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Chapter 5 – Parallel Circuits More Than one Current Path Node Branch Node IT = I1 + I2 + I3 + . . . . Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Voltages across parallel branches are the same. Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. The same voltage appears across each resistor in parallel. Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. FIGURE 5-11 V1 = V2 = VS Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Kirchhoff’s current law: The current into a node equals the current out of that node Current Dividers – Node A IT in = Node A Out (I1 + I2 + I3) IT = I1 + I2 + I3 Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. FIGURE 5-13 Generalized circuit node illustrates Kirchhoff’s current law. Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Examples of circuits with two parallel paths. Sometimes it’s not too Obvious what’s in Parallel with What (Need to Redraw Circuit) Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Total Parallel Resistance • When resistors are connected in parallel, the total resistance of the circuit decreases • The total resistance of a parallel circuit is always less than the value of the smallest resistor Connecting resistors in parallel reduces total resistance and increases total current. 1/R = Conductance More Parallel Resistors/Circuits Increases Conductance And Therefore Total Current Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Formula for Total Parallel Resistance 1/R = Conductance Two Resistors in Parallel • The total resistance for two resistors in parallel is equal to the product of the two resistors divided by the sum of the two resistors RT = R1R2/(R1 + R2) Notation for Parallel Resistors • To indicate 5 resistors, all in parallel, we would write: R1||R2||R3||R4||R5 Total current divides between the two branches. Current Dividers Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e IT = I1 + I2 Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. The branch with the lowest resistance has the most current, and the branch with the highest resistance has the least current Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Ohms Law Entire Circuit • IT = VS/RT • VS = RT IT • RT = VS/IT Individual Parallel Resistors • IRX= VRX/RX • VRX = RX IRX Circuit Analysis Steps: 1. Determine RT 2. Determine IT 3. Determine Current Through Individual Branches as Necessary FIGURE 5-29 RT = ? Ohms IT = VS/RT = > ?mA Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. FIGURE 5-29 RT = 35.9 Ohms IT = VS/RT = > 278mA Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. General Current-Divider Formula • This Formula is for any number of Branches • The current (Ix) through any branch equals the total parallel resistance (RT) divided by the resistance (Rx) of that branch, and then multiplied by the total current (IT) FIGURE 5-29 278mA I1 = (35.9 Ohms/100 Ohms)278mA => 99.8mA I2 = (35.9 Ohms/56 Ohms)278mA => 178mA Total 278mA Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Current-Divider Formula for Two Branches • This Formula is for Two Branches Only • The current through one of the branches equals the Other Branch Divided by the two Branch Resistances Added together, and then multiplied by the total current (IT) FIGURE 5-29 278mA I1 = (56 Ohms/156 Ohms)278mA => 99.8mA I2 = (100 Ohms/156 Ohms)278mA => 178mA Total 278mA Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Power in Parallel Circuits PT = VSIT PT = IT2RT PT = VS2/RT PT = P1 + P2 + P3 + . . . + Pn FIGURE 5-29 PT = (10V)(278mA) => 2.78W Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. FIGURE 5-27 Examples of parallel circuits in residential wiring and appliances. Application of Parallel Circuits Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. FIGURE 5-26 shown. Simplified diagram of the exterior light system of an automobile. All lights are off when the switches are in the positions Application of Parallel Circuits Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Correct? Application of Parallel Circuits Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Troubleshooting When a lamp filament opens: 1) Total current decreases by the amount of current in the lamp that opened 2) The other branch currents remain unchanged Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Troubleshooting – Open Branch All other parallel branches have the same voltage and the same current even though total current decreases. Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Troubleshooting Open Branch Measures Full Calculated Current IT = 50V/85 Ohms = > 589mA IR1 = (85 Ohms/560 Ohms)589mA => 89.3mA IR2 = (85 Ohms/100 Ohms)589mA => 500mA Current Decreases by Current Of Open Branch Current Decreases by Current Of Open Branch Troubleshooting Shorted Branch Any Completely Shorted Branch Creates a Short Directly Across the Source Thomas L. Floyd Electronics Fundamentals, 6e Electric Circuit Fundamentals, 6e Copyright ©2004 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved.
