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Inductor L = 100mH , resistance =100 W AND battery = 100 V are connected in series. After long time battery is disconnected current in it after 10 -3 s?? ….….1/e A, A e, 1 A

Inductor L = 100mH , resistance =100 W AND battery = 100 V are connected in series. After long time battery is disconnected current in it after 10-3 s?? ….….1/e A, A e, 1 A

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2 Answers

Sumit Majumdar IIT Delhi
askIITians Faculty 137 Points
7 years ago
Dear student,
The current at any instant of time is given by:
I\left ( t \right )=\frac{V}{R}\left ( 1-e^{\frac{-Rt}{L}} \right )
hence, the current would be given by:
I\left ( t \right )=\frac{V}{R}\left ( 1-e^{\frac{-Rt}{L}} \right )=\frac{100}{100}\left ( 1-e^{\frac{-100\times 10^{-3}}{100\times 10^{-3}}} \right )=\left ( 1-e^{-1} \right ) amp
Regards
Sumit
Saurabh Kumar
24 Points
7 years ago
For the discharging, through R, put the value and find the answer.

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