Chapter 9 Kirchhoffs Laws Topics Covered in Chapter

Chapter 9 Kirchhoff’s Laws Topics Covered in Chapter 9 9 -1: Kirchhoff’s Current Law (KCL) 9 -2: Kirchhoff’s Voltage Law (KVL) 9 -3: Method of Branch Currents 9 -4: Node-Voltage Analysis 9 -5: Method of Mesh Currents © 2007 The Mc. Graw-Hill Companies, Inc. All rights reserved.

9 -1: Kirchhoff’s Current Law (KCL) § The sum of currents entering any point in a circuit is equal to the sum of currents leaving that point. § Otherwise, charge would accumulate at the point, reducing or obstructing the conducting path. § Kirchhoff’s Current Law may also be stated as IIN = IOUT Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. Fig. 9 -1: Current IC out from point P equals 5 A + 3 A into P.

9 -2: Kirchhoff’s Current Law (KCL) Fig. 9 -2: Series-parallel circuit illustrating Kirchhoff’s laws. Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

9 -2: Kirchhoff’s Current Law (KCL) The 6 -A IT into point C divides into the 2 -A I 3 and 4 -A I 4 -5 is the current through R 4 and R 5 IT − I 3 − I 4 -5 = 0 6 A − 2 A − 4 A = 0 At either point C or point D, the sum of the 2 -A and the 4 -A branch currents must equal the 6 A line current. Therefore, Iin = Iout

9 -2: Kirchhoff’s Voltage Law (KVL) § Loop Equations § A loop is a closed path. § This approach uses the algebraic equations for the voltage around the loops of a circuit to determine the branch currents. § Use the IR drops and KVL to write the loop equations. § A loop equation specifies the voltages around the loop.

9 -2: Kirchhoff’s Voltage Law (KVL) § Loop Equations § ΣV = VT means the sum of the IR voltage drops must equal the applied voltage. This is another way of stating Kirchhoff’s Voltage Law.

9 -2: Kirchhoff’s Voltage Law (KVL) Fig. 9 -2: Series-parallel circuit illustrating Kirchhoff’s laws. Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

9 -2: Kirchhoff’s Voltage Law (KVL) In Figure 9 -2, for the inside loop with the source VT, going counterclockwise from point B, 90 V + 120 V + 30 V = 240 V If 240 V were on the left side of the equation, this term would have a negative sign. The loop equations show that KVL is a practical statement that the sum of the voltage drops must equal the applied voltage.

9 -2: Kirchhoff’s Voltage Law (KVL) § The algebraic sum of the voltage rises and IR voltage drops in any closed path must total zero. For the loop CEFDC without source the equation is −V 4 − V 5 + V 3 = 0 − 40 V − 80 V + 120 V = 0 0=0 Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. Fig. 9 -2: Series-parallel circuit illustrating Kirchhoff’s laws.

9 -3: Method of Branch Currents Fig. 9 -5: Application of Kirchhoff’s laws to a circuit with two sources in different branches. VR = I 1 R 1 1 VR = I 2 R 2 2 VR = I 3 R 3 VR = (I 1+I 2)R 3 3 3 Loop equations: V 1 – I 1 R 1 – (I 1+I 2) R 3 = 0 Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display. V 2 – I 2 R 2 – (I 1+I 2) R 3 = 0

9 -3: Method of Branch Currents Fig. 9 -5

9 -3: Method of Branch Currents

9 -3: Method of Branch Currents Also, in loop 2, 2 I − 3 I 2 − 6 (I 1 + I 2) = 0 Multiplying (I 1 + I 2) by 6 and combining terms and transposing, the two equations are 18 I 1 − 6 I 2 = − 84 − 6 I 1 − 9 I 2 = − 21 Divide the top equation by − 6 and the bottom by − 3 which results in simplest and positive terms 3 I 1 + I 2 = 14 2 I 1 + 3 I 2 = 7

9 -3: Method of Branch Currents Solving for currents Using the method of elimination, multiply the top equation by 3 to make the I 2 terms the same in both equations 9 I 1 + 3 I 2 = 42 1 I 1 + 3 I 2 = 7 Subtracting 7 I 1 = 35 I 1 = 5 A To determine I 2, substitute 5 for I 1 2(5) + 3 I 2 = 7 − 10 3 I 2 = − 3 I 2 = − 1 A

