Chapter 31B - Transient Currents and Inductance

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1 Chapter 31B - Transent Currents and Inductance A PowerPont Presentaton by Paul E. Tppens, Professor of Physcs Southern Polytechnc State Unversty 007

2 Objectves: After completng ths module, you should be able to: Defne and calculate nductance n terms of a changng current. Calculate the energy stored n an nductor and fnd the energy densty. Dscuss and solve problems nvolvng the rse and decay of current n capactors and nductors.

3 Self-Inductance Consder a col connected to to resstance and voltage V.. When swtch s s closed, the rsng current I I ncreases flux, producng an nternal back emf n n the col. Open swtch reverses emf. Increasng I Lenz s s Law: The back emf (red arrow) must oppose change n flux: Decreasng I

4 Inductance The back emf E nduced n a col s proportonal to the rate of change of the current I/ I/t. E L t ; Lnductance Increasng / t An nductance of one henry (H) means that current changng at the rate of one ampere per second wll nduce a back emf of one volt. 1 V 1 H 1 A/s

5 Example 1: A col havng 0 turns has an nduced emf of 4 mv when the current s changng at the rate of A/s.. What s the nductance? / t = A/s 4 mv L E E L ; L t / t ( V) A/s L =.00 mh Note: We are followng the practce of of usng lower case for transent or or changng current and upper case I for steady current.

6 Calculatng the Inductance ecall two ways of fndng E: E N t E L t Settng these terms equal gves: N t L t Thus, the nductance L can be found from: Increasng / t Inductance L N L I

7 B Inductance of a Solenod Solenod l Inductance L Combnng the last two equatons gves: The B-feld created by a current I for length l s: B 0NI 0 NIA and = BA L L 0NA N I

8 Example : A solenod of area 0.00 m and length 30 cm,, has 100 turns.. If the current ncreases from 0 to A n 0.1 s, what s the nductance of the solenod? Frst we fnd the nductance of the solenod: L NA (4 x 10 )(100) (0.00 m ) -7 Tm 0 A l A m L = 8.38 x H Note: L does NOT depend on current,, but on physcal parameters of of the col.

9 Example (Cont.): If the current n the H solenod ncreased from 0 to A n 0.1 s, s what s the nduced emf? l A L = 8.38 x H E L t E -5 (8.38 x 10 H)( A - 0) s E 1.68 mv

10 Energy Stored n an Inductor At an nstant when the current s changng at / /t,, we have: E L ; PE L t t Snce the power P = Work/t, Work = P t.. Also the average value of L s L/ durng rse to the fnal current I. Thus, the total energy stored s: Potental energy stored n nductor: U 1 L

11 Example 3: What s the potental energy stored n a 0.3 H nductor f the current rses from 0 to a fnal value of A? A L = 0.3 H U 1 L U 1 (0.3 H)( A) J I = A U = J Ths energy s equal to the work done n reachng the fnal current I; ; t s returned when the current decreases to zero.

12 Energy Densty (Optonal) l A The energy densty u s the energy U per unt volume V 0N A 1 L ; U LI ; V A Substtuton gves u = U/V : N AI 0 1 0N A U ; U I u V A u 0N I

13 Energy Densty (Contnued) l A Energy densty: u 0N I ecall formula for B-feld: B NI NI B 0 B 0 u 0 NI 0 B 0 u B 0

14 Example 4: The fnal steady current n a solenod of 40 turns and length 0 cm s 5 A. What s the energy densty? NI B u -7 0 (4 x 10 )(40)(5 A) B = 1.6 mt 0.00 m -3 B (1.6 x 10 T) (4 x 10 ) -7 Tm 0 A u = 0.68 J/m 3 l A Energy densty s mportant for the study of electro- magnetc waves.

15 The -L L Crcut An nductor L and resstor are connected n seres and swtch 1 s closed: V E = E L t V L t S 1 S E V L Intally, /t s s large, makng the back emf large and the current small. The current rses to to ts maxmum value II when rate of of change s s zero.

