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The cell in the circuit shown in figure is ideal. The coil has an inductance of 4 mH and a resistance of $2 m Ω$ . The switch is closed at t = 0. The amount of energy stored in the inductor at t=2 s is (take e=3)

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$\frac{4}{3} J$

$\frac{8}{9} \times 10^{3} J$

$2 \times 10^{3} J$

$\frac{8}{3} \times 10^{- 3} J$

answer is B.

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Detailed Solution

$U s e i = \frac{E}{R} [1 - e^{- t / τ}] a n d U = \frac{1}{2} L i^{2}$

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