TECH TIP # 37 SOLVING SERIES/PARALLEL CIRCUITS THREE LAWS --- SERIES CIRCUITS LAW # 1 --- THE SAME CURRENT FLOWS THROUGH ALL PARTS OF THE CIRCUIT



Similar documents
Chapter 5. Parallel Circuits ISU EE. C.Y. Lee

Electrical Fundamentals Module 3: Parallel Circuits

13.10: How Series and Parallel Circuits Differ pg. 571

Experiment NO.3 Series and parallel connection

Tristan s Guide to: Solving Series Circuits. Version: 1.0 Written in Written By: Tristan Miller Tristan@CatherineNorth.com

Series and Parallel Circuits

Student Exploration: Circuits

ELECTRICAL CIRCUITS. Electrical Circuits

6 Series Parallel Circuits

Parallel Circuits. Objectives After studying this chapter, you will be able to answer these questions: 1. How are electrical components connected

Cornerstone Electronics Technology and Robotics I Week 15 Combination Circuits (Series-Parallel Circuits)

AP1 Electricity. 1. A student wearing shoes stands on a tile floor. The students shoes do not fall into the tile floor due to

Kirchhoff's Current Law (KCL)

Objectives 200 CHAPTER 4 RESISTANCE

Series and Parallel Resistive Circuits

Parallel DC circuits

Resistors in Series and Parallel

= (0.400 A) (4.80 V) = 1.92 W = (0.400 A) (7.20 V) = 2.88 W

Experiment 4 ~ Resistors in Series & Parallel

Ohm's Law and Circuits

PHYSICS 111 LABORATORY Experiment #3 Current, Voltage and Resistance in Series and Parallel Circuits

Series and Parallel Circuits

CURRENT ELECTRICITY - I

Module 2. DC Circuit. Version 2 EE IIT, Kharagpur

AP Physics Electricity and Magnetism #4 Electrical Circuits, Kirchoff s Rules

Experiment #5, Series and Parallel Circuits, Kirchhoff s Laws

People s Physics Book

W03 Analysis of DC Circuits. Yrd. Doç. Dr. Aytaç Gören

Circuits. The light bulbs in the circuits below are identical. Which configuration produces more light? (a) circuit I (b) circuit II (c) both the same

DC Circuits (Combination of resistances)

Kirchhoff s Laws Physics Lab IX

Series-Parallel Circuits. Objectives

Series and Parallel Resistive Circuits Physics Lab VIII

Series and Parallel Circuits

Voltage Drop (Single-Phase)

Chapter 7 Direct-Current Circuits

FB-DC3 Electric Circuits: Series and Parallel Circuits

OHM S LAW AND RESISTANCE

Lesson Plan. Parallel Resistive Circuits Part 1 Electronics

Parallel and Series Resistors, Kirchoff s Law

Tristan s Guide to: Solving Parallel Circuits. Version: 1.0 Written in Written By: Tristan Miller Tristan@CatherineNorth.com

Unit: Charge Differentiated Task Light it Up!

Basic Laws Circuit Theorems Methods of Network Analysis Non-Linear Devices and Simulation Models

PS-6.2 Explain the factors that determine potential and kinetic energy and the transformation of one to the other.

Chapter 1. Fundamental Electrical Concepts

7. What is the current in a circuit if 15 coulombs of electric charge move past a given point in 3 seconds? (1) 5 A (3) 18 A (2) 12 A (4) 45 A

Introduction to Electricity & Magnetism. Dr Lisa Jardine-Wright Cavendish Laboratory

Resistors in Series and Parallel

CHAPTER 28 ELECTRIC CIRCUITS

STUDY MATERIAL FOR CLASS Physics- CURRENT ELECTRICITY. The flow of electric charges in a particular direction constitutes electric current.

Analysis of a single-loop circuit using the KVL method

Σ I in = Σ I out E = IR 1 + IR 2 FXA 2008 KIRCHHOFF S LAWS 1. Candidates should be able to : LAW 1 (K1)

Measuring Electric Phenomena: the Ammeter and Voltmeter

Eðlisfræði 2, vor 2007

DC mesh current analysis

Electrical Circuit Theory

Resistors in Series and Parallel Circuits

Resistors. Some substances are insulators. A battery will not make detectible current flow through them.

HOW ACCURATE ARE THOSE THERMOCOUPLES?

GenTech Practice Questions

Preamble. Kirchoff Voltage Law (KVL) Series Resistors. In this section of my lectures we will be. resistor arrangements; series and

After completing this chapter, the student should be able to:

Electrical Fault Level Calculations Using the MVA Method

Electronics. Basic Concepts. Yrd. Doç. Dr. Aytaç GÖREN Yrd. Doç. Dr. Levent ÇETİN

Current Electricity Lab Series/Parallel Circuits. Safety and Equipment Precautions!

