Solutions-_-Tutorial-11_AC2.pdf

Solutions-_-Tutorial-11_AC2.pdf - Manora Caldera Electrical...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Manora Caldera Electrical Systems Tutorial 11 – Solutions 1. (a) (b) Reactance 2 (i) For L = 0.5H, and f =60Hz, (ii) For L = 0.5H, and f =1kHz, (iii) For L = 0.5H, and =500rad/s, For 100 sin Using rms values, 2 2 2 2 60 0.5 188.5Ω 1000 0.5 3.14 Ω 500 0.5 250Ω 30° 30° √ 70.7 30° . 90°Ω and . (i) . 22.52√2 sin 2 ° 0.2828√2 sin 500 100 sin 377 377 200 L = 200mH and 75.4 /√ . Then, (b) Then, if 10sin 2 2 ° ° 60 60° 60° 1000 0.2828 ° Then, 60° 22.52 ° . (iii) 2. (a) ° . Then, 0.375 ° 0.375√2 sin 2 Then, (ii) ° . 0.530 sin 377 60° 31.8 sin 6283 60° A 60° mA 60° 60° 0.4 sin 500 60° A and hence 377rad/s 75.4Ω 10 90°Ω and 0.938 0.938√2 sin 377 90° 90° 1.33 sin 377 90° A 400 60° , and hence 400 2 400 200 10 502.65Ω 502.65 90°Ω and 10/√2 60° 502.65 90° 3.554 30° 3.554√2 sin 2 400 30° 5.03 sin 2 400 30° V 1 Manora Caldera (c) For 40 sin Reactance 30° 30° 364 sin and Using rms values, √ 28.28 2 10 257.39 60° √ 60° and for L= 2mH Ω 60° and 30° 90°Ω and 28.28 30° 257.39 10 3 60° 109.87 90°Ω j109.87Ω Then, 2 10 Ω 109.87Ω This gives, 2 3. (a) 54.935 8.743 / and Reactance (b) (i) For C= 5F, and f =60Hz, 530.52Ω (ii) For C= 5F, and f =1kHz, 31.83Ω (iii) For C= 5F, and =500rad/s, For 100 sin Using rms values, 400Ω 30° √ 30° 70.7 30° . 90°Ω and . (i) ° . ° . ° . ° ° Then, 2.221 2.221√2 sin 2 Then, (iii) 0.133 0.133√2 sin 2 Then, (ii) ° . 0.1768 120° 60 120° 0.188 sin 377 120° A 120° 1000 120° 3.14 sin 6283 120° A 120°A 0.1768√2 sin 500 120° 0.25 sin 500 120° A 2 Manora Caldera For Reactance 4. (a) and for C= 50F 100 sin 377 1 377 5 10 Using rms values, 0° √ 530.52Ω 70.7 0° . 90°Ω and . ° . 0.133√2 sin 377 Then, For Reactance (b) 0.133 ° 10 sin 2 400 90°A 90° 60° 0.188 sin 377 and for C= 50F 1 2 Using rms values, √ 90° A 79.58Ω 400 5 10 60° 7.07 60° . 90°Ω and and 7.07 0.562√2 sin 2 Then, (c) For 362 sin Reactance 33° Using rms values, √ 33° √ 255.97 10 400 60° 79.58 150° 57° 0.562 400 150° 150° V and for C= 2.2F 1 Ω 2.2 10 66.47 57° and 57° 33° 90°Ω and 255.97 33° 3.85 3 57° 66.47 10 Then, 0.795 sin 2 94 sin and 90° 1 2.2 10 Ω 90° Ω 3.85 j3.85kΩ 10 Ω This gives, 2 118.06 / and 18.8 3 Manora Caldera 5. For the circuit in Figure (a), 30Ω, j35Ω 30 Thus, 35 35 25 90°Ω and 30 10Ω j25Ω 31.62 25 90°Ω 18.43°Ω For the circuit in Figure (b), 3 Ω, 4.2 Ω j1.5kΩ 3 Thus, 4.2 j3.3kΩ 1.5 3.3 3.3 90°kΩ and 90°kΩ and 1.5 5.9 7.2 j5.9kΩ 4.1 Ω 8.29 5.9 90°kΩ 29.70°kΩ 6. For the circuit in the Figure, 40Ω, and j20Ω 20 90°Ω j50Ω 50 90°Ω Thus, 40 As 20 50 40 30Ω ° , 2.4 2.4 . ° 36.87° 36.87° 2.4 20 50 50 36.87°Ω 36.87° 90° 90° 48 126.87°V 120 53.13°V 4 Manora Caldera 2.4 36.87° 40 0° 96 36.87°V Now you can show that In this case, you need to convert the voltages into rectangular form first. Phasor diagrams: Note that these are not drawn to the scale -53.13 VC = 120<-53.13 7. (a) 1 (b) 10√2 sin 4000 √ √ 4000 5 10 1 4000 12.5 10 0° 10 20Ω 20Ω 0° 5 Manora Caldera (c) 10 ° (d) 1 ° 0° and 1 (e) 10 20 20=10 Ω √2 sin 4000 0° 10 10 1 0° 20 90° 20 1 0° 20 90° 20 8. For Figure (a), and the total admittance 600 j900Ω 600 900 90°Ω and ° 90°Ω and j1800Ω 1800 j j ° 90°Ω and j j ° . Hence, giving 900Ω 900 90°Ω For Figure (b), and the total admittance 50 Ω and j3kΩ 3 90°kΩ and j6kΩ 6 90°kΩ and j Hence, j . . . ° . j giving ° 0.02 . . j 0.5 . . . . . . . 0.08 2 Ω 6 Manora Caldera 9. 1 1 1 1 3 1 3 8 3 3 Voltage 3 30 4 3 4 4 3 4 3 6 4 1 8 6 4 4 3 3 8 41 6 294 6 92 4 0° 8 8 12 6 4 4 6 4 5.65 5.92 1.77 Ω 17.39° 5.92 177.6 17.39°kΩ 17.39° 10. In this circuit, 100 and the –j40 elements are in series giving a total impedance of 100 40Ω. This total impedance is in parallel with the j80 inductor giving a total impedance of the parallel combination as 100 40 80 3200 8000 800 4 10 100 40 // 80 Ω 100 40 80 100 40 20 5 2 Now the total impedance of the circuit will be 40 4 10 40 4 10 5 2 40 40 42 50 50 50 5 2 5 2 5 2 25 4 50 Thus, ° 40 40 42 29 ° . ° 500 105.17 57.93 120 28.85°Ω 28.85° 7 Manora Caldera 11. (a) 4000 5 10 1 4000 12.5 10 1 10√2 sin 4000 (b) √ 0° √ 10 20Ω 20Ω 0° (c) 10Ω Therefore, ° (d) 1 ° ° (e) 0° and √2 sin 4000 1 ° 0.5 ° 10 20 0° 90° 90° 0.5 0.5 90° 0.5 12. (a) Using the current divider rule, 72 72 96 96 72 72 72 96 96 72 72 (b) And = 599.76 96 96 10 0° 72 96 144 10 0° 10 0° 72 96 144 10 0° 120 120 72 8.33 53.13° 599.76 53.13° 96 96 90° 8.33 53.13° 799.68 53.13° 799.68 36.87° 360 480 53.13° 10 144 53.13° 10 144 0° 0° 8.33 8.33 53.13° 53.13° 36.87° 640 480 280 8 ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern