Solutions-_-Tutorial-11_AC2.pdf

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

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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 ...
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