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Q.

The self induced emf of a coil is $25$ volts. When i the current in it is changed at uniform rate from $10 A t o 25 A$ in $0.5 s,$ the change in the energy of the inductance is

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a

637.5 J

b

525 J

c

437.5 J

d

740 J

answer is B.

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Detailed Solution

$ε = 60 V, I_{1} = 10 A, I_{2} = 25 A, t = 0.5 s, ∆ E = ? ε = L \frac{∆ I}{∆ t} o r 60 = L (\frac{25 - 10}{0.5}); L = \frac{60}{30} = 2 H ∆ E = \frac{1}{2} L (I_{2}^{2} - I_{1}^{2}) = \frac{1}{2} \times 2 \times (25^{2} - 10^{2}) = 525 = 525 J$

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