1. 1 SECOND ORDER Examples

2. 2 What Circuit do we require and why? 1. Circuit at t = 0 - This circuit is required to find the initial values of i(0 - ) and/or v(0 - ). These values are used to form the first equation of finding A 1 and A 2. 3. Circuit for t > 0 This circuit is required to find the initial values of di(0 + )/dt and/or dv(0 + )/dt. These values are used to form the second equation of finding A1 and A2. 2. Circuit at t = 0 + a) Source free circuit We can divide this circuit into two: This circuit is used to find the natural response of the circuit. b) Circuit at t = infinity This circuit is used to find the steady-state response of the circuit. Superposition theorem allows us to solve separately and then combine the solution in (a) and (b) together and to get the complete solution easily.

3. 3 Example 1 (i)Find v(t) for t > 0. (ii) Find i(t) for t >0. v i 24 V R = 6  1 H 0.25 F + - 1  t = 0 The switch has been closed for a long time and it is open at t = 0.

4. 4 v i 24V 5  1 H 0.25 F + - 1  t = 0 Example 2 t = 0 (i) Find i(t) for t > 0. (ii) Find v(t) for t >0. The switches have been opened for a long time and they are closed at t = 0.

5. 5 (i)Find i(t) for t > 0. (ii) Find i R (t) for t >0. Example 3 v i 24 V R = 1  1 H 0.25 F + - 1  t = 0 iRiR The switch has been opened for a long time and it is closed at t = 0.

6. 6 Example 1 : Solution (i)Find v(t) for t > 0. (ii) Find i(t) for t >0. v i 24 V R = 6  1 H 0.25 F + - 1  t = 0 The switch has been closed for a long time and it is open at t = 0.

7. 7 1. Draw the circuit for t = 0 - This statement means that the circuit is in steady state at t = 0 -. Therefore, C is opened and L is shorted. “The switch has been closed for a long time and it is open at t = 0” 24 V 6  11 i(0) = 3.429 A v (0) = 3.429 V + - Find i(0) and v(0) Steps to solve this problem

8. 8 2. Draw the circuit for t = 0 + This is a starting point for the circuit to experience transient. Therefore, C is not opened and L is not shorted. We know that i(0 - ) = i(0 + ) and v(0 - ) = v(0 + ) 24 V 6  i(0 + ) = 3.429 A 0.25 F + - v(0 + ) = 3.429 V Find dv(0 + )/dt or/and di(0 + )/dt 1 H

9. 9 24 V 6  i(0 + ) = 3.429 A 0.25 F + - v(0 + ) = 3.429 V Find dv(0 + )/dt or/and di(0 + )/dt 1 H + - v L (0 + ) -24+ 6(3.429)+v L (0 + )+3.429 = 0 v L (0 + ) = 0 KVL around the loop

10. 10 3. Draw the circuit for t = ∞ At t = ∞ the circuit has reached the steady state again. Therefore, C is open and L is shorted. 24 V 6  + - Find v(∞) or/and i(∞) v(∞) = 24 V i(∞) = 0

11. 11 4. Draw the source free circuit for t >0 1H 6  + - Voltage source is shorted and current source is opened. i v 0.25F Generally for this step, we have to find the differential equation for the source free circuit, then its characteristic equation. Since the circuit is a series RLC, we can directly write down its characteristic equation, and determine the type of the response.

12. 12 For second order circuit:For second order circuit, its characteristics equation is General form Series RLC circuit rad/s the response is overdamped the roots are real and distinct

13. 13 5. Write the general solution for the circuit for t > 0. 24 V 6  i 0.25 F + - v 1H or

14. 14 6. Find A 1 and A 2 24 V 6  i 0.25 F + - v 1H 1 2

15. 15 7. Find other circuit quantities for t > 0. 24V 6  i 0.25 F + - v 1H

16. 16 Verification Using PSpice 24V 6  i 0.25 F + - v 1H 0 1 2 3 - + 1h 2h E1h G2h 1  Hand Calculation Verification: Example 1 **Overdamped response: Case 1.Param R=6 L=1 C=0.25 ************************************************************************************** E1h 1h 0 value={24-21.019*EXP(-0.764*TIME)+0.448*EXP(-5.236*TIME)} R1h 1h 0 1 ; V(1h) = output voltage ************************************************************************************** G2h 0 2h value={4.015*EXP(-0.764*TIME)-0.586*EXP(-5.236*TIME)} R2h 2h 0 1 ; V(2h) = output current *************************************************************************************** V1 1 0 DC 24V R1 1 2 {R} L1 2 3 {L} IC=3.429A C1 3 0 {C} IC=3.429V.Tran 5m 5s 0 5m UIC.Probe V(1h) V(3).Probe I(R2h) I(L1).End

17. 17 (i)Find i(t) for t > 0. (ii) Find v(t) for t >0. 1  Example 2 : Solution v i 24 V 5  1 H 0.25 F + - t = 0 The switches have been opened for a long time and they are closed at t = 0.

