Question

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 1

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).