Question

The R–L circuit with $R=10\,\text{k}\Omega$ and $L=1\,\text{mH}$ is excited by a step current $I_0u(t)$. At $t=0^-$, the inductor current is $i_L=I_0/5$. The minimum time for $i_L(t)$ to reach $99%$ of its final value is _____ $\mu$s (round off to two decimal places)

Hint:

For a current source feeding parallel $R$–$L$, the inductor obeys $di_L/dt+(R/L)i_L=(R/L)I_0$ with $\tau=L/R$; use the exact initial condition to get the {minimum} time to a given percentage.

Answer 1

Correct Answer: 0.43

With a current source feeding the parallel $R$–$L$, KCL at the node gives \[ I_0 = i_R + i_L = \frac{v}{R} + i_L,\qquad v=L\,\frac{di_L}{dt}. \] Hence \[ \frac{di_L}{dt}+\frac{R}{L}\,i_L=\frac{R}{L}\,I_0 \Rightarrow i_L(t)=I_0+\big(i_L(0^+)-I_0\big)e^{-t/\tau}, \] where $\tau=\dfrac{L}{R}=\dfrac{1\,\text{mH}}{10\,\text{k}\Omega}=0.1\,\mu\text{s}$ and $i_L(0^+)=I_0/5$. Set $i_L(t)=0.99\,I_0$: \[ 0.99I_0=I_0-\frac{4}{5}I_0\,e^{-t/\tau} \ \Rightarrow\ e^{-t/\tau}=0.0125 \ \Rightarrow\ t=\tau\ln(80)=0.1\,\mu\text{s}\times 4.382=0.438\,\mu\text{s}\approx 0.44\,\mu\text{s}. \]