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Chapter 5—Electrical Network 5–10 ✲ I ✲ I 3 ☞ > ❄ I 1 ❄ I 2 < ☞✌ > V = 100̸ 0 o < 4Ω j8 Ω > ☞✌ < ✌ > < > < 6Ω > < −j11 Ω Figure 5: Parallel RLC circuit. The effort required may be smaller or greater for one approach as compared to the other, depending on the structure of the circuit. The student should consider the relative difficulty of both techniques before solving the problem, to minimize the total effort. Example Find the apparent power and power factor of the circuit in Fig. 5. One solution technique is to first find the input impedance. Z = 1 1/4+1/j8+1/(6 − j11) = 1 0.25 − j0.125 + 1/(12.53/ − 61.39 o ) = 1 0.25 − j0.125 + 0.080/61.39 o = 1 0.25 − j0.125 + 0.038 + j0.070 = 1 0.288 − j0.055 = 1 0.293/ − 10.79 o =3.41/10.79o Ω The input current is then I = V Z = 100/0o 3.41/10.79 o =29.33/ − 10.79o A The apparent power is |S| = |V ||I| = 100(29.33) = 2933 VA The power factor is pf = cos θ = cos 10.79 o =0.982 lag Another solution technique is to find the individual component powers. We have to find the current I 3 to find the real and reactive powers supplied to that branch. Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001
Chapter 5—Electrical Network 5–11 I 3 = The capacitive reactive power is then V 6 − j11 = 100/0 o 12.53/ − 61.39 o =7.98/61.39o The inductive reactive power is Q C = −|I 3 | 2 X C = −(7.98) 2 (11) = −700 var Q L = V 2 = (100)2 = 1250 var X L 8 The real power supplied to the circuit is P = V 2 4 + |I 3| 2 (6) = (100)2 +(7.98) 2 (6) = 2500 + 382 = 2882 W 4 The complex power is then S = P + jQ = 2882 + j1250 − j700 = 2882 + j550 = 2934/10.80 o var so the apparent power is 2934 var and the power factor is pf = cos /10.80 o =0.982 lag The total effort by a person proficient in complex arithmetic may be about the same for either approach. A beginner is more likely to get the correct result from the second approach, however, because it reduces the required complex arithmetic by not requiring the determination of the input impedance. We now turn our attention to three-phase circuits. We are normally interested in a balanced set of voltages connected in wye as shown in Fig. 6. If we select E a , the voltage of point a with respect to the neutral point n, as the reference, then E a = |E a |/0 o V E b = |E a |/ − 120 o V (20) E c = |E a |/ − 240 o V This set of voltages is said to form an abc sequence, sinceE b lags E a by 120 o ,andE c lags E b by 120 o . WeusethesymbolE rather than V to indicate that we have a source voltage. Wind Energy Systems by Dr. Gary L. Johnson November 21, 2001
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WIND ENERGY SYSTEMS Electronic Edit
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