First slide

Inductance

Question

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

Moderate

A

41.5 J
B

62.50 J
C

125 J
D

250 J

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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