CIRCUITS - MilitaryNewbie
SUBCOURSEIT 0334EDITIONCUS ARMY INTELLIGENCE CENTERCIRCUITS
CIRCUITSSubcourse Number IT0334EDITION CUS ARMY INTELLIGENCE CENTERFORT HUACHUCA, AZ 85613-60004 Credit HoursEdition Date: December 1995SUBCOURSE OVERVIEWThis subcourse is designed to teach you to calculate values in DCseries circuits and to solve for unknown values of voltage, currentand resistance in series-parallel circuits. There are twoprerequisites for this subcourse. You should already have completedsubcourses IT 0332 and IT 0333. If this material is strictly reviewfor you, you may be able to take the subcourses out of order. Inthat case, remember that this subcourse builds upon material learnedin the preceding subcourses. If you encounter any difficulty, goback to those subcourses.This Subcourse replaces SA0703 and SA0704.TERMINAL LEARNING OBJECTIVE.ACTION:You will calculate resistance, current, voltage, andpower in DC series circuits and solve for unknown valuesof voltage, current and resistance in series-parallelcircuits.CONDITION:You will use the information provided in this subcourse.STANDARD:To demonstrate competency of this task, you must achievea minimum of 70% on the subcourse examination.iIT0334
TABLE OF CONTENTSSectionPageSubcourse OverviewiLesson 1:DC Series Circuits1-1Practice Exercise1-29Answer Key and Feedback1-34Series-Parallel Circuits2-1Part A:Series Circuits Connected in Parallel2-2Part B:Parallel Circuits Connected in Series2-15Lesson 2:IT0334ii
LESSON 1DC SERIES CIRCUITSOVERVIEWLESSON DESCRIPTION:Upon completion of this lesson, you will be able to calculate valuesin DC series circuits.This lesson replaces SA 0703.TERMINAL LEARNING OBJECTIVE:ACTION:Calculate resistance, current, voltage, and power in DCseries circuits.CONDITION:Given the information provided in this lesson.STANDARD:To demonstrate competency of this task, you must achievea minimum of 70 percent on the subcourse examination.INTRODUCTIONIn any type of work that uses the effects of electron flow, aknowledge of series or parallel circuits is desirable. None of theeffects which accompany electron flow--for example, heating,lighting, or magnetic effects--are possible without the use ofelectric circuits, and many electric devices can be more effectivelyused if the operator has a knowledge of how the basic circuits work.An electric circuit is a complete path through which electrons flowfrom the negative terminal of the voltage source, through theconnecting wires, through the load, and back to the positive terminalof the voltage source. A circuit is thus made up of a voltagesource, the necessary connecting wires, and the load.A series circuit is defined as two or more component parts connectedend to end to form only one path for current flow.1-1IT0334
In the electric circuit shown above, current flows from the negativeterminal of the battery, through the resistor, the ammeter, and backinto the positive terminal of the battery.The electron flow in a series circuit may be compared to the fluidflow in a hydraulic system as illustrated below.In each circuit, the fluid or current has only one path to follow.The battery in the electric circuit compares in action to the pump inthe hydraulic system. The resistors in the electric circuit compareto the physical restrictions shown in the hydraulic system.IT03341-2
The series circuit above contains three lamps. The current flow fromthe battery must flow through each lamp in succession in order tocomplete the electrical path. When one lamp burns out, theelectrical path is interrupted, and no current flows in the circuit;therefore, none of the lamps will light.1-3IT0334
