ELEC273 Lecture Notes Set 4, Mesh Analysis

Transcription

1 ELEC273 Lecture Notes Set 4, Mesh Analysis The course web site is: The list of homework problems is in the course outline. For this week: Do these problems! 2

2 ELEC 273 Mid-Term Exam Saturday October 21, :00 to 18:30 Section R H407 H411 Section D H415 H435 Topic areas for the midterm test: Chapter 1 Basic Concepts Chapter 2 Basic Laws Chapter 3 Methods of Analysis (Node and Mesh) Chapter 4 Circuit Theorems The test lasts 90 minutes and there will be four questions.

3 Parts of a Circuit: branch, node, mesh node mesh branch mesh node branch mesh branch A branch is a resistor, inductor, capacitor. A node is a connection point. A mesh is the simplest closed path that encloses nothing.

4 Mesh Analysis VV ss,, and II 3 are mesh currents. We can find,, and II 3 using three mesh equations which are KVL equations for the mesh paths. Idea Define mesh currents which flow around closed mesh paths in the circuit. Mesh currents automatically satisfy KCL: when a mesh current flows into a node it also flows out of the node. Mesh currents are related in a very simple way to branch currents. Write one KVL equation for each mesh in the circuit to get a set of independent equations that can be solved for the mesh currents. RR 1 RR 2 RR 3 II 3 RR 4 RR 5

5 Procedure for writing mesh equations VV ss vv 1 _ vv 2 RR 1 vv 3 RR 2 RR 3 II 3 RR 4 RR 5 vv 4 _ vv 5,, and II 3 are mesh currents. vv 1, vv 2, vv 3, vv 4 and vv 5 are branch voltages. 1. Identify the meshes and assign mesh currents. 2. Write the voltage across each branch in terms of the mesh currents. 3. Write a KVL equation for each mesh path. 4. Solve the equations.

6 How are the branch voltages related to the mesh currents? VV ss vv 1 _ vv 2 _ RR 1 vv 3 RR 2 RR 3 II 3 RR 4 RR 5 vv 4 _ vv 5 _,, and II 3 are mesh currents. vv 1, vv 2, vv 3, and vv 4 are branch voltages. vv 1 _ vv 1 = ( )RR 1 vv II 3 2 vv 2 _ II 3 II 3 vv 4 _ vv 5 _ vv 2 = ( II 3 )RR 2 vv 3 = (II 3 )RR 3 vv 4 = RR 4 vv 5 = II 3 RR 5

7 KVL for the mesh paths: VV ss B RR 1 ( ) D RR 2 ( II 3 ) C RR 1 RR 3 (II 3 ) RR 2 RR 3 II 3 RR 4 RR 5 F G RR 4 RR 5 II 3 A E H Mesh ABCDEA: VV ss ( )RR 1 ( II 3 )RR 2 = 0 Mesh DCFGD: RR 1 RR 4 II 3 RR 3 = 0 Mesh EDGHE: II 3 RR 2 II 3 RR 3 II 3 RR 5 = 0

8 Collect terms: V ( )RR 1 ( II 3 )RR 2 = 0 RR 1 RR 4 II 3 RR 3 = 0 II 3 RR 2 II 3 RR 3 II 3 RR 5 = 0 VV ss RR 1 RR 2 RR 3 RR 4 RR 5 RR 1 RR 2 RR 1 RR 2 II 3 = VV ss RR 1 (RR 1 RR 3 RR 4 )RR 2 RR 3 II 3 = 0 RR 2 RR 3 RR 2 RR 3 RR 5 II 3 = 0 II 3 Write the mesh equations as a 3x3 matrix: RR 1 RR 2 RR 1 RR 2 RR 1 RR 1 RR 3 RR 4 RR 3 RR 2 RR 3 RR 2 RR 3 RR 5 II 3 = VV ss 0 0

9 Write the mesh equations by inspection: ZZ iiii II ii = VV ii 1.The diagonal entry ZZ iiii is the sum of the resistances around the mesh. 2.The off-diagonal entry ZZ iiii is the negative of the resistance common to mesh i and mesh j. 3.The right-hand side VV ii is the voltage generator in mesh i. VV ss RR 1 RR 3 RR 2 II 3 RR 4 RR 5 RR 1 RR 2 RR 1 RR 2 RR 1 RR 1 RR 3 RR 4 RR 3 RR 2 RR 3 RR 2 RR 3 RR 5 II 3 = VV 0 0 Tip: the inspection rules are handy for checking your mesh equations on exams!

