2Objectives: 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.
3The 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:
4InductanceThe 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.
5Example 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.
6Calculating the Inductance Recall two ways of finding E:Increasing Di/ DtRInductance LSetting these terms equal gives:Thus, the inductance L can be found from:
7Inductance of a Solenoid RInductance LlBSolenoidThe B-field created by a current I for length l is:and F = BACombining the last two equations gives:
8First 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.
9Example 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
10Energy 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:
11Example 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.
12Energy Density (Optional) The energy density u is the energy U per unit volume VRlASubstitution gives u = U/V :
13Energy Density (Continued) lARecall formula for B-field:
14Example 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
15The 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.
16The 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.
17The 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.
18The 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.
19Example 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
20The 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.
21Rise 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.
22The 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.
23The 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.
24Current 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.
25Example 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
26Potential Energy Energy Density: SummaryRlAPotential Energy Energy Density:
27Summary 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.
28Summary (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.
29Summary (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
30CONCLUSION: Chapter 31B Transient Current - Inductance