Question

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 $e^{-1} = 0.37$)

Hint:

After one time constant (\(t = \tau\)), any decaying quantity in an RC or LR circuit drops to exactly 37% of its initial value.

Answer 1

Correct Answer: 74

Step 1: Understanding the Concept:
When the current in an LR circuit is switched off, the energy stored in the inductor decays through the resistor, following an exponential decay.
Step 2: Key Formula or Approach:
1. Maximum current (steady state): \(I_0 = \frac{V}{R}\).
2. Decay equation: \(I = I_0 e^{-t/\tau}\), where \(\tau = \frac{L}{R}\).
Step 3: Detailed Explanation:
Calculate \(I_0\):
\[ I_0 = \frac{20 \text{ V}}{10^4 \text{ }\Omega} = 2 \times 10^{-3} \text{ A} = 2 \text{ mA} \]
Calculate time constant \(\tau\):
\[ \tau = \frac{10 \times 10^{-3} \text{ H}}{10^4 \text{ }\Omega} = 10^{-6} \text{ s} = 1 \mu\text{s} \]
Current at \(t = 1 \mu\text{s}\):
Since \(t = \tau\):
\[ I = I_0 e^{-1} = 2 \text{ mA} \times 0.37 = 0.74 \text{ mA} \]
According to the question:
\[ I = \frac{x}{100} \text{ mA} = 0.74 \implies x = 74 \]
Step 4: Final Answer:
The value of \(x\) is 74.