An inductor of inductance $$L = 400 mH$$ and resistors of resistance $${R_1} = 2\Omega $$ and $${R_2} = 2\Omega $$ are connected to a battery of emf $$12\,V$$ as shown in the figure. The internal resistance of the battery is negligible. The switch $$S$$ is closed at $$t = 0.$$ The potential drop across $$L$$ as a function of time is:
A.
$$\frac{{12}}{t}{e^{ - 3t}}V$$
B.
$$6\left( {1 - {e^{ - \frac{t}{{0.2}}}}} \right)V$$
C.
$$12{e^{ - 5t}}V$$
D.
$$6{e^{ - 5t}}V$$
Answer :
$$12{e^{ - 5t}}V$$
Solution :
Growth in current in $$L{R_2}$$ branch when switch is closed is given by
$$i = \frac{E}{{{R_2}}}\left[ {1 - {e^{ - \frac{{{R_2}t}}{L}}}} \right] \Rightarrow \frac{{di}}{{dt}} = \frac{E}{{{R_2}}}.\frac{{{R_2}}}{L}.{e^{ - \frac{{{R_2}t}}{L}}} = \frac{E}{L}{e^{ - \frac{{{R_2}t}}{L}}}$$
Hence, potential drop across
$$\eqalign{
& L = \left( {\frac{E}{L}{e^{ - \frac{{{R_2}t}}{L}}}} \right)L = E{e^{ - \frac{{{R_2}t}}{L}}} = 12{e^{ - \frac{{2t}}{{400 \times {{10}^{ - 3}}}}}} \cr
& = 12{e^{ - 5t}}V \cr} $$
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