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 complete solution of current.

(a) i = e^-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t + π/4 + 88.5⁰)

(b) i = e^-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t – π/4 + 88.5⁰)

(c) i = e^-37.5t(c1cos1290t + c2sin1290t) – 0.7cos(500t – π/4 + 88.5⁰)

(d) i = e^-37.5t(c1cos1290t + c2sin1290t) – 0.7cos(500t + π/4 + 88.5⁰)

Answer 1

Correct answer is (a) i = e^-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t + π/4 + 88.5⁰)

The best I can explain: The complete solution is the sum of the complementary function and the particular integral. So  i = e^-37.5t(c1cos1290t + c2sin1290t) + 0.7cos(500t + π/4 + 88.5⁰).