Circuit Analysis using the Node and Mesh Methods

Chapter 13: Electric Circuits

The Membrane Equation

V out. Figure 1: A voltage divider on the left, and potentiometer on the right.

Experiment 8 Series-Parallel Circuits

GT Sensors Precision Gear Tooth and Encoder Sensors

PROCEDURE: 1. Measure and record the actual values of the four resistors listed in Table 10-1.

LAB2 Resistors, Simple Resistive Circuits in Series and Parallel Objective:

NAME: DATE: (Please Print) RESISTOR NETWORK AND OHMS LAW PRACTICE EXAM (See formulas at the back of this exam) 120 Volts AC 35 Ohms

Power Supplies. 1.0 Power Supply Basics. Module

BJT AC Analysis. by Kenneth A. Kuhn Oct. 20, 2001, rev Aug. 31, 2008

Experiment #3, Ohm s Law

Lab 3 - DC Circuits and Ohm s Law

Beware of Simplistic Voltage Drop Calculations. By Aaron Hesse, PE, Coffman Engineers, Spokane, Washington INTRODUCTION

Current, Resistance and Electromotive Force. Young and Freedman Chapter 25

CURRENT ELECTRICITY INTRODUCTION TO RESISTANCE, CAPACITANCE AND INDUCTANCE

ELECTRICAL INSPECTION BULLETIN (Effective )

Chapter 19 Operational Amplifiers

2 A bank account for electricity II: flows and taxes

Fig. 1 Analogue Multimeter Fig.2 Digital Multimeter

Chapter 11. Inductors ISU EE. C.Y. Lee

THE BREADBOARD; DC POWER SUPPLY; RESISTANCE OF METERS; NODE VOLTAGES AND EQUIVALENT RESISTANCE; THÉVENIN EQUIVALENT CIRCUIT

Physics, Chapter 27: Direct-Current Circuits

Lecture - 4 Diode Rectifier Circuits

MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

ES250: Electrical Science. HW7: Energy Storage Elements

WHAT DESIGNERS SHOULD KNOW ABOUT DATA CONVERTER DRIFT

101 BASICS SERIES LEARNING MODULE 2: FUNDAMENTALS OF ELECTRICITY. Cutler-Hammer

Equations, Inequalities & Partial Fractions

COMPOUND CIRCUITS. Voltage (volts) Across Bulb B. Across the Battery. Across Bulb A. Current (amperes) Between A & B. Between Battery & B

Nodal and Loop Analysis

Lecture Notes: ECS 203 Basic Electrical Engineering Semester 1/2010. Dr.Prapun Suksompong 1 June 16, 2010

3.- What atom s particle moves through a conductor material? 4.- Which are the electric components of an elemental electric circuit?

RTD and thermocouple circuits, with millivolt calculations

Energy, Work, and Power

Transcription:

TECH TIP # 37 SOLVING SERIES/PARALLEL CIRCUITS Please study this Tech Tip along with assignment 4 in Basic Electricity. Parallel circuits differ from series circuits in that the current divides into a number of separate, independent branches. In other words, if the current has more than one path to flow through the circuit, it is a parallel circuit. Moreover, these branches may have different resistances, and therefore, the value of current in each branch may be different. The loads that are added in parallel result in additional paths for current flow. This, in effect, decreases resistance. In the same manner, the more outlets a water pipe has, the smaller the total resistance and the greater the water flow. There are electrical rules or laws for series and parallel circuits. These should be committed to memory. THREE LAWS --- SERIES CIRCUITS LAW # 1 --- THE SAME CURRENT FLOWS THROUGH ALL PARTS OF THE CIRCUIT When working with series circuits, the current through each individual component is the same through every other part of the circuit. This means if you had three 15 ohm resistors wired in series, the current flow through the first one would be the same through the second and the third. Knowing the current value of one component would help you determine the other unknown values. LAW #2 -- RESISTANCES ARE ADDITIVE One of a series of dealer contractor technical advisories prepared by HARDI wholesalers as a customer service. The formula for resistors wired in series is R T = R 1 + R 2 + R 3.. Therefore, to find the total equivalent resistance of three 15 ohm resistors wired in series, just add them up. R T = 15 + 15 + 15 = 45 ohms. LAW #3 -- VOLTAGE DROP ACROSS EACH RESISTANCE EQUALS APPLIED VOLTAGE If you had three equal resistors wired in series with an applied voltage of 120 volts, each resistor would have a voltage drop of 40 volts. Voltage Total = 40 + 40 + 40 = 120 volts. Published by the Independent Study Institute, a division of the Heating, Airconditioning & Refrigeration Distributors International. The Institute offers accredited, industry training courses in HVAC/R technology. Direct inquiries to HARDI 3455 Mill Run Drive, Ste. 820, Columbus, OH 43026. Phone 888/253-2128 (toll free) 614/345-4328 Fax 614/345-9161 www.hardinet.org

