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CHAPTER 2: DC Circuit Analysis and AC Circuit Analysis Methods of Circuit Analysis and Circuit Theorems Introduction Motivation(1) & Motivation(2) Nodal Analysis (Node-Voltage Method) Mesh Analysis (Mesh-Current Method) Thevenin’s Theorem Motivation (1) If you are given the following circuit, how can we determine (1) the voltage across each resistor, (2) current through each resistor. (3) power generated by each current source, etc. What are the things which we need to know in order to determine the answers? Motivation (2) Things we need to know in solving any resistive circuit with current and voltage sources only: • Kirchhoff’s Current Laws (KCL) • Kirchhoff’s Voltage Laws (KVL) • Ohm’s Law How should we apply these laws to determine the answers? Introduction (Continued…) 1. 2. 3. 4. There are four ways of solving simultaneous equations: Cramer’s rule Calculator (real numbers only) Normal substitution and elimination (not more than two equations) Computer program packages: mathcad, maple, mathematica etc. Circuit Analysis Method NODAL ANALYSIS (Node-Voltage Method) Concept Developed based on the systematic approach of Kirchhoff’s current law (KCL) to find all circuit variables without having to sacrifice any of the elements. General procedure which is making use of node voltages in circuit analysis as key solutions. Assumptions KCL is performed with current going out from a node as positive (+ve) while current entering a node as negative (ve). in – negative (subtract) out – positive (add) All unknown currents assumed to be leaving a particular node. Step to determine Node Voltages: 1. 2. 3. Mark all essential nodes and assign proper voltage designations except for the appointed reference node. Apply KCL to each non-reference nodes. Use Ohm’s law to formulate the equation in terms of node voltages. Assume all unknown currents are directing out of the nodes. Solve the resulting simultaneous equations to obtain the unknown node voltages. Applying Nodal Analysis on Simple Circuit Example 1 (3 unknowns) Is2 R1 Is1 R2 R3 R4 Solution: Step 1: Mark all essential nodes V1 Assign unknown I node voltages Indicate the reference node. s1 Is2 R1 V2 R3 R2 V3 R4 Solution (continued…) Step 2: Perform KCL at each marked non-reference nodes using Ohm’s law to formulate the equations in terms of the node voltages. Solution (continued…) KCL V1: or I s1 I s 2 V1 V2 R1 (1) I s 1 I s 2 G1 (V1 V2 ) KCL V2: V2 0 V2 V3 V2 V1 0 (2) R1 R3 R2 KCL V3: I s2 V3 V2 V 0 3 0 R2 R4 (3) Solution (continued…) Step 3: Solve the resulting simultaneous equations from step 2 above. 3 mA 10 k V1 2 mA V2 5 k 4 k V3 2 k Solution (continued…) KCL V1: 2m 3m Simplify to KCL V2: Simplify to KCL V3: Simplify to V1 V2 10k V1 - V2 = 50 (1) V2 V1 V2 0 V2 V3 0 10k 4k 5k -2V1 + 11V2 - 4V3 = 0 V3 V2 V3 0 3m 0 5k 2k -2V2 + 7V3 = -30 (2) (3) Solution (continued…) Cramer’s rule: Put the equations in matrix forms. Col-1 a1 b1 c1 V1 d1 a b c V d 2 2 2 2 2 a3 b3 c3 V3 d3 = Col-2 Col-3 1 1 0 V1 50 2 11 4 V 0 2 0 2 7 V3 30 Left Matrix: Col-1: coefficients of V1 i.e. a1, a2 and a3 Col-2: coefficients of V2 i.e. b1, b2 and b3 Col-3: coefficients of V3 i.e. c1, c2 and c3 Middle Matrix: Right Matrix: Unknown variables i.e. V1, V2 and V3 Constants i.e. d1, d2 and d3 Solution (continued…) For third-order determinants, we use shorthand methods of expansion solution. Shorthand method consists of repeating the first two columns of the determinant to the right of the determinant and then summing the products along the specific diagonals as shown below. Solution (continued…) Use determinants to solve for each variable: 1 1 0 1 c2 2 11 4 -2 0 2 7 0 c3 a1 b1 c1 a 2 b2 a3 b3 -1 11 -2 (1)(11)(7) (1)( 4)(0) (0)( 2)( 2) (0)(11)(0) (2)(4)(1) (7)(2)(1)} 77 (22) 55 Solution (continued…) Determinant 1 when coefficients for V1 is replaced by the constants. 50 1 0 50 -1 c2 0 11 4 0 11 -30 -2 c3 30 2 7 d1 b1 c1 1 d 2 b 2 d 3 b3 1 (50)(11)(7) (1)( 4)( 30) (0)(0)( 2) (30)(11)(0) (2)(4)(50) (7)(0)(1)} 3730 (400) 3330 Solution (continued…) Determinant 2 when coefficients for V2 is replaced by the constants. a1 d1 c1 2 a2 d 2 a3 d 3 1 50 0 c2 2 0 4 0 30 7 c3 1 50 -2 0 0 -30 2 (1)(0)(7) (50)( 4)(0) (0)( 2)( 30) (0)(0)(0) (30)(4)(1) (7)(2)(50)} (580) 580 Solution (continued…) Determinant 3 when coefficients for V3 is replaced by the constants. 