10 Mesh analysis example 1 2 Ω 4 Ω 10 3 Ω 5 Find the branch currents using mesh analysis. Mesh Analysis: 1. Identify the meshes and assign mesh currents. 2. Write the voltage across each branch in terms of the mesh currents. 3. Write a KVL equation for each mesh path. 4. Solve the equations.

11 1.Identify the meshes and assign mesh currents. 2 Ω 4 Ω 3 Ω Write the voltage across each branch in terms of the mesh currents ( ) 10 5

12 3.Write a KVL Equation for each mesh path. B 2 3( ) C E A D F Mesh path ABDCA: = 0 Mesh path DCEFD: = 0

13 4.Solve the equations: start by collecting the terms = = = = 5 2 Ω 4 Ω 3 Ω = = = 10 5 It is customary to make the diagonal elements positive: 5 3 =

14 Inspection Rules: 1. The diagonal matrix element rr iiii is the sum of the resistances going around mesh #i. 2. The off diagonal matrix element rr iijj is minus the resistance that is common to meshes #i and #j. 3. The right hand side vv ii is the algebraic sum of the voltage generators going around mesh #i. Mesh equations by inspection: 2 Ω 4 Ω 3 Ω = 10 5 rr 11 rr 12 rr 21 rr 22 II2 = vv 1 vv 2

15 4.Solve the equations: = 10 5 Cramer s Rule DD = DD 1 = 5 7 DD 2 = = DD 1 = 55 = A DD 26 = DD 2 = 5 = A DD 26 = (3) = 26 = (5) = 55 = 5(5) (3) 10 = 5 2 Ω 4 Ω 3 Ω 10 5

16 5.Find the Branch Currents = A = A = Ω 4 Ω 3 Ω 10 5 In the 2 ohm resistor: Ω = = A In the 3 ohm resistor: II 3Ω = ( ) = = A In the 4 ohm resistor: II 4Ω = = A

17 Example #2: Write a set of mesh equations for this circuit. 3 Ω 5 Ω 17 Ω 10 volts 15 volts 2 volts 17 volts 11 Ω 7 Ω 13 Ω Mesh Analysis: 1. Identify the meshes and assign mesh currents. 2. Write the voltage across each branch in terms of the mesh currents. 3. Write a KVL equation for each mesh path. 4. Solve the equations.

18 1. Assign mesh currents: 3 Ω 5 Ω 17 Ω 10 volts volts 11 Ω 7 Ω II 4 13 Ω Mesh Analysis: 1. Identify the meshes and assign mesh currents. 2. Write the voltage across each branch in terms of the mesh currents. 3. Write a KVL equation for each mesh path. 4. Solve the equations. II 3

19 2. Write the voltage across each branch in terms of the mesh currents: 3 3 Ω 5 5 Ω 17 Ω 17 II 4 Mesh Analysis: 1. Identify the meshes and assign mesh currents. 2. Write the voltage across each branch in terms of the mesh currents. 3. Write a KVL equation for each mesh path. 4. Solve the equations. 10 volts volts 11(II II 3 ) 1 11 Ω II 4 7 Ω 13 Ω 7 II 3 13 II 3 II 4 II 3

20 10 3. Write a KVL equation for each mesh: 3 3 Ω 5 5 Ω 17 Ω (II 3 ) 17 II 4 11 Ω II 4 7 Ω 13 Ω 7 II 3 13 II 3 II 4 II II 3 = II II 3 = II 3 11 II 3 13 II 3 II 4 = 0 13 II 3 II II 4 17 = 0 Mesh Analysis: 1. Identify the meshes and assign mesh currents. 2. Write the voltage across each branch in terms of the mesh currents. 3. Write a KVL equation for each mesh path. 4. Solve the equations.

