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Step-by-Step Solution
Step 1: Identify the Maximum Current in the Circuit
When the circuit has been connected to the battery for a long time (steady state), the inductor acts like a simple connecting wire, and the current reaches its maximum value. This maximum current is given by:
I_{\max} = \frac{V}{R} .
Here, V = 20\,\text{V} and R = 10\,\text{k}\Omega = 10 \times 10^3 \,\Omega . So:
I_{\max} = \frac{20}{10 \times 10^3} = 2 \times 10^{-3}\,\text{A} = 2\,\text{mA}.
Step 2: Use the Exponential Decay Formula
Immediately after the switch is opened, the current in an LR circuit decays according to the formula:
I(t) = I_{\max} \, e^{-\frac{R\,t}{L}},
where
R is the resistance,
L is the inductance,
t is the time elapsed after the circuit is opened.
Step 3: Substitute the Known Values
Given:
R = 10 \times 10^3\,\Omega ,
L = 10\,\text{mH} = 10 \times 10^{-3}\,\text{H} ,
t = 1\,\mu\text{s} = 1 \times 10^{-6}\,\text{s} ,
I_{\max} = 2\,\text{mA} .
So,
I(1 \times 10^{-6}\,\text{s}) = 2\,\text{mA} \cdot e^{-\frac{(10 \times 10^3)\times(1 \times 10^{-6})}{10 \times 10^{-3}}}.
Step 4: Simplify the Exponential
Inside the exponential, we have:
\frac{R\,t}{L} = \frac{(10 \times 10^3)\times(1 \times 10^{-6})}{10 \times 10^{-3}} = \frac{10^{4} \times 10^{-6}}{10^{-2}} = \frac{10^{-2}}{10^{-2}} = 1.
Thus,
I(1 \times 10^{-6} \,\text{s}) = 2\,\text{mA} \cdot e^{-1}.
We are given e^{-1} \approx 0.37 . Hence,
I(1 \times 10^{-6} \,\text{s}) = 2\,\text{mA} \times 0.37 = 0.74\,\text{mA}.
In the form given by the problem statement, this is \frac{74}{100}\,\text{mA} . Therefore, x = 74.
Step 5: Final Answer
The required value of x is 74.
