ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Lecture 3 Circuit Laws, Voltage & Current Dividers
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
• Alternatively, the sum of the currents entering a node equals the sum of the currents leaving a node.
• The net current entering a node is zero.
KIRCHHOFF’S CURRENT LAW
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
The algebraic sum of the voltages equals zero for any closed path (loop) in an electrical circuit.
KIRCHHOFF’S VOLTAGE LAW
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Using KVL, KCL, and Ohm’s Law to Solve a Circuit
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ix = 2 A
ix + 0.5ix = i y
15 V iy = =3A 5Ω
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Vs = 35 V
Vs = v x + 15
v x = 10ix = 20 V
ix = 2 A
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
3. Solve circuits by the node-voltage technique.
2. Apply the voltage-division and currentdivision principles.
1. Solve circuits (i.e., find currents and voltages of interest) by combining resistances in series and parallel.
Chapter 2 Resistive Circuits
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
7. Draw the circuit diagram and state the principles of operation for the Wheatstone bridge.
6. Apply the superposition principle.
5. Find Thévenin and Norton equivalents and apply source transformations.
4. Solve circuits by the mesh-current technique.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
v = v1 + v2 + v3 = iR1 + iR2 + iR3 = i ( R1 + R2 + R3 ) = iReq
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
1 1 1 v v v v i = i1 + i2 + i3 = + + = v + + = R1 R2 R3 R1 R2 R3 Req
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
1 1 1 1 1 1 1 1 2 3 = + + = + + = + + =1 Req R2 R3 R4 6 3 2 6 6 6
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Rtotal = R1 + 1Ω = 3Ω
1Ω
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Req = 2.0Ω
1 1 1 2 1 3 1 = + = + = = Req 3 6 6 6 6 2
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Req = 8Ω + 2Ω = 10Ω
1.5Ω
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
1 1 1 2 = + = Ω → Req = 5Ω Req 10 10 10
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
1. Begin by locating a combination of resistances that are in series or parallel. Often the place to start is farthest from the source. 2. Redraw the circuit with the equivalent resistance for the combination found in step 1.
Circuit Analysis using Series/Parallel Equivalents
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
4. Solve for the currents and voltages in the final equivalent circuit.
3. Repeat steps 1 and 2 until the circuit is reduced as far as possible. Often (but not always) we end up with a single source and a single resistance.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
v1 = R1i1 = (10Ω )(3 A) = 30V
v2 60V i3 = = = 1A R3 60Ω
v2 60V i2 = = = 2A R2 30Ω
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Of the total voltage, the fraction that appears across a given resistance in a series circuit is the ratio of the given resistance to the total series resistance.
R2 v 2 = R2 i = v total R1 + R2 + R3
R1 v1 = R1i = v total R1 + R2 + R3
Voltage Division
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
1000 = ×15 1000 + 1000 + 2000 + 6000 = 1.5V
R1 vtotal v1 = R1 + R2 + R3 + R4
Application of the VoltageDivision Principle
R1 R2 RR → v = Req itotal = 1 2 itotal R1 + R2 R1 + R2
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Req =
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
For two resistances in parallel, the fraction of the total current flowing in a resistance is the ratio of the other resistance to the sum of the two resistances.
R1 v i2 = = itotal R2 R1 + R2
R2 v i1 = = itotal R1 R1 + R2
Current Division
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
R2 30 (1.25) = 0.417 A i3 = is = R2 + R3 30 + 60
vs 100 is = = = 1.25 A R1 + R x 60 + 20
Find i3
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Find i1
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
20 i1 = is = 15 = 10A R1 + Req 10 + 20
Req
R2 R3 30 × 60 Req = = = 20Ω R2 + R3 30 + 60
Application of the CurrentDivision Principle
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Use the voltage division principle to find the unknown voltages
R1 5 vs = (120V ) = 120V = 10V 5 + 10 + 15 + 30 12 R1 + R2 + R3 + R4
R3 15 vs = (120V ) = 120V = 30V R1 + R2 + R3 + R4 5 + 10 + 15 + 30 4
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
R4 30 120V (120V ) = v4 = vs = = 60V R1 + R2 + R3 + R4 5 + 10 + 15 + 30 2
v3 =
R2 10 120V (120V ) = v2 = vs = = 20V 5 + 10 + 15 + 30 6 R1 + R2 + R3 + R4
v1 =
1 1 + 7 5
= 2.92Ω
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
R4 4 ( 4)(20V ) v4 = vs = (20V ) = = 8.06V R1 + Req + R4 3 + 2.92 + 4 9.92
Req
2.92 ( 2.92 )(20V ) (20V ) = v2 = vs = = 5.89V R1 + Req + R4 3 + 2.92 + 4 9.92
R1 ( 3 3)(20V ) v1 = vs = (20V ) = = 6.05V R1 + Req + R4 3 + 2.92 + 4 9.92
Req =
1
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Use the current division principle to find the currents
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
R1 + R2 30 (3) = 2 A i3 = it = R1 + R2 + R3 10 + 20 + 15
R3 15 (3) = 1A it = i1 = R1 + R2 + R3 10 + 20 + 15
Req
5 i3 = is = (3) = 1A Req + R1 5 + 10
Req
5 (3) = 1A i2 = is = Req + R1 5 + 10
Req
5 (3) = 1A i1 = is = Req + R1 5 + 10
1 1 + 10 10
=5
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Req =
1
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Third Edition, by Allan R. Hambley, ©2005 Pearson Education, Inc.
Resistor Cube
