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GATE 2023 IN – Question 47

Electrical Circuits and Machines · Transients and two-port networks · 2 marks · Numerical answer

The R–L circuit shown has $R=10$ kΩ, $L=1$ mH, and step-current excitation $I_0u(t)$. At $t=0^-$, inductor current is $I_L=I_0/5$. The minimum time for inductor current to reach 99% of its final value is ______ μs (two decimal places).

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Correct answer: 0.43 to 0.45

Explanation

The time constant is $\tau=L/R=10^{-3}/10^4=0.1$ μs. Including the initial current gives $i_L/I_0=1-0.8e^{-t/\tau}$. Set this to 0.99: $t=\tau\ln80=0.4382$ μs. Answer: **0.44 μs**.