$(a)\;6\;e^{-5t}\;V \\ (b)\;\frac{12}{t} e^{-3t} \;V \\(c)\;6 [1-e^{t/.2}]V \\(d)\;12\;e^{-10\;t}V $

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Let Emf of battery be E.

Current in the inductance branch of the circuit is

$i =\large\frac{E}{R_2} \bigg[1- e^{-R_2 t/l}\bigg]$

$\large\frac{di}{dt} =\frac{E}{R_2} \frac{R_2}{L} \bigg[e^{-R_2 t/l}\bigg]$

$\qquad= \large\frac{E}{L} e^{-R_2 t/l}$

Drop across $L\bigg[ \large\frac{E}{L} e^{-R_2 t/L}\bigg]$$\times L$

$\qquad= 12 e^{\Large\frac{-2t}{200 \times 10^{-3} }}$

$\qquad=12\;e^{-10\;t}V $

Hence d is the correct answer.

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