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(B) Just after closing the key,the inductor acts as an open circuit (infinite resistance). The circuit simplifies to a series combination of the $10 \

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$\frac{5}{3} 	ext{ A}, 0 	ext{ A}$

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(B) Just after closing the key,the inductor acts as an open circuit (infinite resistance). The circuit simplifies to a series combination of the $10 	ext{ V}$ battery,the $2 \Omega$ resistor,and the $4 \Omega$ resistor. The current $I$ is given by $I = \frac{V}{R_{eq}} = \frac{10}{2 + 4} = \frac{10}{6} 	ext{ A} = \frac{5}{3} 	ext{ A}$.$(b)$ $A$ long time after closing the key,the inductor reaches a steady state and acts as a short circuit (zero resistance). Since the inductor is in parallel with the $4 \Omega$ resistor,all the current will flow through the inductor,and the potential difference across the $4 \Omega$ resistor becomes zero. Thus,the current in the $4 \Omega$ resistor is $0 	ext{ A}$.

Find the current in the $4 \Omega$ resistance shown in the figure in the following cases:
$(a)$ Just after the closing of the key.
$(b)$ $A$ long time after the closing of the key.

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Find the current in the $4 \Omega$ resistance shown in the figure in the following cases:$(a)$ Just after the closing of the key.$(b)$ $A$ long time after the closing of the key.