Chapter06-mutualinductance

Chapter06-mutualinductance - 210 CHAPTER 6. Inductance,...

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Unformatted text preview: 210 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 16 dig = 125:44e¡4t dt Test: 371:20e¡4t ¡ 480e¡5t + 8 + 232e¡4t ¡ 240e¡5t + 63:36e¡4t ¡ 80e¡5t ? ¡8 ¡ 792e¡4t + 800e¡5t = ¡125:44e¡4t (8 ¡ 8) + (800 ¡ 480 ¡ 240 ¡ 80)e¡5t ? +(371:20 + 232 + 63:36 ¡ 792)e¡4t = ¡125:44e¡4t ? (800 ¡ 800)e¡5t + (666:56 ¡ 792)e¡4t = ¡125:44e¡4t ¡125:44e¡4t = ¡125:44e¡4t DE 6.8 [a] L2 = à N1 = N2 [b] P1 = s ! = L1 = L2 (0:09)2 = 50 mH (0:75)2 (0:288) s 288 = 2:4 50 L1 0:288 = = 0:2 £ 10¡6 Wb/A 2 N1 (1200)2 P2 = DE 6.9 P1 = M2 k 2 L1 (OK) L2 0:05 = = 0:2 £ 10¡6 Wb/A 2 2 N2 (500) L1 = 2 nWb/A; 2 N1 P12 = P21 = P2 = L2 = 2 nWb/A 2 N2 M = 1:2 nWb/A N1 N2 P11 = P1 ¡ P21 = 0:8 nWb/A DE 6.10 [a] W = (0:5)L1 i2 + (0:5)L2 i2 + Mi1 i2 1 2 q M = 0:85 (18)(32) = 20:4 mH W = [9(36) + 16(81) + 20:4(54)] = 2721:6 mJ [b] W = [324 + 1296 + 1101:6] = 2721:6 mJ [c] W = [324 + 1296 ¡ 1101:6] = 518:4 mJ [d] W = [324 + 1296 ¡ 1101:6] = 518:4 mJ Problems q DE 6.11 [a] M = 1:0 (18)(32) = 24 mH; i1 = 6 A Therefore 16i2 + 144i2 + 324 = 0; 2 µ¶ 9 Therefore i2 = ¡ § 2 211 s µ ¶2 9 2 i2 + 9i2 + 20:25 = 0 2 ¡ 20:25 = ¡4:5 § p 0 Therefore i2 = ¡4:5 A [b] No, setting W equal to a negative value will make the quantity under the square root sign negative. Problems P 6.1 p = vi = 40t[e¡10t ¡ 10te¡20t ¡ e¡20t ] W= Z 1 p dx = 0 Z 1 0 40x[e¡10x ¡ 10xe¡20x ¡ e¡20x ] dx = 0:2 J This is energy stored in the inductor at t = 1: P 6.2 0·t<1 103 Z t e¡3x 30 £ 10¡3 e¡3x dx + 2:5 = 7:5 iL = 40 ¡3 = 5 ¡ 2:5e¡3t A; 0·t·1 ¯t ¯ ¯ + 2:5 ¯ 0 234 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance [e] w ¡3 = 5 £ 10 Z t 0 e¡500x dx = 104 (1 ¡ e¡500t ) nJ :¢: 104 (1 ¡ e¡500t ) = 5000; Thus, t = P 6.32 e¡500t = 0:5 ln 2 = 1:39 ms. 500 dio dt ¡3 = (20 £ 10 )(50 £ 10¡3 )fe¡8000t [¡6000 sin 6000t + 12;000 cos 6000t] v2 (t) = 20 £ 10¡3 +(¡8000e¡8000t )[cos 6000t + 2 sin 6000t]g = e¡8000t f4 cos 6000t ¡ 22 sin 6000tg V :¢: v2 (0) = 4 V i0 (0) = 50 mA vR (0) = 320(50 £ 10¡3 ) = 16 V v1 (0) = 16 + 4 = 20 V P 6.33 vc = ¡106 Z t ¡80x e sin 60x dx ¡ 300 20 0 = 5e¡80t [80 sin 60t + 60 cos 60t] + 300 ¡ 300 vL = 400e¡80t sin 60t + 300e¡80t cos 60t V dio =5 dt = 5[¡80e¡80t sin 60t + 60e¡80t cos 60t] = ¡400e¡80t sin 60t + 300e¡80t cos 60t V vo = vc ¡ vL = (300e¡80t cos 60t ¡ 300e¡80t cos 60t + 400e¡80t sin 60t+ 400e¡80t sin 60t) = 800e¡80t sin 60t V P 6.34 [a] Yes, vo = 20(i2 ¡ i1 ) + 60i2 Problems = 20(1 ¡ 52e¡5t + 51e¡4t ¡ 4 ¡ 64e¡5t + 68e¡4t )+ [b] vo 60(1 ¡ 52e¡5t + 51e¡4t ) = 20(¡3 ¡ 116e¡5t + 119e¡4t ) + 60 ¡ 3160e¡5t + 3060e¡4t = ¡5440e¡5t + 5440e¡4t V vo = L2 [c] vo d di1 (ig ¡ i2 ) + M dt dt d d (15 + 36e¡5t ¡ 51e¡4t ) + 8 (4 + 64e¡5t ¡ 68e¡4t ) dt dt ¡5t ¡4 t ¡5t = ¡2880e + 3264e ¡ 2560e + 2176e¡4t = 16 = ¡5440e¡5t + 5440e¡4t V vo P 6.35 = 5(ig ¡ i1 ) + 20(i2 ¡ i1 ) + 60i2 [a] vg = 5(16 ¡ 16e¡5t ¡ 4 ¡ 64e¡5t + 68e¡4t )+ 20(1 ¡ 52e¡5t + 51e¡4t ¡ 4 ¡ 64e¡5t + 68e¡4t )+ 60(1 ¡ 52e¡5t + 51e¡4t ) = 60 + 5780e¡4t ¡ 5840e¡5t V [b] vg (0) = 60 + 5780 ¡ 5840 = 0 V [c] pdev = vg ig = 960 + 92;480e¡4t ¡ 94;400e¡5t ¡ 92;480e¡9t + 93;440e¡10t W [d] pdev (1) = 960 W [e] i1 (1) = 4 A; i2 (1) = 1 A; p5− = (16 ¡ 4)2 (5) = 720 W ig (1) = 16 A; p20− = 32 (20) = 180 W p60− = 12 (60) = 60 W X X :¢: P 6.36 [a] ¡2 pabs = 720 + 180 + 60 = 960 W pdev = X pabs = 960 W dig di2 + 16 + 32i2 = 0 dt dt 16 di2 dig + 32i2 = 2 dt dt 235 236 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance [b] i2 = e¡t ¡ e¡2t A di2 = ¡e¡t + 2e¡2t A=s dt ig = 8 ¡ 8e¡t A dig = 8e¡t A=s dt :¢: ¡16e¡t + 32e¡2t + 32e¡t ¡ 32e¡2t = 16e¡t [c] v1 =4 dig di2 ¡2 dt dt = 4(8e¡t ) ¡ 2(¡e¡t + 2e¡2t ) = 34e¡t ¡ 4e¡2t V; t¸0 [d] v1 (0) = 34 ¡ 4 = 30 V; Also dig di2 v1 (0) = 4 (0) ¡ 2 (0) dt dt = 4(8) ¡ 2(¡1 + 2) = 32 ¡ 2 = 30 V Yes, the initial value of v1 is consistent with known circuit behavior. [e] P 6.37 µ ¶µ ¶ µ 1 P11 P22 P11 [a] 2 = 1 + 1+ = 1+ k P12 P12 P21 Therefore P12 P21 k2 = (P21 + P11 )(P12 + P22 ) ¶µ P22 1+ P12 ¶ Now note that Á1 = Á11 + Á21 = P11 N1 i1 + P21 N1 i1 = N1 i1 (P11 + P21 ) Problems and similarly Á2 = N2 i2 (P22 + P12 ) It follows that (P11 + P21 ) = and (P22 + P12 ) = Á1 N1 i1 à Á2 N2 i2 ! Therefore (Á12 =N2 i2 )(Á21 =N1 i1 ) Á12 Á21 k2 = = (Á1 =N1 i1 )(Á2 =N2 i2 ) Á1 Á2 or và !à ! u u Á21 Á12 k=t Á1 Á2 [b] The fractions (Á21 =Á1 ) and (Á12 =Á2 ) are by de¯nition less than 1.0, therefore k < 1: P 6.38 [a] vab = L1 di di di di di + L2 + M + M = (L1 + L2 + 2M ) dt dt dt dt dt It follows that Lab = (L1 + L2 + 2M ) [b] vab = L1 di di di di di ¡ M + L2 ¡ M = (L1 + L2 ¡ 2M ) dt dt dt dt dt Therefore Lab = (L1 + L2 ¡ 2M ) P 6.39 23:8 M =p = 0:85 [a] k = p L1 L2 784 [b] M = 28 mH µ 2 L1 N1 P1 N1 [c] =2 = L2 N2 P2 N2 :¢: µ N1 N2 ¶2 = ¶2 196 = 49 4 N1 =7 N2 P 6.40 q p [a] M = k L1 L2 = 0:4 9 £ 104 = 120 mH P1 = L1 400 £ 10¡3 = = 100 nWb/A 2 N1 4 £ 106 237 238 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance dÁ11 P11 = = 0:25; dÁ21 