Answer :
Explanation:
It is given that,
Inductance of the inductor, [tex]L=10\ mH=10\times 10^{-3}\ H=10^{-2}\ H[/tex]
Resistance of the resistor, R = 10 ohms
(a) Let [tex]\tau[/tex] is the time constant of the circuit. It is given by :
[tex]\tau=\dfrac{L}{R}[/tex]
[tex]\tau=\dfrac{10^{-2}}{10}[/tex]
[tex]\tau=0.001\ s[/tex]
[tex]\tau=1\ ms[/tex]
(b) The current equation in RL circuit is given by :
[tex]I'=I(1-e^{\dfrac{-t}{\tau}})[/tex]
I' = 0.99 I
[tex]0.99=1-e^{\dfrac{-t}{\tau}}[/tex]
[tex]e^{\dfrac{-t}{\tau}}=0.01[/tex]
[tex]e^{\dfrac{-t}{1\ ms}}=0.01[/tex]
t = 4.6 ms
Hence, this is the required solution.