chap13

# chap13 - Chapter 13 Solution 1 For coil 1 L1 M12 M13 = 6 4...

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Chapter 13, Solution 1. For coil 1, L 1 – M 12 + M 13 = 6 – 4 + 2 = 4 For coil 2, L 2 – M 21 – M 23 = 8 – 4 – 5 = – 1 For coil 3, L 3 + M 31 – M 32 = 10 + 2 – 5 = 7 L T = 4 – 1 + 7 = 10H or L T = L 1 + L 2 + L 3 – 2M 12 – 2M 23 + 2M 12 L T = 6 + 8 + 10 = 10H Chapter 13, Solution 2. L = L 1 + L 2 + L 3 + 2M 12 – 2M 23 2M 31 = 10 + 12 +8 + 2x6 – 2x6 –2x4 = 22H Chapter 13, Solution 3. L 1 + L 2 + 2M = 250 mH (1) L 1 + L 2 – 2M = 150 mH (2) Adding (1) and (2), 2L 1 + 2L 2 = 400 mH But, L 1 = 3L 2, , or 8L 2 + 400, and L 2 = 50 mH L 1 = 3L 2 = 150 mH From (2), 150 + 50 – 2M = 150 leads to M = 25 mH k = M/ 150 x 50 / 5 . 2 L L 2 1 = = 0.2887

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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, L eq = L 1 + L 2 + 2M L 2 L 1 L 2 L 1 I 1 I 2 I s V s + L eq (a) (b) (b) For the parallel coil, consider Figure (b). I s = I 1 + I 2 and Z eq = V s /I s Applying KVL to each branch gives, V s = j ω L 1 I 1 + j ω MI 2 (1) V s = j ω MI 1 + j ω L 2 I 2 (2) or ω ω ω ω = 2 1 2 1 s s I I L j M j M j L j V V = ω 2 L 1 L 2 + ω 2 M 2 , 1 = j ω V s (L 2 – M), 2 = j ω V s (L 1 – M) I 1 = 1 / , and I 2 = 2 / I s = I 1 + I 2 = ( 1 + 2 )/ = j ω (L 1 + L 2 – 2M)V s /( – ω 2 (L 1 L 2 – M)) Z eq = V s /I s = j ω (L 1 L 2 – M)/[j ω (L 1 + L 2 – 2M)] = j ω L eq i.e., L eq = (L 1 L 2 – M)/(L 1 + L 2 – 2M)
Chapter 13, Solution 5. (a) If the coils are connected in series, = + + = + + = 60 x 25 ) 5 . 0 ( 2 60 25 M 2 L L L 2 1 123.7 mH (b) If they are connected in parallel, = + = + = mH 36 . 19 x 2 60 25 36 . 19 60 x 25 M 2 L L M L L L 2 2 1 2 2 1 24.31 mH Chapter 13, Solution 6. V 1 = (R 1 + j ω L 1 )I 1 – j ω MI 2 V 2 = –j ω MI 1 + (R 2 + j ω L 2 )I 2 Chapter 13, Solution 7. Applying KVL to the loop, 20 30 ° = I(–j6 + j8 + j12 + 10 – j4x2) = I(10 + j6) where I is the loop current. I = 20 30 ° /(10 + j6) V o = I(j12 + 10 – j4) = I(10 + j8) = 20 30 ° (10 + j8)/(10 + j6) = 22 37.66 ° V Chapter 13, Solution 8. Consider the current as shown below. j 6 j 4 1 -j3 j2 10 + I 2 I 1 4 + V o

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For mesh 1, 10 = (1 + j6)I 1 + j2I 2 (1) For mesh 2, 0 = (4 + j4 – j3)I 2 + j2I 1 0 = j2I 1 +(4 + j)I 2 (2) In matrix form, + + = 2 1 I I j 4 2 j 2 j 6 j 1 0 10 = 2 + j25, and 2 = –j20 I 2 = 2 / = –j20/(2 + j25) V o = –j3I 2 = –60/(2 + j25) = 2.392 94.57 ° Chapter 13, Solution 9. Consider the circuit below. 2 -j1 2 For loop 1, -j2V + j 4 j 4 8 30 o + I 2 I 1 8 30 ° = (2 + j4)I 1 – jI 2 (1) For loop 2, ((j4 + 2 – j)I 2 – jI 1 + (–j2) = 0 or I 1 = (3 – j2)i 2 – 2 (2) Substituting (2) into (1), 8 30 ° + (2 + j4)2 = (14 + j7)I 2 I 2 = (10.928 + j12)/(14 + j7) = 1.037 21.12 ° V x = 2I 2 = 2.074 21.12 °
Chapter 13, Solution 10. Consider the circuit below. I o j ω L j ω L 1/j ω C I 2 I 1 j ω M I in 0 o 2 2 1 L L L k M = = = L, I 1 = I in 0 ° , I 2 = I o I o (j ω L + R + 1/(j ω C)) – j ω LI in – (1/(j ω C))I in = 0 I o = j I in ( ω L – 1/( ω C)) /(R + j ω L + 1/(j ω C)) Chapter 13, Solution 11.

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