2 Objectives: After completing this module, you should be able to: Define and calculate inductance in terms of a changing current.Calculate the energy stored in an inductor and find the energy density.Discuss and solve problems involving the rise and decay of current in capacitors and inductors.
3 The back emf (red arrow) must oppose change in flux: Self-InductanceConsider a coil connected to resistance R and voltage V. When switch is closed, the rising current I increases flux, producing an internal back emf in the coil. Open switch reverses emf.RIncreasing IRDecreasing ILenz’s Law:The back emf (red arrow) must oppose change in flux:
4 InductanceThe back emf E induced in a coil is proportional to the rate of change of the current DI/Dt.RIncreasing Di/ DtAn inductance of one henry (H) means that current changing at the rate of one ampere per second will induce a back emf of one volt.
5 Example 1: A coil having 20 turns has an induced emf of 4 mV when the current is changing at the rate of 2 A/s. What is the inductance?RDi/ Dt = 2 A/s4 mVL = 2.00 mHNote: We are following the practice of using lower case i for transient or changing current and upper case I for steady current.
6 Calculating the Inductance Recall two ways of finding E:Increasing Di/ DtRInductance LSetting these terms equal gives:Thus, the inductance L can be found from:
7 Inductance of a Solenoid RInductance LlBSolenoidThe B-field created by a current I for length l is:and F = BACombining the last two equations gives:
8 First we find the inductance of the solenoid: Example 2: A solenoid of area m2 and length 30 cm, has 100 turns. If the current increases from 0 to 2 A in 0.1 s, what is the inductance of the solenoid?First we find the inductance of the solenoid:RlAL = 8.38 x 10-5 HNote: L does NOT depend on current, but on physical parameters of the coil.
9 Example 2 (Cont. ): If the current in the 83 Example 2 (Cont.): If the current in the 83.8-mH solenoid increased from 0 to 2 A in 0.1 s, what is the induced emf?RlAL = 8.38 x 10-5 H
10 Energy Stored in an Inductor At an instant when the current is changing at Di/Dt, we have:RSince the power P = Work/t, Work = P Dt. Also the average value of Li is Li/2 during rise to the final current I. Thus, the total energy stored is:Potential energy stored in inductor:
11 Example 3: What is the potential energy stored in a 0 Example 3: What is the potential energy stored in a 0.3 H inductor if the current rises from 0 to a final value of 2 A?L = 0.3 HI = 2 ARU = JThis energy is equal to the work done in reaching the final current I; it is returned when the current decreases to zero.
12 Energy Density (Optional) The energy density u is the energy U per unit volume VRlASubstitution gives u = U/V :
13 Energy Density (Continued) lARecall formula for B-field:
14 Example 4: The final steady current in a solenoid of 40 turns and length 20 cm is 5 A. What is the energy density?RlAB = 1.26 mTEnergy density is important for the study of electro-magnetic waves.u = J/m3
15 The R-L CircuitRLS2S1VAn inductor L and resistor R are connected in series and switch 1 is closed:EiV – E = iRInitially, Di/Dt is large, making the back emf large and the current i small. The current rises to its maximum value I when rate of change is zero.
16 The Rise of Current in L t i I At t = 0, I = 0 Current Rise Time, tIiCurrent RiseAt t = 0, I = 0t0.63 IAt t = ¥, I = V/RThe time constant t:In an inductor, the current will rise to 63% of its maximum value in one time constant t = L/R.
17 The R-L Decay V Now suppose we close S2 after energy is in inductor: E = iRFor current decay in L:Initially, Di/Dt is large and the emf E driving the current is at its maximum value I. The current decays to zero when the emf plays out.
18 The Decay of Current in L Time, tIiCurrent DecayAt t = 0, i = V/RAt t = ¥, i = 0t0.37 IThe time constant t:In an inductor, the current will decay to 37% of its maximum value in one time constant t.
19 Example 5: The circuit below has a 40-mH inductor connected to a 5-W resistor and a 16-V battery. What is the time constant and what is the current after one time constant?5 WL = 0.04 H16 VRTime constant: t = 8 msAfter time t:i = 0.63(V/R)i = 2.02 A
20 The R-C CircuitVClose S1. Then as charge Q builds on capacitor C, a back emf E results:S1ES2iV – E = iRRCInitially, Q/C is small, making the back emf small and the current i is a maximum I. As the charge Q builds, the current decays to zero when Eb = V.
21 Rise of Charge t q Capacitor Qmax t = 0, Q = 0, I = V/R Time, tQmaxqIncrease in ChargeCapacitort = 0, Q = 0, I = V/Rt0.63 It = ¥ , i = 0, Qm = C VIn a capacitor, the charge Q will rise to 63% of its maximum value in one time constant t.The time constant t:Of course, as charge rises, the current i will decay.
22 The Decay of Current in C Time, tIiCurrent DecayCapacitorAt t = 0, i = V/RAt t = ¥, i = 0t0.37 IThe time constant t:As charge Q increasesThe current will decay to 37% of its maximum value in one time constant t; the charge rises.
23 The R-C Discharge V Now suppose we close S2 and allow C to discharge: E = iRFor current decay in L:Initially, Q is large and the emf E driving the current is at its maximum value I. The current decays to zero when the emf plays out.
24 Current Decay t i Capacitor I Current Decay At t = 0, I = V/R Time, t As the current decays, the charge also decays:In a discharging capacitor, both current and charge decay to 37% of their maximum values in one time constant t = RC.
25 Example 6: The circuit below has a 4-mF capacitor connected to a 3-W resistor and a 12-V battery. The switch is opened. What is the current after one time constant t?3 WC = 4 mF12 VRt = RC = (3 W)(4 mF)Time constant: t = 12 msAfter time t:i = 0.63(V/R)i = 2.52 A
26 Potential Energy Energy Density: SummaryRlAPotential Energy Energy Density:
27 Summary Time, t I i Current Rise t 0.63I Inductor In an inductor, the current will rise to 63% of its maximum value in one time constant t = L/R.The initial current is zero due to fast-changing current in coil. Eventually, induced emf becomes zero, resulting in the maximum current V/R.
28 Summary (Cont.) Time, t I i Current Decay t 0.37I Inductor The initial current, I = V/R, decays to zero as emf in coil dissipates.The current will decay to 37% of its maximum value in one time constant t = L/R.
29 Summary (Cont.)When charging a capacitor the charge rises to 63% of its maximum while the current decreases to 37% of its maximum value.Time, tQmaxqIncrease in ChargeCapacitort0.63 ITime, tIiCurrent DecayCapacitort0.37 I
30 CONCLUSION: Chapter 31B Transient Current - Inductance