Chapter 3_Circuit Analysis Techniques

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

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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

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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 )
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• Fall '15
• Resistor, Thévenin's theorem, Design of Linear Circuits, Solution Manual Chapter

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