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Chapter 1 techniques of dc circuit analysis

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Chapter 1 techniques of dc circuit analysis

  1. 1. Menuntut ilmu adalah TAQWA Menyampaikan ilmu adalah IBADAH Mengulang-ulang ilmu adalah ZIKIR Mencari ilmu adalah JIHADChapter 1 : Techniques of DC Circuit Analysis Engr. Mohd Riduwan bin Ghazali (Grad. IEM)
  2. 2. 1.1 Review Circuit Analysis I: Most Important thing that should remember: Ohm Laws,V = IR (Kirchoff’s) KCL/KVL Independent/Dependent Source Nodal/SuperNode Mesh/SuperMesh Source Transformation
  3. 3. OHM’S LAWOhm’s Law state that the voltage, v across a resistor is directlyproportional to the current, i flowing through the resistor. V=iRCurrent and voltage are linearly proportional
  4. 4. KIRCHHOFF’S LAW Kirchhoff current law (KCL) • States that the sum of currents entering a node (or closed boundary) is zero. i1 + (−i2 ) + i3 + i4 + ( −i5 ) = 0 i1 + i3 + i4 = i2 + i5 Total current in = Total current out
  5. 5. KIRCHHOFF’S LAW Kirchhoff voltage law (KVL) • States that the sum of voltages around a closed path (or loop) is zero −v1 + v2 + v3 − v4 + v5 = 0 v2 + v3 + v5 = v1 + v4
  6. 6. CIRCUIT ELEMENTS Active Elements Passive Elements • voltage source comes with polarities (+-) in its symbol • current source comes with an arrow Independent Dependant Sources Sources (round shape) (diamond shape)
  7. 7. NODAL ANALYSISProvides a general procedure for analyzing circuits using node voltagesas the circuit variables.• Steps: i) select a node as a reference node (ground) ii) assign voltage designations to nonreference nodes iii) redraw the circuit to avoid too much information on the same circuit iv) apply KCL for each nonreference node - at node 1: I1 = I2 + i1 + i2 -at node 2: I2 = i3 – i2 v) apply Ohm’s law (current flows from a higher potential to a lower potential) vhigher − vlower v1 − 0 v1 − v2 v2 − 0 i= i1 = i2 = i3 = R R1 R2 R3
  8. 8. Nodal Analysis With Voltage SourcesWe now consider how voltage sources affect nodal analysis:CASE 1: If a voltage source is connected between the reference node, wesimply set the voltage at the nonreference node equal to the voltage of thevoltage source E.g.: v1 = 10VCASE 2: If the voltage source (dependent or independent) is connectedbetween two nonreference nodes, the two nonreference nodes form ageneralized node or supernode; apply both KCL and KVL to determinethe node voltagesE.g.: nodes 2 and 3 form a supernode
  9. 9. MESH ANALYSISSteps 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 thevoltages in terms of the mesh currents• solve the resulting simultaneous equations to get the meshcurrents A p p l y i n g K V L t o m e s h 1, − V 1 + R 1 i1 + R 3 ( i1 − i 2 ) = 0 ( R 1 + R 3 ) i1 − R 3 i 2 = V 1 A p p ly in g K V L to m e s h 2 , R 2 i 2 + V 2 + R 3 ( i 2 − i1 ) = 0 − R 3 i1 + ( R 2 + R 3 ) i 2 = − V 2 R1+ R3 − R3  i1   V1   − R3 R2 + R3 i  = −V    2  2
  10. 10. MESH ANALYSIS WITH CURRENT SOURCESCASE 1 When a current source exists only in one mesh, we set i2 = -5A and write a mesh equation for the other mesh in the usual way: -10 + 4i1 + 6(i1 – i2) = 0 i1 = -2A
  11. 11. MESH ANALYSIS WITH CURRENT SOURCESCASE 2 When a current source exists between two meshes, we create a supermesh by excluding the current source and any elements connected in series with it Applying KVL in (b), -20 + 6i1 + 10i2 + 4i2 = 0 6i1 + 14i2 = 20 ………..(i) Applying KCL to a node where the two meshes intersect, i2 = i1 + 6 ………… (ii) Solve (i) and (ii), i1 = -3.2 A, i2 = 2.8A
  12. 12. SOURCE TRANSFORMATION• series parallel combination and wye-delta transformation help simplifycircuits• source transformation is another tool for simplifying circuits• is the process of replacing a voltage source, Vs in series with a resistor by acurrent source in parallel with a resistor, or vice versa• basic to these tools is the concept of equivalence v s = is R vs is = R
  13. 13. Superposition theorem For circuit network have more than one independent source voltage or current produced by a source acting in isolation can be determined by assuming other sources do not work, where the resources should be switched off in the following manner: - Independent voltage source – short circuit (0 V) or internal resistance if have Independent current source – open circuit (0 A)
  14. 14. Superposition theorem cont. Example for superposition theorem. IBSolution.1. Current source IB work ( voltage source VB off- short circuit) I’ Get the value of I’
  15. 15. Superposition theorem cont.2. Voltage source VB work (current source IB off – open circuit) I” Get the value of I”3. So get the value of current flowing at resistance R2 with I = I’ +I”Record:- Various methods can be used to obtain the value of I and I “, such as current divider or mesh analysis or nodal analysis or node.