The series circuit above contains three resistors and two batteries.Each resistor is labeled with an "R," the symbol for resistance. Inaddition, each "R" is followed with a numeral to identify thespecific resistor (R1, R2, etc.). The batteries are identified in asimilar manner. When performing calculations in electric circuitscontaining two or more components, it becomes necessary to usesubscripts to identify specific values. Subscripts are numbers orletters written below and to the right of the original circuitfunction letter, as in IR1 or ER3 (I R1 is read "I sub R1"). Totalcircuit values are normally identified with a subscript "t." Totalvoltage, total current, and total resistance are identified as Et, It,and Rt.PROBLEM:How many paths for current flow are there in a series circuit?A.OneC.ThreeB.TwoD.FourIT03341-4
ANSWER:A.There is only one path for current flow in a series circuit.Referring to the circuit above, note that the physical components areidentified as R1 R2, R3, E1 and E2. The electrical values areassociated with the physical components by the use of subscripts.Voltage drops across the resistors are identified as ER1, ER2 and ER3.The current flowing through each resistor is identified as I R1 I 2 andIR3. Each value, such as ER1 or IR2, is specifically associated with acomponent by the proper use of subscripts (R1 and R2). The totalcurrent flow in the circuit is measured by the indicated meter andidentified as It.RPROBLEM:What is the definition of a series circuit?1-5IT0334
ANSWER:A series circuit is defined as a circuit having two or morecomponents connected end to end to form only one path for currentflow.In the hydraulic system shown above, when fluid is flowing out of thepump at the rate of three gallons per minute, fluid must be flowinginto the other side of the pump at the rate of three gallons perminute. Therefore, the rate of flow of fluid throughout the systemmust be constant. Similarly, this same concept is true regardingcurrent flow in the series circuit shown below. When current isflowing out of the negative terminal of the battery at the rate ofthree amperes, the current flowing into the positive terminal to thebattery is at the same rate, or three amperes.(CONTINUED)IT03341-6
Since the current flow in the series circuit is the same throughoutthe circuit, the value of current flowing through any resistor (IR1,IR2, and IR3) is equal to the value of total current (It). The formulawhich expresses this relation isIt IR1 IR2 IR3PROBLEMS:a. The total current in a series circuit is to thecurrent through any resistor in the circuit.b. Select the formula used to express the relationship ofcurrent throughout a series circuit.A.It IR1 IR3 IR2B.It IR1 IR2 IR3C.It IR1 IR2 IR3 .c. Match each symbol in column A with the function it representsin column B.AB(1)Ita.Current flow through R1.(2)Rtb.Total resistance.(3)IR1c.Total current.(4)ER3d.Voltage drop across R3.d.What is the definition of a series circuit?1-7IT0334
ANSWERS:a.equalb.Cc.(1)c.(2)b.(3)a.(4)d.d. Two or more components connected end to end to form only onepath for current flow.The total resistance of a series circuit is calculated by adding theindividual resistances of the resistors in the circuit. The sum ofthe individual resistances in the circuit below is equal to the totalresistance. This relation is expressed asRt R1 R2 R3IT03341-8
PROBLEMS:a. The sum of the individual resistances in a series circuit isequal to the resistance of the circuit.b. Select the formula used to solve for total resistance in aseries circuit.A.Rt R1 R2 R3B.Rt R1 R2 R3C.Rt R1 R2 R3D.Rt R1 R2 R3c. Match each symbol in column A with the circuit function itrepresents in column B.AB(1)IR1a.Current through R1.(2)Itb.Total resistance.(3)Rtc.Resistor R3.(4)Etd.Total voltage.(5)ER1e.Total current.(6)R3f.Voltage drop across R1.1-9IT0334
ANSWERS:a. totalb.B.c.(1)a.(2)e.(3)b.(4)d.(5)f.(6)c.The total voltage in a series circuit is equal to the sum of theindividual voltage drops across the resistors in that circuit.The series circuit containing three resistors (R1, R2, and R3)has three voltage drops (ER1, ER2 and ER3) . The sum of these threevoltage drops is equal to the total voltage. Expressedmathematically:Et ER1 ER2 ER3IT03341-10
