Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Regenerative circuit wikipedia , lookup
Index of electronics articles wikipedia , lookup
Valve RF amplifier wikipedia , lookup
Mathematics of radio engineering wikipedia , lookup
Standing wave ratio wikipedia , lookup
Nanofluidic circuitry wikipedia , lookup
Zobel network wikipedia , lookup
0-2 Operations with Complex Numbers Simplify. 1. i–10 SOLUTION: 2. i2 + i8 SOLUTION: 3. i3 + i20 SOLUTION: 4. i100 SOLUTION: 5. i77 SOLUTION: 6. i4 + i–12 SOLUTION: eSolutions Manual - Powered by Cognero 7. i5 + i9 Page 1 SOLUTION: 0-2 Operations with Complex Numbers 6. i4 + i–12 SOLUTION: 7. i5 + i9 SOLUTION: 8. i18 SOLUTION: Simplify. 9. (3 + 2i) + (–4 + 6i) SOLUTION: 10. (7 – 4i) + (2 – 3i) SOLUTION: 11. (0.5 + i) – (2 – i) SOLUTION: 12. (–3 – i) – (4 – 5i) SOLUTION: eSolutions Manual - Powered by Cognero Page 2 SOLUTION: 0-2 Operations with Complex Numbers 12. (–3 – i) – (4 – 5i) SOLUTION: 13. (2 + 4.1i) – (–1 – 6.3i) SOLUTION: 14. (2 + 3i) + (–6 + i) SOLUTION: 15. (–2 + 4i) + (5 – 4i) SOLUTION: 16. (5 + 7i) – (–5 + i) SOLUTION: 17. ELECTRICITY Engineers use imaginary numbers to express the two-dimensional quantity of alternating current, which involves both amplitude and angle. In these imaginary numbers, i is replaced with j because engineers use I as a variable for the entire quantity of current. Impedance is the measure of how much hindrance there is to the flow of the charge in a circuit with alternating current. The impedance in one part of a series circuit is 2 + 5j ohms, and the impedance in another part of the circuit is 7 – 3j ohms. Add these complex numbers to find the total impedance in the circuit. SOLUTION: Simplify. 18. (–2 – i)2 SOLUTION: eSolutions Manual - Powered by Cognero Page 3 in the circuit. SOLUTION: 0-2 Operations with Complex Numbers Simplify. 18. (–2 – i)2 SOLUTION: 19. (1 + 4i)2 SOLUTION: 20. (5 + 2i)2 SOLUTION: 21. (3 + i)2 SOLUTION: 22. (2 + i)(4 + 3i) SOLUTION: eSolutions Manual - Powered by Cognero Page 4 0-2 Operations with Complex Numbers 22. (2 + i)(4 + 3i) SOLUTION: 23. (3 + 5i)(3 – 5i) SOLUTION: 24. (5 + 3i)(2 + 6i) SOLUTION: 25. (6 + 7i)(6 – 7i) SOLUTION: Simplify. 26. SOLUTION: eSolutions Manual - Powered by Cognero Page 5 0-2 Operations with Complex Numbers Simplify. 26. SOLUTION: 27. SOLUTION: 28. SOLUTION: eSolutions Manual - Powered by Cognero Page 6 0-2 Operations with Complex Numbers 28. SOLUTION: 29. SOLUTION: 30. SOLUTION: eSolutions Manual - Powered by Cognero Page 7 0-2 Operations with Complex Numbers 30. SOLUTION: 31. SOLUTION: 32. SOLUTION: eSolutions Manual - Powered by Cognero Page 8 0-2 Operations with Complex Numbers 32. SOLUTION: 33. SOLUTION: eSolutions Manual - Powered by Cognero Page 9 ELECTRICITY The voltage E, current I, and impedance Z in a circuit are related by E = I · Z. Find the voltage (in volts) in each of the following circuits given the current and impedance. 0-2 Operations with Complex Numbers 33. SOLUTION: ELECTRICITY The voltage E, current I, and impedance Z in a circuit are related by E = I · Z. Find the voltage (in volts) in each of the following circuits given the current and impedance. 34. I = 1 + 3j amps, Z = 7 – 5j ohms SOLUTION: Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit. Therefore, the voltage of the circuit is 22 + 16j volts. 35. I = 2 – 7j amps, Z = 4 – 3j ohms SOLUTION: Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit. eSolutions Manual - Powered by Cognero Page 10 0-2 Therefore, Operations with ofComplex the voltage the circuit isNumbers 22 + 16j volts. 35. I = 2 – 7j amps, Z = 4 – 3j ohms SOLUTION: Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit. Therefore, the voltage of the circuit is –13 – 34j volts. 36. I = 5 – 4j amps, Z = 3 + 2j ohms SOLUTION: Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit. Therefore, the voltage of the circuit is 23 – 2j volts. 37. I = 3 + 10j amps, Z = 6 – j ohms SOLUTION: Substitute the given values for I and Z into the equation E = I · Z to find the voltage E of the circuit. Therefore, the voltage of the circuit is 28 + 57j volts. Solve each equation. 38. 5x2 + 5 = 0 SOLUTION: eSolutions Manual - Powered by Cognero Page 11 0-2 Operations with Complex Numbers Therefore, the voltage of the circuit is 28 + 57j volts. Solve each equation. 