THREE LAWS --- PARALLEL CIRCUITS LAW #1 --- THE SAME VOLTAGE DROP IS ACROSS EACH ELEMENT This law is perhaps the most abused of the electrical laws. The fallacious concept that prevails in the mind of many is that differing resistances result in differing voltage drops in a branch circuit. This is completely in error, and Figure 15 corroborates this. Four parallel branches or elements are connected to a 24 volt battery. Although all branches have different resistors, they are connected across the same voltage source. The battery potential between the minus and positive electrodes is 24 volts. Each branch, in effect, is an independent connection from the minus electrode to the positive electrode. Therefore, regardless of branch resistance, each branch or element must have a 24 volt drop. LAW #2 --- THE TOTAL CURRENT IS EQUAL TO THE SUM OF THE CURRENTS FLOWING IN THE INDIVIDUAL BRANCHES If a river divides into a number of branches, and later they rejoin, no change in water volume occurs. The initial volume of water merely divides according to the resistance offered by the various branches. When the branches rejoin into one stream, the leaving volume of water is the same as the starting volume. The sum of the branch volumes, however, would have to equal the total volume. There is no point where water could be gained or lost. A parallel electrical circuit is similar. In Figure 15, current flow in branch #1 would be the 24 volts impressed across the branch divided by the branch resistance of 3 ohms or E 1 24 volts I = = = 8 amperes R 1 3 ohms In branch #2, it would be: In branch #3, it would be: In branch #4, it would be: 24 volts = 6 amperes 4 ohms 24 volts = 4 amperes 6 ohms 24 volts = 6 amperes 4 ohms Grand total = 8 + 6 + 4 + 6 = 24 amperes Therefore the sum of the branches would be 24 amperes. 24 amperes would originate at the negative terminal and flow until it divided at the branches. After the branches rejoin into one conductor the branch currents combine again into a 24 ampere flow. 2

LAW #3 --- TOTAL RESISTANCE IN A PARALLEL CIRCUIT IS EQUAL TO THE APPLIED VOLTAGE DIVIDED BY THE TOTAL CURRENT Hence, the effective or total resistance in Figure 15 would be: E T 24 volts or = 1 ohm I T 24 amperes It may seem unusual for a circuit with branch resistances of 3 ohms, 4 ohms, 6 ohms and 4 ohms to combine into a total resistance of only 1 ohm. The 1 ohm is less than the resistance of the smallest branch resistor. This is logical when it is realized that each branch represents another outlet for electrical energy or in the case of piping for water flow. Each outlet contributes to a lessening resistance. In effect, the multiple outlets are equivalent to one large conductor or water pipe with their resultant decreases in resistance. Mini Quiz #15 (see answers on last pages) If the above circuit has the values of R 1 = 12 ohms, R 2 = 8 ohms, R 3 = 4 ohms and R 4 = 24 ohms, then the total resistance would equal. Underline the correct answer. a. 2 ohms b. 4 ohms c. 3 ohms d. 1.5 ohms SHORTCUT METHODS FOR COMBINING PARALLEL RESISTANCES Three rules also apply to the shortcut methods for computing total branch resistances. RULE #1 If all branch resistors are identical, the total resistance is simply the resistance of one branch divided by the number of branches, e.g., five 10 ohm resistors would be: 3

10 ohms/5 branches = 2 ohms total resistance Rule #1 is illustrated in Figure 16. Computing resistance by the long method necessitates the determination of total current which is 30 amperes. Then dividing 60 volts by 30 amps yields an effective resistance of 2 ohms. In contrast merely taking the value of one resistor, 10 ohms, and dividing this by the number of branches 5, yields the same answer -- 2 ohms. Moreover, the answer is instantaneously computed. RULE #2 This is called the product over sum method. Here the product of each pair of resistances is divided by the sum of the two resistances. The long method and the product over sum method are illustrated by Figure 17. In the conventional method, current is computed for each branch. Then current is totaled as a value of 12 amperes. 24 volts are divided by 12 amperes to yield 2 ohms as total resistance. In the product over sum method, the first pair is computed: 3 ohms x 6 ohms 18 ohms = = 2 ohms 3 ohms + 6 ohms 9 ohms The 2 ohms are then combined with the remaining 2 ohm branch. 2 ohms x 2 ohms 4 ohms = = 1 ohms 2 ohms + 2 ohms 4 ohms Mini Quiz # 16 (see answers on last pages) Using the above diagram, find total current and total resistance if all branch resistors were 15 ohms each. Underline the correct answer. a. 3 amperes -- 20 ohms b. 15 amperes -- 4 ohms c. 4 amperes -- 15 ohms d. 20 amperes -- 3 ohms 4