21 4. Solve the equations: 3 Ω From the previous page: 5 Ω 17 Ω II 3 = II II 3 = Ω 17 7 II 3 11 II 3 13 II 3 II 4 = 0 13 II 3 II II 4 17 = 0 7 Ω 13 Ω II 4 II 3 Collect the terms: II 3 = Mesh Analysis: 1.Identify the meshes and assign mesh currents. 2.Write the voltage across each branch in terms of the mesh currents. 3.Write a KVL equation for each mesh path. 4.Solve the equations II 4 11II 3 11 = II 3 11II II 3 13II 4 = 0 13II 3 13II II 4 = 2 17

22 4. Solve the equations write a matrix: 3 Ω II 3 = Ω 17 Ω II 4 11II 3 11 = II 3 11II II 3 13II 4 = 0 13II 3 13II II 4 = Ω 7 Ω 13 Ω II 4 17 II II 3 II 4 = Multiply by -1 so that the diagonal elements are positive: II 3 II 4 =

23 Check the matrix with the inspection rules: 1. The diagonal matrix element rr iiii is the sum of the resistances going around mesh #i. 2. The off diagonal matrix element rr iijj is minus the resistance that is common to meshes #i and #j. 3. The right hand side vv ii is the algebraic sum of the voltage generators going around mesh #i. 3 Ω 5 Ω 17 Ω Ω 7 Ω 13 Ω II II 3 II 4 = II 3

24 Homework: write mesh equations for this circuit: Alexander and Sadiku Problem 3.41 Mesh Analysis: 1.Identify the meshes and assign mesh currents. 2.Write the voltage across each branch in terms of the mesh currents. 3.Write a KVL equation for each mesh path. - Arrange you mesh equations into a matrix - Check your mesh equations using the inspection rules. 4.Solve the equations. 5.Solve the circuit with Spice and check your answers.

25 Mesh analysis including a current generator Write mesh equations for this circuit: The path including the 6 ohm resistor, the 3 ohm resistor and the 5 ohm resistor is called a supermesh because it encloses a current generator. 2 Ω 2 Ω 4 Ω 3 Ω 3 Ω 10 v 6 Ω 2 A 5 Ω Mesh Analysis: 1.Identify the meshes and assign mesh currents. 2.Write the voltage across each branch in terms of the mesh currents. 3a.Write a KVL equation for each mesh path, including the supermesh. 3b.Write a constraint equation for the current generator. 4.Solve the equations.

26 1. Identify mesh paths and assign mesh currents. 2 Ω 2 Ω II 3 4 Ω 3 Ω 3 Ω 10 v 2 A 6 Ω II 4 II 5 5 Ω Mesh Analysis: 1.Identify the meshes and assign mesh currents. 2.Write the voltage across each branch in terms of the mesh currents. 3a.Write a KVL equation for each mesh path, including the supermesh. 3b.Write a constraint equation for the current generator. 4.Solve the equations.

27 2. Find the branch voltages from the mesh currents. 2 Ω 2 Ω 2 2II 3 4( II 3 ) 4 Ω II 3 3 Ω 3 Ω 3 3 II 4 II 3 10 v 6( II 4 ) 2 A II 5 6 Ω II 4 5 Ω 5II 5 2.Write the voltage across each branch in terms of the mesh currents. Label each of the resistors with the voltage in terms of the mesh currents.

28 3a.Write a KVL Equation for each Mesh Path C B 2 Ω 2 Ω 2 2II 3 4( II 3 ) 4 Ω II 3 3 Ω 3 Ω 3 3 II 4 II 3 10 v 6( II 4 ) 2 A II 5 6 Ω II 4 A F D E I G H 5 Ω K J 5II 5 Path BCFEB 2 4 II 3 3( ) = 0 Path ABEDA II 4 = 0 Path EFIHE 4( II 3 ) 2II 3 3(II 4 II 3 ) = 0 Supermesh Path DEHKJGD 6 II 4 3 II 4 II 3 5II 5 = 0 3b.Write a constraint equation for the current generator: II 55 II 44 = 22

29 4.Collect the equations into a matrix: 2 4 II 3 3 = II 4 = 0 4( II 3 ) 2II 3 3(II 4 II 3 ) = 0 6 II 4 3 II 4 II 3 5II 5 = 0 II 4 II 5 = II 3 II 4 II 5 =

