Topic 12 Mesh Analysis (4.5 - 4.8)

Mesh Currents R1 R2 R3 R4 R5 R6 R7 R8 vs Remember our planar circuit example? R7 R1 R2 R3 R4 R5 R6 R8 vs M4 Mesh analysis can only be done on planar circuits M2 (the same restriction does not apply to node analysis) M1 M3 How many meshes? 2/22/2019 Mesh Method

Mesh Currents R7 R1 R2 R3 R4 R5 R6 R8 vs im1 im2 im3 im4 M1 M2 M3 M4 Imagine each of these meshes as having a circulating current Here we have counterveiling currents im1 im2 In this region we just have im1 In this region we just have im2 Net current down is im1-im2 2/22/2019 Mesh Method

Branch Currents R7 R1 R2 R3 R4 R5 R6 R8 vs i7 i1 im1 im2 im3 im4 i4 i8 Runs through vs, R1 & R2 i6 i5 Once we know all the mesh currents, we know all the branch currents (and we can find them by inspection) Runs through R7 Runs through R4 Runs through R5 Runs through R6 Runs through R3 Runs through R8 2/22/2019 Mesh Method

Mesh Current Method ia ib 10 5 2k 5k 8k Label the mesh currents including their direction Write KVL around the meshes in terms of the mesh currents 1 2 1 2 1 2 2/22/2019 Mesh Method

Method Identify Law In terms of Method Comparison Method Identify Law In terms of Node-Voltage critical nodes KCL at crit. nodes node voltages Mesh-Current meshes KVL around meshes mesh currents 2/22/2019 Mesh Method

Prelab 3, part 2.2b v1 + - v3 - + R1 R2 R3 R4 R5 vs1 vs2 i1 i3 Find the voltages and currents indicated using the mesh-current method ib ic ia v4 - + i4 Define the mesh currents Write KVL around each mesh 1 vs1=12 v; vs2=6 v; R1=820Ω; R2=330Ω; R3=100Ω; R4=680Ω; R5=120Ω; 2 3 1 2 3 2/22/2019 Mesh Method

1 2 3 4 from 1 5 from 3 4 5 into 2 2/22/2019 Mesh Method

Find the voltages and currents indicated using the mesh-current method Prelab 3, part 2.2b v1 + - v3 - + R1 R2 R3 R4 R5 vs1 vs2 i1 i3 Find the voltages and currents indicated using the mesh-current method 9.3 v4 - + -3.8 10.7 i4 2/22/2019 Mesh Method

5 26 90 8 80 30 Assessment 4.7 ib ia ic Use the mesh method to find the power delivered by the source and the power absorbed by the 8Ω resistor Define the mesh currents Write KVL around each mesh collecting terms 1 2 3 2/22/2019 Mesh Method

1 2 3 25 x 1 + 3 4 25 x 2 +18 x 3 5 & 4 5 + 5 & from 1 2/22/2019 Mesh Method

5 26 90 8 80 30 Assessment 4.7 ib ia ic Use the mesh method to find the power delivered by the source and the power absorbed by the 8Ω resistor ib ia Current out of source is ia Current through the 8Ω resistor is ib 2/22/2019 Mesh Method

Dependent Sources (Assessment 4.2) 5 3 1 25 14 10 -3vφ vφ + - Use mesh currents to find how much power is being delivered to the dependent source ib ia ic Write KVL around each mesh in terms of the mesh currents 1 2 3 1 4 4 2 into 5 5 3 2/22/2019 Mesh Method

Dependent Sources (Assessment 4.2) 5 3 1 25 14 10 -3vφ vφ + - Dependent Sources (Assessment 4.2) Use mesh currents to find how much power is being delivered to the dependent source ib ia ic -1 12 1 5 4 3 3 It is straightforward to solve the algebra, giving ia = 4 A, ib = -1 A, ic = 3 A & vφ = 12 v ##*#@!!! triple negative! ib is defined to go uphill but power absorbed is downhill power 2/22/2019 Mesh Method

3 5 2 50 100 10 6 4 ic ia ib The Supermesh The current source gives us a problem (like a floating voltage source in the node method) Notice that the emphasis is on being able to write KVL around the super mesh in terms of the mesh currents and voltage sources by inspection. When we write KVL around meshes a and b we don’t know the voltage across the current source But we do know Since KVL works around any loop And write KVL around the bigger loop (still using ia and ic)… We mentally remove the current source …well, actually just ia since ic =5+ia 2/22/2019 Mesh Method

3 5 2 50 100 10 6 4 ic ia ib The Supermesh So now we have ic in terms of ia and one equation in ia and ib We get our last equation from mesh b 1 2 Adding a resistor in series with the current source would not change the analysis at all Gather terms… 3 2 equation 1 would be the same 3 KVL around the supermesh and mesh b would be the same 3 3x + 2 4 2/22/2019 Mesh Method

Assessment 4.10 3 2 8 16 30 6 4 5 ib ia ic Use the mesh method to find the power dissipated in the 2Ω resistor Identify the mesh currents No. Since ic is the only mesh current through the branch with the current source, we know ic=16 Do we have a supermesh? Leaves us only two to work with 1 1 2 2 These are readily solved as & The current through the 2Ω resistor is: So 2/22/2019 Mesh Method

Node vs Mesh Analysis How many essential nodes are there? How many meshes are there? Are there any supernodes which will reduce the no. of equations? Are there any supermeshes which will reduce the no. of equations? What exactly is it you need to solve? Will one of the methods give you that more directly or quickly? 2/22/2019 Mesh Method

Assesment 4.13: Node or Mesh? Find the power delivered by the 2A current source 15Ω 10Ω 20v 2A 25v v1 20 25 two essential nodes one node equation two meshes one supermesh but One node equation a little easier than one supermesh equation? Current is uphill relative to the voltage, so… 2/22/2019 Mesh Method

Assesment 4.14: Node or Mesh? Find the power delivered by the 4A current source 4 4Ω 3Ω four essential nodes three node equations ix 30ix 6Ω ib 128v ia three meshes one known so two equations 2Ω 5Ω but KVL around mesh a around mesh b Dependent current Solving Voltage across 4A 2/22/2019 Mesh Method