An inductor of inductance L=400mH and resistors of resistances

Question:

An inductor of inductance $\mathrm{L}=400 \mathrm{mH}$ and resistors of resistances $\mathrm{R}_{1}=2 \Omega$ and $\mathrm{R}_{2}=2 \Omega$ are connected to a battery of emf $12 \mathrm{~V}$ as shown in the figure. The internal resistance of the battery is negligible. The switch $\mathrm{S}$ is closed at $\mathrm{t}=0$. The potential drop across $\mathrm{L}$ as a function of time is:-

  1. $6\left(1-\mathrm{e}^{-t / 0.2}\right) \mathrm{V}$

  2. $12 \mathrm{e}^{-5 \mathrm{t}} \mathrm{V}$

  3. $6 e^{-5 t} \mathrm{~V}$

  4. $\frac{12}{t} e^{-3 t} V$


Correct Option: , 2

Solution:

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now