1 1 50 d 2 2 11 0 0 2 30 d3 a1 b1 d1 3 a2 b2 a3 b3 1 -2 0 -1 11 -2 3 (1)(11)(30) (1)(0)(0) (50)(2)(2) (0)(11)(50) (2)(0)(1) (30)(2)(1)} 130 (60) 70 Solution (continued…) V1 = 1/ = 3330/55 = 60.55 V V2 = 2/ = 580/55 = 10.55 V V3 = 3/ = -70/55 = -1.27 V You should verify your answers using calculator for three unknowns. Solution (continued…) Knowing all the node voltages, we can obtain the element voltages, element currents and even power dissipated by the passive elements such as: VR1= V1 – V2 IR1 = (V1 – V2)/R1 PR1 = IR12R1 = VR12/R1 VR2= V2 – V3 IR1 = (V2– V3)/R2 PR2 = IR22R2 = VR22/R2 **VR3= V2 IR1 = V2/R3 PR3 = IR32R3 = VR32/R3 **VR4= V3 IR4 = V32/R4 PR4 = IR42R3 = VR42/R4 **In these two cases, the element voltages identical to node voltages because one of its terminals is at reference node. Can you find the power dissipated by the 10 k resistor? Need to find the element voltage of 10-k resistor because power calculation formula uses element voltage. P10k = (V1 – V2)2/10k = (60.55 –10.55)2/10k = 502/10k = 0.25 W Applying Nodal Analysis on Circuit with Voltage Sources Three different effects depending on placement of voltage source in the circuit. Does the presence of a voltage source complicate or simplify the analysis? Case 1: Voltage source between two non-reference essential nodes. R E P U S Nonreference essential node V1 Supernode Equation: E D NO Vs V2 Nonreference essential node Vs V1 V2 Case 2: Voltage source between a reference essential node and a non-reference essential node. E D NO N GE W O LTA N K VO Nonreference essential node V1 Known node voltage: Vs 0V Reference essential node V1 Vs Case 3: Voltage source between an essential node and a nonessential node. Vs Nonreference essential node V1 FIN NO D N N- OD ES E TE SE VO RM NT LT S O IAL AG VO N E LT F NO OD AT Va AG D E IN E E Non-essential node Node voltage at non-essential node: Va V1 Vs Solution Case 2 (Known node voltage) Step 1: Mark essential nodes and assign unknown node voltages and indicate the reference node. 10 V i 5 8 50 30 3A V2 KNOWN NODE VOLTAGE V1 40 Checklist: 3 essential nodes – 1 ref node – 1 known node voltage = 1 KCL Eqn. Solution (Continued…) Step 2: Perform KCL at each marked nonreference nodes using Ohm’s law to formulate the equations in terms of node voltages. Immediately known node voltage at V1: V1 10 V V2 10 V2 KCL V2: 3 80 8 11V2 230 (1) Solution (Continued…) Step 3: Solve the resulting simultaneous equations which have been simplified in step 2 above. Solving Eqn. (1) yields, 230 V2 20.91 V 11 Finding current through the voltage source, KCL at V1: V1 V1 V2 i0 40 80 10 10 20.91 i 0.636 A 40 80 Solution (Continued…) Hence, P10-V = Vi= (10)(-0.636) = -6.36 W. (Delivering energy) Circuit Analysis Method MESH ANALYSIS (Mesh-Current Method) Concept Similar to nodal analysis. Developed based on the systematic approach of Kirchhoff’s voltage law (KVL) to find all circuit variables without having to sacrifice any of the elements. General procedure which is making use of mesh current in circuit analysis as key solutions. Importance terms Mesh Current: Assigned unknown current flows around the perimeter of the particular mesh/loop. Element Current: Actual current through any element or branch in the circuit. Assumptions KVL is performed in clockwise direction. Voltage rise – negative (subtract) Voltage drop – positive (add) Mesh Analysis Procedures: 1. 2. 3. Label all independent meshes and assign proper unknown mesh currents in clockwise direction. Do the checklist. Formulate KVL/Supermesh/Constraint Equation. Solve the resulting simultaneous equations to obtain the unknown mesh current. Applying Mesh Analysis on Simple Circuit Example (2 unknowns) Question: Find power dissipated in 12resistor and 3-resistor using mesh analysis. 2 9 12 12 V 4 3 8V Solution: 1. Label all independent meshes and assign proper unknown mesh currents in clockwise direction. 