Chapter 3_Circuit Analysis Techniques

Chapter 3_Circuit Analysis Techniques - The Analysis and...

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

View Full Document Right Arrow Icon
The Analysis and Design of Linear Circuits Seventh Edition 3 Circuit Analysis Techniques 3.1 Exercise Solutions Exercise 3–1. The reference node and node voltages in the bridge circuit of Figure 3–3 are v A = 5 V, v B = 10 V, and v C = 3 V. Find the element voltages. Apply the node-voltage definitions to find the voltages as follows: v 1 = v B 0 = 10 0 = 10 V v 2 = 0 v C = 0 ( 3) = 3 V v 3 = v B v C = 10 ( 3) = 13 V v 4 = v A v C = 5 ( 3) = 8 V v 5 = v A v B = 5 10 = 5 V Exercise 3–2. First find the node voltages in the following circuit and then find v X and v Y . Figure 3–4 defines the reference node as node D with v D = 0 V. Apply the node-voltage definitions to find the node voltages as follows: v A v D = 5 V v A = 5 V v C v D = 6 V v C = 6 V v C v B = 10 V v B = 4 V Apply the node-voltage definitions again to find the voltages v X and v Y . v X = v A v B = 5 ( 4) = 9 V v Y = v A v C = 5 6 = 1 V Exercise 3–3. For the circuit in Figure 3–6 replace the current source i S2 with a resistor R 5 . (a). Using the same node designations and reference node, formulate node-voltage equations for the modified circuit. Place the result in matrix for Ax = b . The reference node, node voltages, and element currents are the same as those in Example 3–1. In addition, the KCL equations are the same: Node A : i 0 i 1 i 2 = 0 Node B : i 1 i 3 + i 5 = 0 Node C : i 2 i 4 i 5 = 0 The element equations remain the same except for the current i 5 , which now involves a resistor: i 0 = i S1 i 3 = 1 R 3 v B i 1 = 1 R 1 ( v A v B ) i 4 = 1 R 4 v C i 2 = 1 R 2 ( v A v C ) i 5 = 1 R 5 ( v C v B ) Solution Manual Chapter 3 Page 3-1
Image of page 1

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

View Full Document Right Arrow Icon
The Analysis and Design of Linear Circuits Seventh Edition Substitute the element equations into the KCL constraints and arrange the result in standard form to get the following: Node A : parenleftbigg 1 R 1 + 1 R 2 parenrightbigg v A 1 R 1 v B 1 R 2 v C = i S1 Node B : 1 R 1 v A + parenleftbigg 1 R 1 + 1 R 3 + 1 R 5 parenrightbigg v B 1 R 5 v C = 0 Node C : 1 R 2 v A 1 R 5 v B + parenleftbigg 1 R 2 + 1 R 4 + 1 R 5 parenrightbigg v C = 0 Write the results in matrix form: parenleftbigg 1 R 1 + 1 R 2 parenrightbigg 1 R 1 1 R 2 1 R 1 parenleftbigg 1 R 1 + 1 R 3 + 1 R 5 parenrightbigg 1 R 5 1 R 2 1 R 5 parenleftbigg 1 R 2 + 1 R 4 + 1 R 5 parenrightbigg v A v B v C = i S1 0 0 (b). Is the resulting A matrix symmetrical? The matrix is symmetrical because entry ( i, j ) is the same as entry ( j, i ) for all off-diagonal terms. Exercise3–4. Formulate node-voltage equations for the circuit of Figure 3–7 and place the results in matrix form Ax = b . Is the resulting matrix A symmetrical? The figure is labeled with a reference node, node voltages, and element currents. The KCL constraints at the two nonreference nodes are: Node A : i 1 i 2 i 3 = 0 Node B : i 3 i 4 + i 5 = 0 The element equations are as follows: i 1 = 20 mA i 4 = 1 2000 v B i 2 = 1 1000 v A i 5 = 50 mA i 3 = 1 1500 ( v A v B )
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