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ECE 221 Electric Circuit Analysis I Chapter 9 Mesh Current Method

The Mesh Current Method is an alternative to the substitution method for computing electric units in circuits. It reduces equations to a smaller number of unknown currents and is applicable only to planar circuits. This text provides examples of solving circuits using the Mesh Current Method.

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ECE 221 Electric Circuit Analysis I Chapter 9 Mesh Current Method

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  1. ECE 221Electric Circuit Analysis IChapter 9 Mesh Current Method Herbert G. Mayer, PSU Status 10/15/2015

  2. Syllabus Definition Circuit1 for Mesh Current First Solve Via Substitution Example 4.4 Example 4.5 Conclusion

  3. Definition The Mesh Current Method is an alternative to the substitution method for computing electric units in circuits It expresses voltages as a function of fictitious currents along a mesh Why have yet another method? The Mesh Current Method is simpler due to a smaller number of equations needed But the method is applicable only to planar circuits Mesh Currents are not identical to branch currents They are fictitious, thus not always measurable with an instrument (Amp meter)

  4. Definition A Mesh Current is associated with a complete path through a mesh, obeying Kirchhoff’s law Valid only in a mesh, i.e. any loop with no interior loops enclosed Valid only in the perimeter of a mesh Has a defined direction, indicated via an arrow, to be used consistently across mesh If a basic element is affected by multiple mesh currents, they all have to be accounted for in computing currents Example:

  5. Circuit 1 for Mesh Current

  6. First Solve Via Substitution Number of essential nodes is 2 Number of essential branches is 3 To compute the currents i1, i2, and i3, using substitution, there is just one independent KCL equation So we need 2 more independent voltage equations for a solution by substitution When complete, compare the result with the Mesh Current Method

  7. First Solve Via Substitution (1) KCL: i2 + i3 = i1 (2) KVL: R1*i1 + R3*i3 - v1 = 0 (3) KVL: R2*i2 - R3*i3 + v2 = 0 Solve (1) for i3 and substitute into (2) and (3): (2)’ v1 = i1*(R1 + R3) - i2*R3 (3)’ v2 = -i2*(R2 + R3) + i1*R3 Solve Via Substitution

  8. Then Solve Via Mesh Current (1) R1*ia + R3*(ia – ib) – v1 = 0 (2) R2*ib + v2 + R3*(ib – ia) = 0 (1)” v1 = ia*(R1 + R3) – ib*R3 (2)” v2 = -ib*(R2 + R3) + ia*R3

  9. Substitution vs. Mesh Current i1 == ia // == stands for “identical to” i2 == ib i3 = ia - ib

  10. Conclusion Mesh Current Method is simpler than Substitution Equations are reduced to a smaller number of unknown currents Similar to Node Voltage Method, which reduces equations to a smaller number of unknown voltages

  11. Example 4.4

  12. Example 4.4 via Mesh Current

  13. Example 4.4 via Mesh Current Find the power associated with each voltage source: First find the two interesting currents, the ones through the constant voltage sources Their voltages are known, hence their two currents must be found to compute power There are b = 7 branches with unknown currents And n = 5 nodes Hence we need b-(n-1) = 7-(5-1) = 3 equations

  14. Example 4.4 via Mesh Current (1) 2*ia + 8*(ia - ib) - 40 = 0 (2) 6*ib + 6*(ib - ic) + 8*(ib - ia) = 0 (3) 4*ic + 20 + 6*(ic - ib) = 0

  15. Example 4.4 via Mesh Current (1) 2*ia + 8*(ia - ib) - 40 = 0 (2) 6*ib + 6*(ib - ic) + 8*(ib - ia) = 0 (3) 4*ic + 20 + 6*(ic - ib) = 0 . . . (1’) ia = 4 + 8*ib / 10 (3’) ic = -2 + 6*ib / 10

  16. Example 4.4 via Mesh Current (1) 2*ia + 8*(ia - ib) - 40 = 0 (2) 6*ib + 6*(ib - ic) + 8*(ib - ia) = 0 (3) 4*ic + 20 + 6*(ic - ib) = 0 . . . (1’) ia = 4 + 8*ib / 10 (3’) ic = -2 + 6*ib / 10 Substitute both (1’) and (3’) in (2) . . . ia = 5.6 A ib = 2.0 A ic = -0.8 A

  17. Example 4.5

  18. Next: Example 4.5 via Mesh Current

  19. Example 4.5 via Mesh Current Use the Mesh Current Method to compute the power P4 dissipated in the 4 Ω resistor To this end, compute the currents i2, and i3 in the 3 meshes; also find i1 With 3 unknowns we need 3 equations And need to express the branch current that controls the dependent voltage source as a function of other currents Then we can express the power P4 consumed in the 4 Ω resistor

  20. Example 4.5 via Mesh Current 20*(i1 - i3) - 50 + 5*(i1 - i2) = 0 5*(i2 - i1) + i2 + 4*(i2 - i3) = 0 4*(i3 - i2) + 15*iα + 20*(i3 - i1) = 0 express iα as a function of the other currents: iα = i1 - i3

  21. Example 4.5 via Mesh Current 20*(i1 - i3) - 50 + 5*(i1 - i2) = 0 5*(i2 - i1) + i2 + 4*(i2 - i3) = 0 4*(i3 - i2) + 15*iα + 20*(i3 - i1) = 0 express iα as a function of the other currents: iα = i1 – i3 (1’) 50 = 25*i1 - 5*i2 - 20*i3 (2’) 0 = -5*i1 + 10*i2 - 4*i3 (3’) 0 = -5*i1 - 4*i2 + 9*i3

  22. Example 4.5 via Mesh Current 20*(i1 - i3) - 50 + 5*(i1 - i2) = 0 5*(i2 - i1) + i2 + 4*(i2 - i3) = 0 4*(i3 - i2) + 15*iα + 20*(i3 - i1) = 0 express iα as a function of the other currents iα = i1 – i3 (1’) 50 = 25*i1 - 5*i2 - 20*i3 (2’) 0 = -5*i1 + 10*i2 - 4*i3 (3’) 0 = -5*i1 - 4*i2 + 9*i3 i1 = 29.6 Ai2 = 26 Ai3 = 28 A Current in 4 Ω resistor = 28 A - 26 A = 2 A P4 = 2*2*4 = 16 W

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