1. Network Analysis Week 10 RLC Circuits

2. Similar to RL and RC circuits, RLC circuits has two parts of responses, namely: natural response and forced response. Force Response: Similar to RL and RC, a step input causes a step output. Natural Response: Different and more difficult than RL, RC.

3. Source-free RLC Circuits We study the natural response by studying source-free RLC circuits. Parallel RLC Circuit Second-order Differential equation

4. This second-order differential equation can be solved by assuming solutions The solution should be in form of If the solution is good, then substitute it into the equation will be true. which means s=??

5. Use quadratic formula, we got Both and are solution to the equation Therefore, the complete solution is

6. From Define resonant frequency Damping factor Therefore, in which we divide into 3 cases according to the term inside the bracket

7. Solution to Second-order Differential Equations 1. α > ω 0 (inside square root is a positive value) Overdamped case 2.α = ω 0 (inside square root is zero) Critical damped case 3.α < ω 0 (inside square root is a negative value) Underdamped case

8. 1. Overdamped case, α > ω 0 Initial condition v c (0) = 0, i L (0) = -10A Find v(t) α > ω 0,therefore, this is an overdamped case s1 = -1, s2 = -6

9. Therefore, the solution is in form of Then, we will use initial conditions to find A1, A2 From v c (0) = 0 we substitute t=0 From KCL At t=0

10. Solve the equation and we got A1 = 84 and A2 = -84 and the solution is t v(t)

11. 2. Critical damped case, α = ω 0 Initial condition v c (0) = 0, i L (0) = -10A Find v(t) α = ω 0, this is an critical damped case s1 = s2 = -2.45 The complete solution of this case is in form of

12. Then, we will use initial conditions to find A1, A2 From v c (0) = 0 we substitute t=0 Find A1 from KCL at t=0 Therefore A2 =0 and the solution is reduced to

13. Solve the equation and we got A1 = 420 and the solution is t v(t)

14. 3. underdamped case, α < ω 0 from The term inside the bracket will be negative and s will be a complex number define Then and

15. Use Euler’s Identity

16. 3. Underdamped case, α < ω 0 Initial condition v c (0) = 0, i L (0) = -10A Find v(t) α < ω 0,therefore, this is an underdamped case and v(t) is in form

17. Then, we will use initial conditions to find B1, B2 From v c (0) = 0 we substitute t=0 Find B2 from KCL at t=0 Therefore B1 =0 and the solution is reduced to

18. we got and then the solution is t v(t)

19. Natural response at different time Mechanical systems are similar to electrical systems A pendulum is an example of underdamped second-order systems in mechanic. t displacement(t)

20. Series RLC

21. Parallel RLC

22. Response time Period 1 How to Solve RLC Problems? Period 2 Period 3 Divide in to several periods (3 periods as shown below) Period 1, 3 have constant V, I -> Use DC circuit analysis Period 2 is transient.

23. Procedure for Solving Transient in RLC Circuits 1.Decide that it is a series or parallel RLC circuit. Find α and ω0. Then, decide which case it is (overdamped, critical damped, underdamped). 2. Assume the solution in form of (natural response+ forced response) for i L (t) or v C (t) Overdamped Critical damped Underdamped 3. Find A, B, Vf using initial conditions and stable conditions

24. Example Find vc(t) This is overdamped case, so the solution is in form

25. Consider the circuit we found that the initial conditions will be v C (0) = 150 V and i L (0) = 5 A and the condition at stable point will be v C (∞) = 150 V and i L (∞) = 9 A Using v C (0) = 150 V, we got Using v C (∞) = 150 V, we got Therefore, Vf = 150, A1+A2 = 0 ------------(1)

26. Use the initial condition i L (0) = 5 A, we have to change vc(t) to i L (t) From Next, use KCL on the circle below or

27. From From (1), (2)A1 = 13.5, A2 = -13.5 Therefore, Substitute i L (0) = 5A ------------------(2)