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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0.5 amp/sec

2.0 amp/sec

2.5 amp/sec

0.25 amp/sec

answer is C.

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Detailed Solution

$i = i_{0} (1 - e^{\frac{- Rt}{L}}) \Rightarrow \frac{di}{dt} = \frac{d}{dt} i_{0} - \frac{d}{dt} i_{0} e^{\frac{- Rt}{L}}$

$\Rightarrow \frac{di}{dt} = 0 - i_{0} (- \frac{R}{L}) e^{- \frac{Rt}{L}} = \frac{i_{0} R}{L} e^{- \frac{Rt}{L}}$

$Initially, t = 0 \Rightarrow \frac{di}{dt} = \frac{i_{0} \times R}{L} = \frac{E}{L} = \frac{5}{2} = 2 .5 amp / \sec .$

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