In the circuit shown here, the point 'C' is kept connected to point 'A' till the current flowing through the circuit becomes constant. Afterward, suddenly, point 'C' is disconnected from point 'A' an connected to point 'B' at time t =0. Ratio of the voltage across resistance and the inductor at `t =L//R` will be equal to:

Options

A. -1
B. `(1 - e)/(e)`
C. `(e)/(1 - e)`
D. 1

Answer

D

Explanation

When C is connected to A let `I_(0)` be the current set up in the circuit. When C is disconnected to `A` and is connected to `B`, then current in `L-R` circuit decays form `I_(0)`. At time `t`, current is given by
`I = I_(0) e^((-R//L)t), (dI)/(dt) = - (R )/(L) I_(0) e^(-(R )/(L)) = - (R )/(L) I`
When `t = (L)/(R ) , I = I_(0) e^(-(R//L) xx (L//R)) = (I_(0))/(e)`
`(dI)/(dt) = - (R )/(L) xx (I_(0))/(e)`
Voltage across `L`,
`V_(L) = - L (dI)/(dt) = - L [-(R )/(L) (I_(0))/(e)] = (RI_(0))/(e)`
Voltage across `R`, `V_(R ) = IR = (I_(0))/(e) R :. (V_(L))/(V_(R )) = 1`