Respuesta :
Answer:
(A) [tex]9.14\times 10^{-9}sec[/tex]
(B) [tex]6.20\times 10^{-3}A[/tex]
Explanation:
We have given inductance [tex]L=5.41\mu H=5.41\times 10^{-6}H[/tex]
Resistance [tex]R=0.949kohm=0.949\times 10^3ohm[/tex]
Time constant of RL circuit is equal to [tex]\tau =\frac{L}{R}[/tex]
[tex]\tau =\frac{5.41\times 10^{-6}}{0.949\times 10^3}=5.70\times 10^{-9}sec[/tex]
Battery voltage e = 16 volt
(a) It is given current becomes 79.9% of its final value
Current in RL circuit is given by
[tex]i=i_0(1-e^{\frac{-t}{\tau }})[/tex]
According to question
[tex]0.799i_0=i_0(1-e^{\frac{-t}{\tau }})[/tex]
[tex]e^{\frac{-t}{\tau }}=0.201[/tex]
[tex]{\frac{-t}{\tau }}=ln0.201[/tex]
[tex]{\frac{-t}{5.7\times 10^{-9} }}=-1.6044[/tex]
[tex]t=9.14\times 10^{-9}sec[/tex]
(b) Current at [tex]t=\tau sec[/tex]
[tex]i=i_0(1-e^{\frac{-t}{\tau }})[/tex]
[tex]i=\frac{16}{0.949\times 10^3}(1-e^{\frac{-\tau }{\tau }})[/tex]
[tex]i=6.20\times 10^{-3}A[/tex]
