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CE2004 : Chapter 2 Nodal & Loop Analysis Techniques

CE2004 : Chapter 2 Nodal & Loop Analysis Techniques. Week 2, 3 : Lecture 3, 4, 5. Nodal and Loop Analysis Techniques. Nodal Analysis Mesh Analysis Loop Analysis In Chapter 2: KCL, KVL and Ohm’s Law single-loop (series) and single-node-pair (//) circuits

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CE2004 : Chapter 2 Nodal & Loop Analysis Techniques

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  1. CE2004 : Chapter 2Nodal & Loop Analysis Techniques Week 2, 3 : Lecture 3, 4, 5

  2. Nodal and Loop Analysis Techniques • Nodal Analysis • Mesh Analysis • Loop Analysis In Chapter 2: • KCL, KVL and Ohm’s Law • single-loop (series) and single-node-pair (//) circuits Complex circuits – need a more systematic approach

  3. Nodal Analysis Chapter 2 : Nodal & Loop Analysis Techniques

  4. Nodal Analysis NOTE: Complex Cct – where you need more than one KCL equations. • variables are node voltages • uses KCL and Ohm’s Law •  one node in the circuit is chosen as reference (ground) and all other node voltages are defined with respect to that node: if there are N nodes in the circuit, there will be N -1 KCL equations • reference node is often the one to which the largest number of branches are connected (also often at one end of an independent voltage source)   • resistor current – put in terms of node voltages via Ohm’s law.

  5. NOTE: If necessary, use Ohm Law to work out the current in each branch Nodal Analysis Resistor Current Ohm’s law equation: ib ia Rb ic Ra Rc

  6. Ccts Containing Only Independent Current Sources Example 2-1 Write the node equations for the cct below. No. of node eqns = 3 – 1 = 2 Node 1: Node 2:

  7. I4 I1 I3 I2 I5 I6 Ccts Containing Only Independent Voltage Sources • Example 2-2 - Find all the currents in the circuit. The presence of Independent Voltage Sources results in certain nodes with a known voltage  Need fewer node eqns. Since N = 4 in this cct, in general we need N-1 = 4 -1 = 3 node eqns. NOTE: Fewer Nodes.

  8. Now that voltages at node 1 & node 3 are known, the no. of node eqns needed is reduced by 2. i.e. only 3 - 2 = 1 node eqn is needed. I3 I1 Node 2: I2 All multiply by 12k (remove common denominator) : V2 – 12 + 2V2 + V2 + 6 = 0 V2 = 1.5V I1 = (V1 - V2)/(12k) = (12 – 1.5)/(12k) = 0.875 mA I2 = V2/(6k) = 1.5/(6k) = 0.25 mA

  9. I4 I1 I3 I2 I5 I6 I3 = [V2-(-6)]/(12k) = (1.5 + 6)/(12k) = 0.625 mA I4 = (V1- V3)/(9k) = [12 – (-6)]/(9k) = 2 mA I5 = I1 + I4 = 2.875 mA I6 = I3 + I4 = 2.625 mA

  10. Sometimes certain nodes in a cct may have a voltage that can be readily obtained from the voltage of another node. This can also helps reduce the no. of node eqns needed. Example 2-3 Using nodal analysis, find I1 and I2. I1 I2 1 V2 can be readily obtained once V1 is known & vice versa, hence no. of node eqns required = 3 -1 -1 = 1.

  11. *Note: The book uses the concept of super node to solve this problem

  12. Node 1: I1 I2 but V1- V2 = 6 • All multiply by 12k (remove common denominator) : -72 + 2V1+ V1 - 6 + 48 = 0 3V1 = 30 V1 = 10V V2 = V1 - 6 = 10 – 6 = 4V I1 = V1/(6k) = 10/(6k) = 5/3 mA I2 = V2/(12k) = 4/(12k) = 1/3 mA

  13. Mesh Analysis Chapter 2 : Nodal & Loop Analysis Techniques

  14. Mesh Analysis & Loop Analysis Recall from chapter 2:  • Loop - simply any closed path through the circuit in which no node is encountered more than once. • Mesh - A loop that has no branches cutting across the path. (i.e. a mesh is a special type of loop)

  15. Mesh Analysis & Loop Analysis Fictitious mesh currents or loop currents can be assigned to solve cct problems. For the given cct on the right, there are 3 loops of which 2 are meshes. Loop (not mesh) Loop (not mesh)

  16. Mesh Analysis NOTE: Complex Cct – where you need more than one KVL equations. • variables are mesh currents • uses KVL and Ohm’s Law • directions of mesh currents are arbitrarily selected • every branch must be ‘covered’ by at least one mesh • No. of independent meshes (KVL equations) = No. of branches - No. of nodes + 1

