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The complete solution of the current in the sinusoidal response of R-L circuit is?(a) i = e^-t(R/L)[V/√(R^2+(ωL)^2) cos(θ-tan^-1)(ωL/R))]+V/√(R^2+(ωL)^2) cos(ωt+θ-tan^-1)(ωL/R))(b) i = e^-t(R/L)[-V/√(R^2+(ωL)^2) cos(θ-tan^-1)(ωL/R))]-V/√(R^2+(ωL)^2) cos(ωt+θ-tan^-1)(ωL/R))(c) i = e^-t(R/L)[V/√(R^2+(ωL)^2) cos(θ-tan^-1)(ωL/R))]-V/√(R^2+(ωL)^2) cos(ωt+θ-tan^-1)(ωL/R))(d) i = e^-t(R/L)[-V/√(R^2+(ωL)^2) cos(θ-tan^-1)(ωL/R))]+V/√(R^2+(ωL)^2) cos(ωt+θ-tan^-1)(ωL/R))The question was asked in final exam.This intriguing question originated from Sinusoidal Response of an R-L Circuit topic in portion Transients of Network Theory |
Answer» CORRECT ANSWER is (d) i = e^-t(R/L)[-V/√(R^2+(ωL)^2) cos(θ-tan^-1)(ωL/R))]+V/√(R^2+(ωL)^2) cos(ωt+θ-tan^-1)(ωL/R)) Explanation: The complete solution for the CURRENT becomesi = e^-t(R/L)[-V/√(R^2+(ωL)^2) cos(θ-tan^-1)(ωL/R))]+V/√(R^2+(ωL)^2)cos(ωt+θ-tan^-1)(ωL/R)). |
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