Time -Varying current in L-R circuit

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Question

An inductor of 2 henry and a resistance of 10 ohms are connected in series with a battery of 5 volts. The initial rate of change of current is

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Solution

$i = {i_0}\left( {1 - {e^{\frac{{ - Rt}}{L}}}} \right) \Rightarrow \frac{{di}}{{dt}} = \frac{d}{{dt}}{i_0} - \frac{d}{{dt}}{i_0}{e^{\frac{{ - Rt}}{L}}}$
$\Rightarrow \frac{{di}}{{dt}} = 0 - {i_0}\left( { - \frac{R}{L}} \right)\;{e^{ - \frac{{Rt}}{L}}} = \frac{{{i_0}R}}{L}{e^{ - \frac{{Rt}}{L}}}$
$Initially\,\,t = 0 \Rightarrow \frac{{di}}{{dt}} = \frac{{{i_0} \times R}}{L} = \frac{E}{L} = \frac{5}{2} = 2.5\;amp/\sec .$

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