DeCarlo Ch12 Solutions

Linear Circuit Analysis: The Time Domain and Phasor Approach

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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 .
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern