1. 0 ECE 222 Electric Circuit Analysis II Chapter 3 Inductor Herbert G. Mayer, PSU Status 2/12/2016 For use at CCUT Spring 2016

2. 1 Syllabus Inductor Voltage Current Pulse, Circuit 1 [1] Inductor Current Voltage Pulse, Circuit 2 [1] Review: Power & Energy Exercise Bibliography

3. 2 Inductor Voltage

4. 3 A current creates a magnetic field around the conductor it flows through; DC as well as AC Magnetic field can create a current in another conductor nearby, named induction Phenomenon is magnified when conductor is arranged into coil via multiple windings And this can be amplified with certain materials inside the core of such windings; e.g. ferrite Either way, phenomenon is called induction Coil and its electric symbol in some circuit is referred to as an inductor Inductance symbol is L, measured in Henry [H] Unit of H is: [ H ] = [ V s A -1 ]

5. 4 Inductor Voltage In the Passive Sign Convention we view the reference direction of the current through inductor to be the same as the direction of the voltage drop That voltage is proportional to the time rate change of the current; key equation: v(t) = L di / dt Voltage drop is created at the terminals only if the current changes Under DC conditions (except for initial moment) an inductor is a short circuit: di / dt = 0 Hence v(t) = 0, i.e. no voltage drop across the terminals of an ideal inductor with DC!

6. 5 Inductor Voltage Voltage across inductor terminals can change instantaneously; of interest for AC circuits Yet the current only changes gradually; it first builds up (or tears down) the magnetic field around the inductor When a circuit with inductivities is interrupted, the magnetic field breaks down, since there is no more current to support or reverse it This break down of the magnetic field creates a current, so strong that it can cause a spark, AKA arcing; that arc can be visible, an indicator of very high voltage

7. 6 Serial, Parallel Inductors

8. 7 Duality simplifies life for EEs Recall that R i resistances in series, i = 1.. n, can be replaced by one single, equivalent resistor R eq, with R eq = Σ R i Also, multiple parallel resistors R i can be replaced by an equivalent resistor R eq, where the following holds: 1 / R eq = 1 / R1 + 1 /R 2... simply stating that conductances add up We see similar relations with inductors and dually with capacitors

9. 8 Serial Inductances With n inductances L n in series, the question is, what is their equivalent single induction L eq ? Clearly, all n currents through L n are identical, the same charge has to travel through them all The voltages add up; if we view n = 3 inductors, then the total voltage v along all of them is: v = v 1 + v 2 + v 3 v 1 = L 1 di/dt, v 2 = L 2 di/dt, v 3 = L 3 di/dt

10. 9 Serial Inductances With di/dt factored out we get from v = v 1 + v 2 + v 3 v = L 1 di/dt + L 2 di/dt L 3 di/dt v = ( L 1 + L 2 + L 3 ) di/dt v = Σ L n di/dt, for n = 1.. 3 As the original n inductors carry some initial current i 0 before the AC current starts at t 0, that same current also must flow through the equivalent L eq at t 0

11. 10 Parallel Inductances Parallel inductors L n all have the same voltage v at their terminals The partial currents through all L n add up; for example, with n = 3 here it follows : i = i 1 + i 2 + i 3 Starting at time t 0 to time t: i 1 = 1 / L 1 v * dt + i 1 (t 0 ) i 2 = 1 / L 2 v * dt + i 2 (t 0 ) i 3 = 1 / L 3 v * dt + i 3 (t 0 ) Factoring out the identical integrals, we get i = ( 1 / L 1 + 1 / L 2 + 1 / L 3 ) v * dt + Σ i n (t 0 ), for n = 1..3

12. 11 Parallel Inductances We compute the equivalent inductance, using the analog rule from resistor conductance: 1 / L eq = 1 / L 1 + 1 / L 2 + 1 / L 3 i(t 0 ) = Σ i n (t 0 )

13. 12 Circuit 1 with Inductivity

14. 13 Current Pulse, Circuit 1 Analyze the following simple circuit 1, with: Independent current source i(t) of decreasing pulse, where i(t)=0for t <= 0 i(t)=10 t e -5t for t > 0 Only other component in circuit 1 is an inductivity L 1 L1=100 mH No other component in circuit, e.g. no resistor to protect current surges Can this pose a problem in the circuit handled here?

