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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 fnd V BC . To do this, write a KVL equation to fnd V AB , and use the known phase angle relationship between V AB and V BC to fnd 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 fnd 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 fnd V AB . To do this, write a KVL equation to fnd V BC , and use the known phase angle relationship between V AB and V BC to fnd 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 fnd 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 fnd 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 fnd 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 = 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 fnd V ab , use the a-phase circuit to fnd V AN , and use the relationship between phase voltages and line voltages For a y-connection (see ±ig. 11.9[b]). ±rom the a-phase circuit, use KVL: V an = V aA + V AN =(0 . 1+ j 0 . 8) I aA + 2400/0 . j 0 . 8)(96 j 72) + 2400/0 = 2467 . 2+ j 69 . 6 2468 . 18/1 . 62 V ±rom ±ig. 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 0 n = V a 0 a + V an =(0 . 2+ j 0 . 16) I aA + V an . 02 + j 0 . 16)(96 j 72) + (2467 . j 69 . 9) = 2480 . 64 + j 83 . 52 = 2482 .
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ch11_ssm - 11 Balanced Three-Phase Circuits Assessment...

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