}{3}$ H. Change in energy $\Delta E = \frac{1}{2}L(i_2^2 - i_1^2) = \frac{1}{2} \times \frac{5}{3} \times (625 - 100) = \frac{5}{6} \times 525 = 437.5$ J.
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