Analysis_Notes

# Analysis_Notes - Defining Voltages The mesh current’s direction is positive while adjacent currents are negative entering a positive terminal

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Nodal Analysis Goal: Determine node voltages Steps: 1. Name the nodes (V 1 , V 2 , … V n-1 ) and place a ground reference 2. Apply KCL to each node to obtain n-1 equations 3. Solve the simultaneous equations for the node voltages 4. If currents are needed, use Ohm’s law to calculate them Defining Currents: I = ( V origination – V destination ) / R Workaround: If a current cannot be defined (e.g., in the case of a voltage source) then create a supernode that surrounds the undefined region. Apply KCL to the supernode, and make up for the lost equation using KVL through a loop that includes the supernode. When to use: 1. When there are fewer nodes than meshes 2. Lots of parallel connections 3. Large supernodes 4. Current source heavy Mesh Analysis Goal: Determine mesh currents Steps: 1. Assign mesh currents to each mesh (loop with no sub-loops) 2. Apply KVL to all meshes using Ohm’s Law 3. Solve the simultaneous equations for the currents 4. If voltages are need, use other relations to calculate
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Unformatted text preview: Defining Voltages: The mesh current’s direction is positive, while adjacent currents are negative; entering a positive terminal first is a positive contribution to loop voltage. Workaround: If voltage cannot be defined (e.g., current sources are present) then create a supermesh that surrounds the undefined region. Apply KVL to the supermesh, and make up for lost equations with KCL within the mesh. When to use: 1. When there are more nodes than meshes 2. Lots of series connections 3. Large supermeshes 4. Voltage source heavy Superposition Key Concept: The voltage across (or current through) an element in a linear circuit equals the sum of the voltages (or currents) due to each independent source acting alone Source Transformation Key concept: A voltage source in series with a resistance can be transformed into a current source in parallel with the same resistance, and vice versa, with a current equal to the voltage source divided by the resistance....
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## This note was uploaded on 11/02/2010 for the course ECE 3710 taught by Professor Haris during the Spring '07 term at Georgia Institute of Technology.

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