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Chapter 13 Problem 3, Two coils connected in series aiding fashion have a total inductance of 250 mH When. connected in a series opposing configuration the coils have a total inductance of 150. mH If the inductance of one coil L1 is three times the other find L1 L2 and M What is. the coupling coefficient,Chapter 13 Solution 3,L1 L2 2M 250 mH 1. L1 L2 2M 150 mH 2,Adding 1 and 2,2L1 2L2 400 mH,But L1 3L2 or 8L2 400 and L2 50 mH. L1 3L2 150 mH,From 2 150 50 2M 150 leads to M 25 mH. k M L1L 2 25 50 x150 0 2887, PROPRIETARY MATERIAL 2007 The McGraw Hill Companies Inc All rights reserved No part.
of this Manual may be displayed reproduced or distributed in any form or by any means without the prior. written permission of the publisher or used beyond the limited distribution to teachers and educators. permitted by McGraw Hill for their individual course preparation If you are a student using this Manual. you are using it without permission,Chapter 13 Problem 4. a For the coupled coils in Fig 13 74 a show that,Leq L1 L2 2M. b For the coupled coils in Fig 13 74 b show that,Figure 13 74. For Prob 13 4, PROPRIETARY MATERIAL 2007 The McGraw Hill Companies Inc All rights reserved No part. of this Manual may be displayed reproduced or distributed in any form or by any means without the prior. written permission of the publisher or used beyond the limited distribution to teachers and educators. permitted by McGraw Hill for their individual course preparation If you are a student using this Manual. you are using it without permission,Chapter 13 Solution 4.
a For the series connection shown in Figure a the current I enters each coil from. its dotted terminal Therefore the mutually induced voltages have the same sign as the. self induced voltages Thus,Leq L1 L2 2M,b For the parallel coil consider Figure b. Is I 1 I2 and Zeq Vs Is,Applying KVL to each branch gives. Vs j L1I1 j MI2 1,Vs j MI1 j L2I2 2,Vs j L1 j M I1. or V j M j L 2 I 2,2L1L2 2M2 1 j Vs L2 M 2 j Vs L1 M. I1 1 and I2 2,Is I1 I2 1 2 j L1 L2 2M Vs 2 L1L2 M2.
L1 L2 2M Vs j L1L2 M2,Zeq Vs Is j L1L2 M2 L1 L2 2M j Leq. i e Leq L1L2 M2 L1 L2 2M, PROPRIETARY MATERIAL 2007 The McGraw Hill Companies Inc All rights reserved No part. of this Manual may be displayed reproduced or distributed in any form or by any means without the prior. written permission of the publisher or used beyond the limited distribution to teachers and educators. permitted by McGraw Hill for their individual course preparation If you are a student using this Manual. you are using it without permission,Chapter 13 Problem 5. Two coils are mutually coupled with L1 25 mH L2 60 mH and k 0 5 Calculate the. maximum possible equivalent inductance if,a the two coils are connected in series. b the coils are connected in parallel,Chapter 13 Solution 5.
a If the coils are connected in series,L L1 L 2 2M 25 60 2 0 5 25x 60 123 7 mH. b If they are connected in parallel,L1 L 2 M 2 25x 60 19 36 2. L mH 24 31 mH,L1 L 2 2M 25 60 2x19 36, PROPRIETARY MATERIAL 2007 The McGraw Hill Companies Inc All rights reserved No part. of this Manual may be displayed reproduced or distributed in any form or by any means without the prior. written permission of the publisher or used beyond the limited distribution to teachers and educators. permitted by McGraw Hill for their individual course preparation If you are a student using this Manual. you are using it without permission,Chapter 13 Problem 6. The coils in Fig 13 75 have L1 40 mH L2 5 mH and coupling coefficient k 0 6. Find i1 t and v2 t given that v1 t 10 cos t and i2 t 2 sin t 2000 rad s. Figure 13 75,For Prob 13 6, PROPRIETARY MATERIAL 2007 The McGraw Hill Companies Inc All rights reserved No part.
of this Manual may be displayed reproduced or distributed in any form or by any means without the prior. written permission of the publisher or used beyond the limited distribution to teachers and educators. permitted by McGraw Hill for their individual course preparation If you are a student using this Manual. you are using it without permission,Chapter 13 Solution 6. M k L1 L2 0 6 40 x5 8 4853 mH,j L j 2000 x 40 x10 j80. j L j 2000 x5 x10 j10,8 4853mH j M j 2000 x8 4853 x10 j16 97. We analyze the circuit below,V1 j80 I1 j16 97 I 2 1. V2 16 97 I1 j10 I 2 2, But V1 10 0o and I 2 2 90o j 2 Substituting these in eq 1 gives.
