An inductor of inductance L = 400 mH and resistors of R1 = 2Ω and R2 = 2Ω are connected to a battery of emf 12 V as shown in the figure. The internal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is?

Consider the branch containing L and R2

Image pending


di/dt = E/R2e<sup>-R2t/L</sup>.R2/l

di/dt = E/L e<sup>-R2t/L</sup>

Therefore, V<sub>L</sub> = l di/dt = E e<sup>-R2t/L</sup> = 12e<sup>-5t</sup>V

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