Chapter 8 – Second-Order Circuits

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Presentation transcript:

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-)

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.

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

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

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

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