PROBLEMS:a. Total voltage is a series circuit equal to theof the individual voltage drops.b. Select the formula used to solve for total voltage in aseries circuit.A.Et ER1 ER2 ER3B.Et ER1 ER2 ER3C.Et ER1 ER2 ER3D.Et ER1 ER2 ER3c. Write the formula used to solve for total current (It) in thecircuit below.d. Write the formula used to solve for total resistance (Rt) inthe circuit below.e. Write the formula used to solve for total voltage (Et) in thecircuit below.1-11IT0334
ANSWERS:a.sumb.D.C.It IR1 IR2 IR3d.Rt R1 R2 R3e.Et ER1 ER2 ER3IT03341-12
The solution of a typical series circuit problem requires theapplication of Ohm's law and the series circuit formulas contained inthis lesson. The formulas used are determined by analyzing theinformation given in the problem and the answers required.Refer to the circuit above. When the current flowing through anyresistor is known or can be calculated, the total current isdetermined by using the formulaIt IR1 IR2 IR3It IR2It 2 AThe total current in the circuit above may also be calculated byusing the Ohm's law formulaIt EtRt 100 volts50 ohmsIt 2 A1-13IT0334
PROBLEMS:a.Solve for It in the circuit below.b.Select the value of total current (It) in the circuit below.A.0.5 AC.1.5 AB.1.0 AD.2.0 AIT03341-14
ANSWERS:a.5 Ab.B.(1.0 amps)The total resistance in the circuit above may be calculated by usingthe formulaRt R1 R2 R3 10 ohms 10 ohms 5 ohmsRt 25 ohmsThe total resistance in the circuit above may also be calculated byusing the formula derived from Ohm's law:Rt EtIt 100 volts4 ampsRt 25 ohms1-15IT0334
PROBLEM:Solve for Rt in the circuit below.IT03341-16
ANSWER:40 ohmsSince Rt R1 R2 R3we have:Rt 20 ohms 5 ohms 15 ohmsRt 40 ohmsWhen the voltage drops across the resistors in a series circuit areknown, the total voltage may be calculated by using the formulaEt ER1 ER2 ER3 20 volts 30 volts 50 voltsEt 100 voltsThe total voltage in a circuit may also be calculated by using theformula derived from Ohm's law:Et It x Rt 1 amp X 100 ohmsEt 100 volts1-17IT0334
PROBLEM:Solve for Et in the circuit below.IT03341-18
ANSWER:50 voltsSince Et It X Rt,we have: E .25 amps X 200 ohmsE 50 voltsPROBLEMS:a. Select the value of total voltage (Et) applied to the circuitbelow.b.A.5 VC.25 VB.10 VD.50 VCalculate the total current (It) in the circuit below.1-19IT0334
ANSWERS:a.C.25 Vb.3 ampsMost unknown values in a series circuit may be calculated a number ofdifferent ways with various formulas. The shortest method ofcalculation is determined by analyzing the information given in theproblem. Recognizing the shortest possible way to solve a problemcomes with practice and experience.In the circuit above, the values of ER1 ER2, ER3, IR, IR2, and IR3 are tobe calculated. Referring to the chart below, analyze the knownvalues.IT03341-20
(CONTINUED)Since the rate of current is constant throughout a series circuit,the value of 2 amperes (It) is designated as the value of currentthrough R1, R2, and R3, as shown in the chart below.Total resistance of the circuit is computed in the following manner:Rt EtIt 200 volts2 ampsRt 100 ohmsSince Rt is equal to 100 ohms and Rt R1 R2 R3, the value of R3,the value of R3 may be calculated by subtracting the known resistancesfrom Rt.This difference is then equal to the value of the unknown resistor.Expressed mathematically:Rt R1 R2 R3R3 Rt - R1 - R2R3 100 ohms - 50 ohms - 15 ohmsR3 35 ohms1-21IT0334
(CONTINUED)All values are now known, as shown in the chart below, except thevoltage drop across each resistor.The calculations for the voltage drops are performed by using theOhm's law formulas.ER1 IR1 X R1 2 A X 50ER1 100 voltsER2 IR2 X R2ER3 IR3 X R3 2 A X 15 2 A X 35ER2 30 voltsER3 70 voltsWhen the values of ER1, ER2, and ER3 are inserted into the chart below,it is readily seen that Et ER1 ER2 ER3.IT03341-22
PROBLEMS:a.Solve for ER1, ER2, and ER3 in the circuit below.b.Solve for ER1, ER2, and ER3 in the circuit below.1-23IT0334
