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Lecture - 6 Mesh analysis  

Lecture - 6 Mesh analysis  . Outline. Terms of describing circuits. The Mesh-Current method. The concept of supermesh . The Node-Voltage method Versus the Mesh-Current method. Terms of describing circuits . Example 1. For the circuit in the figure, identify a) all nodes.

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Lecture - 6 Mesh analysis  

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  1. Lecture - 6 Mesh analysis  

  2. Outline • Terms of describing circuits. • The Mesh-Current method. • The concept of supermesh. • The Node-Voltage method Versus the Mesh-Current method

  3. Terms of describing circuits

  4. Example 1 • For the circuit in the figure, identify a) all nodes. b) all essential nodes. c) all branches. d) all essential branches. e) all meshes. f) two paths that are not loops or essential branches. g) two loops that are not meshes.

  5. Example 1 • The nodes are a, b, c, d, e, f, and g. b) The essential nodes are b, c, e, and g. c) The branches are v1, v2, R1, R2, R3, R4, R5, R6, R7 and I. d) The essential branches are: v1 – R1 , R2 – R3 , v2 – R4 , R5, R6, R7 and I

  6. Example 1 e) The meshes are: v1 – R1– R5 – R3 – R2, v2–R2 – R3 – R6 –R4, R5– R7 – R6 and R7 –I f) The two paths that are not loops or essential branches are R1 – R5 – R6 and v2 - R2 , because they do not have the same starting and ending nodes), nor are they an essential branch (because they do not connect two essential nodes). g)The two loops that are not meshes are v1 — R1 — R5 – R6 - R4 - v2 and I — R5 —R6 , because there are two loops within them.

  7. Mesh-Current method • Steps to Determine Mesh Currents: • Assign mesh currents i1, i2, .., in to the n meshes. • Apply KVL to each of the n meshes. Use Ohm’s law to express the voltages in terms of the mesh currents. • Solve the resulting n simultaneous equations to get the mesh currents.

  8. Example 2 a) Use the mesh-current method to determine the power associated with each voltage source in the circuit shown. b) Calculate the voltage voacross the 8 Ω resistor. ----------------------------------------

  9. Example 2 • a) The three meshes equations are: • Therefore, the three mesh currents are • ia = 5.6 A, ib= 2.0 A, ic= -0.80 A. • The power associated with each voltage source: • P40V = -40ia = -224 W, P20V = 20ic = -16 W. • b) vo = 8(ia- ib) = 8(3.6) = 28.8 V.

  10. Example 3 Use the mesh-current method of circuit analysis to determine the power dissipated in the 4Ωresistorin the circuit shown. --------------------------------------------------------------

  11. Example 3 • The three mesh-current equations are: • We now express the branch current controlling the dependent voltage source in terms of the mesh currents as: iΦ= i1 – i3 • Therefore, the mesh currents are: i1= 29.6 i2 = 26 A i3 = 28 A and P4Ω= (i3 - i2)2(4) = (2)2(4) = 16 W.

  12. The Concept of a Supermesh • A supermesh results when two meshes have a (dependent , independent) current source in common. • If a supermesh consists of two meshes, two equations are needed; one is obtained using KVL and Ohm’s law to the supermesh and the other is obtained by relation regulated due to the current source.

  13. Example 4 Use the mesh-current method to find the power dissipated in the 1 Ω resistor in the circuit shown. ---------------------------------------------

  14. Example 4 • The 2 A current source is shared by the meshes ia and ib. Thus we combine these meshes to form a supermesh and write the following equation: −10 + 2ib + 2(ib − ic) + 2(ia − ic) = 0 • The other mesh current equation is −6 + ic + 2(ic − ia) + 2(ic − ib) = 0 • The supermesh constraint equation is ia − ib = 2

  15. Example 4 • Place these three equations in standard form: 2ia + 4ib − 4ic = 10 −2ia − 2ib + 5ic = 6 ia− ib+ 0ic = 2 • Solving, ia = 7 A ib= 5 A ic = 6 A • Thus, P1Ω = ic2(1) = (6)2(1) = 36 W

  16. The Node-Voltage method Versusthe Mesh-Current method • Asking a number of questions, may help you identify the more efficient method before plunging into the solution process: • Does one of the methods result in fewer simultaneous equations to solve? • Does the circuit contain supernodes? If so, using the node-voltage method will permit you to reduce the number of equations to be solved. • Does the circuit contain supermeshes? If so, using the mesh-current method will permit you to reduce the number of equations to be solved. • Will solving some portion of the circuit give the requested solution? If so, which method is most efficient for solving just the pertinent portion of the circuit?

  17. Summary • The mesh-current method works only with planar circuits • Mesh currents are assigned to each mesh, and Kirchhoff's voltage law is used to write one equation per mesh. • The number of equations is b — (n — 1), where b is the number of branches in which the current is unknown, and n is the number of nodes. • The mesh currents are used to find the branch currents.

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