REVIEW-KCL KVL NV ML & Transforms

REVIEW-KCL KVL NV ML & Transforms - ENGR 210 REVIEW...

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THINGS YOU SHOULD KNOW Assign voltages to each node and apply KCL at nodes where voltage is unknown. The problems are voltage sources. There are three ways to deal with them: 1. Ground one end of a voltage source. Doesn’t work with multiple sources unless they are connected to each other. 2. Transform a voltage source (with a series resistance) to a current source. 3. Define a supernode which encloses the voltage source and write a second equation which is simply the voltage difference between the nodes. ABOUT MESH CURRENT (Lectures 10 Write mesh currents and apply KVL around each loop. The problems are current sources. There are three ways to deal with them: 1. A current in an external loop is no problem. It becomes the mesh current. 2. Transform a current source (with a parallel resistance) to a voltage source. 3. Define a supermesh of two meshes which enclose the current source and write a second equation which is simply the current difference between the two meshes. PAGE NODE-VOLTAGE EXAMPLES 2-13 NODE-VOLTAGE SPECIAL CASES 7 SUPERNODE 12-13 MESH-CURRENT EXAMPLES 15-22 MESH-CURRENT SPECIAL CASES 17 SUPERMESH 21-22 SOURCE TRANSFORMATIONS 23-27 ENGR 210 REVIEW Page 1
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ENGR 210 HW3 Solution Page 3 Problem 2 HINT: Check out Example 4.4 The circuit shown below is a DC model of a residential power distribution circuit. 110V + 24 ± 3 ± 2 ± i 5 - i 6 16 ± i 2 8 ± i 4 110V + - 2 ± i 1 i 3 (a) Use the node-voltage method to find the branch currents i 1 -i 6 . (b) Test your solution for the branch currents by showing that the total power dissipated (what is it?) equals the total power developed (what is it?)/ SOLUTIONS There are 4 essential nodes and so we will need 4-1=3 node-voltage equations. To reduce the number of variables let’s choose the node between the two voltage sources as the reference node (v=0) and define three other nodes as 1-at the top of the 8 ± resistor; 2- between the 8 ± and 24 ± resistors; and 3 at the bottom of the 24 ± resistor and label the node voltages v1,v2,v3 respectively. We can write the KCL equations at each node as node 1: + 110 ± v 1 2 ± v 1 ± v 2 8 ± v 1 ± v 3 16 = 0 ² 11 v 1 ± 2 v 2 ± v 3 = 880 node 2: + v 1 ± v 2 8 ± v 2 ± 0 3 ± v 2 ± v 3 24 = 0 ²± 3 v 1 + 12 v 2 ± v 3 = 0 node 3: - v 3 ±± 110 () 2 + v 2 ± v 3 24 + v 1 ± v 3 16 = 0 ²± 3 v 1 ± 2 v 2 + 29 v 3 = ± 2640 ENGR 210 REVIEW Page 2
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ENGR 210 HW3 Solution Page 4 We can solve these three equations in three unknowns to get
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REVIEW-KCL KVL NV ML & Transforms - ENGR 210 REVIEW...

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