V j16 97 I 2 10 j16 97 x j 2,I1 1 0 5493 90o,i1 t 0 5493sin t A. V2 16 97 x 0 j 5493 j10 x j 2 20 j 9 3216 22 0656 24 99o. v2 t 22 065cos t 25o V, PROPRIETARY MATERIAL 2007 The McGraw Hill Companies Inc All rights reserved No part. of this Manual may be displayed reproduced or distributed in any form or by any means without the prior. written permission of the publisher or used beyond the limited distribution to teachers and educators. permitted by McGraw Hill for their individual course preparation If you are a student using this Manual. you are using it without permission,Chapter 13 Problem 7. For the circuit in Fig 13 76 find Vo,Figure 13 76,For Prob 13 7. Chapter 13 Solution 7, We apply mesh analysis to the circuit as shown below.
12 I1 j6 j4 I2 1 Vo,For mesh 1,12 I1 2 j 6 jI 2 1,For mesh 2. 0 jI1 2 j1 j 4 I 2,0 jI1 2 j 3 I 2 2,In matrix form. 12 2 j 6 j I1,0 j 2 j 3 I 2,I 2 0 4381 j 0 3164,Vo I2x1 540 5 144 16 mV. PROPRIETARY MATERIAL 2007 The McGraw Hill Companies Inc All rights reserved No part. of this Manual may be displayed reproduced or distributed in any form or by any means without the prior. written permission of the publisher or used beyond the limited distribution to teachers and educators. permitted by McGraw Hill for their individual course preparation If you are a student using this Manual. you are using it without permission,Chapter 13 Problem 8. Find v t for the circuit in Fig 13 77,Figure 13 77.
For Prob 13 8,Chapter 13 Solution 8,j L j 4 x 2 j8. j L j 4 x1 j 4,Consider the circuit below,2 0o I1 j8 j4 2 V t. 2 4 j8 I1 j 4 I 2 1,0 j 4 I1 2 j 4 I 2 2,In matrix form these equations become. 2 4 j8 j 4 I1,0 j 4 2 j 4 I,Solving this leads to,I2 0 2353 j0 0588. V 2I2 0 4851 14 04o,v t 0 4851cos 4t 14 04o V, PROPRIETARY MATERIAL 2007 The McGraw Hill Companies Inc All rights reserved No part.
of this Manual may be displayed reproduced or distributed in any form or by any means without the prior. written permission of the publisher or used beyond the limited distribution to teachers and educators. permitted by McGraw Hill for their individual course preparation If you are a student using this Manual. you are using it without permission,Chapter 13 Problem 9. Find Vx in the network shown in Fig 13 78,Figure 13 78. For Prob 13 9,Chapter 13 Solution 9,Consider the circuit below. For loop 1,8 30 2 j4 I1 jI2 1,For loop 2 j4 2 j I2 jI1 j2 0. or I1 3 j2 i2 2 2,Substituting 2 into 1 8 30 2 j4 2 14 j7 I2.
I2 10 928 j12 14 j7 1 037 21 12,Vx 2I2 2 074 21 12. PROPRIETARY MATERIAL 2007 The McGraw Hill Companies Inc All rights reserved No part. of this Manual may be displayed reproduced or distributed in any form or by any means without the prior. written permission of the publisher or used beyond the limited distribution to teachers and educators. permitted by McGraw Hill for their individual course preparation If you are a student using this Manual. you are using it without permission,Chapter 13 Problem 10. Find vo in the circuit of Fig 13 79,Figure 13 79,For Prob 13 10. Chapter 13 Solution 10,2H j L j 2 x 2 j 4,0 5H j L j 2 x0 5 j. 2 j C j 2 x1 2,Consider the circuit below,24 0 I1 I2 Vo j.
