Behavior long after closing the switch

For a DC circuit, long after the switch is closed, the current becomes steady.

• The inductor behaves like a short circuit in steady state.
• So the branch containing the inductance has effectively zero resistance.

Simplify the circuit

Since the two resistors are identical, each has resistance $R=5\,\Omega$

In the given arrangement, after the inductor becomes a short circuit, one of the resistors gets bypassed and the effective resistance seen by the battery becomes just: $R_{\text{eq}}=5+\left(5\parallel 0\right)$

But $5\parallel 0 = 0$

Hence, $R_{\text{eq}}=5\,\Omega$

Current through the battery

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

Given, $V=15\,\text{V}$

Therefore, $I=\frac{15}{5}=3\,\text{A}$

Check options

• A: $6\,\text{A}$
• B: $7.5\,\text{A}$
• C: $3\,\text{A}$
• D: $5.5\,\text{A}$

So the correct option is: $\boxed{\text{C: }3\,\text{A}}$

Comparison with stored answer

The stored correct answer is A (6 A), but from steady-state DC analysis of an inductor, the battery current is clearly $\boxed{3\,\text{A}}$

Hence, I disagree with the stored answer.