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

• A. 0.5 amp/sec ​
• B. 2.0 amp/sec
• C. 2.5 amp/sec
• D. 0.25 amp/sec ​

Answer 1

The correct option is C 2.5 amp/sec

$i=i_0(1-e^{\frac{-Rt}{L}})\Rightarrow \frac{di}{dt}=\frac{d}{dt}{i}_0-\frac{d}{dt}{i}_0{e}^{\frac{-Rt}{L}}$
$\Rightarrow \frac{di}{dt}=0-i_0(-\frac{R}{L})e^{-\frac{Rt}{L}}=\frac{i_{0}R}{L}{e}^{-\frac{Rt}{L}}$
Initially, $t=0\Rightarrow \frac{di}{dt}=\frac{i_0\times R}{L}=\frac{E}{L}=\frac{5}{2}=2.5amp/sec$