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METHODS OF ANALYSIS

METHODS OF ANALYSIS. Nodal analysis. Mesh Analysis. 6 . 10 . 2 . 20V. 4 . + . 6 A. i 1. i 2. Mesh Analysis 2. Supermesh. Solve the mesh currents. Step 1. Assign mesh currents to the meshes. Step 2.

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METHODS OF ANALYSIS

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  1. METHODS OF ANALYSIS • Nodal analysis • Mesh Analysis

  2. 6 10 2 20V 4 +  6A i1 i2 Mesh Analysis 2 Supermesh Solve the mesh currents Step 1 Assign mesh currents to the meshes Step 2 For every mesh, apply KVL – Using Ohm’s law, write down the equations in terms of mesh currents We cannot write equation for the meshes since we cannot write the voltage drop across the 6A source in terms of mesh currents!

  3. 6 10 2 20V 4 +  6A i1 i2 Mesh Analysis 2 Supermesh Solve the mesh currents We form a supermesh by EXCLUDING the branch containing the current source and elements connected in series with it We cannot write equation for the meshes since we cannot write the voltage drop across the 6A source in terms of mesh currents!

  4. 6 10 2 20V 4 +  6A i1 i2 Mesh Analysis 2 Supermesh SUPERMESH Solve the mesh currents We cannot write equation for the meshes since we cannot write the voltage drop across the 6A source in terms of mesh currents! We form a supermesh by EXCLUDING the branch containing the current source and elements connected in series with it KVL for the supermesh : -20 + 6i1 + 10i2 + 4i2 = 0

  5. 6 10 2 20V 4 +  6A i1 i2 Mesh Analysis 2 Supermesh Solve the mesh currents Since we have combined meshes 1 and 2, we lost one equation. We need one more equation to solve the mesh current We obtain the extra equation by using KCL on the branch current that we have removed. -20 + 6i1 + 10i2 + 4i2 = 0 i2 - i1 = 6

  6. 6 10 2 20V 4 +  6A i1 i2 Mesh Analysis 2 Supermesh Solve the mesh currents Step 3 Solve mesh currents in equations obtained in step 2, simultaneously -20 + 6i1 + 10i2 + 4i2 = 0 i2 - i1 = 6

  7. 6 10 2 20V 4 +  6A i1 i2 -20 + 6i1 + 10i2 + 4i2 = 0 i2 - i1 = 6 Mesh Analysis 2 Supermesh Solve the mesh currents Step 3 Solve mesh currents in equations obtained in step 2, simultaneously i1 = - 3.2 A, i2 = 2.8 A

  8. Mesh Analysis 2 Supermesh To summarize …. We form a supermesh when there is a current source in a branch sharing 2 mesh currents Supermesh has no current of its own. To solve the mesh currents, we need extra equation which can be obtained by applying KCL in the branch containing the current source.

  9. 2 2 6V + - 4 1 8 3A i1 i3 i2 Mesh Analysis 2 Supermesh Example 2 Step 1 Assign mesh currents to the meshes

  10. 2 2 6V + - 4 1 8 3A i1 i3 i2 Mesh Analysis 2 Supermesh Example 2 Step 2 For every mesh, apply KVL – Using Ohm’s law, write down the equations in terms of mesh currents Step 3 Solve mesh currents in equations obtained in step 2, simultaneously

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