15. 14 Current Pulse, Circuit 1 Circuit 1: Independent Current Source, Pulsed

16. 15 Current Pulse, Circuit 1 Analyze circuit 1, and solve the following: 1. Compute current i(t) from 0 to 10.0 [s] 2. Plot i(t) 3. At what moment of time t 1 is i at its maximum i max ? 4. What is the value of i max ? 5. Compute voltage v(t) at the terminals of L1 6. At what instant t 2 is the voltage 0 [V]? 7. At what moment t 3 is voltage v(t) minimal? 8. At what moment t 4 does voltage changes instantaneously? 9. Plot v(t) 10. Plot the power p(t) 11. Plot the energy w(t)

17. 16 1. Current i(t) from 0 to 10 for Circuit 1

18. 17 2. Plot i(t) for Circuit 1, f(x) = i(t)

19. 18 3. When is i(t 1 ) max for Circuit 1 i(t)=10 t e -5 t i is max, when di / dt = 0 at time t 1 Compute i’ by product rule, then set = 0 [A] di / dt=i’=10 e -5 t + (-5) 10 t e -5 t 0=10 e -5 t - 50 t e -5 t 10 e -5 t =50 t e -5 t -- divide by 10 e -5 t 1=5 t t=1 / 5-- this is the time t 1 we seek! t 1 =0.2 [s]

20. 19 4. What is Value of i max for Circuit 1 i max = i(t)for t = 0.2 i(t)=10 t e -5 t Substitute t = t 1 = 0.2 i max =10 * 0.2 e -5 * 0.2 i max =2 e -1 i max =0.735758882 [A]

21. 20 5. Compute Voltage v(t) for Circuit 1 v(t)=L di / dt v(t)=L d( 10 t e -5 t ) / dt v(t)=0.1 * 10 ( -5 t e -5 t + e -5 t ) v(t)=0.1 * 10 e -5 t ( 1 – 5 t ) v(t)=e -5 t ( 1 - 5 t ) v(t)=e -5 t - 5 t e -5 t

22. 21 6. Voltage v(t) = 0 for Circuit 1 Goal to find time t 2, when voltage v(t 2 ) will be 0 v(t)=e -5 t - 5 t e -5 t -- find time t 2 for voltage v(t 2 ) = 0 0=e -5 t - 5 t e -5 t -- same as: e -5 t =5 t 2 e -5 t 1=5 t-- that is the time t t 2 =0.2 [s] At time t 2 v(t) changes polarity, goes negative after t 2 !

23. 22 7. When is Voltage v(t 3 ) minimal? v(t)=e -5 t - 5 t e -5 t -- find v(t 3 ) when v(t)’ = 0 Compute t 3 Find first derivative v(t)’ then set to 0 v(t)=e -5 t - 5 t e -5 t v(t)’=-5 e -5 t - 5 (-5) t e -5 t - 5 e -5 t v(t)’=-10 e -5 t + 25 t e -5 t -- now set to 0 10e -5 t =25 t e -5 t 10=25 t t=0.4-- that is time t 3, when v is minimal t 3 =0.4 [s] -- see plot in section 9.

24. 23 8. Instantaneous Voltage in Circuit 1? At which moment t 4 does the voltage change instantaneously? Current through an inductor cannot change instantaneously, would require ∞ voltage But voltage can change instantaneously Circuit 1 is inactive at time t <= 0, thus voltage v(t) is zero for t <= 0 When the switch for the circuit closes, the independent current source starts pushing i(t), and the voltage at terminals of L1 immediately rises to its start value That happens at t = 0, thus t 4 = 0 [s]

25. 24 9. Plot v(t) for Circuit 1, f(x) = v(t) min v(t)

26. 25 10. Power p(t) for Circuit 1 We know the following for Circuit 1: i(t)=10 t e -5 t v(t)=e -5 t ( 1 - 5 t )-- so we can compute p(t) p(t)=v(t) i(t)-- substitute v(t) and i(t) p(t)=10 t e -5 t e -5 t ( 1 - 5 t ) p(t)=10 t e -10 t ( 1 - 5 t ) for t=0  p = 0 for t=1/5  p = 0 When is power max? When p(t)’ = 0 p(t)’=10 t e -10 t - 200 t e -10 t + 500 t 2 e -10 t Set to 0, solve for t, will have 2 solutions...

27. 26 10. Plot Power f(x) = p(t) for Circuit 1 power p(t) change power p(t) max power p(t) min

28. 27 11. Energy w(t) for Circuit 1 i(t)=10 t e -5 t v(t)=L di / dt p(t)=i(t) v(t)=L i(t) di / dt w(t)= p(t) dt=i(t) 2 L / 2 w(t)=0.5 * 0.1 * 100 t 2 e -10 t w(t)=5 t 2 e -10 t for t=0it follows:w = 0 w(t) is max when curve for w(t)’ = 0 w(t)’=2 * 5 t e -10 t - 50 t 2 e -10 t w(t)’=0for t = 0.2 10 t e -10 t = 50 t 2 e -10 t t = 0.2 energy is max at 0.2 [s]

29. 28 11. Plot Energy w(t) for Circuit 1 max w(t)

30. 29 11. Energy Questions in Circuit 1 a. In which time interval is energy being stored in the inductor of circuit 1? b. In which time interval is energy extracted from the inductor? c. What is the maximum energy stored in the inductor?

