Mesh Equatons Introducton The crcuts n ths problem set consst of resstors and ndependent sources. We wll analyze these crcuts by wrtng and solvng a set of mesh equatons. To wrte mesh equatons, we. Express the element currents and voltages n terms of the mesh currents.. Apply KVL to the meshes of the crcut. In partcular, we wll express resstor voltages n terms of the mesh currents. Consder the blue resstor n each of these two stuatons. (a) (b) In (a) the blue resstor s located on the nsde of the crcut. Consequently, t s n two meshes, mesh and mesh wth correspondng mesh currents and, n ths case. Apply KCL n (a) to get = + = Next, apply Ohm s law to get v = = In (b) the blue resstor s located on the outsde of the crcut. Consequently, t s n only one mesh, mesh wth correspondng mesh current, n ths case. Apply KCL n (b) to get Next, apply Ohm s law to get = v = =
To summarze:. Consder a resstor havng resstance and located on the nsde of the crcut, n both mesh and mesh. Label the resstor current and voltage v to adhere to the passve conventon. Suppose that the drecton of mesh current s the same as the drecton of but that the drecton of mesh current s opposte the drecton to. The resstor voltage s gven by v = + opposte.) of, the resstor voltage s gven by v =.. ( + corresponds to same and corresponds to. Consder a resstor havng resstance and located on the outsde of the crcut, n mesh. Label the resstor current and voltage v to adhere to the passve conventon. When the drecton of mesh current s the same as the drecton of, the resstor voltage s = gven by v. When the drecton of mesh current s the opposte to the drecton We wll also express the currents of current sources currents n terms of mesh currents. Consder the blue current source n each of these two stuatons. (a) (b) In (a) the current source s located on the nsde of the crcut. Consequently, t s n two meshes, mesh and mesh wth correspondng mesh currents and, n ths case. Apply KCL n (a) to get = s + s = There s no easy way to express the current source voltage n terms of the mesh currents. In (b) the current source s located on the outsde of the crcut. Consequently, t s n only one mesh, mesh wth correspondng mesh current, n ths case. Apply KCL to get = s
Agan, there s no easy way to express the current source voltage n terms of the mesh currents. To summarze:. Consder a current source havng current I and located on the nsde of the crcut, n both mesh and mesh. Suppose that the drecton of mesh current s the same as the drecton of I but that the drecton of mesh current two mesh currents are related by the equaton corresponds to opposte.) s opposte the drecton of I. The I = +. (+ corresponds to same and. Consder a current source havng current I and located on the outsde of the crcut, n mesh. When the drecton of mesh current s the same as the drecton of I, the mesh current s gven by I. On the other hand, when the drecton of mesh current s opposte as the drecton of I, the mesh current s gven by = = I.. There s no easy way to express the current source voltage n terms of the mesh currents. Example: Worked Examples Analyze ths crcut by wrtng and solvng a set of mesh equatons. Soluton: Label the label the mesh currents. Then, label the element currents n terms of the mesh currents:
Notce that the A source on the outsde of the crcut s n mesh and that the currents A and have the same drecton. Consequently Apply KVL to mesh to get = A 5 + 9 + 8 = 0 s the voltage across the 5 Ω resstor (+ on the left), 9( ) s the voltage across the 9 Ω resstor (+ on top) and 8 s the voltage across the 8 Ω resstor (+ on In ths equaton 5( ) bottom). Substtutng = A and dong a lttle algebra gves 4 9 = 50 Next, apply KVL to mesh to get In ths equaton 4( ) source voltage and = A ( ) ( ) 4 + 4 9 = 0 s the voltage across the 4 Ω resstor (+ on the left), 4 s the voltage ) s the voltage across the 9 Ω resstor (+ on top). Substtutng 9( and dong a lttle algebra gves 9 + = 4+ 4 =4 The smultaneous equatons can be wrtten n matrx form
4 9 = 50 4 9 50 = 9+ = 4 9 4 5 We can use MATLAB to solve the matrx equaton: Then That s, the mesh currents are.40 = 0.698 =.40 A and = 0.698 A.
Example: Ths crcut s represented by the smultaneous equatons: a + a = 4 a + a = 0 Determne the values of the coeffcents,, and a. a a a Soluton: Label the label the mesh currents. Then, label the element currents n terms of the mesh currents: Notce that the 0.4 A source on the nsde of the crcut s n both mesh and mesh. Mesh current s drected n the same way as current source current but the mesh current s drected opposte to the current source current. Consequently = 0.4 A
7 The current source s n both mesh and mesh so we apply KVL to the supermesh correspondng to the current source (.e. the permeter of meshes and ). The result s 0 + 9 + 0 = 0 ( ) In ths equaton 0 s the voltage across the horzontal 0 Ω resstor (+ on the left), 9 s the voltage across the 9 Ω resstor (+ on top) and 0 s the voltage across the vertcal 0 Ω resstor (+ on bottom). Substtutng = 0.4 and dong a lttle algebra gves 9 9 = 4 Next, apply KVL to mesh to get 0 9 = 0 In ths equaton s the voltage across the Ω resstor (+ on the left), 0 s the voltage ( ) source voltage and 9 s the voltage across the 9 Ω resstor (+ on top). Dong a lttle algebra gves 9 + 4 = 0 To summarze, the crcut s represented by the smultaneous equatons: 9 9 = 4 9 + 4 = 0 Comparng these equatons to the gven equatons shows a = 9, a = 9, a = 9 and a = 4.