An inductor of
Question:

An inductor of $10 \mathrm{mH}$ is connected to a $20 \mathrm{~V}$ battery through a resistor of $10 \mathrm{k} \Omega$ and a switch. After a long time, when maximum current is set up in the circuit, the current is switched off. The current in the circuit after $1 \mu \mathrm{s}$ is $\frac{\mathrm{x}}{100} \mathrm{~mA}$. Then $\mathrm{x}$ is equal to __________ (Take $\left.\mathrm{e}^{-1}=0.37\right)$

Solution:

$\mathrm{I}_{\max }=\frac{\mathrm{V}}{\mathrm{R}}=\frac{20 \mathrm{~V}}{10 \mathrm{~K} \Omega}=2 \mathrm{~mA}$

For LR – decay circuit

$\mathrm{I}=2 \mathrm{~mA} \mathrm{e} \mathrm{e}^{-1}$

$\mathrm{I}=2 \times 0.37 \mathrm{~mA}$

$\mathrm{I}=\frac{74}{100} \mathrm{~mA}$

$x=74$

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