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

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TestHub · Accepted Answer

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

$\frac{1}{R_{e q}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1 \Rightarrow R_{e q} = 1 Ω$

so, the circuit becomes,

$\Rightarrow I = \frac{30}{3} = 10 A$

Physics - EMI/AC Question with Solution | TestHub

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