For the given circuit the current i through the battery when the...

In a steady state, the inductor behaves as a conducting wire. So, the equivalent circuit becomes,

$\dfrac{1}{{R}_{eq}} = \dfrac{1}{{R}_{1}} + \dfrac{1}{{R}_{2}} + \dfrac{1}{{R}_{3}} = \dfrac{1}{3} + \dfrac{1}{3} + \dfrac{1}{3} = 1 ⇒ {R}_{eq} = 1Ω$

so, the circuit becomes,

$⇒ I = \dfrac{30}{3} = 10A$