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Chapter 31B - Transient Currents and Inductance A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University © 2007 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. Self-Inductance Consider 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. Increasing I R Lenz’s Law: The back emf (red arrow) must oppose change in flux: Decreasing I R Inductance The back emf E induced in a coil is proportional to the rate of change of the current DI/Dt. Di E L ; Dt L inductance An 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. Increasing Di/ Dt R 1V 1 H 1 A/s 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? Di/ Dt = 2 A/s 4 mV R Di E L ; Dt (0.004 V) L 2 A/s E L Di / Dt L = 2.00 mH Note: We are following the practice of using lower case i for transient or changing current and upper case I for steady current. Calculating the Inductance Recall two ways of finding E: D E N Dt Di E L Dt Increasing Di/ Dt R Setting these terms equal gives: D Di N L Dt Dt Thus, the inductance L can be found from: Inductance L N L I Inductance of a Solenoid Solenoid B l The B-field created by a current I for length l is: B 0 NI and = BA R Inductance L Combining the last two equations gives: 0 NIA L N L I 0 N 2 A Example 2: A solenoid of area 0.002 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: 0 N A (4 x 10 L 2 l -7 Tm A 2 2 )(100) (0.002 m ) 0.300 m L = 8.38 x 10-5 H A R Note: L does NOT depend on current, but on physical parameters of the coil. Example 2 (Cont.): If the current in the 83.8H solenoid increased from 0 to 2 A in 0.1 s, what is the induced emf? l L = 8.38 x 10-5 H A R Di E L Dt (8.38 x 10-5 H)(2 A - 0) E 0.100 s E 1.68 mV Energy Stored in an Inductor At an instant when the current is changing at Di/Dt, we have: R Di P Ei Li Dt Since 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: Di EL ; Dt Potential energy stored in inductor: U 12 Li 2 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? U 12 Li 2 L = 0.3 H R I=2A U 12 (0.3 H)(2 A)2 0.600 J U = 0.600 J This energy is equal to the work done in reaching the final current I; it is returned when the current decreases to zero. Energy Density (Optional) The energy density u is the energy U per unit volume V l A R L 0 N 2 A ; U LI ; V A 1 2 2 Substitution gives u = U/V : 0 N A 2 U I ; 2 1 2 0 N 2 AI 2 2 U u V A u 0 N 2 I 2 2 2 Energy Density (Continued) Energy u density: l A R 2 2 Recall formula for B-field: B 0 NI 2 0 NI 0 B u 2 2 2 0 2 0 N 2 I 2 NI 2 B u 20 B 0 Example 4: The final steady current in a solenoid of 40 turns and length 20 cm is 5 A. What is the energy density? B 0 NI (4 x 10-7 )(40)(5 A) 0.200 m B = 1.26 mT 2 A R -3 2 B (1.26 x 10 T) u -7 Tm 20 2(4 x 10 A ) u = 0.268 J/m3 l Energy density is important for the study of electromagnetic waves. The R-L Circuit An inductor L and resistor R are connected in series and switch 1 is closed: Di V – E = iR EL Dt Di V L iR Dt V S1 S2 R i L E Initially, 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. The Rise of Current in L V ( R / L )t i (1 e ) R At t = 0, I = 0 At t = , I = V/R The time constant t: L t R i I 0.63 I Current Rise t Time, t In an inductor, the current will rise to 63% of its maximum value in one time constant t = L/R. The R-L Decay Now suppose we close S2 after energy is in inductor: Di EL E = iR Dt For current L Di iR Dt decay in L: V S1 S2 R i L E 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. The Decay of Current in L V ( R / L )t i e R i I At t = 0, i = V/R At t = , i = 0 The time constant t: L t R Current Decay 0.37 I t Time, t In an inductor, the current will decay to 37% of its maximum value in one time constant t. 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? L 0.040 H t R 5W 16 V 5W L = 0.04 H After time t: i = 0.63(V/R) R Time constant: t = 8 ms V i (1 e ( R / L )t ) R 16V i 0.63 5W i = 2.02 A The R-C Circuit Close S1. Then as charge Q builds on capacitor C, a back emf E results: Q V – E = iR E C Q V iR C V S1 S2 C R i E Initially, 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. Rise of Charge Qmax q Q t = 0, Q = 0, V iR 0.63 I I = V/R C t = , i = 0, Qm = C V Q CV (1 e t / RC ) The time constant t: t RC Capacitor Increase in Charge t Time, t In a capacitor, the charge Q will rise to 63% of its maximum value in one time constant t. Of course, as charge rises, the current i will decay. The Decay of Current in C V t / RC i e R i Capacitor I At t = 0, i = V/R At t = , i = 0 The time constant t: t RC Current Decay 0.37 I t Time, t As charge Q increases The current will decay to 37% of its maximum value in one time constant t; the charge rises. The R-C Discharge Now suppose we close S2 and allow C to discharge: Q E E = iR C Q For current iR C decay in L: V S1 S2 C R E 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. i Current Decay V t / RC i e R I i Capacitor t RC At t = 0, I = V/R At t = , I = 0 As the current decays, the charge also decays: Current Decay 0.37 I t Q CVe Time, t t / RC In a discharging capacitor, both current and charge decay to 37% of their maximum values in one time constant t = RC. Example 6: The circuit below has a 4-F 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? 12 V 3W C = 4 F After time t: i = 0.63(V/R) t = RC = (3 W)(4 F) R Time constant: t = 12 s V i (1 e t / RC ) R 12V i 0.63 3W i = 2.52 A Summary Di E L ; Dt L l L inductance 0 N A 2 Potential Energy Energy Density: A N L I R U Li 1 2 2 2 B u 20 Summary V ( R / L )t i (1 e ) R L t R I i Inductor Current Rise 0.63I t Time, t 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. Summary (Cont.) V ( R / L )t i e R The initial current, I = V/R, decays to zero as emf in coil dissipates. I i Inductor Current Decay 0.37I t Time, t The current will decay to 37% of its maximum value in one time constant t = L/R. Summary (Cont.) When charging a capacitor the charge rises to 63% of its maximum while the current decreases to 37% of its maximum value. Qmax q Capacitor I Increase in Charge 0.63 I t Capacitor Current Decay 0.37 I Time, t Q CV (1 et / RC ) i t RC t Time, t V t / RC i e R CONCLUSION: Chapter 31B Transient Current - Inductance