24 j 4 I1 jI 2 1,0 jI1 j 4 j I 2 0 I1 3I 2 2,In matrix form. 24 j 4 j I1,Solving this,I 2 j 2 1818 Vo jI 2 2 1818. vo 2 1818cos2t V, PROPRIETARY MATERIAL 2007 The McGraw Hill Companies Inc All rights reserved No part. of this Manual may be displayed reproduced or distributed in any form or by any means without the prior. written permission of the publisher or used beyond the limited distribution to teachers and educators. permitted by McGraw Hill for their individual course preparation If you are a student using this Manual. you are using it without permission,Chapter 13 Problem 11. Use mesh analysis to find ix in Fig 13 80 where,is 4 cos 600t A and vs 110 cos 600t 30.
Figure 13 80,For Prob 13 11,Chapter 13 Solution 11. j L j 600 x800 x10 j 480,j L j 600 x600 x10 j 360,1200mH j L j 600 x1200 x10 j 720. 12 F j138 89,j C 600x12x10 6, PROPRIETARY MATERIAL 2007 The McGraw Hill Companies Inc All rights reserved No part. of this Manual may be displayed reproduced or distributed in any form or by any means without the prior. written permission of the publisher or used beyond the limited distribution to teachers and educators. permitted by McGraw Hill for their individual course preparation If you are a student using this Manual. you are using it without permission, After transforming the current source to a voltage source we get the circuit shown below. 200 j480 j138 89 150,j360 j720 110 30,For mesh 1,800 200 j 480 j 720 I1 j 360 I 2 j 720 I 2.
800 200 j1200 I1 j 360 I 2 1,For mesh 2,110 30 150 j138 89 j720 I2 j360I1 0. 95 2628 j 55 j 360 I1 150 j 581 1 I 2 2,In matrix form. 800 200 j1200 j 360 I1,95 2628 j 55 j 360 150 j 581 1 I 2. Solving this using MATLAB leads to,Z 200 1200i 360i 360i 150 581 1i. 0 2000 1 2000i 0 0 3600i,0 0 3600i 0 1500 0 5811i,V 800 95 26 55i.
0 9526 0 5500i,0 1390 0 7242i,0 0609 0 2690i,Ix I1 I2 0 0781 j0 4552 0 4619 80 26. Hence ix 461 9cos 600t 80 26 mA, PROPRIETARY MATERIAL 2007 The McGraw Hill Companies Inc All rights reserved No part. of this Manual may be displayed reproduced or distributed in any form or by any means without the prior. written permission of the publisher or used beyond the limited distribution to teachers and educators. permitted by McGraw Hill for their individual course preparation If you are a student using this Manual. you are using it without permission,Chapter 13 Problem 12. Determine the equivalent Leq in the circuit of Fig 13 81. Figure 13 81,For Prob 13 12,Chapter 13 Solution 12. Applying KVL to the loops,1 j8 I 1 j 4 I 2 1,0 j 4 I 1 j18 I 2 2.
Solving 1 and 2 gives I1 j0 1406 Thus,Leq 7 111 H, We can also use the equivalent T section for the transform to find the equivalent. inductance, PROPRIETARY MATERIAL 2007 The McGraw Hill Companies Inc All rights reserved No part. of this Manual may be displayed reproduced or distributed in any form or by any means without the prior. written permission of the publisher or used beyond the limited distribution to teachers and educators. permitted by McGraw Hill for their individual course preparation If you are a student using this Manual. you are using it without permission,Chapter 13 Problem 13. For the circuit in Fig 13 82 determine the impedance seen by the source. Figure 13 82,For Prob 13 13,Chapter 13 Solution 13. Z in 4 j 2 5 4 j7 4 308 j6 538,j5 4 j j2 4 j6, PROPRIETARY MATERIAL 2007 The McGraw Hill Companies Inc All rights reserved No part.
of this Manual may be displayed reproduced or distributed in any form or by any means without the prior. written permission of the publisher or used beyond the limited distribution to teachers and educators. permitted by McGraw Hill for their individual course preparation If you are a student using this Manual.

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