1. Chapter 8 – Second-Order Circuits
Has 2 storage elements: 2 C’s, or 2 L’s, or 1 C and 1 L
Involves solving a 2nd order ODE – needs 2 initial conditions
Finding Final Values:
Steady State:
C  open circuit; L  short circuit.
Finding Initial Values:
From “no jump” condition: Continuity of Energy
vc(0+) = vc(0-)
iL(0+) = iL(0-)

2. c) Initial values of derivatives of: vC, iL, and vR.
Example1: Find: a) vC(∞), iL(∞) and vR(∞); b) vC(0+), iL(0+) and vR(0+),
c) Initial values of derivatives of: vC, iL, and vR.

5. Source-Free Series RLC
Given: iL(0) = Io, vC(0) = Vo

10. Example 2.
Given: R = 4Ω , L = 1H, F = 1/3 F
iL(0) = 0, vC(0) = Vo = 5V
Find: i(t)

12. Source-Free Parallel RLC
Given: iL(0) = Io, vC(0) = Vo

15. Example 3.
Given: Switch open for a long time, and is closed at t=0.
Find: v(t) for t>0.