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

i=E/R2(1-e<sup>-R2t/L</sup>)

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

Was this answer helpful?

  
   

0 (0)

Upvote (0)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

BOOK

Free Class

Ask
Question