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As shown in the figure, a battery of emf...

As shown in the figure, a battery of emf is connected to an inductor L and resistance R in series. The switch is closed at . The total charge that flows from the battery, between `t=t_e` and `t=2t_e` (`t_e` is the time constant of the circuit) is:

A

`(epsiL)/(R^2)(1-1/e)`

B

`(epsiL)/R^2 (1+1/e^2 -1/e)`

C

`(epsiL)/(eR^2) (1- 2/e^2 - 1/e)`

D

`(epsiL)/(eR^2)`

Text Solution

Verified by Experts

The correct Answer is:
B
`i = i_0 ( 1-e^(-R//L))=i=_0 (1-e^(-t//t_(C)))`
`q = underset(t_C)overset(2t)(int)i" "dt " "underset(t_C)overset(2t)(int)epsi/R(1-e^(-t//t_C))dt = (epsiL)/R^2(1+1/e^2 -1/e)`
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