30 Can we solve this circuit smarter? 10 v 10 v 2 Ω 2 Ω 4 Ω 3 Ω 3 Ω 2 Ω 2 Ω 4 Ω 3 Ω 3 Ω II 3 2 A 6 Ω II 4 5 Ω II 5 II 3 II 55 = II A 6 Ω II 4 5 Ω II4 2 By smarter I mean with fewer unknown currents. We used five mesh currents:, ii 2, II 3, II 4, II 5 We had to solve a 5x5 matrix! We can solve the circuit smarter by using the constraint equation directly in the mesh equations. II 5 II 4 = 2 So II 5 = II 4 2 Then the KVL equation for the Supermesh Path changes from 6 II 4 3 II 4 II 3 5II 5 = 0 to 6 II 4 3 II 4 II 3 5(II 4 2) = 0 The other three KVL equations don t change: 2 4 II 3 3 = II 4 = 0 4( II 3 ) 2II 3 3(II 4 II 3 ) = 0 There are only four unknowns :, ii 2, II 3, II 4. The matrix is only 4x4.

31 Homework: solve this circuit using mesh analysis. Alexander and Sadiku problem Identify the meshes and assign mesh currents. 2.Write the voltage across each branch in terms of the mesh currents. 3a.Write a KVL equation for each mesh path including the supermesh. 3b.Write a constraint equation for the current generator. 4.Solve the equations. 5.Find ii oo. 6.Check your answer with SPICE

32 Mesh analysis with controlled sources 10 volts 2 Ω 3 Ω vv 7 Ω 5 Ω 2vv 6 Ω Use mesh analysis to find the branch currents. vv 2vv This circuit contains a voltage-controlled voltage source (VCVS). The voltage v across the 3 ohm resistor controls the voltage across the controlled source. Mesh Analysis: 1. Identify the meshes and assign mesh currents. 2. Write the voltage across each branch in terms of the mesh currents. 3. Write a KVL equation for each mesh path. 4. Solve the equations.

33 1.Identify the meshes and assign mesh currents. 10 volts 2 Ω 3 Ω vv 7 Ω 5 Ω 2vv 6 Ω 10 2 Ω 3 Ω 7 Ω 5 Ω vv II 3 2vv 6 Ω 2.Write the voltage across each branch in terms of the mesh currents II 3 5( II 3 ) vv = 3( ) 6II 3 2vv = 2 3( ) = 6( )

34 3.Write a KVL equation for each mesh path II 3 5( II 3 ) vv = 3( ) 6II 3 2vv = 2 3( ) = 6( ) Mesh 1: = 0 Mesh 2: II 3 6( ) = 0 Mesh 3: 6( ) 5 II 3 6II 3 = 0

35 Collect terms and write a matrix equation: Mesh 1: = 0 Mesh 2: II = 0 Mesh 3: II 3 6II 3 = 0 Collect terms: = II = II 3 6II 3 = 0 Collect terms: 5 3 = II 3 = II 3 = II 3 = 0

36 Practice Problem: Alexander and Sadiku prob Write mesh equations for ii 1 and ii 2. 2.Arrange your equations into a 2x2 matrix. 3.Solve the equations. 4.Find the branch currents.

37 ELEC273 Lecture Notes Set 4, Mesh Analysis The course web site is: The list of homework problems is in the course outline. For this week: Do these problems! 68

38 ELEC 273 Mid-Term Exam Saturday October 21, :00 to 18:30 Section R H407 H411 Section D H415 H435 Topic areas for the midterm test: Chapter 1 Basic Concepts Chapter 2 Basic Laws Chapter 3 Methods of Analysis (Node and Mesh) Chapter 4 Circuit Theorems The test lasts 90 minutes and there will be four questions.

39 Controlled Sources Example #2 ii 11 Ω We solved this circuit by node analysis! 2ii 3 Ω vv 2 Ω 7 Ω 12 volts 1 Ω 8vv Write a set of mesh equations for this circuit. Mesh Analysis: 1. Identify the meshes and assign mesh currents. 2. Write the voltage across each branch in terms of the mesh currents. 3. Write a KVL equation for each mesh path. 4. Solve the equations.