P21 P21 = 4P11 P1 = P11 + P21 = 5P11 1 P11 = P1 = 20 nWb/A 5 P21 = 4P11 = 80 nWb/A N2 = M 120 £ 10¡3 = = 750 turns N1 P21 (2000)(80 £ 10¡9 ) L2 225 £ 10¡3 = = 400 nWb/A 2 N2 (0:75)2 £ 106 [c] P11 = 20 nWb/A [see part (a)] Á22 P2 400 [d] ¡1=4 = ¡1= Á12 P12 80 [b] P2 = P 6.41 [a] Dot terminal 1; the °ux is up in coil 1-2, and down in coil 3-4. Assign the current into terminal 3; the °ux is down in coil 3-4. Therefore, dot terminal 3. Hence, 1 and 3 or 2 and 4. [b] Dot terminal 2; the °ux is up in coil 1-2, and right-to-left in coil 3-4. Assign the current into terminal 4; the °ux is right-to-left in coil 3-4. Therefore, dot terminal 4. Hence, 2 and 4 or 1 and 3. [c] Dot terminal 2; the °ux is up in coil 1-2, and right-to-left in coil 3-4. Assign the current into terminal 4; the °ux is right-to-left in coil 3-4. Therefore, dot terminal 4. Hence, 2 and 4 or 1 and 3. [d] Dot terminal 1; the °ux is down in coil 1-2, and down in coil 3-4. Assign the current into terminal 4; the °ux is down in coil 3-4. Therefore, dot terminal 4. Hence, 1 and 4 or 2 and 3. P 6.42 When the switch is opened the induced voltage is negative at the dotted terminal. Since the voltmeter kicks upscale, the induced voltage across the voltmeter must be positive at its positive terminal. Therefore, the voltage is negative at the negative terminal of the voltmeter. Thus, the lower terminal of the unmarked coil has the same instantaneous polarity as the dotted terminal. Therefore, place a dot on the lower terminal of the unmarked coil. P 6.43 [a] vab = L1 0 = L1 d(i1 ¡ i2 ) di2 +M dt dt d(i2 ¡ i1 ) di2 di2 d(i1 ¡ i2 ) ¡M +M + L2 dt dt dt dt Problems Collecting coe±cients of [di1 =dt] and [di2 =dt]; the two mesh-current equations become vab = L1 di2 di1 + (M ¡ L1 ) dt dt and di1 di2 + (L1 + L2 ¡ 2M ) dt dt Solving for [di1 =dt] gives 0 = (M ¡ L1 ) di1 L1 + L2 ¡ 2M = vab dt L1 L2 ¡ M 2 from which we have vab = à L1 L2 ¡ M 2 L1 + L2 ¡ 2M :¢: Lab = !à di1 dt ! L1 L2 ¡ M 2 L1 + L2 ¡ 2M [b] If the magnetic polarity of coil 2 is reversed, the sign of M reverses, therefore L1 L2 ¡ M 2 Lab = L1 + L2 + 2M P 6.44 When the button is not pressed we have C2 dv d = C1 (vs ¡ v ) dt dt or (C1 + C2 ) dvs dv = C1 dt dt C1 dv dvs = dt (C1 + C2 ) dt 239 ...
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This note was uploaded on 03/29/2012 for the course EE 215 taught by Professor Unknown during the Spring '05 term at University of Washington.

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