Q 1339178912.     In the given circuit shown : `L= 200 mH` Emp of battery `E= 12 VR_1=R_2=2 Omega` the internal resistance of battery is negligible . The potential drop across the inductance `L` after time `t`.



A

`6 e^{-5t} V`

B

`frac{12}{t} e^{-3t} V`

C

`6 [1-e^{t//.2}]V`

D

`12 e^{-10 t}V`

HINT

`i = frac{E}{R_2} [1- e^{-R_2 t//l}]`

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