31. 30 11. Energy Answers in Circuit 1 a. Energy is being stored in the inductor as long as the energy curve rises; i.e. increasing value for w(t). That happens in the interval 0.. 0.2 [s]. Note that at 0.2 s the tangent changes from positive to negative! Also note that this corresponds to p(t) = 0 b. Energy is extracted from the inductor, when the energy curve decreases. That happens in the interval 0.2.. ∞ [s]. This corresponds to p(t) < 0 a. Maximum energy is stored in the inductor, when the state changes from storing to extracting, i.e. when the curve for w(t) changes direction, or w(t)’ = 0. That max amount is about 27 mJ, according to the energy plot

32. 31 Inductor Current

33. 32 Inductor Current Voltage in an inductor is proportional to the time rate change of the current: v(t) = L di / dt Transform, to express the current as a function of the voltage: v dt = L di L di = v dt i(t) = 1/L v dt + i 0 --from time 0 to t Where i(t) is the current at time t, i 0 the current thru inductor at the start of integration, and the time often starting at t = 0

34. 33 Circuit 2 with Inductivity

35. 34 Voltage Pulse, Circuit 2 Analyze simple circuit 2, with: Independent voltage source v(t) v(t) = 0for t <= 0 v(t) = 20 t e -10t for t > 0 Only other component is inductivity L 2 = 100 mH in circuit Analyze circuit 2!

36. 35 Voltage Pulse, Circuit 2

37. 36 Voltage Pulse, Circuit 2 Analyze circuit 2, and solve the following: 1. Compute v = v(t) as a function of time from 0 to 1.0 [s] 2. Plot v(t) 3. At what time t 1 is v(t) at max value, named v max ? 4. What is that voltage v max at t 1 ? 5. What is the current i(t) at the terminals of L 2 ? 6. Plot i(t)

38. 37 1. Voltages v(t) Table for Circuit 2

39. 38 2. Voltage f(x)=v(t) Plot for Circuit 2

40. 39 3. When is Voltage v max for Circuit 2 v(t)=20 t e -10t v(t) is max when v(t)’ = 0 So compute first derivative, using product rule v(t)’=20 e -10t + 20 t (-10) e -10t 0=20 e -10t – 200 t e -10t 20 e -10t =200 t e -10t 1=10 t-- that is the t 1 we seek t 1 =1/10 =0.1[s] v(t) is max at t 1 = 0.1 seconds

41. 40 4. Value of v max for Circuit 2 v(t)=20 t e -10t v(t) is max when v(t)’ = 0 t 1 =0.1 v(0.1)=20 * 0.1 e -10 * 0.1 v(0.1)=2 e -1 v(0.1)=0.735758882 [V]

42. 41 5. Current i(t) at Terminals for Circuit 2 v(t)=L di / dt-- here v(t) is: v(t)=20 t e -10t di=1 / L v(t) dt-- integrate di=1 / L 20 t e -10t dt-- see rule for x e c*x i =1 / L * 20 t e -10t dt-- find from 0 to t i =10 * 20 ( t / (-10) e -10t - 1/100 e -10t - 0 +1/100 e 0 ) i =2 ( -10 t e -10t - e -10t + 1 ) [ A ] For very large t (t >> 1 s) the current is 2 A, as solely the part 2 (... + 1 ) contributes; the other negative powers of e -10t all converge toward 0

43. 42 6. Plot of Current i(t), Circuit 2

44. 43 Resistor Needed in Circuit 2? When a current is stable (e.g. DC) in some circuit, an ideal inductor in that circuit poses no resistance In our example the sole component aside from the current source is an inductor Does this pose a problem by causing a short- circuit? Generally yes, since an inductivity simply conducts DC current Luckily, here we have a voltage source that converges toward 0 V

45. 44 Review: Power & Energy

46. 45 Power in Inductor p(t)=i v Voltage in inductor is proportional to the time rate change of the current i(t), and equal, when multiplied by L. So power = p(t)=i L di / dt-- function of current Or power expressed as a function of the voltage, with i 0 being the current at time t=0: p(t)=v(t) 1/L v dt + i 0 -- function of voltage

47. 46 Energy in Inductor p(t)=i(t) v(t) Power is also the time rate of expending energy w: p(t)=dw / dt p(t)=L i di / dt dw/dt=L i di/dt dw=L i di Integrate, when energy w=0, corresponding to current i=0: w= L i di w=i(t) 2 L / 2

48. 47 Exercises with Current Pulse in Inductor

49. 48 Current Pulse with L 1 Problem 6.1 from[1] Given a circuit with single current source and one inductivity L 1, compute the following: 1. Compute formula for v(t) 2. Compute v(t 0 ) at t 0 = 0 [s] 3. Moment of time t 1, when voltage v 1 is = 0 V 4. Formula for the power p(t) delivered to L 1 5. Others t.b.d. See circuit next page

50. 49 Current Pulse with L 1 Circuit with current pulse and inductivity L 1

51. 50 1. Compute Formula for v(t) for L 1 Compute formula for v(t) We know i(t) = 8 ( e -300t - e -1200t ) v(t)= L di/dt v(t)= 4 * 10 -3 * 8 * d( e -300t - e -1200t ) / dt v(t)= 32 10 -3 d( e -300t - e -1200t ) / dt v(t)= 32 10 -3 ( -300 e -300t + 1200 e -1200t ) v(t)= 9.6 ( -e -300t + 4 e -1200t )

52. 51 2. Compute v(t 0 ) at t 0 = 0 for L 1 We know the formula for v(t): v(t) = 9.6 ( -e -300t + 4 e -1200t ) At time t=0 both expression e become 1 v(t)= 9.6 ( -1+ 4 ) v(t)= 28.8 [V]

53. 52 3. At Which time t 1 is Voltage 0 in L 1 We know the formula for v(t): v(t) = 9.6 ( -e -300t + 4 e -1200t ) At time t1 voltage is 0 V 0= 9.6 ( -e -300t + 4 e -1200t ) 0= ( -e -300t + 4 e -1200t ) e -300t =4 e -1200t -- take ln() -300 t=ln 4 - 1200 t 900 t=ln 4 t=1.5403[ms]

54. 53 4. Power p(t) Delivered to L 1 We know the formula for v(t) and i(t): i(t)=8 ( e -300t - e -1200t ) v(t)=9.6 ( -e -300t + 4 e -1200t ) p(t)=i(t) * v(t) p(t)=8 ( e -300t - e -1200t ) * 9.6 ( -e -300t + 4 e -1200t ) p(t)=-76.8 e -600t - 307.2 e -2400t + 384 e -1500t

55. 54 Exercises with Parallel Inductors

56. 55 Parallel Inductors L 1 and L 2 Use Assessment Problem 6.4 from [1], with two parallel inductors L 1 = 60 mH, and L 2 = 240 mH Initial values of the currents in the inductors are i 1 = 3 A, i 2 = -5 A Voltage at the terminals, and thus at both inductors is v(t) = -30 e -5t [mV] Compute the following units: 1. Equivalent inductivity L eq if L 1 and L 2 are replaced by one equivalent L eq 2. Initial current and its reference direction 3. i(t) using solely L eq 4. i 1 (t) and i 2 (t) and confirm KCL

57. 56 Parallel Inductors L 1 and L 2 Circuit with voltage pulse v(t) = -30 e -5t and parallel inductivities L 1 and L 2

58. 57 1. Find Equivalent Inductor L eq We know L 1 = 60 mH and L 2 = 240 mH L eq = L 1 * L 2 / (L 1 + L 2 ) L eq = 60 * 240 / 300 L eq = 48[mH]

59. 58 2. Initial Current i 0 Initial currents in L 1 and L 2 are given to be 3 [A] and -5 [A] Hence initial current i = i 1 + i 2 = -2 [A] Or 2 amp pointing up in the circuit

60. 59 3. Compute i(t) Using L eq We know: v(t) = -30 e -5t We know: i(t) = 1/L v(t) dt + c 0 i(t)=1/L eq -30 e -5t dt + c 0 i(t)=1/48 10 -3 10 3 1/-5 ( -30 ) e -5t + c 0 i(t)=0.125 e -5t - 2 - 0.125[A]

61. 60 4. Find i1(t) and i2(t), Confirm i(t) by KCL We know: i(t) = 1/L v(t) dt + c 0 Hence step 1: i 1 (t) = 1/L 1 v(t) dt + c 1 i 1 (t)=1/60 10 -3 10 3 1/-5 ( -30 ) e -5t + 2.9 i 1 (t)=0.1 e -5t + 2.9 [A] Hence step 2: i 2 (t) = 1/L 2 v(t) dt + c 1 i 2 (t)=1/240 10 -3 10 3 1/-5 ( -30 ) e -5t – 5.025[A] q.e.d.: i(t) =i 1 (t) + i 2 (t)

62. 61 Bibliography  Differentiation rules: http://www.codeproject.com/KB/recipes/Differentiati on.aspx  Electric Circuits, 10 nd edition, Nilsson and Riedel, Pearsons Publishers, © 2015 ISBN-13: 978-0-13- 376003-3  Table of integrals: http://integral- table.com/downloads/single-page-integral-table.pdf