Download Chapter 13-Electrical Circuit Analysis-Problem Solutions and more Exercises Electrical Circuit Analysis in PDF only on Docsity! Chapter 13, Problem 1. For the three coupled coils in Fig. 13.72, calculate the total inductance. Figure 13.72 For Prob. 13.1. Chapter 13, Solution 1. For coil 1, L1 – M12 + M13 = 6 – 4 + 2 = 4 For coil 2, L2 – M21 – M23 = 8 – 4 – 5 = – 1 For coil 3, L3 + M31 – M32 = 10 + 2 – 5 = 7 LT = 4 – 1 + 7 = 10H or LT = L1 + L2 + L3 – 2M12 – 2M23 + 2M12 LT = 6 + 8 + 10 = 10H Chapter 13, Problem 2. Determine the inductance of the three series-connected inductors of Fig. 13.73. Figure 13.73 For Prob. 13.2. Chapter 13, Solution 2. L = L1 + L2 + L3 + 2M12 – 2M23 –2M31 = 10 + 12 +8 + 2x6 – 2x6 –2x4 = 22H docsity.com 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/ 150x50/25LL 21 = = 0.2887 docsity.com =++=++= 60x25)5.0(26025M2LLL 21 123.7 mH (b) If they are connected in parallel, = −+ − = −+ − = mH 36.19x26025 36.1960x25 M2LL MLLL 2 21 2 21 24.31 mH docsity.com 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 ω t and i2(t) = 2 sin ω t, ω = 2000 rad/s. Figure 13.75 For Prob. 13.6. docsity.com Chapter 13, Solution 6. 1 2 0.6 40 5 8.4853 mHM k L L x= = = 3 40 2000 40 10 80mH j L j x x jω − ⎯⎯→ = = 3 5 2000 5 10 10mH j L j x x jω − ⎯⎯→ = = 3 8.4853 2000 8.4853 10 16.97mH j M j x x jω − ⎯⎯→ = = We analyze the circuit below. 1 1 280 16.97V j I j I= − (1) 2 1 216.97 10V I j I= − + (2) But 1 210 0 and 2 90 2 o oV I j= < = < − = − . Substituting these in eq.(1) gives 1 2 1 16.97 10 16.97 ( 2) 0.5493 90 80 80 oV j I j x jI j j + + − = = = < − 1( ) 0.5493sin Ai t tω= From (2), 2 16.97 ( 0. 5493) 10 ( 2) 20 9.3216 22.0656 24.99 oV x j j x j j= − − + − = + = < 2 ( ) 22.065cos( 25 ) V ov t tω= + • • + _ V2 I1 I2 + _ V1 16.77 Ω j80 Ω j10 Ω docsity.com 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° 2 Ω + – o 2 Ω j 4 j 4 -j1 + – docsity.com Find vo in the circuit of Fig. 13.79. Figure 13.79 For Prob. 13.10. Chapter 13, Solution 10. 2 2 2 4H j L j x jω⎯⎯→ = = 0.5 2 0.5H j L j x jω⎯⎯→ = = 1 11 2 2 1/ 2 F jj C j xω ⎯⎯→ = = − Consider the circuit below. 1 224 4j I jI= − (1) 1 2 1 20 ( 4 ) 0 3jI j j I I I= − + − ⎯⎯→ = − + (2) In matrix form, 1 2 24 4 0 1 3 Ij j I − ⎡ ⎤⎡ ⎤ ⎡ ⎤ = ⎢ ⎥⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦ Solving this, 2 22.1818, 2.1818oI j V jI= − = − = − vo = –2.1818cos2t V + _ 24 ∠ 0° • • + _ Vo j4 j –j I1 I2 j4 docsity.com
