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(D) Just after closing the switch $S$ $(t = 0)$,the inductor opposes any change in current,so it behaves as an open circuit (infinite resistance).Ther

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(D) Just after closing the switch $S$ $(t = 0)$,the inductor opposes any change in current,so it behaves as an open circuit (infinite resistance).Therefore,the branch containing the inductor carries no current.The circuit simplifies to the $6\,V$ battery in series with the $2\,\Omega$ resistor and the $4\,\Omega$ resistor.The total resistance of the circuit is $R_{eq} = 2\,\Omega + 4\,\Omega = 6\,\Omega$.Using Ohm's law,the current through the battery is $I = \frac{V}{R_{eq}} = \frac{6\,V}{6\,\Omega} = 1\,A$.

For the given circuit,the current through the battery of $6\,V$ just after closing the switch $S$ will be $.......A$.

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