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

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Its time constant is 0.25s

In steady state, current through the inductance will be equal to zero

In steady state, current through the battery will be equal to 0.75A

None of these

answer is A, C.

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Detailed Solution

In the circuit shown in the figure in problem, $6 Ω a n d 12 Ω$ resistances are in parallel with each other and
their parallel combination is in series with $4 Ω$ and the inductance of 2H. Hence, equivalent resistance of these three resistances is equal to $8 Ω$ .Therefore, the time constant for the circuit is

$λ = \frac{L}{R} = \frac{2}{8} = 0.25 s$
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.