Steady-state behavior of the inductor

When the key is closed and a long time has passed, the circuit reaches steady state for DC. In steady state, an ideal inductor behaves like a short circuit because $V_L=L\frac{di}{dt}=0$ since the current becomes constant.

Interpretation of the circuit

In the given circuit, after the inductor becomes a short circuit, the network reduces to the battery connected to an equivalent resistance.

On simplifying the resistor combination in the circuit, the equivalent resistance seen by the battery comes out to be $R_{\text{eq}}=1\,\Omega$

Current through the battery

Using Ohm’s law, $i=\frac{V}{R_{\text{eq}}}$

With battery voltage $V=10\text{ V}$, $i=\frac{10}{1}=10\text{ A}$

Matching with the options

Therefore, the current through the battery is $\boxed{10\text{ A}}$

So the correct option is C.