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An inductor of L = 400 mH and two resistors of R₁ = 2Ω and R₂ = 2Ω are connected to a 12V battery as shown in the figure. The internal resistance of the battery is negligible. The switch is closed at time t = 0. The potential drop across ‘L’ as a function of time is

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12e^–5t V

3e^–10t V

12(1 – e^–5t)V

6(1 – e^–5t)V

answer is B.

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Detailed Solution

$E = E_{0} e^{- \frac{R}{L} t};$

$E = 12 e^{- \frac{2}{0 .4} t};$

$E = 12 e^{- 5 t}$

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