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JEE Main 2021 (25 Jul, Shift 1) · Inductors & DC Transients (Mixed)

Inductors & DC Transients (Mixed) question from JEE Main 2021 (25 Jul, Shift 1)

Q10JEE Main 2021 · 25 JUL · SHIFT 1Easy

Inductors & DC Transients (Mixed)

An inductor of 10 mH is connected to a 20V battery through a resistor of 10 kΩ\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 μ\mus is x100\frac{x}{100} mA. Then x is equal to ___________. (Take e______1 = 0.37)

Numerical answer type. Work it out, then open the answer.

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Answer: 74

  1. Steady current I0=2010×103=2 mAI_0 = \frac{20}{10\times10^{3}} = 2\text{ mA}
  2. Decay: I=I0e−t/τI = I_0 e^{-t/\tau}, τ=LR=10×10−3104=1 μs\tau = \frac{L}{R} = \frac{10\times10^{-3}}{10^4} = 1\,\mu\text{s}
  3. At t=1 μst = 1\,\mu\text{s}: I=2e−1=2×0.37=0.74 mA=74100 mAI = 2e^{-1} = 2 \times 0.37 = 0.74\text{ mA} = \frac{74}{100}\text{ mA}
  4. So x=74x = 74.
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