JEE MAIN - Physics (2019 - 10th January Evening Slot - No. 21)
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 1s, the change in the energy of the inductance is -
740 J
637.5 J
540 J
437.5 J
Explanation
$$L{{di} \over {dt}} = 25$$
$$L \times {{15} \over 1} = 25$$
$$L = {5 \over 3}H$$
$$\Delta E = {1 \over 2} \times {5 \over 3} \times ({25^2} - {10^2})$$
$$ = {5 \over 6} \times 525 = 437.5$$ J
$$L \times {{15} \over 1} = 25$$
$$L = {5 \over 3}H$$
$$\Delta E = {1 \over 2} \times {5 \over 3} \times ({25^2} - {10^2})$$
$$ = {5 \over 6} \times 525 = 437.5$$ J
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