Mini Quiz # 17 (see answers on last pages) If R 3 = 15 ohms, R 2 = 10 ohms and R 1 = 12 ohms, what is the total resistance in the circuit. Underline the correct answer. a. 2 ohms b. 6 ohms c. 4 ohms d. 8 ohms RULE #3 In the third method of finding the total resistance, the joint resistance of a parallel circuit is equal to the reciprocal of the sum of the reciprocals of the individual resistances or stated as a formula: 1 R j = 1 1 1 + + R 1 R 2 R 3 Substituting the values in Figure 17, this would be: 1 R j = or 1 1 1 + + 2 3 6 Converting the fractions to the common denominator 6 1 1 1 R j = = = or 1 ohm 3 2 1 6 1 + + 6 6 6 6 5

This could also be converted to decimal form: 1 1 = or 1 ohm.5 +.333 +.167 1 All methods achieve the same solution. PARALLEL CIRCUIT CHARACTERISTICS The predominating characteristic of a series circuit is that it divides the applied voltage in direct proportion to circuit resistance. In parallel circuits the predominating characteristic is that it divides current in direct proportion to circuit resistance. In control technology, these are very important assumptions. They serve to make even the most sophisticated solid state circuits understandable. KIRCHHOFF S LAW APPLIED TO PARALLEL CIRCUITS Kirchhoff s Law for parallel circuits states that at any junction of conductors the algebraic sum of the currents is zero. This means that as many electrons leave the junction as enter it. Mathematically stated: L T (or total current) ± I 1 ± I 2 ± I 3 = 0 This merely presents a mathematical relationship which has been obvious from logical deduction. Further stated, it also applies to power in a parallel circuit. Total power is equal to the sum of the power consumed in the individual resistances. SERIES-PARALLEL CIRCUITS As the title indicates, series-parallel circuits are composite or compound circuits whose loads are connected in both series and parallel configurations. These circuits are extremely important to the refrigeration and air conditioning technician or application engineer because they are the basis for modulating or proportional controls. These controls must deliver a variable response to a variable signal. The on-of f or snap switch control is incapable of delivering this response. Even the single pole double throw (SPDT) or floating control is ineffective for modulation when precise performance is required. The bridge or Wheatstone bridge type circuits have the capabilities required for proportional control. Therefore, these controls predominate in the commercial-industrial applications where electrical control systems are utilized. Fundamentally, bridge circuits are a series-parallel circuit. 6

SERIES-PARALLEL CIRCUIT SOLUTIONS Series-parallel circuits may be solved by application of the rules already given for simple series and simple parallel circuits. To do this, the series-parallel circuit is reduced to an equivalent, simplified circuit. Each group of parallel resistors is first replaced by it equivalent single resistance, and then the entire circuit is treated as a simple series circuit. This method is illustrated in Figures 18A, B, C and D. In Figure 18A, the original circuit shows resistors R 3 and R 4 in series with each other. These resistors are in parallel with R 2. All of these resistors are in series with R 1. 7

Mini Quiz # 18 (see answers on last pages) If in 18A, R 2 is changed to 15 ohms and R 3 to 6 ohms with the other values unchanged, the total resistance would be. Underline the correct answer. a. 8 ohms b. 10 ohms c. 11 ohms d. 12 ohms The recommended procedure is to start at the circuit s right side. 1. Resistor R 3 and R 4 are in series and additive so they are the equivalent of one 8 ohm resistor. Figure 18A is combined into 18B. 2. This results in an 8 ohm resistor in parallel with R 2 or another 8 ohm resistor (Figure 18B). 3. Reduce these two resistors down to an equivalent resistance. Hence rule #1 for combination circuits is applicable --- with equal resistors, the value for one resistor is divided by the number of branches. Resistance 8 ohms = = 4 ohms (Figure 18C) # of branches 2 branches 4. Since R 1 and R 2 are in series and additive, the simple series equivalent now consists of one 8 ohm resistor as shown in Figure 18D. Mini Quiz # 15 solution 1 1 1 = = = 2 ohms 1/12 + 1/8 + 1/4 + 1/24 0.0833 + 0.125 + 0.25 + 0.04166.5000 Mini Quiz # 16 solution Resistance of individual branch 15 R t = = = 3Ω # of Branches 5 and I t = 60 volts/3ω = 20 amps 8

Mini Quiz # 17 solution Using the product over sum formula, find the equivalent value of resistors R 2 and R 3. 10 x 15 150 R E = = = 6 Ω 10 + 15 25 Using the formula again for R 1 12 x 6 72 R T = = = 4 Ω 12 + 6 18 Mini Quiz # 18 solution 1. R 3 (6 ohms) + R 4 (4 ohms) = 10 ohms 2. 15 x 10 150 R E = = 6 ohms 15 + 10 25 3. R T = 6 ohms + 4 ohms = 10 ohms 9