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 complementary current.

(a) ic = e^-37.5t(c1cos1290t + c2sin1290t)

(b) ic = e^-37.5t(c1cos1290t – c2sin1290t)

(c) ic = e^37.5t(c1cos1290t – c2sin1290t)

(d) ic = e^37.5t(c1cos1290t + c2sin1290t)

Answer 1

Correct answer is (a) ic = e^-37.5t(c1cos1290t + c2sin1290t)

For explanation I would say: The roots of the charactesistic equation are D1 = -37.5+j1290 and D2 = -37.5-j1290. The complementary current obtained is ic =  e^-37.5t(c1cos1290t + c2sin1290t).