1.

In the circuit shown below, the switch is closed at t = 0. Applied voltage is v (t) = 400cos (500t + π/4). Resistance R = 15Ω, inductance L = 0.2H and capacitance = 3 µF. Find the particular solution.(a) ip = 0.6cos(500t + π/4 + 88.5⁰)(b) ip = 0.6cos(500t + π/4 + 89.5⁰)(c) ip = 0.7cos(500t + π/4 + 89.5⁰)(d) ip = 0.7cos(500t + π/4 + 88.5⁰)I got this question in unit test.Enquiry is from Sinusoidal Response of an R-L-C Circuit topic in section Transients of Network Theory

Answer»

The correct option is (d) ip = 0.7cos(500T + π/4 + 88.5⁰)

BEST EXPLANATION: PARTICULAR SOLUTION is ip = V/√(R^2+(1/ωC-ωL)^2) cos⁡(ωt+θ+tan^-1)⁡((1/ωC-ωL)/R)). ip = 0.7cos(500t + π/4 + 88.5⁰).



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