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

Options

A. `(EL)/(2.7R^(2))`
B. `(2.7EL)/(R^(2))`
C. `(EL)/(7.3R^(2))`
D. `(7.3EL)/(R^(2))`

Answer

A

Explanation

`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`