For the circuit shown in figure, which of the following statements is/are correct?

 

In the circuit shown in the figure in problem, $6\Omega and12\Omega$ resistances are in parallel with each other and 
 their parallel combination is in series with $4\Omega$ and the inductance of 2H. Hence, equivalent resistance of  these three resistances is equal to $8\Omega$.Therefore, the time constant for the circuit is 

$\lambda =\frac{L}{R}=\frac{2}{8}=0.25s$                                                                    
Hence (1) is correct.
In steady state, no e.m.f. will be induced in the inductance. Hence current through the circuit will be equal to $\frac{E}{R}$ where R is the equivalent resistance. Hence, the steady state current will be equal to 6/8=0.75A.

Hence (3) is correct and (2) is wrong.