An inductor of 10 mH is connected to a 20 V battery through a resistor of 10 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 $$\mu$$s is $$\frac{x}{100}$$ mA. Then $$x$$ is equal to ___. (Take $$e^{-1} = 0.37$$)

For an R-L circuit we first recall the basic facts. When a constant emf is suddenly applied, the current grows according to

$$I(t)=I_{0}\left(1-e^{-t/\tau}\right),$$

and when the source is removed after reaching the steady value $$I_{0},$$ the current decays as

$$I(t)=I_{0}\,e^{-t/\tau}.$$

Here $$\tau$$ is the time constant of the circuit, given by the well-known formula

$$\tau=\frac{L}{R},$$

where $$L$$ is the self-inductance and $$R$$ is the resistance.

Now let us insert the numerical data. We have

$$L = 10\ \text{mH}=10\times10^{-3}\ \text{H},\qquad R = 10\ \text{k}\Omega = 10\times10^{3}\ \Omega.$$

Therefore

$$\tau = \frac{L}{R} = \frac{10\times10^{-3}}{10\times10^{3}} = \frac{10}{10}\times\frac{10^{-3}}{10^{3}} = 1\times10^{-6}\ \text{s} = 1\ \mu\text{s}.$$

Before the switch is opened, the current has reached its maximum steady value. Ohm’s law for the dc source gives this value as

$$I_{0}=\frac{V}{R} =\frac{20\ \text{V}}{10\times10^{3}\ \Omega} =\frac{20}{10}\times10^{-3}\ \text{A} =2\times10^{-3}\ \text{A} =2\ \text{mA}.$$

Immediately after the battery is disconnected, the current starts to fall. After a time $$t$$ the current is given by the exponential decay formula

$$I(t)=I_{0}\,e^{-t/\tau}.$$

We are asked to find the current at $$t = 1\ \mu\text{s}.$$ Substituting $$t=1\ \mu\text{s}$$ and $$\tau = 1\ \mu\text{s},$$ we get

$$I(1\ \mu\text{s}) = I_{0}\,e^{-1} = 2\ \text{mA}\times e^{-1}.$$

The problem statement provides the numerical value $$e^{-1}=0.37.$$ Hence

$$I(1\ \mu\text{s}) = 2\ \text{mA}\times 0.37 = 0.74\ \text{mA}.$$

We can express $$0.74\ \text{mA}$$ as a fraction of a milliampere:

$$0.74\ \text{mA} = \frac{74}{100}\ \text{mA}.$$

By comparison with the form given in the question, $$\displaystyle I=\frac{x}{100}\ \text{mA},$$ we identify

$$x = 74.$$

So, the answer is $$74$$.