2 12 V I1 12 4 9 I2 3 Checklist: 2 meshes = 2 KVL 8 V Eqns. Solution (Continued…) 2. Formulate KVL/Supermesh/Constraint Eq. KVL I1: 18I1 – 12I2 = 12 KVL I2: -12I1 + 24I2 = -8 (1) (2) Solution (Continued…) 3. Solve the resulting simultaneous equations to obtain the unknown mesh current. I1 = 1/ 12 12 1 I2 = 2/ 8 24 18 12 12 24 (18)( 24) (12)( 12) 288 (12)( 24) (12)( 8) 192 18 12 2 12 8 (18)( 8) (12)(12) 0 Solution (Continued…) Using calculator/Cramer’s rule we obtain: I1 = 0.667 A and I2 = 0 A P12 = (I1 -I2)2(12) = 5.33 W P3 = I22(3) = 0 W Notice that the branch (3-resistor) forming the outer most boundary of the circuit will have mesh current = element current. Case 1: Current source located at the outer most boundary Connecting mesh current immediately known. No need to apply KVL around that loop/mesh. Mesh Current = Element Current = Current Source Value R1 R2 Is I2 I1 Immediately known mesh current, R3 12 V I3 I3 = -Is R4 Case 2: Current source located at the boundary between 2 meshes Enclose the current source and combine the two meshes to form a SUPERMESH. KVL is performed around the supermesh – do not consider voltage across current source. Formulate supermesh equation – express the relationship between mesh currents that form the supermesh and current source that it encloses. TIPS: Nodal versus Mesh Analysis To select the method that results in the smaller number of equations. For example: 1. Choose nodal analysis for circuit with fewer nodes than meshes. *Choose mesh analysis for circuit with fewer meshes than nodes. *Networks that contain many series connected elements, voltage sources, or supermeshes are more suitable for mesh analysis. *Networks with parallel-connected elements, current sources, or supernodes are more suitable for nodal analysis. 2. If node voltages are required, it may be expedient to apply nodal analysis. If branch or mesh currents are required, it may be better to use mesh analysis. Circuit Theorem SOURCE TRANSFORMATION What benefits from source transformation? Another tool to simplify circuit – the simpler the circuit, the easier will be the solution. How to simplify? – rearrange the resistors/sources by Source Transformation so that they end up with series/parallel connections. Definition An equivalent circuit is one whose v-i characteristics are identical with the original circuit. A Source Transformation is the process of replacing a voltage source Vs in series with resistor Rs by a current source is in parallel with the same resistor Rs or vice versa. Equivalent Circuits The connections of each case should be between the same terminals before and after transformation. Vs Is Rs (a) Independent source transform Vs I s Rs (b) Dependent source transform • The arrow of the current source is directed toward the positive terminal of the voltage source. • The source transformation is not possible when R = 0 for voltage source and R = ∞ for current source. Circuit Theorem THEVENIN’S THEOREM Purpose Used when we are interested ONLY in the terminal behavior of the circuit particularly where a variable load is connected to. Provides a technique to replaced the fixed part of the circuit by a simple equivalent circuit. Avoid the re-do on the analysis of the entire circuit except for the changed load. Thevenin’s Theorem It states that a linear two-terminal circuit (Fig. a) can be replaced by an equivalent circuit (Fig. b) consisting of a voltage source VTH in series with a resistor RTH, where • VTH is the open-circuit voltage at the terminals. • RTH is the input or equivalent resistance at the terminals when the independent sources are turned off. Procedures to obtain VTh and RTh Step 1: Preliminary – Omitting load resistor RL (Not applicable if no load resistor) Step 2: Find RTh – setting all independent sources to zero. Find the resultant resistance between the marked terminals. Voltage source – short circuit (s.c.) Current sorce – open circuit (o.c.) Step 3: Find VTh – calculate VTh by returning all sources back to their original positions. Find the open circuit voltage between the marked terminals using the method which takes least effort. Exercise 1: Find the Thevenin equivalent between terminal a-b. (Ans: VTh=32V, RTh=8) 5 4 a 25 V 20 3A b Exercise 2: Find the Thevenin equivalent circuit at the terminal a-b of the circuit below. (Ans: VTh=- 4.8V, RTh=2.4) 4 a 6 3 b 8V 2