Mesh and Node Equations: More Circuits Containing Dependent Sources

Transcription

1 Mesh nd Node Equtions: More Circuits Contining Dependent Sources Introduction The circuits in this set of problems ech contin single dependent source. These circuits cn be nlyzed using mesh eqution or using node equtions. When doing so, it is useful to express the controlling current or voltge of the dependent source s function of the mesh currents or node voltges. Node equtions re discussed in Sections 4.3, 4.4 nd 4.5 of Introduction to Electric Circuits by R.C. Dorf nd J.A Svobod. Section 4.5 considers circuits tht contin dependent sources. Mesh equtions re discussed in Sections 4.6 nd 4.7. Section 4.7 considers circuits tht contin dependent sources. Worked Exmples Exmple 1: Consider the circuit shown in Figure 1. Find the vlue of the gin, A, of the VCVS. Figure 1 The circuit considered in Exmple 1. Solution: Figure 2 shows the circuit from Figure 1 fter replcing the voltmeter by n equivlent open circuit nd lbeling the voltge mesured by the voltmeter. We will nlyze this circuit by writing nd solving node equtions. Figure 3 shows the circuit fter selecting reference node nd numbering the other nodes. Let v 1, v 2 nd v 3 denote the node voltges t nodes 1, 2 nd 3, respectively.

2 Figure 2 The circuit from Figure 1 fter replcing the voltmeter by n open circuit. Figure 3 The circuit from Figure 2 fter lbeling the nodes. The voltge cross the dependent source is represented in three wys. It is A*v with the + of reference direction t the top, 5.14 V with the + t the bottom nd v 3 0 = v 3 with the + t the top. Consequently v3 = Av = 5.14 V The voltge of the 12 V voltge source cn be expressed in terms of the node voltges s 12 = v 0 v = 12 V 1 1 The controlling voltge of the dependent source, v, is the voltge cross n open circuit. This voltge cn be expressed in terms of the node voltges t the nodes of the open circuit. Hence Apply KCL to node 2 to get v = v 0 = v 2 2 Finlly, v1 v2 v2 v3 12 v2 v2 ( 5.14) = = v2 = 1.71 V Av Av 5.14 A = = = = 3 V/V v v

3 Exmple 2: Consider the circuit shown in Figure 4. Find the vlue of the gin, A, of the VCCS. Figure 4 The circuit considered in Exmple 2. Solution: Figure 5 shows the circuit from Figure 4 fter replcing the mmeter by n equivlent short circuit nd lbeling the current mesured by the mmeter. We will nlyze this circuit by writing nd solving node equtions. Figure 6 shows the circuit fter selecting reference node nd numbering the other nodes. Let v 1, v 2 nd v 3 denote the node voltges t nodes 1, 2 nd 3, respectively. Figure 5 The circuit from Figure 4 fter replcing the mmeter by short circuit. Figure 6 The circuit from Figure 5 fter lbeling the nodes.

4 The voltge of the 36 V voltge source cn be expressed in terms of the node voltges s 36 = v 0 v = 36 V 1 1 The controlling voltge of the dependent source, v, is the voltge cross n open circuit. This voltge cn be expressed in terms of the node voltges t the nodes of the open circuit. Hence v = v 0 = v 2 2 Node 3 is connected to the reference node by the short circuit tht replced the mmeter. The voltges cross short circuit is 0 V. Consequently Apply KCL t node 2 to get v 0= 0 V v = 0 V 3 3 v1 v2 v2 v3 36 v2 v2 0 = = v2 = 10.3 V Apply KCL t node 3 to get v2 v3 = Av + ( 39.9) Av = 41.2 V 8 Finlly, Av Av 41.2 A = 4 A/V v = v = 10.3 = 2

5 Exmple 3: Consider the circuit shown in Figure 7. Find the vlue of the gin, A, of the CCCS. Figure 7 The circuit considered in Exmple 3. Solution: Figure 8 shows the circuit from Figure 7 fter replcing the voltmeter by n equivlent open circuit nd lbeling the voltge mesured by the voltmeter. We will nlyze this circuit using mesh equtions. Figure 9 shows the circuit fter numbering the meshes. Let i 1 nd i 2 denote the mesh currents in meshes 1 nd 2, respectively. Figure 8 The circuit from Figure 7 fter replcing the voltmeter by n open circuit. Figure 9 The circuit from Figure 8 fter lbeling the meshes.

6 The controlling current of the dependent source, i, is the current in short circuit. This short circuit is common to meshes 1 nd 2. The short circuit current cn be expressed in terms of the mesh currents s i = i i The dependent source is in only one mesh, mesh 2. The reference direction of the dependent source current does not gree with the reference direction of i 2. Consequently 1 2 Apply KVL to mesh 1 to get Ai = i i1 24 = 0 i1 = A 4 Apply KVL to mesh 2 to get i2 (30) = 0 i2 = A 16 Finlly, A Ai i i i i 2 = = = = A/A

7 Exmple 4: Consider the circuit shown in Figure 10. Find the vlue of the gin, A, of the CCVS. Figure 10 The circuit considered in Exmple 4. Solution: Figure 11 shows the circuit from Figure 10 fter replcing the voltmeter by n equivlent open circuit nd lbeling the voltge mesured by the voltmeter. We will nlyze this circuit using mesh equtions. Figure 11 shows the circuit fter numbering the meshes. Let i 1 nd i 2 denote the mesh currents in meshes 1 nd 2, respectively. Figure 11 The circuit from Figure 10 fter replcing the voltmeter by n open circuit. Figure 12 The circuit from Figure 11 fter lbeling the meshes.

8 The voltge cross the dependent source is represented in two wys. It is A*v with the + of reference direction t the bottom nd -7.2 V with the + t the top. Consequently ( ) Av = 7.2 = 7.2 V The controlling current of the dependent source, i, is the current in short circuit. This short circuit is common to meshes 1 nd 2. The short circuit current cn be expressed in terms of the mesh currents s i = i i Apply KVL to mesh 1 to get Apply KVL to mesh 2 to get i 36 = 0 i = 3.6 A i + ( 7.2) = 0 i = 1.8 A 2 2 Finlly, Ai Ai 7.2 A = 4 V/A i = i i = = 1 2