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Transcript
Chapter 31B - Transient Currents and Inductance AA PowerPoint PowerPoint Presentation Presentation by by Paul Paul E. E. Tippens, Tippens, Professor Professor of of Physics Physics Southern Southern Polytechnic Polytechnic State State University 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 Consider aa coil coil connected connected to to resistance resistance RR and and voltage voltage VV.. When When switch switch isis closed, closed, the the rising rising current current II increases increases flux, flux, producing producing an an internal internal back back emf emf in in the the coil. coil. Open Open switch switch reverses reverses emf. 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 I/t. i E L ; t 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 i/ t 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? i/ t = 2 A/s 4 mV R i E L ; t (0.004 V) L 2 A/s E L i / t 2.00 mH mH LL == 2.00 Note: We We are are following following the the practice practice of of using using Note: lower case case ii for for transient transient or or changing changing current current lower and upper upper case case II for for steady steady current. current. and Calculating the Inductance Recall two ways of finding E: E N t i E L t Increasing i/ t R Setting these terms equal gives: i N L t t Thus, Thus, the the inductance inductance LL can can be be found found from: 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 R Inductance L Combining the last two equations gives: 0 NI 0 NIA L and = BA 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: L 0 N A (4 x 10 2 l -7 Tm A 2 2 )(100) (0.002 m ) 0.300 m 8.38 xx 10 10-5-5 HH LL == 8.38 A R Note: LL does does NOT NOT depend depend Note: on current, current, but but on on physical physical on parameters of of the the coil. coil. parameters 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 -5 H -5 L = 8.38 x 10 L = 8.38 x 10 H A R i E L t (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 i/t, we have: R i P Ei Li t Since the power P = Work/t, Work = P t. Also the average value of Li is Li/2 during rise to the final current I. Thus, the total energy stored is: i EL ; t 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 B 0 2 B u 2 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 2 0 2(4 x 10 A ) uu == 0.268 0.268 J/m J/m33 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: i V – E = iR EL t i V L iR t V S1 S2 R i L E Initially, i/t isis large, Initially, i/t large, making making the the back back emf emf large large and small. The The current current rises rises to to its its and the the current current ii small. maximum maximum value value II when when rate rate of of change change isis zero. 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 L R i I 0.63 I Current Rise Time, t In In an an inductor, inductor, the the current current will will rise rise to to 63% 63% of of its its maximum . == L/R L/R. maximum value value in in one one time time constant constant The R-L Decay Now suppose we close S2 after energy is in inductor: i EL E = iR t For current L i iR t decay in L: V S1 S2 R i L E Initially, i/t isis large Initially, i/t large and and the the emf emf EE driving driving the the current current isis at at its its maximum maximum value value II.. The The current current decays decays to to zero zero when when the the emf emf plays plays out. 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 L R Current Decay 0.37 I Time, t In In an an inductor, inductor, the the current current will will decay decay to to 37% 37% of of its its maximum maximum value value in in one one time time constant constant Example 5: The circuit below has a 40-mH inductor connected to a 5- resistor and a 16-V battery. What is the time constant and what is the current after one time constant? L 0.040 H R 5 16 V 5 L = 0.04 H After time i = 0.63(V/R) R Time Time constant: constant: = = 88 ms ms V i (1 e ( R / L )t ) R 16V i 0.63 5 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, Initially, Q/C Q/C isis small, small, making making the the back back emf emf small small and . As and the the current current ii isis aa maximum maximum II. As the the charge charge QQ builds, = V. V. builds, the the current current decays decays to to zero zero when when EEbb = Rise of Charge Qmax q t = 0, Q = 0, Q V iR 0.63 I I = V/R C t = , i = , Qm = C V Q CV (1 e t / RC ) The time constant RC Capacitor Increase in Charge Time, t In In aa capacitor, capacitor, the the charge charge QQ will will rise rise to to 63% 63% of of its its maximum maximum value value in in one one time time constant constant 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 RC Current Decay 0.37 I Time, t As charge Q increases The The current current will will decay decay to to 37% 37% of of its its maximum maximum value the the charge charge rises. rises. value in in one one time time constant constant 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, Initially, QQ isis large large and and the the emf emf EE driving driving the the current current isis at at its its maximum maximum value value II.. The The current current decays decays to to zero zero when when the the emf emf plays plays out. out. i Current Decay V t / RC i e R I i Capacitor RC At t = 0, I = V/R At t = , I = 0 As the current decays, the charge also decays: Current Decay 0.37 I Q CVe Time, t t / RC In In aa discharging discharging capacitor, capacitor, both both current current and and charge charge decay decay to to 37% 37% of of their their maximum maximum values values in in one one time time constant constant == RC. RC. Example 6: The circuit below has a 4-F capacitor connected to a 3- resistor and a 12-V battery. The switch is opened. What is the current after one time constant ? 12 V 3 C = 4 F After time i = 0.63(V/R) = RC = (3 )(4 F) R Time Time constant: constant: = = 12 12 s s V i (1 e t / RC ) R 12V i 0.63 3 i = 2.52 A Summary i E L ; t 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 2 0 Summary V ( R / L )t i (1 e ) R L R I i Inductor Current Rise 0.63I Time, t In In an an inductor, inductor, the the current current will will rise rise to to 63% 63% of of its its maximum . == L/R L/R. maximum value value in in one one time time constant constant The The initial initial current current isis zero zero due due to to fast-changing fast-changing current current in in coil. coil. Eventually, Eventually, induced induced emf emf becomes becomes zero, zero, resulting resulting in in the the maximum maximum current current V/R. V/R. Summary (Cont.) V ( R / L )t i e R The The initial initial current, current, II = , decays = V/R V/R, decays to to zero zero as as emf emf in in coil coil dissipates. dissipates. I i Inductor Current Decay 0.37I Time, t The The current current will will decay decay to to 37% 37% of of its its maximum maximum value == L/R. L/R. value in in one one time time constant constant Summary (Cont.) When When charging charging aa capacitor capacitor the the charge charge rises rises to to 63% 63% of of its its maximum maximum while while the the current current decreases decreases to to 37% 37% of of its its maximum maximum value. value. Qmax q Capacitor I i Capacitor Current Decay Increase in 0.37 I Charge 0.63 I Time, t Q CV (1 e t / RC ) RC Time, t V t / RC i e R CONCLUSION: Chapter 31B Transient Current - Inductance