40 1.Identify the meshes and assign mesh currents. 11 Ω ii ii 3 Ω vv 2 Ω 7 Ω 12 volts 1 Ω 8vv Use one mesh current for each window in the circuit diagram. The current source forces themesh current in the mesh at left to be 2ii. Then we need three unknown mesh currents: ii 1, ii 2,, ii 3. There are two dependencies: ii = ii 2 and vv = 3(2ii ii 1 ) 2ii 2ii 3 Ω vv ii 1 2 Ω 1 Ω ii 7 Ω 11 Ω ii 2 ii 3 8vv 12

41 2.Write the voltage across each branch in terms of the mesh currents 2ii 2ii 3 Ω 11ii 2 ii 11 Ω 2 Ω ii 1 2(ii 1 ii 2 ) ii 2 vv = 3(2ii ii 1 ) 7 Ω 7(ii 3 ii 2 ) 1 Ω 1(ii ii 1 ii 3 ) 3 8vv 12 volts 3.Write a constraint equation for the current generator a KVL equation for each mesh path: Mesh #1: 3 2ii ii 1 2(ii 1 ii 2 ) 1(ii 1 ii 3 ) = 0 Mesh #2: 2(ii 1 ii 2 ) 11ii (ii 3 ii 2 ) = 0 Mesh #3: 1(ii 1 ii 3 ) 7 ii 3 ii 2 8vv = 0 Also ii = ii 2 And vv = 3(2ii ii 1 )

42 4.Solve the equations: eliminate vv and ii and arrange into a matrix equation. Mesh #1: 3 2ii ii 1 2(ii 1 ii 2 ) 1(ii 1 ii 3 ) = 0 Mesh #2: 2(ii 1 ii 2 ) 11ii (ii 3 ii 2 ) = 0 Mesh #3: 1(ii 1 ii 3 ) 7 ii 3 ii 2 8vv = 0 Also ii = ii 2 And vv = 3(2ii ii 1 ) Simplify by eliminating vv and ii from the mesh equations: Since ii = ii 2 we can write 2ii = 2ii 2 for mesh #1 3 2ii ii 1 2(ii 1 ii 2 ) 1(ii 1 ii 3 ) = 0 as 3 2ii 2 ii 1 2 ii 1 ii 2 1 ii 1 ii 3 = 0 Since vv = 3 2ii ii 1 = 3(2ii 2 ii 1 ) we can write for mesh #3 1 ii 1 ii 3 7 ii 3 ii 2 8vv = 0 as 1 ii 1 ii 3 7 ii 3 ii 2 24(2ii 2 ii 1 ) = 0

43 Simplify: 3 2ii 2 ii 1 2 ii 1 ii 2 1 ii 1 ii 3 = 0 6ii 2 3ii 1 2ii 1 2ii 2 ii 1 ii 3 = 0 6ii 1 4ii 2 ii 3 = 0 2(ii 1 ii 2 ) 11ii (ii 3 ii 2 ) = 0 2ii 1 2ii 2 11ii 2 7ii 3 7ii 2 = 12 2ii 1 20ii 2 7ii 3 = 12 1 ii 1 ii 3 7 ii 3 ii 2 24(2ii 2 ii 1 ) = 0 ii 1 ii 3 7ii 3 7ii 2 48ii 2 24ii 1 = 0 23ii 1 41ii 2 8ii 3 = 0 Hence: 6ii 1 4ii 2 ii 3 = 0 2ii 1 20ii 2 7ii 3 = 12 23ii 1 41ii 2 8ii 3 = 0

44 Arrange into a matrix equation: 6ii 1 4ii 2 ii 3 = 0 2ii 1 20ii 2 7ii 3 = 12 23ii 1 41ii 2 8ii 3 = ii 1 ii 2 ii 3 = ii 2ii 3 Ω 11ii 2 ii 11 Ω 2 Ω ii 1 2(ii 1 ii 2 ) ii 2 vv = 3(2ii ii 1 ) 7 Ω 7(ii 3 ii 2 ) 1 Ω 1(ii ii 1 ii 3 ) 3 8vv 12 volts

45 Practice Problem: Alexander and Sadiku prob Write mesh equations for ii 1, ii 2 and ii 3. 2.Arrange your equations into a 3x3 matrix. 3.Solve the equations. 4.Find the branch currents. 5.What is the value of vv 0?