18. 18 1. Draw the circuit for t = 0 - This statement means that the circuit is in steady state at t = 0 -. Therefore, C is opened and L is shorted. “The switch have been opened for a long time and it is closed at t = 0” 24 V 5  1  v(0 - ) = 24 V + - Find i(0) and v(0) Steps to solve this problem i(0 - ) = 0 A

19. 19 2. Draw the circuit for t = 0 + This is a starting point for the circuit to experience transient. Therefore, C is not open and L is not shorted. We know that i(0 - ) = i(0 + ) and v(0 - ) = v(0 + ) 24 V i(0 + ) = 0 0.25 F + - v(0 + ) = 24 V Then we can find dv(0 + )/dt or/and di(0 + )/dt 1  1H

20. 20 i(0 + ) = 0 0.25 F + - v(0 + ) = 24 V 1  1H i R (0 + ) = 24/1 = 24 A i C (0 + ) i C (0 + ) = -i R (0+) = -24 A and from the circuit

21. 21 3. Draw the circuit for t = ∞ At t = ∞ the circuit reaches steady state again. Therefore, C is open and L is shorted. Find v(∞) or/and i(∞) 24 V i( ∞ ) = 24 A + - v( ∞ ) = 24 V 1 

22. 22 4. Draw the source free circuit for t >0 Voltage source is shorted and current source is open. i(t) 0.25 F + - v(t) 1  1H Generally for this step, we have to find the differential equation for the source free circuit, then its characteristic equation. Since the circuit is a parallel RLC, we can directly write down its characteristic equation, and determine the type of the response.

23. 23 For second order circuit:For second order circuit, its characteristics equation is General form Parallel RLC circuit rad/s the response is critically damped The roots are real and equal

24. 24 5. Write the general solution for the circuit for t > 0. 24 V i(t) 0.25 F + - v(t) 1  1H

25. 25 6. Find A 1 and A 2 24 V i(t) 0.25 F + - v(t) 1  1H

26. 26 7. Find other circuit quantities for t > 0. 24 V i(t) 0.25 F + - v(t) 1  1H

27. 27 (i)Find i(t) for t > 0. (ii) Find i R (t) for t >0. v i 24 V R = 1  1 H 0.25 F + - 1  t = 0 Example 3: Solution iRiR The switch has been opened for a long time and it is closed at t = 0.

28. 28 1. Draw the circuit for t = 0 - This statement means that the circuit is in steady state at t = 0 -. Therefore, C is opened and L is shorted. “The switch has been opened for a long time and it is closed at t = 0” 24 V 1  i(0 - ) = 0 v(0 - ) = 24 V + - Find i(0) and v(0) Steps to solve this problem

29. 29 2. Draw the circuit for t = 0 + This is a starting point for the circuit to experience transient. Therefore, C is not opened and L is not shorted. We know that i(0 - ) = i(0 + ) and v(0 - ) = v(0 + ) 24 V i(0 + ) = 0 0.25 F + - v(0 + ) = 24 Then we can find dv(0 + )/dt or/and di(0 + )/dt 1  1H 1  i R (0 + ) = 24A

30. 30 24 V i(0 + ) = 0 0.25 F + - v(0 + ) = 24 1  1H 1  i R (0 + ) = 24A i C (0 + ) i C (0 + ) = -i R (0 + )

31. 31 3. Draw the circuit for t = ∞ At t = ∞ the circuit reaches steady state again. Therefore, C is open and L is shorted. Find v(∞) or/and i(∞) 24 V i( ∞ ) = 12 A + - v( ∞ ) = 12V 1 

32. 32 4. Draw the source free circuit for t >0. Voltage source is shorted and current source is opened. Since this circuit is neither a parallel nor a series RLC. This circuit must be a general second order. So, we have to find the differential equation in terms of i(t) or v(t). Is this a series RLC circuit? Is this a parallel RLC circuit? i(t) 0.25 F + - v(t) 1  1H 1 

33. 33 i(t) 0.25 F + - v(t) 1  1H 1  Applying KCL at node a: a Applying KVL to the left mesh:

34. 34 For second order circuit:For second order circuit, its characteristics equation is General form The circuit under-study rad/s the response is underdamped The roots are complex conjugate

35. 35 5. Write the general solution for the circuit for t > 0. 24 V i(t) 0.25 F + - v(t) 1  1H 1 

36. 36 6. Find A 1 and A 2 24 V i(t) 0.25 F + - v(t) 1  1H 1 

37. 37 7. Find other circuit quantities for t > 0. 24 V i(t) 0.25 F + - v(t) 1  1 H 1  i C (t) i R (t)

38. 38 Example 4 t = 0 5 H 3  8 A 0.05 F i(t) - + v(t) 1  Find v (t) for t >0

39. 39 Example 5 6  4u(t) 0.1 F 1 H v (t) + - i (t) Find i (t) for t >0

40. 40 Example 1a A series RLC circuit has R = 10 k , L = 0.1 mH, and C = 10  F. What type of damping is exhibited by the circuit. Example 2a A parallel RLC circuit has R = 10 k , L = 0.1 mH, and C = 10  F. What type of damping is exhibited by the circuit.

41. 41 Example 4a The responses of a series RLC circuit are Determine the values of R, L, and C. V mA (a) Overdamped(b) Critically damped (c) Underdamped If R = 20 , L = 0.6 H, what value of C will make an RLC series circuit: Example 3a

42. 42 Example 5a Find i(t) in the circuit of Fig. 8.10. Assume that the circuit has reached steady state at t = 0 -. Fig 8.10

43. 43 Example 8.9 Find v(t) for t > 0 in Fig. 8.29.

44. 44 Example 8.9 Find v(t) for t > 0 in Fig. 8.29.

45. 45 Problem 8.56 + v - iRiR i 0.75  0.25 H 4  0.04 F 20 V Find i R (t), v(t), and i(t) for t > 0. Given v(0) = 0 and i(0) = 0