Question

Consider a series circuit including a 4Ω, a 16mH-inductor, and a power supply of 9 volts.

How many times constant does it take for the current to reach 99.9% of its final value?

Answer 1

R=4 $\Omega$

L=16mH

V=9V

$I={0.999I_0}$

$\tau=\frac{L}{R}$

$\tau=\frac{16\times10^3}{4}=4\times10^{-3}$ Sec

$I=I_0(1-e^{(\frac{t}{\tau})})$

$0.999I_0 =I_0(1-e^{(\frac{-t}{\tau})})$

$0.999=(1-e^{(\frac{-t}{\tau})})$

$1\times10^{-3}=e^{(\frac{-t}{\tau})}$

$10^3=e^{\frac{t}{\tau}}$

$ln1000=\frac{t}{\tau}$

$t=\tau ln1000$

$t=27.63\times10^{-3}=27.63msec$ $t=27.63 msec$ times constant does it take for the current to reach 99.9% of its final value