  16. 16. Superposition theorem cont. Example 1.10 For the circuit in the figure below, find the value of I, the voltage across the resistor 2Ω and the power absorbed by the resistor 0.5 Ω 0.5 Ω I 5V 10 V 2Ω V2Ω 0.5 Ω 0.5 Ω
  17. 17. Superposition theorem cont.Solution1. Voltage source 5V work ( voltage source 10 V off – short circuit) Find the value of I’ RT = 1 + ( 2 //1) = 1.67Ω  5  IT =   = 3A  1.67  1 I= xIT = 1A 2 +1
  18. 18. Superposition theorem cont.2. Voltage source 10 V work ( voltage source 5 V off – short circuit)Find the value of I” RT = 1 + ( 2 //1) = 1.67Ω  10  IT =   = 6A  1.67  1 I"= xIT = 2 A 2 +13. So the value of current I (make sure the direction of I’ and I”) I = I + (− I ") = −1A V2 Ω = IR = (−1)(2) = −2V P2 Ω = I 2 ( R) = (−1) 2 (2) = 2W
  19. 19. Superposition theorem cont. Example 1.11Refer to the figure below, calculate current flowing and voltage across resistance 4Ω. 1Ω 8Ω I 4Ω 2Ω 5A 1Ω
  20. 20. Superposition theorem cont. Example 1.12 Refer to the figure below, determine voltage across resistance 4 Ω. 5Ω 2Ω 4Ω - 3Ω Vx 3A + 4 Vx 1Ω
  21. 21. Thevenin theorem This theorem is in use to facilitate a complex circuit network to a simple circuit called the Thevenin equivalent circuit. The equivalent circuit contains a voltage source Vth in series with a resistor Rth a Complex Circuit b
  22. 22. Thevenin theorem cont. I=0 A a a Complex OFF VTH RTH Circuit b Circuit b The steps to get the Thevenin equivalent circuit:- a. Remove section of the network where to find the thevenin equivalent circuit and mark clearly the two terminals as a-b b. Determine the Thevenin equivalent resistance seen from the terminal a-b with independent sources is turn off c. Get the values of Thevenin voltage on the voltage across the terminal a-b when the terminal at open circuit. (various method can be used to obtain Vth, whether to used loop analysis/nod analysis) d. Draw the Thevenin equivalent circuit and connect the back portion removed from the (a) above
  23. 23. Thevenin theorem cont. Example 1.13For circuit below, sketch the Thevenin equivalent circuit at terminals a-b. 6Ω 2Ω a 20 V 5A RL 4Ω b
  24. 24. Thevenin theorem cont. Solutiona. Remove RL from circuitb. Determine RTH seen from terminal a-b with all independent sources are turn off.
