An inductor of 10 mH is connected to a 20V battery through a resistor of 10 kΩ 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 μs is x/100 mA. Then x is equal to ___________. (Take e⁻¹ = 0.37)

Solution

Physics sir · 3 steps

Find the steady current, find L/R, and let the current decay by one factor of e in one time constant.

1. Steady current. After a long time the inductor acts like a plain wire, so only the resistor limits the current. I₀ = V/R = 20/(10 × 10³) = 2 × 10⁻³ A = 2 mA.
2. Time constant. τ = L/R = (10 × 10⁻³)/(10 × 10³) = 10⁻⁶ s = 1 μs. So the time asked about is exactly one time constant.

   I = I₀ e^(-t/τ) = 2 mA × e⁻¹ = 2 × 0.37 mA = 0.74 mA

3. So. 0.74 mA = 74/100 mA, so x = 74.

The common slip: 126. Using the growth formula I₀(1 - e⁻¹) instead of the decay formula gives 1.26 mA, that is x = 126. The current is falling here, not rising.

The idea to keep. When an LR circuit is switched off, the current decays as I₀ e^(-t/τ) with τ = L/R. After one time constant only about 37% of the current is left.

Answer from the NTA final answer key. Solution by PrepWiser.