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In the circuit shown in figure, the switch S is closed at t = 0, when reading of the voltmeter is 8 volt, the energy stored in the inductor is [Given $R = 4 Ω and L = 10 mH$ ]

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20 mJ

16 mJ

25 mJ

50 mJ

answer is A.

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Detailed Solution

$V = iR \Rightarrow i = \frac{8}{4} A = 2 A$

$∴$ U = Magnetic energy stored in the inductor $= \frac{1}{2} {Li}^{2} = \frac{1}{2} \times 10 \times 10^{- 3} \times {(2)}^{2} J = 20 mJ$

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