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Consider the LR circuit shown in the fig...

Consider the LR circuit shown in the figure. If the switch S is closed at t = 0 then the amount of charge that passes through the battery between `t=0` and `t=L/R` is

A

`(EL)/(2.7R^(2))`

B

`(2.7EL)/(R^(2))`

C

`(EL)/(7.3R^(2))`

D

`(7.3EL)/(R^(2))`

Text Solution

Verified by Experts

The correct Answer is:
A
`q=int_(0)^(tau)E/R(1-e^(-(tR)/L))dt`
`=[E/Rt-[E/Re^(-(tR)/L)]L/R]_(0)^(tau)=E/R tau+(EL)/(R^(2))e^(-1)-(EL)/(R^(2))=(EL)/(eR^(2))-(EL)/(R^(2))=(EL)/(eR^(2))=(EL)/((2.7)R^(2))e=2.7`
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