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. `(E L)/(eR^(2))`
D. `(e L)/(E R)`

Answer

C

Explanation

The current at time `t` is given by `i = i_(0) (1-e^(-t//tau))`
Here `i_(0) = (E)/(R )` and `tau = (L)/(R )`
`:. q = int_(0)^(tau)i d t = int_(0)^(t)i_(0)(1-e^(-t//tau))dt`
`= (i_(0) tau)/(e) = (((E)/(R ))((L)/(R )))/(e) = (EL)/(eR^(2))`