The circuit shown in the figure with the switch S open, is in steady state. After the switch S is closed, the time constant of the circuit in seconds is

The correct option is 1.25.

Step-by-Step Explanation:

To find the time constant of the circuit after the switch S is closed, we analyze the circuit parameters and simplify the components as follows:

1. Analyze the given circuit diagram:
From the schematic, we can identify:

• An independent current source of $1\text{A}$ on the leftmost branch.
• A parallel branch containing a $1\Omega$ resistor connected in series with a $1\text{H}$ inductor.
• A switch S that, when closed, connects the left part of the circuit to the right part.
• The branch after the switch S has a series combination of a $1\Omega$ resistor and a $1\text{H}$ inductor, which then feeds into two parallel-connected inductors, each having a value of $1\text{H}$.

2. Calculate the equivalent inductance of the right-hand section:
First, we find the equivalent inductance of the two parallel $1\text{H}$ inductors on the far right:
$$L_p=1\text{H}\parallel 1\text{H}=\frac{1\times 1}{1+1}=0.5\text{H}$$
This parallel combination is in series with the other $1\text{H}$ inductor in the same branch:
$$L_{\text{right}}=1\text{H}+0.5\text{H}=1.5\text{H}$$

3. Determine the equivalent parameters of the circuit with the switch closed:
To determine the time constant, we find the equivalent resistance ($R_{\text{eq}}$) and equivalent inductance ($L_{\text{eq}}$) looking from the reactive elements. We do this by deactivating the independent current source (which becomes an open circuit):

• With the current source open-circuited and switch S closed, the left-hand branch and the right-hand branch form a single closed loop in series.
• The total equivalent resistance of the loop is:
  $$R_{\text{eq}}=1\Omega +1\Omega =2\Omega$$
• The total equivalent inductance of the loop is:
  $$L_{\text{eq}}=1\text{H}+1.5\text{H}=2.5\text{H}$$

4. Calculate the time constant ($\tau$):
The time constant of an RL circuit is given by:
$$\tau =\frac{L_{\text{eq}}}{R_{\text{eq}}}$$
Substituting the calculated values:
$$\tau =\frac{2.5}{2}=1.25\text{seconds}$$