16 The se of Current n L V ( / L) t (1 ) e I At t = 0, I = 0 At t =,, I = V/ The tme constant L 0.63 I Current se Tme, t In an nductor, the current wll rse to to 63% of of ts maxmum value n n one tme constant = = L/.

17 The -L L Decay Now suppose we close S after energy s n nductor: E = E L t For current decay n L: L t S 1 S E V L Intally, /t s s large and the emf E drvng the current s s at at ts maxmum value I. I.. The current decays to to zero when the emf plays out.

18 The Decay of Current n L V e ( / L) t I At t = 0, = V/ At t =, = 0 The tme constant L 0.37 I Current Decay Tme, t In an nductor, the current wll decay to to 37% of of ts maxmum value n n one tme constant

19 Example 5: The crcut below has a 40-mH nductor connected to a 5- resstor and a 16-V battery. What s the tme constant and what s the current after one tme constant? 16 V 5 L = 0.04 H After tme = 0.63(V/) L H 5 Tme constant: = 8 ms 16V V ( / L) t (1 ) e =.0 A

20 The -C C Crcut V Close S 1. Then as charge Q bulds on capactor C,, a back emf E results: Q V E = E C Q V C S 1 S C E Intally, Q/C s s small, makng the back emf small and the current s s a maxmum I. I. As the charge Q bulds, the current decays to to zero when E b = V. V.

21 se of Charge V Q C t = 0, Q = 0, I = V/ t =, =, Q m = C V QCV e t/ C (1 ) Q max 0.63 I q Capactor Increase n Charge Tme, t The tme constant C In a capactor, the charge Q wll rse to to 63% of of ts maxmum value n n one tme constant Of course, as charge rses, the current wll decay.

22 The Decay of Current n C V e t/ C At t = 0, = V/ At t =, = 0 The tme constant C I 0.37 I Capactor Current Decay Tme, t As charge Q ncreases The current wll decay to to 37% of of ts maxmum value n n one tme constant the charge rses.

23 The -C C Dscharge V Now suppose we close S and allow C to dscharge: Q E = E C For current decay n L: Q C S 1 S C E Intally, Q s s large and the emf E drvng the current s s at at ts maxmum value I. I.. The current decays to to zero when the emf plays out.

24 Current Decay V e t/ C At t = 0, I = V/ At t =,, I = 0 C As the current decays, the charge also decays: 0.37 I I Capactor Q CVe Current Decay Tme, t t/ C In a dschargng capactor, both current and charge decay to to 37% of of ther maxmum values n n one tme constant = = C.

25 Example 6: The crcut below has a 4-F capactor connected to a 3- resstor and a 1-V battery. The swtch s opened. What s the current after one tme constant? 1 V = C = (3 )(4 F) 3 Tme constant: = 1 s C = 4 F V e t/ C (1 ) After tme = 0.63(V/) 1V =.5 A

26 E L t L 0NA Summary ; Lnductance L N I l A Potental Energy Energy Densty: U 1 L u B 0

27 Summary V e ( / L) t (1 ) I 0.63I Inductor Current se L Tme, t In an nductor, the current wll rse to to 63% of of ts maxmum value n n one tme constant = = L/. The ntal current s s zero due to to fast-changng current n n col. Eventually, nduced emf becomes zero, resultng n n the maxmum current V/.

28 Summary (Cont.) V e ( / L) t The ntal current, I = V/,, decays to to zero as emf n n col dsspates. I 0.37I Inductor Current Decay Tme, t The current wll decay to to 37% of of ts maxmum value n n one tme constant = = L/.

29 Summary (Cont.) When chargng a capactor the charge rses to to 63% of of ts maxmum whle the current decreases to to 37% of of ts maxmum value. Q max q Capactor I Capactor 0.63 I Increase n Charge 0.37 I Current Decay Tme, t Tme, t Q CV e t/ C (1 ) C V e t/ C

30 CONCLUSION: Chapter 31B Transent Current - Inductance

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