An inductor of inductance L = 400 mH and resistors of resistances R_1 = 2 Ω and R_2 = 2 Ω are connected to a b

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An inductor of inductance L = 400 mH and resistors of resistances R_1 = 2 Ω and R_2 = 2 Ω are connected to a battery of emf 12 V . 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: ![image](https://pub-7faa45

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![image](https://pub-7faa453bd1ec4bf0a04cc031c6f058de.r2.dev/images/dbe92d2d-be62-4768-b14c-71b48aeda489.png) The current in the branch with the inductor L and resistor R_2 follows the equation: I_2(t) = E/R_2 (1 - e^-R_2 t / L) The potential drop across the inductor is V_L = L dI_2/dt . dI_2/dt = E/R_2 ≤ft( R_2/L e^-R_2 t / L ) = E/L e^-R_2 t / L V_L = L × E/L e^-R_2 t / L = E e^-R_2 t / L Given E = 12 V , R_2 = 2 Ω , L = 0.4 H . V_L = 12 e^-2t/0.4 = 12 e^-5t V

An inductor of inductance $L = 400 mH$ and resistors of resistances $R_{1} = 2 Ω$ and $R_{2} = 2 Ω$ are connected to a battery of emf $12 V$ . 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:

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An inductor of inductance L=400mH and resistors of resistances R1​=2Ω and R2​=2Ω are connected to a battery of emf 12V. 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: