1. ECE 222 Electric Circuit Analysis II
Chapter 3 The Inductor
Herbert G. Mayer, PSU
Status 4/28/2016
For use at CCUT Spring 2016

2. Syllabus Inductor Voltage Never Forget Serial, Parallel Inductors
Current Pulse, Circuit 1 [1]
Inductor Current
Voltage Pulse, Circuit 2 [1]
Review: Power & Energy
Exercise
Bibliography

3. Inductor Voltage

4. Inductor Voltage
A current creates & maintains magnetic field around conductor it flows through; DC as well as AC
Magnetic field can create a current in another conductor nearby, via physical phenomenon 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. Inductor Voltage v(t) = L di / dt
In the Passive Sign Convention we view the positive reference direction of a 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 terminals of an ideal inductor with DC!

6. Inductor Voltage
Voltage across inductor terminals can change instantaneously; of interest for AC circuits
Yet current cannot change instantaneously in an inductor; the magnetic field resists the current via magnetic field, being built
Current first builds up (or tears down) the magnetic field around the inductor, given the voltage causing this
When an active 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 it can cause sparks, AKA arcing; arc can be visible: indicator of high voltage

7. v(t) = L di / dt Never Forget
If v(t) is the instantaneous voltage across terminals of an inductor
L is the inductivity of the inductance with v(t)
di / dt is the instantaneous current through L
Then “Ohm’s”  Law for Inductor is:
v(t) = L di / dt

8. Serial, Parallel Inductors

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

10. v1 = L1 di/dt, v2 = L2 di/dt, v3 = L3 di/dt
Serial Inductances
With n inductances Ln in series, the question is, what is their equivalent single induction Leq?
Clearly, all n currents through Ln are identical, the same charge has to travel through them all
The voltages add up; if we view n = 3 inductors, the total voltage v along all n of them is:
v = v1 + v2 + v3
v1 = L1 di/dt, v2 = L2 di/dt, v3 = L3 di/dt

11. v = L1 di/dt + L2 di/dt L3 di/dt
Serial Inductances
With di/dt factored out we get from v = v1 + v2 + v3
v = L1 di/dt + L2 di/dt L3 di/dt
v = ( L1 + L2 + L3 ) di/dt
v = Σ Ln di/dt for n =
Leq = Σ Ln for n =
If the original n inductors carry initial current i0 before the AC current starts at t0, that same current also flows through the equivalent Leq

12. i = ( 1 / L1 + 1 / L2 + 1 / L3 ) v * dt + Σ in(t0), for n = 1..3
Parallel Inductances
Parallel inductors Ln all have the same voltage v at their terminals
The partial currents through all Ln add up; for example, with n = 3 here, it follows:
i = i1 + i2 + i3
Starting at time t0 to time t, integrate over time t0 to t:
i1 = 1 / L1 v * dt + i1(t0)
i2 = 1 / L2 v * dt + i2(t0)
i3 = 1 / L3 v * dt + i3(t0)
Factoring out the identical integrals, we get
i = ( 1 / L1 + 1 / L2 + 1 / L3 ) v * dt + Σ in(t0), for n = 1..3

13. Parallel Inductances
We compute the equivalent inductance, using the analog rule from resistor conductance:
1 / Leq = 1 / L1 + 1 / L2 + 1 / L3
i(t0) = Σ in(t0)

14. Circuit 1 with Inductivity

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

16. Circuit 1: Independent Current Source, Pulsed
Current Pulse, Circuit 1
Circuit 1: Independent Current Source, Pulsed

17. Current Pulse, Circuit 1 Analyze circuit 1, and solve the following:
Compute table of values for current i(t) from 0 to 10.0 [s]
Plot the current i(t)
At what moment of time t1 is i at its maximum imax?
What is the value of imax?
Compute voltage v(t) at the terminals of L1
At what instant t2 is the voltage 0 [V]?
At what moment t3 is voltage v(t) minimal?
At what moment t4 does voltage change instantaneously?
Plot the voltage v(t)
Plot the power p(t) and compute, when power p(t) is 0
Plot the energy w(t); when is energy stored, when is max stored, when is energy extracted?

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

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

20. 3. When is i(t1) max for Circuit 1
See [4] for free function plotter on the web!
i(t) = 10 t e-5 t
i is max, when di / dt = 0 at time t1
Compute i’ by product rule, then set = 0 [A]
di / dt = i’ =
Students, do it in class now! First derivatives are easy!

21. 3. When is i(t1) max for Circuit 1
See [4] for free function plotter on the web!
i(t) = 10 t e-5 t
i is max, when di / dt = 0 at time t1
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 / this is the time t1 we seek!
t1 = 0.2 [s]

22. 4. What is Value of imax for Circuit 1
imax = i(t) for t = 0.2
i(t) = 10 t e-5 t
Substitute t = t1 = 0.2
imax = 10 * 0.2 e-5 * 0.2
imax = 2 e-1 = 2 / e
imax = [A]

23. 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 ( t )
v(t) = e-5 t - 5 t e-5 t
For a plot of function v(t), see sub-problem 9 below

24. 6. Voltage v(t) = 0 for Circuit 1
Goal to find time t2, when voltage v(t2) will be 0
v(t) = e-5 t - 5 t e-5 t -- find time t2 for voltage v(t2) = 0
Students, compute time t2!

25. 6. Voltage v(t) = 0 for Circuit 1
Goal to find time t2, when voltage v(t2) will be 0
v(t) = e-5 t - 5 t e-5 t -- find time t2 for voltage v(t2) = 0
0 = e-5 t - 5 t e-5 t -- same as:
e-5 t = 5 t2 e-5 t
1 = 5 t that is the time t2
t2 = 0.2 [s]
At time t2 v(t) changes polarity, goes negative after t2!

26. 7. When is Voltage v(t3) minimal?
v(t) = e-5 t - 5 t e-5 t -- find v(t3) when v(t)’ = 0
Compute t3
The method: Find first derivative v(t)’ then set to 0
v(t) = e-5 t - 5 t e-5 t
Students, generate first derivative v’(t) now in class and thus find t3

27. 7. When is Voltage v(t3) minimal?
v(t) = e-5 t - 5 t e-5 t -- find v(t3) when v(t)’ = 0
Compute t3
The method: 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 = that is time t3, where v is minimal
t3 = 0.4 [s] -- see plot in subproblem 9.

28. 8. Instantaneous Voltage in Circuit 1?
At which moment t4 does the voltage change instantaneously? (careful: Instantaneous!!)
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 t4 = 0 [s]

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

30. 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 ( 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 ( t )
p(t) = 10 t e-10 t ( t )
for t = 0  p = 0
for t = 1/5  p = 0
When is power max? When p(t)’ = 0
Students, compute first derivative p(t)’ in class now!

31. 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 ( 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 ( t )
p(t) = 10 t e-10 t ( t )
for t = 0  p = 0
for t = 1/5  p = 0
When is power min or max? When p(t)’ = 0
p(t)’ = 10 e-10 t t e-10 t t2 e-10 t
Set to 0, solve for t, will have 2 solutions . . .

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

33. 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) = i(t) L di / dt
w(t) = p(t) dt = i(t)2 L / 2
w(t) = 0.5 * 0.1 * 100 t2 e-10 t
w(t) = 5 t2 e-10 t
for t = 0 it follows: w = 0
w(t) is max when curve for w(t)’ = 0
w(t)’ = 2 * 5 t e-10 t - 50 t2 e-10 t
w(t)’ = 0 for t = 0.2
10 t e-10 t = 50 t2 e-10 t
t = 0.2 energy is max at 0.2 [s]

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

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

36. 11. Energy Answers in Circuit 1
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 [s]. Note that at s the tangent changes from positive to negative! Also note that this corresponds to p(t) = 0
Energy is extracted from the inductor, when the energy curve decreases. That happens in the interval ∞ [s]. This corresponds to p(t) < 0
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

37. Inductor Current

38. i(t) = 1/L v dt + i0 --from time 0 to t
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 + i0 --from time 0 to t
Where i(t) is the current at time t, i0 the current through inductor at the start of integration, the time starting at t = t0 = 0

39. Circuit 2 Also With Inductivity

40. Voltage Pulse, Circuit 2 Analyze simple circuit 2, with:
Independent voltage source v(t)
v(t) = 0 for t <= 0
v(t) = 20 t e-10t for t > 0
Only other component is inductivity L2 = 100 mH in circuit
Analyze circuit 2!

41. Voltage Pulse, Circuit 2

42. Voltage Pulse, Circuit 2 Analyze circuit 2, and solve the following:
Compute v = v(t) as a function of time from 0 to 1.0 [s]
Plot voltage v(t)
At what time t1 is v(t) at max value, named vmax?
What is that voltage vmax at t1?
What is the current i(t) at the terminals of L2?
Plot i(t)

43. 1. Voltages v(t) Table for Circuit 2

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

45. 3. When is Voltage vmax for Circuit 2
v(t) = 20 t e-10t
v(t) is max when v(t)’ = 0
So compute first derivative, using product rule
Students, compute v(t)’ now!

