For the given circuit the current $$i$$ through the battery when the key is closed and the steady state has been reached is:

We need to determine the total steady-state current $$i$$ flowing through the battery after the key is closed.

1. Core Concept: Inductor Behavior in Steady State

When a DC circuit containing inductors reaches a steady-state condition (a long time after the switch is closed):

• The current through the circuit components stabilizes and becomes constant ($$\frac{di}{dt} = 0$$).
• Because the induced electromotive force in an inductor is given by $$e = -L\frac{di}{dt}$$, the voltage across each inductor drops to zero.
• Therefore, in a steady-state DC circuit, every ideal inductor behaves simply as a short-circuit (a zero-resistance connecting wire).

2. Simplify the Circuit Structure

Based on the layout of this standard problem:

• The power supply consists of a $$30\text{ V}$$ battery connected in series with a main-line $$2\ \Omega$$ resistor.
• Following this series resistor, the circuit splits into three parallel branches.
• Each of these three parallel branches contains a $$3\ \Omega$$ resistor (with two of the branches also containing inductors of values $$0.5\text{ mH}$$ and $$0.2\text{ H}$$ that are now replaced with direct shorting wires).

3. Calculate the Total Equivalent Resistance ($$R_{\text{eq}}$$)

First, find the equivalent resistance ($$R_p$$) of the three parallel branches, where each active branch presents an individual resistance of $$3\ \Omega$$:

$$\frac{1}{R_p} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = \frac{3}{3} = 1\ \Omega^{-1}$$

$$R_p = 1\ \Omega$$

Next, add the main line $$2\ \Omega$$ resistor that is connected in series with this entire parallel group to get the total circuit resistance:

$$R_{\text{eq}} = R_{\text{series}} + R_p = 2\ \Omega + 1\ \Omega = 3\ \Omega$$

4. Calculate the Total Current ($$i$$)

Using Ohm’s Law ($$i = \frac{V}{R_{\text{eq}}}$$), we can compute the steady-state current leaving the battery:

$$i = \frac{30\text{ V}}{3\ \Omega} = 10\text{ A}$$

Conclusion

The current through the battery once the steady state has been reached is 10 A, which corresponds to Option A.