Mesh and Node Equations Simple Resistive Circuits

Mesh and Node Equations Simple Resistive Circuits - Mesh...

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Mesh and Node Equations: Simple Resistive Circuits Introduction The circuits in this set of problems are small circuits that consist of independent sources, resistors and a meter. In particular, these circuits do not contain dependent sources. These circuits can be analyzed using mesh equation or using node equations. Node equations are discussed in Sections 4.3 and 4.4 of Introduction to Electric Circuits by R.C. Dorf and J.A Svoboda. Mesh equations are discussed in Section 4.6. Worked Examples Example 1: Consider the circuit shown in Figure 1. Find the value of the current source current, I a . Figure 1 The circuit considered in Example 1. Solution: Figure 2 shows the circuit from Figure 1 after replacing the ammeter by an equivalent short circuit and labeling the current measured by the ammeter. This circuit can be analyzed using mesh equations or using node equations. We will do both. Figure 2 The circuit from Figure 1 after replacing the ammeter by a short circuit. 1

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Figure 3 The circuit from Figure 2 after labeling the mesh currents. First, consider analyzing the circuit in Figure 2 using mesh equations. Figure 3 shows the circuit after labeling the mesh currents, i 1 and i 2 . The mesh current i 2 is equal to the current in the short circuit that replaced the ammeter so 2 3 A i = The current in the current source can be expressed in terms of the mesh currents as 21 3 a 1 I ii i = −=− (1) Apply KVL to the supermesh to get ( )( ) 11 22 3 4 0 1 +− = = A Substituting this value into Equation 1 gives 3 ( 1) 4 A a I = −− = . Next, consider analyzing the circuit in Figure 2 using node equations. Figure 4 shows the circuit after selecting a reference node and numbering the other nodes. Let v 1 and v 2 denote the node voltages at node 1 and node 2, respectively. The voltage of the voltage source can be expressed in terms of the node voltages as 40 4 vv V = −⇒ = The current in the short circuit that replaced the ammeter is equal to the current in one of the 2 resistors. This current can be expressed in terms of the node voltages as 2 2 0 36 2 v v V =⇒ = Apply KCL at node 2 to get 64 33 a I 4 A =+ = + = 2
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This note was uploaded on 09/30/2010 for the course ECE 2025 taught by Professor Juang during the Spring '08 term at Georgia Institute of Technology.

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Mesh and Node Equations Simple Resistive Circuits - Mesh...

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