DeCarlo Ch12 Solutions

# Linear Circuit Analysis: The Time Domain and Phasor Approach

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9/26/01 P12-1 @ DeCarlo & P. M. Lin CHAPTER 12. PROBLEM SOLUTIONS S OLUTION P12.1. By conservation of energy, the instantaneous power consumed by each load when summed together is equivalent to the total power consumed by the three phase load. Thus writing out p tot ( t ) = p AB ( t ) + p BC ( t ) + p CA ( t ) = v AB ( t ) i AB ( t ) + v BC ( t ) i BC ( t ) + v CA ( t ) i CA ( t ) = V L 2 cos( t + v ) V L 2 Z cos( t + i ) + V L 2 cos( t + v - 120 o ) V L 2 Z cos( t + i - 120 o ) + V L 2 cos( t + v + 120 o ) V L 2 Z cos( t + i + 120 o ) = V L 2 Z cos( v - i ) + cos(2 t + v + i ) ( 29 + V L 2 Z cos( v - i ) + cos(2 t + v + i + 120 o ) ( 29 + V L 2 Z cos( v - i ) + cos(2 t + v + i - 120 o ) ( 29 = 3 V L 2 Z cos( v - i ) = 3 V L 2 Z × pf S OLUTION P12.2. To justify the point of this problem we equate the following two equations: (i) For the 3 phase system: P loss ' = 3 × I L 2 R ' = 3 × 3 9 × P L 2 V L 2 R ' = P L 2 V L 2 R ' (ii) For the single phase system: P loss = 2 × I L 2 R ( 29 = P L 2 V L 2 2 R It follows that R' = 2R. Since both systems have the same distance of transmission and the resistance of a wire is inversely proportional to the cross sectional area, the condition R' = 2R implies that the cross section A' of each wire in the three-phase system need only be half of the area A of the wire in the single phase system. But there are two wires in the single phase system and three wires in the three-phase system. Therefore the ratio of the materials used is: material in 3 - phase system material in 1 - phase system = 3 A ' 2 A = 3 2 × 1 2 = 75% S OLUTION P12.3. For row 1 of table 12.1, the impedance in (a) seen between each pair of terminal is Z ik = Z || 2 Z = 2 Z 3 . In (b) the impedance seen between any two terminals is Z jk = Z 3 + Z 3 = 2 Z 3 .

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9/26/01 P12-2 @ DeCarlo & P. M. Lin In row 2, the impedance between any two terminals is Z jk = Z Y + Z Y = 2 Z Y for (c), and Z ik = 3 Z Y || 6 Z Y = 2 Z Y for (d). S OLUTION P12.4 . Consider the -Y relationship of the figures below (row 3 of table 12.1): Let us consider the terminal pair (1,2): (i) For the Y-connected case, Z th = 2 × Z 3 and V oc = V p 3 - 30 o ( 29 - V p 3 - 150 o ( 29 = V p 3 1 ∠- 30 o - 1 ∠ - 150 o ( 29 = V p (ii) For the D-connected case, Z th = Z / / 2 Z ( 29 = 2( Z ) 2 3 Z = 2 3 Z Now note that there is no load connected to the -configuration. Applying KVL to the indicated loop implies that: 0 = 3 Z I loop + V p + V p - 120 o + V p + 120 o = 3 Z I loop Hence I loop = 0 . Finally, V oc = V 12 = Z I loop + V p = V p . Therefore, looking into terminals 1-2, both the -configuration and the Y-configuration have the same Thevenin equivalent. For terminal pairs (1-3) and (2-3), the proof is virtually the same. Hence this establishes the equivalence in row 3 of table 12.1. To establish the equivalence in row 4 of table 12.1, we do all the same computations with the slightly different labeling to obtain the same result, i.e., the circuits are equivalent. S OLUTION 12.5 . For (a) first note the following relationship V N = V 1 Z 1 + V 2 Z 2 Z 1 || Z 2 || Z 3 ( 29 , which is obtained by KCL at the center node, N, with node 3 as the reference node.. Write out KCL at terminal 1, I 1 = V 1 Z 1 - V 1 Z 1 Z 1 + V 2 Z 2 Z 1 Z 1 || Z 2 || Z 3 ( 29 = V 1 Z 3 + Z 2 Z 2 Z 3 + Z 1 Z 3 + Z 1 Z 2
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