9 -3: Method of Branch Currents This solution of − 1 A for I 2 shows that the current through R 2 produced by V 1 is more than the current produced by V 2. The net result is 1 A through R 2 from C to E Calculating the Voltages VR 1 = I 1 R 1 = 5 x 12 = 60 V VR 2 = I 2 R 2 = 1 x 3 = 3 V VR 3 = I 3 R 3 = 4 x 6 = 24 V Note: VR 3 and VR 2 have opposing polarities in loop 2. This results in the − 21 V of V 2

9 -3: Method of Branch Currents Checking the Solution At point C: 5 A = 4 A + 1 A At point D: 4 A + 1 A = 5 A Around the loop with V 1 clockwise from B, 84 V − 60 V − 24 V = 0 Around the loop with V 2 counterclockwise from F, 21 V + 3 V − 24 V = 0 Fig. 9 -6: Solution of circuit 9 -5 with all currents and voltages. Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

9 -4: Node-Voltage Analysis § A principal node is a point where three or more currents divide or combine, other than ground. § The method of node voltage analysis uses algebraic equations for the node currents to determine each node voltage. § Use KCL to determine node currents § Use Ohm’s Law to calculate the voltages. § The number of current equations required to solve a circuit is one less than the number of principal nodes. § One node must be the reference point for specifying the voltage at any other node.

9 -4: Node-Voltage Analysis § Finding the voltage at a node presents an advantage: A node voltage must be common to two loops, so that voltage can be used for calculating all voltages in the loops.

9 -4: Node-Voltage Analysis Fig. 9 -7: Method of node-voltage analysis for the same circuit as in Fig. 9 -5. Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

9 -4: Node-Voltage Analysis

9 -4: Node-Voltage Analysis Fig. 9 -7 VR 1/R 1 + VR 2/R 2= VN/R 3 VR 1/12 + VR 2/3 = VN/6

9 -4: Node-Voltage Analysis VR 1+ VN = 84 or VR 1 = 84 − VN For the loop with V 2 of 21 V, VR 2 + VN = 21 or VR 2 = 21 − VN Substituting values I 1 + I 2 =I 3 Using the value of each V in terms of VN 84 − VN/12 + 21 − VN/3 = VN/6 Fig. 9 -7

9 -4: Node-Voltage Analysis Fig. 9 -7 This equation has only one unknown, VN. Clearing fractions by multiplying each term by 12, the equation is (84 − VN) + 4(21 − VN) = 2 VN 84 - VN + 84 − 4 VN = 2 VN − 7 VN = − 168 VN = 24 V

9 -4: Node-Voltage Analysis Calculating All Voltages and Currents Node Equations § Applies KCL to currents in and out of a node point. § Currents are specified as V/R so the equation of currents can be solved to find a node voltage. Loop Equations § Applies KVL to the voltages in a closed path. § Voltages are specified as IR so the equation of voltages can be solved to find a loop current.

9 -5: Method of Mesh Currents § A mesh is the simplest possible loop. § Mesh currents flow around each mesh without branching. § The difference between a mesh current and a branch current is that a mesh current does not divide at a branch point. § A mesh current is an assumed current; a branch current is the actual current. § IR drops and KVL are used for determining mesh currents.

9 -5: Method of Mesh Currents § The number of meshes is the number of mesh currents. This is also the number of equations required to solve the circuit. Fig. 9 -8: The same circuit as Fig. 9 -5 analyzed as two meshes. Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

9 -5: Method of Mesh Currents § A clockwise assumption is standard. Any drop in a mesh produced by its own mesh current is considered positive because it is added in the direction of the current. § Mesh A: 18 IA − 6 IB = 84 V § Mesh B: 6 IA + 9 IB = − 21 V

9 -5: Method of Mesh Currents § The mesh drops are written collectively here: Mesh A: 18 IA − 6 IB = 84 Mesh B: − 6 IA + 9 IB = − 21 Fig. 9 -8: The same circuit as Fig. 9 -5 analyzed as two meshes. Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

9 -5: Method of Mesh Currents Use either the rules for meshes with mesh currents or the rules for loops with branch currents, but do not mix the two methods. To eliminate IB and solve for IA, divide the first equation by 2 and the second by 3. then 9 IA − 3 IB = 42 − 2 IA + 3 IB = − 7 Add the equations, term by term, to eliminate IB. Then 7 IA = 35 IA = 5 A

9 -5: Method of Mesh Currents Fig. 9 -8: The same circuit as Fig. 9 -5 analyzed as two meshes. To calculate IB, substitute 5 for IA in the second equation: − 2(5) + 3 IB = − 7 + 10 =3 IB = 1 A The positive solutions mean that the electron flow for both IA and IB is actually clockwise, as assumed. Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.