lle = 2+ (5 – 2)e-15t = 2+ 3e-15t A, %3D t > 0 %3D Example At t= 0, switch 1 is closed, and switch 2 is closed 4 s later. Find i(t) for t> 0. Calculate i for 1s 2 s and t= 5 s. t= 0 42 6 2 t=4 40 V 2Ω 5 H 10 V We need to consider the three time intervals t<0, 0St54, and t> 4 separately. For t<0, switches S1 and S2 are open so that i = 0. Since the inductor current cannot change instantly,

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
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le
= 2+ (5 – 2)e-15t
= 2+ 3e-15t
A,
%3D
t >0
Example
At t = 0, switch 1 is closed, and switch 2 is closed 4 s later, Find i(t) for t > 0. Calculate i for a
2 s and t= 5 s.
t=0
42
t=4
40 V
2Ω
5 H
10 V
We need to consider the three time intervals t<0, 0<t54, and t> 4 separately. For t<0,
switches S1 and S2 are open so that i = 0. Since the inductor current cannot change instantly,
Transcribed Image Text:le = 2+ (5 – 2)e-15t = 2+ 3e-15t A, %3D t >0 Example At t = 0, switch 1 is closed, and switch 2 is closed 4 s later, Find i(t) for t > 0. Calculate i for a 2 s and t= 5 s. t=0 42 t=4 40 V 2Ω 5 H 10 V We need to consider the three time intervals t<0, 0<t54, and t> 4 separately. For t<0, switches S1 and S2 are open so that i = 0. Since the inductor current cannot change instantly,
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