  25. 25. Thevenin theorem cont. I=0 Ac. Get VTH at terminal a-b VX − 20 =5 VTH 10 VX = 70V VTH = V5 A = VX = 70 d. Draw the Thevenin equivalent circuit and connect the back portion removed from the (a) above RTH=12Ω a VTH=70 V RL b
  26. 26. Thevenin theorem cont. Example 1.14 Refer to the circuit below, sketch Thevenin equivalent circuit at terminal a-b, next calculate the current flowing, I3Ω and voltage across,V3Ω the resistor 3Ω, 5Ω 1Ω a I3Ω 4Ω V3Ω 3Ω 28 V b
  27. 27. Thevenin theorem cont. Example 1.16 Refer to the circuit below, get the value of V1/3Ω ½Ω 3V ½Ω 2A ¼Ω 1/3 Ω V1/3Ω
  28. 28. Norton theorem This theorem is in use to facilitate a complex circuit network to a simple circuit called the Norton equivalent circuit. This equivalent circuit consists of a current source IN connected in parallel with a resistor RN. a Complex Circuit b
  29. 29. Norton theorem cont. a a Complex OFF IN RN Circuit Circuit b b The steps to get the Norton equivalent circuit:- a. Remove section of the network where to find the Norton equivalent circuit and mark clearly the two terminals as a-b b. Determine the Norton equivalent resistance seen from the terminal a-b with independent sources is turn off c. Get the Norton current value of current flowing through the terminals a-b when a short circuit in the terminal. (various method can be used to obtain IN, whether to used loop analysis/nod analysis) d. Draw the Norton equivalent circuit and connect the back portion removed from the (a) above
  30. 30. Norton theorem cont. Examples Determine Norton equivalent circuit at terminals a-b for circuit below. Next calculate current flowing and voltage across resistance 3Ω 5Ω 1Ω a I3Ω 4Ω V3Ω 3Ω 28 V b
  31. 31. Norton theorem cont. Solution.a. Remove RL from circuitb. Determine RN seen from terminal a-b with all independent sources are turn off. RN=(1+(5//4))=3.22 Ω
  32. 32. Norton theorem cont. ITc. Get IN at terminal a-b (short a-b) IN 4 28 4 I N = IT   =   = 3.86 A  5  (5 + (4 //1))  5  d. Draw the Norton equivalent circuit and connect the back portion removed from the (a) above Current flowing 3Ω, 3.22 I 3Ω = ( 3.86 ) = 2 A (3 + (3.22)) IN=3.86 A RN=3.22Ω 3Ω Voltage across 3Ω V3Ω = I 3Ω ( 3) = 6V
  33. 33. Norton theorem cont. Example 1.19 Get the value of V1/3Ω ½Ω 3V ½Ω 2A ¼Ω 1/3 Ω V1/3Ω
  34. 34. Thevenin and Norton theorem with dependent sources To analyze circuits with independent sources, (IN) and (VTH) may be obtained by using the analysis as before. However, the Thevenin and Norton resistance can not be obtained directly from the network because of dependent sources can not be turned off as an independent source. therefore, to solve the circuit dependent sources, two ways: 1. Determine the value of VTH and IN, so VTH RN = RTH = IN
  35. 35. Thevenin and Norton theorem with dependentsources CONT.2. Introduce an independent voltage source,VT or an independent current source, IT at the root a-b. VT and IT value is any value. However, free resources available on the network must be turned off prior circuit. I a a Off VT Off Vab IT Circuit b Circuit b VT Vab RTH = RN = Ω RTH = RN = Ω I IT
  36. 36. Thevenin and Norton theorem with dependentsources CONT. Example 1.20Sketch the Thevenin equivalent circuit at terminal a-b, next calculate value Iab
  37. 37. Thevenin and Norton theorem with dependentsources CONT. Solution.a) Remove resistance 3 Ω from circuit -+b) Get the value VTH Write equation every loop
  38. 38. Thevenin and Norton theorem withdependent sources CONT. Solve the equation above to get value I2, next find value VTHc) Get resistance value of Thevenin equivalent, RTH RTH can be solve in two way. i) Get value IN -+
  39. 39. Thevenin and Norton theorem withdependent sources CONT. ii) Introduce an independent source. 1. introduce independent voltage source -+
  40. 40. Thevenin and Norton theorem withdependent sources CONT. II. introduce independent current source
  41. 41. Thevenin and Norton theorem with dependent sources CONT. So, sketch the Thevenin equivalent circuitVTH=48.2 V
  42. 42. Relationship between the Thevenin and Norton theorem Thevenin equivalent circuit can be converted into the Norton equivalent circuit or vice versa by changing the concept (super transformation)of supply where: - a) Thevenin resistance (RTH) value is equal to the Norton resistance (RN)
  43. 43. Maximum power transfer a circuit will supply maximum power to the load if the load resistance RL is equal to the equivalent resistance seen by the load Maximum power transfer can be obtained by replace a complex circuit with the Thevenin equivalent circuit or Norton equivalent circuit
  44. 44. Maximum power transfer cont. 2 Power to the load RL,  VTH  PRL = I RL =  2  RL  RTH + RL  Condition maximum power transfer RL = RTH Therefore, maximum power supplied to the load is: VTH 2 VTH 2 PRLmak = = 4 RL 4 RTH
  45. 45. Maximum power transfer cont. 2 V Th R L = R Th p max = 4 R Th
  46. 46. Maximum power transfer cont. ExamplesFrom circuit below, calculate:-a) Value RL when maximum output powerb) Maximum power absorb by load RL
  47. 47. Maximum power transfer cont. Solution.a) Remove RL from circuit and off all source from circuit to find RTHb) Get the thevenin voltage at terminal a-b when RL removed from circuit
  48. 48. Maximum power transfer cont. So, draw Thevenin equivalent circuit when maximum power transfer happen.
  49. 49. Maximum power transfer cont. Examples 1.22From circuit below, calculate:-a) Value RL when maximum output powerb) Maximum power absorb by load RL

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