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(B) When the key $K$ is closed,the inductor opposes the change in current. However,after a long time,the current reaches a steady state.In the steady

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$5$

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(B) When the key $K$ is closed,the inductor opposes the change in current. However,after a long time,the current reaches a steady state.In the steady state,the inductor acts as a simple wire with zero resistance (ideal inductor).The circuit behaves as a simple $DC$ circuit with only the battery and the resistor.Using Ohm's Law: $I = \frac{V}{R}$.Given $V = 10 \ V$ and $R = 2 \ \Omega$.Therefore,$I = \frac{10}{2} = 5 \ A$.

Calculate the current,a long time after the closing of the key $K$ in a circuit containing a battery of $10 \ V$,a resistor of $2 \ \Omega$,and an inductor of $5 \ H$. (in $A$)

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