Basic Electronics Ninth Edition Basic Electronics Ninth Edition ©2002 The McGraw-Hill Companies Grob Schultz

Basic Electronics Ninth Edition Basic Electronics Ninth Edition ©2003 The McGraw-Hill Companies 9 CHAPTER Kirchhoff’s Laws

Topics Covered in Chapter 9  Kirchhoff’s Current Law  Kirchhoff’s Voltage Law  Branch Current Analysis  Node Voltage Analysis  Mesh Current Analysis

Kirchhoff’s Current Law The sum of currents entering any point in a circuit is equal to the sum of currents leaving that point. I1I1 I3I3 I2I2 Example: I 1 + I 3 = I 2 or I 1 - I 2 + I 3 = 0

Kirchhoff’s Voltage Law The algebraic sum of the voltage rises and IR voltage drops in any closed path must total zero. V R1 V R2 VTVT Example: V T - V R1 - V R2 = 0 or V T = V R1 + V R2

Branch Currents 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.

Branch Current Method R1R1 R2R2 R3R3 V1V1 V2V2 I1I1 I2I2 I3I3 V R 1 = I 1 R 1 V R 2 = I 2 R 2 V R 3 = (I 1 +I 2 )R 3 Loop equations: V 1 – I 1 R 1 – (I 1 +I 2 ) R 3 = 0 V R 3 = I 3 R 3 V 2 – I 2 R 2 – (I 1 +I 2 ) R 3 = 0 also

Branch Current Method Applied R1R1 R2R2 R3R3 V1V1 V2V2 I1I1 I2I2 I3I3 V 1 – I 1 R 1 – (I 1 +I 2 ) R 3 = 0V 2 – I 2 R 2 – (I 1 +I 2 ) R 3 =  20  10  15 V13 V 15 – 100I 1 – 10(I 1 +I 2 ) = 0 13 – 20I 2 – 10(I 1 +I 2 ) = 0 15 – 100I 1 – 10I 1 – 10I 2 = 0 13 – 20I 2 – 10I 1 – 10I 2 = 0 15 – 110I 1 – 10I 2 = 0 13 – 10I 1 – 30I 2 = 0

Branch Current Method Applied R1R1 R2R2 R3R3 V1V1 V2V2 I1I1 I2I2 I3I3 100  20  10  15 V13 V 15 – 110I 1 – 10I 2 = 0 13 – 10I 1 – 30I 2 = I I 2 = I 1 = 0 320I 1 = 32 I 1 = 0.1 A 13 – 10I 1 – 30I 2 = 0 } Add the equations -3[ ]

Branch Current Method Applied R1R1 R2R2 R3R3 V1V1 V2V2 I1I1 I2I2 I3I3 100  20  10  15 V13 V 15 – 110I 1 – 10I 2 = 0 13 – 10I 1 – 30I 2 = 0I 1 = 0.1 A 12 – 30I 2 = 0 I 2 = 0.4 A I 3 = = 0.5 A 13 – 1 – 30I 2 = 0 Substitute:

0.5 A Branch Current Method Applied 100  20  10  15 V13 V 0.1 A0.4 A – + – –– + ++ KVL check (outside loop): 15 V – 13 V + 8 V – 10 V = 0 (The 13 V battery is a drop and V 100  is a rise.) – + KVL check (loop 1): 15 V – 5 V – 10 V = 0 1 KVL check (loop 2): 13 V – 5 V – 8 V = 0 2

Node Voltage Analysis A principal node is a point where 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

Node Voltage Method R1R1 R2R2 R3R3 V1V1 V2V2 I1I1 I2I2 I3I3 N At node N:I 1 + I 2 = I 3 or VR1VR1 R1R1 VR2VR2 R2R2 VNVN R3R3 + =

Node Voltage Method Applied R1R1 R2R2 R3R3 V1V1 V2V2 N VR1VR1 R1R1 VR2VR2 R2R2 VNVN R3R3 + = 100  20  15 V13 V VR1VR1 100 VR2VR2 20 VNVN 10 + = 10 

Node Voltage Method Applied R1R1 R2R2 R3R3 V1V1 V2V2 N 100  20  10  15 V13 V VR1VR1 100 VR2VR2 20 VNVN 10 + = V R 1 + V N = 15 or V R 1 = 15 – V N V R 2 + V N = 13 or V R 2 = 13 – V N 15 – V N VNVN 10 = 13 – V N 20 Substitute:

Node Voltage Method Applied R3R3 V1V1 V2V2 N 100  20  10  15 V13 V 15 – V N VNVN 10 = 13 – V N 20 V N = 5 This agrees with the solution found using the branch method. 5 V 8 V10 V

Mesh Current Analysis A mesh is the simplest possible loop. Mesh currents flow around each mesh without branching. IR drops and KVL are used for determining mesh currents.

Mesh Current Method R1R1 R2R2 R3R3 V1V1 V2V2 BA Mesh A: R 1 I A + R 3 I A – R 3 I B = – V 1 Mesh B: R 2 I B + R 3 I B – R 3 I A = V 2 A clockwise assumption is standard. Any drop in a mesh produced by its own mesh current is considered positive.

Mesh Current Method Applied R1R1 R2R2 R3R3 V1V1 V2V2 BA 100  20  10  15 V13 V Mesh A: 110I A – 10I B = – 15 Mesh B: – 10I A + 30I B = 13 The mesh drops are written collectively here:

Mesh Current Method Applied R1R1 R2R2 R3R3 V1V1 V2V2 BA 100  20  10  15 V13 V Multiply equation A by 3 to allow cancellation of an unknown. 330I A – 30I B = – 45 Add the equations. – 10I A + 30I B = I A = – 32 I A = – 0.1 A Mesh A : 110I A – 10I B = – 15Mesh B : – 10I A + 30I B = 13

Mesh Current Method Applied R1R1 R2R2 R3R3 V1V1 V2V2 BA 100  20  10  15 V13 V 110(– 0.1) – 10I B = – 15 I A = – 0.1 A Substitute. I B = 0.4 A The assumed direction of I A was wrong because it solved as negative. The magnitude is correct. Reverse the flow arrow in mesh A. Both mesh currents flow up through R 3. Thus, they add: I R 3 = = 0.5 A 0.5 A Mesh A : 110I A – 10I B = – 15Mesh B : – 10I A + 30I B = 13