  17. Ccts with Independent Voltage Sources Example 2-4 Write the mesh eqns for the given cct. Mesh 1: KVL • VAB + VBE + VEF + VFA = 0 • i1R1 + (i1 - i2)R3 + i1R2 - vs1 = 0 Mesh 2: KVL VBC + VCD + VDE + VEB = 0 vs2 + i2R4 + i2R5 + (i2 - i1) R3 = 0

  18. Example 2-5 Find I0 using mesh analysis. 2 window panes  2 meshes Mesh 1: KVL ----------------- (1)’

  19. Mesh 2: KVL --------------- (2)’ (1)’ + (2)’ gives 2I2 = 1m I2= 0.5mA • I1= 1.25mA • Io= I1 - I2 • = 0.75mA

  20. Example 2-5 Find V0 using mesh analysis. 2 window panes  2 meshes I1 I2 So far we have assigned all mesh currents in the clockwise direction. However, we can actually solve a problem by assigning the direction of mesh currents arbitrarily. Let’s assign I1 anti-clockwise and I2 clockwise. Mesh 1: KVL

  21. -- (1) I1 I2 Mesh 2: KVL ----------------- (2) ----------------- (2)’ (2) x 4 : (2)’- (1): 30I2 = 33m I2 =11/10 mA VO = I2 x6k = 33/5 V

  22. 1Ω 2 Ω 3Ω I1 5Ω 4Ω VX 3A 2V I2 I3 Ccts Containing Independent Current Sources Case 1: The current source is involved in only 1 mesh current fewer mesh eqns needed Example 2-7-1 Using mesh analysis, find Vx. 3 window panes  3 meshes The 3-A current source is involved in only 1 mesh current I3  obviously I3 = 3A. No need to write mesh eqn for mesh 3. Need only 2 mesh eqns to solve the given problem. NOTE: Fewer Mesh

  23. 1Ω 2 Ω 3Ω I1 5Ω 4Ω VX 3A 2V I2 I3 Mesh 2: KVL (I2 - I1)3 + I2 (4) -2 + (I2- I3 )5 = 0 • Mesh 1: KVL I1 (1) + (I1- I2 )3 + (I1 - I3 )2 = 0

  24. 1Ω 2 Ω 3Ω I1 5Ω 4Ω VX 3A 2V I2 I3 • Mesh 1: KVL I1 (1) + (I1- I2 )3 + (I1 - I3 )2 = 0 I1 (1 + 3 + 2) + I2(-3) + I3(-2) = 0 6I1 - 3I2 = 2 x 3 2I1 - I2 = 2 -------------- (1) Mesh 2: KVL (I2 - I1)3 + I2 (4) -2 + (I2- I3 )5 = 0 I1 (-3) + I2(3 + 4 + 5) + I3(-5) = 2 • -3I1 +12I2 = 17 ------------- (2) (1) and (2) gives I2 = 40/21A VX = I2(4) = 7.62V

  25. 1Ω 2Ω I1 3Ω 5Ω VX 4Ω 2V 3A I2 I3 Case 2: The current source is involved in more than 1 mesh current. voltage of the current source needs to be introduced as an extra variable in mesh eqns involving the current source Example 2-7-2 Using mesh analysis, find Vx. 3window panes  3 meshes Mesh 1: KVL I1 (1) + (I1- I2 )3 + (I1 - I3 )2 = 0 I1 (1 + 3 + 2) + I2(-3) + I3(-2) = 0 6I1 - 3I2 - 2I3 = 0 ----- (1)

  26. 1Ω 2Ω I1 3Ω 5Ω VX 4Ω 2V 3A I2 I3 Mesh 2: KVL • Mesh 3: KVL V Mesh 1: KVL (I2 -I1)3 + I2(4) + (-v) + (I2- I3 )5 = 0 I1 (1) + (I1- I2 )3 + (I1 - I3 )2 = 0 We have 3 eqns but 4 unknowns  need an additional eqn. I2 - I3 = 3 (I3 -I1)2 + (I3- I2 )5 + v - 2 = 0