38. 5x2 + 5 = 0 SOLUTION: 39. 4x2 + 64 = 0 SOLUTION: 40. 2x2 + 12 = 0 SOLUTION: 41. 6x2 + 72 = 0 SOLUTION: 2 42. 8x Manual + 120 = 0 eSolutions - Powered by Cognero SOLUTION: Page 12 0-2 Operations with Complex Numbers 42. 8x2 + 120 = 0 SOLUTION: 43. 3x2 + 507 = 0 SOLUTION: 44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance and the reactance due to inductance ohms) for R, , and , , and can be written as a complex number R + (X L – X C)j . The values (in in the first and second parts of a particular series circuit are shown. a. Write complex numbers that represent the impedances in the two parts of the circuit. b. Add your answers from part a to find the total impedance in the circuit. c. The admittance S of a circuit is the measure of how easily the circuit allows current to flow, and is the reciprocal of impedance. Find the admittance (in siemens) in a circuit with an impedance of 6 + 3j ohms. SOLUTION: a. Substitute the given values for R, X C, and The impedance in the first part of the circuit is b. ohms eSolutions Manual - Powered by Cognero c. The reciprocal of a number x is denominator. into the equation Z = R + (X L – X C)j . ohms and in the second of the circuit is , so the reciprocal of is ohms. . Simplify this fraction by rationalizingPage the 13 0-2 Operations with Complex Numbers 44. ELECTRICITY The impedance Z of a circuit depends on the resistance R, the reactance due to capacitance and the reactance due to inductance ohms) for R, , and , , and can be written as a complex number R + (X L – X C)j . The values (in in the first and second parts of a particular series circuit are shown. a. Write complex numbers that represent the impedances in the two parts of the circuit. b. Add your answers from part a to find the total impedance in the circuit. c. The admittance S of a circuit is the measure of how easily the circuit allows current to flow, and is the reciprocal of impedance. Find the admittance (in siemens) in a circuit with an impedance of 6 + 3j ohms. SOLUTION: a. Substitute the given values for R, X C, and The impedance in the first part of the circuit is b. ohms c. The reciprocal of a number x is into the equation Z = R + (X L – X C)j . ohms and in the second of the circuit is , so the reciprocal of is ohms. . Simplify this fraction by rationalizing the denominator. Find the values of x and y to make each equation true. 45. 3x + 2iy = 6 + 10i SOLUTION: To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the equation and then solve for x and y respectively. eSolutions Manual - Powered by Cognero Page 14 0-2 Operations with Complex Numbers Find the values of x and y to make each equation true. 45. 3x + 2iy = 6 + 10i SOLUTION: To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the equation and then solve for x and y respectively. 46. 5x + 3iy = 5 – 6i SOLUTION: To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the equation and then solve for x and y respectively. 47. x – iy = 3 + 4i SOLUTION: To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the equation and then solve for x and y respectively. 48. –5x + 3iy = 10 – 9i SOLUTION: To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the equation and then solve for x and y respectively. 49. 2x + 3iy = 12 + 12i SOLUTION: To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the equation and then solve for x and y respectively. eSolutions Manual - Powered by Cognero 50. 4x – iy = 8 + 7i Page 15 equation and then solve for x and y respectively. 0-2 Operations with Complex Numbers 49. 2x + 3iy = 12 + 12i SOLUTION: To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the equation and then solve for x and y respectively. 50. 4x – iy = 8 + 7i SOLUTION: To find the values of x and y that make this equation true, equate the real and imaginary parts of each side of the equation and then solve for x and y respectively. Simplify. 51. (2 – i)(3 + 2i)(1 – 4i) SOLUTION: 52. (–1 – 3i)(2 + 2i)(1 – 2i) SOLUTION: 53. (2 + i)(1 + 2i)(3 – 4i) eSolutions Manual - Powered by Cognero SOLUTION: Page 16 0-2 Operations with Complex Numbers 53. (2 + i)(1 + 2i)(3 – 4i) SOLUTION: 54. (−5 – i)(6i + 1)(7 – i) SOLUTION: eSolutions Manual - Powered by Cognero Page 17