46. 3. When is Voltage vmax 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 t1 we seek
t1 = 0.1 [s]
v(t) is max at t1 = 0.1 seconds

47. 4. Value of vmax for Circuit 2
Given we know the time t = t1, when the voltage is at maximum, compute it’s momentary value:
v(t) = 20 t e-10t
v(t) is max when v(t)’ = 0
t1 = 0.1
v(0.1) = 20 * 0.1 e-10 * 0.1
v(0.1) = 2 e-1
v(0.1) = [V]

48. 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 /100 e0 )
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 ( ) contributes; the other negative powers of e-10t all converge toward 0

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

50. Resistor Needed in Circuit 2?
When a current is stable (i.e. DC current) 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

51. Review: Power & Energy

52. Power in Inductor p(t) = i v v = L di / dt
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 only
Or students express p(t) as a function of voltage now!

53. Power in Inductor p(t) = i v v = L di / dt
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 only
Or power expressed as a function of the voltage, with i0 being the current at time t=0:
p(t) = v(t) 1/L v dt -- function of voltage only

54. Energy in Inductor w = i(t)2 L / 2 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

55. Exercise Current Pulse in Inductor

56. Current Pulse with L1 Problem 6.1 from[1]
Given a circuit with single current source and one inductivity L1, compute the following:
Derive formula for v(t)
Compute v(t0) at t0 = 0 [s]
Moment of time t1, when voltage v1 is = 0 V
Formula for the power p(t) delivered to L1
Others t.b.d.
See circuit next page

57. Current Pulse with L1
Circuit with current pulse and inductivity L1

58. 1. Compute Formula for v(t) for L1
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) = d( e-300t - e-1200t ) / dt
v(t) = ( -300 e-300t e-1200t )
v(t) = 9.6 ( -e-300t + 4 e-1200t )

59. 2. Compute v(t0) at t0 = 0 for L1 v(t) = 28.8 [V]
We know the formula for v(t):
v(t) = 9.6 ( -e-300t + 4 e-1200t )
At time t = 0 both expressions of e become 1
v(t) = 9.6 ( )
v(t) = [V]

60. 3. At Which time t1 is Voltage 0 in L1
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() on both sides!
-300 t = ln t
900 t = ln 4
t = [ms]

61. 4. Power p(t) Delivered to L1
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) = e-600t e-2400t e-1500t

62. Exercises with Parallel Inductors

63. Parallel Inductors L1 and L2
Use Assessment Problem 6.4 from [1], with two parallel inductors L1 = 60 [mH], and L2 = 240 [mH]
Initial values of the currents in the inductors are i1(0) = 3 A, i2(0) = [A]
Voltage at the terminals, and thus at both inductors is v(t) = -30 e-5t [mV]
Compute the following units:
Equivalent inductivity Leq if L1 and L2 are replaced by one equivalent Leq
Initial current and its reference direction
i(t) using solely Leq
i1(t) and i2(t) and confirm KCL

64. Parallel Inductors L1 and L2
Circuit with voltage pulse v(t) = -30 e-5t [mV] and parallel inductivities L1 and L2

65. 1. Find Equivalent Inductor Leq
We know L1 = 60 mH and L2 = 240 mH
Leq = L1 * L2 / (L1 + L2 )
Leq = 60 * 240 / 300
Leq = 48 [mH]

66. 2. Initial Current i0
Initial currents in L1 and L2 are given to be 3 [A] and -5 [A]
Hence initial current i0(0) = i1(0) + i2(0) = -2 [A]
Equivalently: 2 [A] pointing up in the circuit

67. 3. Compute i(t) Using Leq i(t) = 0.125 e-5t - 2 - 0.125 [A]
We know: v(t) = -30 e-5t
We know: i(t) = 1/L v(t) dt + i0
i(t) = 1/Leq -30 e-5t dt + i0
i(t) = 1/ /-5 ( -30 ) e-5t + i0
i(t) = e-5t [A]

68. 4. Find i1(t) and i2(t), Confirm i(t) by KCL
We know: i(t) = 1/L v(t) dt + i(0)
Compute i1(t) in step 1:
i1(t) = 1/L1 v(t) dt + i1(0)
i1(t) = 1/ /-5 ( -30 ) e-5t + 2.9
i1(t) = 0.1 e-5t [A]
Compute i2(t) in step 2:
i2(t) = 1/L2 v(t) dt + i2(0)
i2(t) = 1/ /-5 ( -30 ) e-5t – 1/40 - 5
i2(t) = e-5t – [A]
q.e.d.:
i(t) = i1(t) + i2(t) = e-5t – [A]

69. Bibliography
Differentiation rules:
Electric Circuits, 10nd edition, Nilsson and Riedel, Pearsons Publishers, © 2015 ISBN-13:
Table of integrals:
Free plotter for circuits: