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Mesh Analysis

Mesh Analysis

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Mesh Analysis

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  1. Mesh Analysis Introducing Supermeshes!!!

  2. Mesh Analysis • A mesh is a loop with no other loops within it; an independent loop. • Mesh analysis provides another general procedure for analyzing circuits, using mesh currents as the circuit variables. • This is more convenient and reduces the number of equations that must be solved simultaneously.

  3. Mesh Analysis (cont.) • Mesh analysis applies KVL to find unknown currents and is not as general because it only applies to a planar circuit. • A planar circuit is one that can be drawn in a plane and no branches that cross.

  4. Steps to Determine Mesh Currents • 1. Assign mesh currents i1, i2, … , in to the n meshes. • 2. Apply KVL to each of the n meshes. Use Ohm’s Law to express the voltages in terms of the mesh currents. • 3. Solve the resulting n simultaneous equations to get the mesh currents.

  5. Applying Mesh Analysis to circuits with current sources is easier because the current sources reduce the number of equations. Case 1: When a current source exists only in one mesh, set that mesh current to the source current, being aware of the direction of that current. Case 2: When a current source exists between two meshes, we create a supermesh. Mesh Analysis with Current Sources

  6. Supermesh • With supermeshes, exclude the current source and any elements connected in series with it. • We treat supermeshes differently because mesh analysis applies KVL and we do not know the voltage across a current source in advance.

  7. Supermesh (cont.) • Note the properties of a supermesh: • 1. The current source provides the constraint equation necessary to solve for mesh currents. • 2. A supermesh has no current of its own. • 3. A supermesh requires KVL and KCL.

  8. Examples Step 1: Assign mesh currents i1, i2, … in to the n meshes. Step 2: Apply KVL to each of the meshes. 1: -15 + 5i1 + 10(i1 – i2) + 10 = 0 2: -10 + 10(i2 – i1) + 10i2 = 0 Step 3: Solve the equations to get the mesh currents 1: 15i1 – 10i2 = 5 2: -10i1 + 20i2 = 10 i1 = 1 A i2 = 1 A I1 = 1 A I2 = 1 A I3 = 0 A NOTE: Remember, currents flowing in opposite directions are subtracted.

  9. Examples Step 1: Assign mesh currents i1, i2, … in to the n meshes. Step 2: Apply KVL to each of the meshes. 1: -12 + 2i1 + 12(i1 – i2) + 4i1 = 0 2: 12(i2 – i1) + 9i2 + 8 + 3i2 = 0 Step 3: Solve the equations to get the mesh currents 1: 18i1 – 12i2 = 12 2: -12i1 + 24i2 = -8 i1 = 667 mA i2 = 0 A NOTE: There are no ground nodes. Therefore, current goes into the positive terminal.

  10. Examples Note: io = i1 – i2 Also, there is a big difference between branch currents and loop currents. They are equal only when the loop current is the only current involved. For example in loop 2, the currents i2 are the same through the 24 ohm resistor, but they are different at the 4 and 10 ohm resistors in the same loop because those resistors are being shared with other loops.

  11. Examples Note: Keep the currents separate when proceeding through the supermesh. The currents do not overlap into the other loops. KVL: -20 + 6i1 + 10i2 + 4i2 = 0 KCL (currents at the bottom node): i2 = i1 + 6 Solve equations as usual: 6i1 + 14i2 = 20 -i1 + i2 = 6 i1 = -3.2 A i2 = 2.8 A