  27. 1Ω 2Ω I1 3Ω 5Ω VX 4Ω 2V 3A I2 I3 (I2 -I1)3 + I2(4) + (-v) + (I2- I3 )5 = 0 -3I1 + 12I2 - 5I3 - v = 0 --- (2) Mesh 2: KVL • Mesh 3: KVL (I3 -I1)2 + (I3- I2 )5 + v - 2 = 0 - 2I1 - 5I2 + 7I3 + v – 2 = 0 ----------- (3) We have 3 eqns but 4 unknowns  need an additional eqn. I2 - I3 = 3 ----------------- (4) (2) + (3): - 5I1 + 7I2 + 2 I3 – 2 = 0 ----------- (5) Solving (1), (4) & (5) gives I2 = 18/29 A VX = I2 (4) = 2.48V

  28. Loop Analysis Chapter 2 : Nodal & Loop Analysis Techniques

  29. Loop Analysis • Similar to mesh analysis – all statements made for mesh analysis also hold true for loop analysis. • In addition, loop analysis • may lead to a simpler solution than mesh analysis • is applicable to both planar and non-planar ccts

  30. Example 2-8 Solve example 2-5 again using loop analysis with the loops chosen as indicated. Cct is planar: 2 window panes  2 loops Loop 1: KVL (I1 + I2)6k + I1 (6k) – 12 = 0 I1(12k) + I2 (6k) = 12 2 I1 + I2 = 2m ------------- (1)

  31. (I1 + I2)6k + I2 (3k) +3 – 12 = 0 I1(6k) + I2 (6k + 3k) = 9 2 I1 + 3I2 = 3m ------------- (2) Solving (1) & (2) gives I1 = 0.75m  IO = I1 = 0.75mA Loop 2: KVL

  32. Sometimes, a cct problem can be solved very easily with a judicious choice of loops. Example 2-9 Using loop analysis, find I0. Cct is planar: 3 window panes  3 loops In this example, we can show that if we choose the loops wisely such that only one loop passes through each independent current source, we will get a very simple solution.

  33. With the loops chosen as indicated, we need only 1 loop equation to solve the problem. Since the 2-mA source is only involved with I1 and the 4-mA source is only involved with I2 I1 = 2mA I2 = 4mA The only loop eqn needed is for loop 3: Loop 3: KVL • – 6 + I3(1k) + (I3 + I2 )2k + (I3 + I2 – I1)2k + (I3 – I1 )1k = 0 • – 6 + I3(1k + 2k + 2k + 1k) + 8 + 8 – 4 –2 = 0 • I3 = –4/(6k) = –2/3 mA • IO = I1 – I2 – I3 = 2m – 4m – (-2/3m) = –4/3 mA

  34. Recap/Summary Chapter 2: Nodal & Loop Analysis Techniques

  35. Nodal Analysis • variables are node voltages • write KCL eqns • need N-1 node eqns in general (can be fewer) • choose the reference node judiciously • node that has the most no. of branches connected • node at one end of an independent voltage source • Ccts containing: • only independent current sources • independent voltage sources • a node whose v can be readily obtained

  36. Mesh Analysis • variables are mesh currents • write KVL eqns • no. of meshes = B - N + 1 (or = no. of window panes) • Ccts containing: • independent voltage sources • independent current sources • current sources involves only 1 mesh i •  fewer mesh eqns needed • current source involves >1 mesh i •  v of current source is introduced as an extra variable

  37. Loop Analysis – similar to mesh analysis • variables are loop currents • write KVL eqns • need B – N + 1 loops • In addition it may give simpler solution, and can handle both planar & non-planar ccts. •  How to choose loops? Try to choose loops such that each current source is involved in only 1 loop current • Planar & non-planar ccts • planar: can be drawn on a plane with no branches crossing over each other. • Ccts containing: • independent voltage sources • independent current sources

  38. Extras Chapter 2: Nodal & Loop Analysis Techniques

  39. Extra Examples E2-1 Find Vo in the given circuit using nodal analysis. [Ans: 4.5V]

  40. E2.2 Using mesh analysis, find the voltages and currents of all elements in the given circuit. Indicate the V & I obtained clearly in the circuit diagram. [Ans: Vo=6.857V, I= 1.143mA, other v & I are: (6V, 2.429mA),(5.143V,1.286mA), (0.857V, 0.429mA), (0.857V, 2mA), (1.714V, 0.857mA)]

  41. 6 7 1 5 8 9 2 4 10 3

  42. E2-3 Using mesh analysis, find Vo in the given circuit. [Ans: 1.6V]

  43. C 1 3A IX I2 I1 2 1 A 2 F D I4 I3 2 VX 1V B 1 (ref) E E2-4 Consider the non-planar circuit given below: • (a) Using loop analysis, show that the branch current IX is given by IX = (43-2VX)/41 A. • (b) If the voltages at node B and node C are equal, determine the value of VX. Verify your answer using nodal analysis, with node E taken as the reference.

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