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Figure shows a circuit contains three identical resistors with resistance R = 9.0 $\Omega$ each, two identical inductors with inductance L = 2.0 mH each, and an ideal battery with emf $\varepsilon$ = 18 V. The current 'i' through the battery just after the switch closed is
Explanation
Just after closing, inductors carry no current, leaving only one resistor in the path.
Detailed Solution
At t = 0 the inductors behave as open circuits, so no current flows through $R_1$ and $R_3$.
$i = \frac{\varepsilon}{R_2} = \frac{18}{9} = 2$ A
Note: the source says the question is not correctly framed, but 2 A is the best option of those given.