Activity 4: By applying Kirchhoff's Voltage Law to a series RL circuit, we obtain the differential equation: di 4+ 2 i = f(t) , t>0 dt where i(t) = electrical current in Amperes, and t = time in seconds. With zero initial condition (i(0) = 0), use Laplace transform to obtain the solutions to the differential equation, assuming: (4-a) f(t) = te-t (4-b) f(t) = 2 sin 5t .. (for this.. evaluate the generated solution in terms of transient and steady-state regions) (4-c) f(t) = 4 u(t) – 3 u(t – 1)

Activity 4: By applying Kirchhoff's Voltage Law to a series RL circuit, we obtain the differential equation: di 4+ 2 i = f(t) , t>0 dt where i(t) = electrical current in Amperes, and t = time in seconds. With zero initial condition (i(0) = 0), use Laplace transform to obtain the solutions to the differential equation, assuming: (4-a) f(t) = te-t (4-b) f(t) = 2 sin 5t .. (for this.. evaluate the generated solution in terms of transient and steady-state regions) (4-c) f(t) = 4 u(t) – 3 u(t – 1)

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Activity 4:
By applying Kirchhoff's Voltage Law to a series RL circuit, we obtain the differential equation:
di
4+ 2i = f(t), t>0
dt
where i(t) = electrical current in Amperes, and t = time in seconds.
With zero initial condition (i(0) = 0), use Laplace transform to obtain the solutions to the
differential equation, assuming:
(4-a) f(t) = te¬t
(4-b) f(t) = 2 sin 5t ... (for this... evaluate the generated solution in terms of transient and
steady-state regions)
(4-c) f(t) = 4 u(t) – 3 u(t – 1)

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