Consider the series RL circuit given in Problem 9, in discharge mode, with voltage source removed. Parameters such as R = 10 Ω and L = 10 mH are the same. The switch has been closed for long period of time, such that the current has developed to the maximum or steady-state level 1.04 A.How much time

Question

Consider the series RL circuit given in Problem 9, in discharge mode, with voltage source removed. Parameters such as R = 10 Ω and L = 10 mH are the same. The switch has been closed for long period of time, such that the current has developed to the maximum or steady-state level 1.04 A. How much time would need to elapse for the current to drop to 0.5 A after the switch is opened.

Accepted Answer

Apply series RL current equation, Eq. 1.31.

[latex]i_L(t)=i_R(t)=i(0)e^{-\frac{R}{L}t}+\frac{V}{R}\left\lgroup 1-e^{-\frac{R}{L}t}\right\rgroup [/latex]

Given:
t = ?
L = 10 × [latex]10^{−3}[/latex]H
R = 10 Ω
V = 0
i(0) = 1.04 A
[latex]i_L(t)[/latex]= 0.5 A

[latex]i_L(t)=(0.5)=(1.04)e^{-\frac{10}{0.01}t}+(0)\left\lgroup 1-e^{-\frac{10}{0.01}t}\right\rgroup [/latex]

0.5=(1.04)[latex]e^{-\frac{10}{0.01}t}[/latex]

0.481=[latex]e^{-\frac{10}{0.01}t}[/latex]

[latex]\ln(0.481)= \ln{\left\lgroup e^{-\frac{10}{0.01}t}\right\rgroup}[/latex]

−0.7324 = −1,000t
t = 0.00073 s or 0.73 ms

Question 1.SA.10: Consider the series RL circuit given in Problem 9, in discha......

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