Introduction to Electric Circuits

Answers 243

14 = 44 mA. 7. (a) 6012, (b) 8 A, 4 A, 4 A. 8. 0.332 fL 66.4 fL 0.664 ~. 9. (a) 2 f~, (b)
112.5 W. 10. (a) 1.4 V, 3.9 12; (b) 1.28 V. 11. (a) 0.456 A flowing from A to B, (b) 11.5 12 in
parallel with R, (c) 2.29 W. 12. 3.79 mA.

Chapter 4

Self-assessment test

2. Hz. 3.f = 1/T. 4. 25.5. 50 Hz. 6. 90 ~ 7. 100 V. 8. 1.11.9. In phase. 10. 150.8 12.


3. 31.83 k12. 12. Ohm. 13. 63.43 ~ 14. 28.28 12. 15. (a) 22.36 12, (b) 20 12. 17. j. 18. Third.


4. 18/_-56.3 ~ 20. 21.65 + j12.5.21. 5/_53.13 ~ 22. (5 + j6) 12. 23. (0.1 + j0.04) S. 24.
   S = P + jQ. 25. (a) 160 VA, (b) 125 W, (c) 100 Var, (d) 0.832 leading, (e) (25 - j20) V.

Problems

1. v = 250 sin (314t + 53.2~ i = 5 sin (314t + 23.6~ -150 V; -4.58 A; 29.6 ~ current
   lagging. 2. 122 V leading Vl by 2.2 ~ 3. (a) 5 12, 5 f~, 7.07 12; 5 ~1, 10 12, 11.2 12.4. (a) 11.1 A,
   (b) 0.555 lagging. 5. (a) 2.47 A, (b) 123.5 V, (c) 157.3 V. 6. (a) 7.98 Hz, (b) 433 12.7. (a) (i)
   2.48 A, (ii) 124 V, 158 V, (iii) 0.618 leading; (b) 36.7 ~xF; (c) 110 IxF. 8. (a) 2 A, 1.57 A; (b)
   2.54 A; (c) 78.8 12; (d) 38 ~ leading. 9. (a) (i) 13.2 A, (ii) 2.62 kW, (iii) 0.991 lagging; (b) (i)
   31.5 A, (ii) 5.9 kW, (iii) 0.937 lagging. 10. (a) 0.118 S, 0.113 S, 0.164 S; (b) 32.8 A; (c)
   4.72 kW. 11. (a) 13.1 A, (b) 0.953 leading. 12. (a) 23.6 A, (b) 5.1 kW.

Chapter 5

Self-assessment test

7. ABC. 9. EMsin (tot - 240~ 11. 25 A. 12. 30 ~ 13. 17.32 A. 14. 12.65 kW. 15. V'3 VLIL.


8. 28 kVar. 17. 340 W. 18. W1 = VAclA COS (angle between VAC and IA);
   W2 = VBclB COS (angle between VBC and IB). 19. COS {tan-l[V'3(P1 - P2)/(Pa + P2)]}. 20. (a)
   190 W, (b) 103.92 W, (c) 216.56 W, (d) 0.877.

Problems

1. (a) 32.44 A, (b) 8.99 kW. 2. (a) 34.6 A, (b) 18.24 kW, 3. (2.287 - j10.79) kVA. 4. (a)
   30 kW, (b) 0.33, (c) 231 A. 5. 0.866. 6. (a) 400 kW, (b) 0.756, (c) 152 A, (d) 360 kW.

Chapter 6

Self-assessment test

1. toL = 1/toC. 2. 1/27rk/(LC). 3. Unity. 4. 250 rad s -1. 5. 2 12.6. Because the current is a
   maximum at resonance. 8. Q is dimensionless. 9. By reducing the resistance in the circuit.


2. Hz. 14. P1 is 10 log10 (P1/P2) dB above P2. 16. 1/2~/(LC). 17. L/CR. 18. It increases.


3. A minimum. 20. to0/Q.

Problems

1. 4 lq, 504 mH. 2. (a) 40 Hz, (b) 20 A, (c) 5 kV. 3. (a) 20.3 IxF, (b) 3.925 kV, 100 V, (c) 39.3.


2. (a) 22.5 kHz, (b) 17.6, (c) 1.28 kHz. 5. 0.233; -12.6 dB. 6. (a) 1/[1 + 1 jtoCR], (b) -40,
   -20, -3; Comment: 1000 rad s -a is a half-power frequency. (c) Add a series inductance,
   the output still being taken across the resistor. 7. (a) 38 I~F, (b) 60.8 W, (c) 3.93 (11.89 dB).