46 Solve a Circuit with Voltage-Dependent Voltage Sources Modelling Operational Amplifiers 1 5 Ω 10 Ω 5 Ω vv 1 AAvv 1 5 Ω 10 Ω vv 2 AAvv 2 5 Ω 50 Ω vv oo This circuit has two voltagedependent voltage sources with gain A=100. They model operational amplifiers or op-amps. The gain of each amplifier is A=100. Use mesh analysis to: Find the branch currents. Find the output voltage. But typically op-amps have a much higher gain, such as 100,000! Mesh Analysis: 1. Identify the meshes and assign mesh currents. 2. Write the voltage across each branch in terms of the mesh currents. 3. Write a KVL equation for each mesh path. 4. Solve the equations.

47 1.Identify the meshes and assign mesh currents. 2.Write the branch voltages in terms of the mesh currents. 1 5 Ω 10 Ω 5 Ω 10 Ω 5 Ω 50 Ω 5 Ω vv 1 AAvv 1 II vv 3 2 AAvv 4 II 5 3. Write the voltage across each branch in terms of the mesh currents vv oo II 3 10II 4 1 5( ) AA 5 5(II 3 II 4 ) AA 5 II 3 II 4 II 1 II2 II 3 II 4 II 5 50II 5

48 4.Write a KVL equation for each mesh path II 3 10II 4 1 5( ) AA 5 5(II 3 II 4 ) AA 5 II 3 II 4 II 1 II2 II 3 II 4 II 5 50II 5 Mesh 1: = 0 Mesh 2: 5 10 AA[5( )] = 0 where AA = 100 Mesh 3: AA[5( )] 5II 3 5(II 3 II 4 ) = 0 Mesh 4: 5(II 3 II 4 ) 10II 4 AA[5(II 3 II 4 )] = 0 Note that equations 1 and 2 use only and so can be solved immediately. Once and have been found we can solve equations 3 and 4 for II 3 and II 4. Mesh 5: AA[5(II 3 II 4 )] 50II 5 = 0

49 Node Analysis or Mesh Analysis? 1 volt 5 Ω 5 Ω 10 Ω 5 Ω 10 Ω 5 Ω 50 Ω vv 1 AAAA 1 vv 2 AAAA 2 vv oo Use node analysis to find the output voltage. 1.Choose a ground node. 2.Assign node voltages. vv 1 AAvv 1 vv 2 AAvv 2 1 volt 5 Ω 10 Ω 5 Ω 10 Ω 5 Ω 5 Ω 50 Ω vv 1 AAAA 1 vv 2 AAAA 2 vv oo = AAvv 2 Ground is 0 volts

50 3.Write the branch currents in terms of the node voltages. 4.Write KCL at each node. vv 1 AAvv 1 vv 2 AAvv 2 1 volt 5 Ω 10 Ω 5 Ω 10 Ω 5 Ω 5 Ω 50 Ω vv 1 AAAA 1 vv 2 AAAA 2 vv oo = AAvv 2 Node #1: 1 vv 1 5 vv 1 5 vv 1 AAvv 1 10 = 0 One unknown: solve for vv 1 Node #2: AAvv 1 vv 2 5 vv 2 5 vv 2 AAvv 2 10 = 0 We know vv 1, so solve for vv 2. Then vv oo = AAvv 2. For this problem node analysis is much simpler than mesh analysis!

51 Node Analysis or Mesh Analysis? Alexander and Sadiku Section 3.7 page 102 You can solve any problem either by node analysis or by mesh analysis? Which method should you choose? 1.Often the method with fewer unknowns is better. 2.For some problems, like the amplifier above, one method is clearly simpler. 3.In some problems, we need to know the node voltages for comparison with a measurement or a SPICE simulation, and then node analysis is more direct. 4.For some problems one method gives more insight into the behavior of the circuit. Toolbox of Methods You are trying to build a toolbox of methods for solving problems. Node analysis and mesh analysis are two of the tools in your toolbox. Knowing which tool to use comes with experience. It is often useful to solve a problem two different ways to be sure the answer is right!