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Q.

A coil resistance 20 $Ω$ and inductance 5H is connected with a 100V battery. Energy stored in the coil will be

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a

62.50 J

b

41.5 J

c

250 J

d

125 J

answer is B.

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Detailed Solution

$U = \frac{1}{2} {Li}^{2} = \frac{1}{2} L {(\frac{E}{R})}^{2} = \frac{1}{2} \times 5 \times {(\frac{100}{20})}^{2} = 62 .50 J$

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