rlc_problem2 - will go to 0. V f = 0 We find that the...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Sample Problem Topic: RLC Circuits/Response Statement of Problem: Given the circuit shown in the figure below At time t = 0, the switch closes. Assume it has been open for a long time prior to this. Find v 0 (t) for t 0. Solution Find the resonant frequency with the equation: ° 0 = 1 ±² = 1 25 6.25 10 6 = 80 Find the attenuation with the equation: = 1 2 ³² = 1 2 800 6.25 10 6 = 100 The next step is to find the poles of the equation. We simply use the quadratic equation. ? = −∝ ± ´∝ 2 − ° 0 2 = 100 ± ´ 10 4 80 2 = 100 ± 60 We then find: S1 = -40 rad/s S2 = -160 rad/s We know that once the circuit reaches a steady state, the capacitor will act like an open, and the inductor will act as a short. This implies that the voltage v
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: will go to 0. V f = 0 We find that the circuit is overdamped, as 2 2 . Thus, the equation takes on the form: = 1 ? 40 + 2 ? 160 We also know that since the switch was closed for a long time, no current is flowing through the capacitor, and we can assume the voltage across the capacitor at time 0 = 30V. V (0) = 30 = A 1 +A 2 Solving the system of equations with the following, = 0 = 40 1 160 2 We get: A 1 = 40V, A 2 = -10V Finally, we can state v (t) as: = 40 ? 40 10 ? 160 , > 0 +...
View Full Document

Ask a homework question - tutors are online