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# ch11_ssm - 11 Balanced Three-Phase Circuits Assessment...

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11 Balanced Three-Phase Circuits Assessment Problems AP 11.1 Make a sketch: We know V AN and wish to find V BC . To do this, write a KVL equation to find V AB , and use the known phase angle relationship between V AB and V BC to find V BC . V AB = V AN + V NB = V AN V BN Since V AN , V BN , and V CN form a balanced set, and V AN = 240/ 30 V, and the phase sequence is positive, V BN = | V AN | // V AN 120 = 240/ 30 120 = 240/ 150 V Then, V AB = V AN V BN = (240/ 30 ) (240/ 150 ) = 415 . 46/0 V Since V AB , V BC , and V CA form a balanced set with a positive phase sequence, we can find V BC from V AB : V BC = | V AB | /(/ V AB 120 ) = 415 . 69/0 120 = 415 . 69/ 120 V Thus, V BC = 415 . 69/ 120 V 11–1

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11–2 CHAPTER 11. Balanced Three-Phase Circuits AP 11.2 Make a sketch: We know V CN and wish to find V AB . To do this, write a KVL equation to find V BC , and use the known phase angle relationship between V AB and V BC to find V AB . V BC = V BN + V NC = V BN V CN Since V AN , V BN , and V CN form a balanced set, and V CN = 450/ 25 V, and the phase sequence is negative, V BN = | V CN | // V CN 120 = 450/ 23 120 = 450/ 145 V Then, V BC = V BN V CN = (450/ 145 ) (450/ 25 ) = 779 . 42/ 175 V Since V AB , V BC , and V CA form a balanced set with a negative phase sequence, we can find V AB from V BC : V AB = | V BC | // V BC 120 = 779 . 42/ 295 V But we normally want phase angle values between +180 and 180 . We add 360 to the phase angle computed above. Thus, V AB = 779 . 42/65 V AP 11.3 Sketch the a-phase circuit:
Problems 11–3 [a] We can find the line current using Ohm’s law, since the a-phase line current is the current in the a-phase load. Then we can use the fact that I aA , I bB , and I cC form a balanced set to find the remaining line currents. Note that since we were not given any phase angles in the problem statement, we can assume that the phase voltage given, V AN , has a phase angle of 0 . 2400/0 = I aA (16 + j 12) so I aA = 2400/0 16 + j 12 = 96 j 72 = 120/ 36 . 87 A With an acb phase sequence, / I bB = / I aA + 120 and / I cC = / I aA 120 so I aA = 120/ 36 . 87 A I bB = 120/83 . 13 A I cC = 120/ 156 . 87 A [b] The line voltages at the source are V ab V bc , and V ca . They form a balanced set. To find V ab , use the a-phase circuit to find V AN , and use the relationship between phase voltages and line voltages for a y-connection (see Fig. 11.9[b]). From the a-phase circuit, use KVL: V an = V aA + V AN = (0 . 1 + j 0 . 8) I aA + 2400/0 = (0 . 1 + j 0 . 8)(96 j 72) + 2400/0 = 2467 . 2 + j 69 . 6 2468 . 18/1 . 62 V From Fig. 11.9(b), V ab = V an ( 3/ 30 ) = 4275 . 02/ 28 . 38 V With an acb phase sequence, / V bc = / V ab + 120 and / V ca = / V ab 120 so V ab = 4275 . 02/ 28 . 38 V V bc = 4275 . 02/91 . 62 V V ca = 4275 . 02/ 148 . 38 V

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11–4 CHAPTER 11. Balanced Three-Phase Circuits [c] Using KVL on the a-phase circuit V a n = V a a + V an = (0 . 2 + j 0 . 16) I aA + V an = (0 . 02 + j 0 . 16)(96 j 72) + (2467 . 2 + j 69 . 9) = 2480 . 64 + j 83 . 52 = 2482 . 05/1 . 93 V With an acb phase sequence, / V b n = / V a n + 120 and / V c n = / V a n 120 so V a n = 2482 . 05/1 . 93 V V b n = 2482 . 05/121 . 93 V V c n = 2482 . 05/ 118 . 07 V AP 11.4 I cC = ( 3/ 30 ) I CA = ( 3/ 30 ) · 8/ 15 = 13 . 86/ 45 A AP 11.5 I aA = 12/(65
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ch11_ssm - 11 Balanced Three-Phase Circuits Assessment...

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