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Question 1.SA.9: Consider the series RL circuit shown in the diagram below. T......

Consider the series RL circuit shown in the diagram below. The source volt-age is 12 V, R = 10 Ω, and L = 10 mH. The switch is closed at t = 0. What would be the magnitude of current flowing through this circuit at t = 2 ms?

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In most series RL cases, the current value at a certain time “t” can be predicted through Eq. 1.31.

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

Given:

t = 2 × 10^{−3} s
L = 10 ×10^{−3} H
R = 10 Ω
V = 12 V
i(0) = 0

i_L(t)=(0)e^{-\frac{10}{0.01}(0.002)}+\frac{12}{10}\left\lgroup 1-e^{-\frac{10}{0.01}(0.002)}\right\rgroup \\ i_L(t)=\frac{12}{10}\left\lgroup 1-e^{-\frac{10}{0.01}(0.002)}\right\rgroup \\ i_L(t)=(1.2)(1-0.135)=1.04  A

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