For the given circuit the current through battery of 6 V just after closing the switch 'S' will be _____ A.

Right after the switch is closed, the inductor behaves like an open circuit (since current through an inductor cannot change instantaneously).

So the branch containing the 2 H inductor carries no current initially → ignore that entire right-side loop.

step 1: simplify the circuit just after closing

Remaining network:

• left branch: 2 Ω
• middle branch: 4 Ω
Both connected between the same top and bottom nodes

So they are in series.

$$Now\ R_{eq}=4+2=6Ω$$

So $$I=\frac{V}{R}$$

$$I=V/R=6/(6)=1A$$