In the circuit shown in figure switch `S` is closed at time `t=0`. The charge which passes through the battery in one time constant is

Options

A. `(eR^(2)E)/(L)`
B. `E((L)/(R ))`
C. `(EL)/(eR^(2))`
D. `(eL)/(ER)`

Answer

C

Explanation

The current at time `T` is given by
`i=i_(0)(1-e^(-t//tau))`
where `i_(0)=(E)/(R )` and `tau=(L)/(R )`
So `dq=I dt=i_(0)(1-e^(-t//tau))dt`
`Q=intdQ=in_(0)^(R )i_(0)(1-e^(t//tau))dt=i0{[in_(0)^(R )dt-int_(0)^(R )e^(-t//tau)dt]}`
`=(i_(0)tau)/(e)impliesQ=((E)/(R ).(L)/(R))/(e)=(EL)/(eR^(2))impliesQ=(EL)/(ER^(2))`