In the circuit shown in figure, the switch `S` is closed at `t=0`. If `V_L` is the voltge induced across the inductor and `i` is the instantaneous current, the correct variation of `V_L` versus is given by

Options

A.
B.
C.
D.

Answer

D

Explanation

`i=0(1-e^(-t//tau_L))=E/R (1-e^(-t//tau_L))`
`=E/R -(Ee^(-t//tau-L))/R=i_0-(V_C/R) (as V_L =Ee^(-t//tau_L))`
`:. V_L=(i_0R)-(R)i`
`V_L` versus i graph is a straight line with positive intercept and negastive slope.