ANSWERS:a.ER1 2.5 V, ER2 2.5 V, ER3 15 Vb.ER1, 22.5 V, ER2 12.5 V, ER3 15 VPROBLEMS:a.Calculate the total resistance (Rt) in the series circuit below.b.Calculate the total voltage (Et) applied to the circuit below.c.Solve for the values of ER1, ER2, and ER3 in the circuit below.IT03341-24
ANSWER:a.80 ohmsb.50 voltsc.ER1 40 V, ER2 30 V, ER3 50 VA practical application of the laws of DC series circuits isdemonstrated in the study of batteries. A battery is made up of twoor more cells which may be connected in series, parallel, or seriesparallel combinations. This lesson is concerned with the seriesconfiguration only.When connecting cells to form a battery, the voltage and currentrequirements for the particular application determine the method ofconnection. The advantage of connecting cells in series is that ahigher output voltage is obtained. The voltage law for a seriescircuit states that the sum of the voltage drops around a closedcircuit is equal to the applied voltage, orEt, E1 E2 E3Therefore, when battery cells are connected in series, the totalvoltage is equal to the sum of the cell voltages, as shown below.1-25IT0334
There are two basic types of battery cells. These are (1) theprimary cell (sometimes referred to as a "dry cell" because theelectrolyte is in paste form) and (2) the secondary cell (sometimesreferred to as a "wet cell" because the electrolyte is in liquidform). Both types of cells are widely used.One of the most popular primary cells in use is the flashlight cell.The output voltage of the primary cell is about 1.5 volts when new.The output voltage of the secondary cell is about 2 volts. Thisvoltage will vary with specific types of cells. For example, thelead-acid cell used in the automobile battery has an output of 2.2volts. The so-called 12-volt automobile battery consists of sixcells connected in series.PROBLEM:Select the advantage of connecting battery cells in series.A.Economy.B.Less weight.C.Higher output current is obtained.D.Higher output voltage is obtained.IT03341-26
ANSWER:D.Higher output voltage is obtained.1-27IT0334
LESSON 1PRACTICE EXERCISEThe following items will test your grasp of the material contained inthis lesson. There is only one correct answer for each item. Whenyou complete this exercise, check your answers with the answer keythat follows. If you answer any item incorrectly, study again thatpart of the lesson which contains the portion involved.1.Define a series circuit.2. Match each symbol in column A with the circuit function itrepresents in column B.AB(1)Rta.Current flow through resistor R2.(2)Itb.Total resistance.(3)Etc.Resistor R3.(4)IR2d.Total voltage.(5)ER1e.Total current.(6)R3f.Voltage drop across resistor R1.1-29IT0334
(CONTINUED)3. Write the formula used to determine total current (It) in thecircuit below.4. Write the formula used to solve for total resistance (Rt) in thecircuit below.5. Write the formula used to solve for total voltage (Et) in thecircuit below.IT03341-30
(CONTINUED)6.Calculate the total current (It) in the circuit below.7.Calculate the total resistance (Rt) in the circuit below.1-31IT0334
(CONTINUED)8.Calculate the total voltage (Et) applied to the circuit below.9. Solve for the values of ER1, ER2, and ER3 in the circuit shownbelow.IT03341-32
10.State the advantage of connecting battery cells in series.11.State the voltages developed by the following cells.a.a single carbon - zinc cell (flashlight battery).b.a single lead-acid cell (automobile battery).1-33IT0334
LESSON 1PRACTICE EXERCISEANSWER KEY AND FEEDBACK1. Two or more components connected end to end to form only one pathfor current flow.2.(1) b.(2) e.(3) d.(4) a.(5) f.(6) c.3.It IR1 IR2 IR34.Rt R1 R2 R35.Et ER1 ER2 ER36.It 2 A7.Rt 40 ohms8.Et 50 V9.ER1 25 VER2 25 VER3 50 V10.A higher output voltage is obtained.11.a.b.IT03341.5 V2.0 V1-34
LESSON 2SERIES-PARALLEL CIRCUITSOVERVIEWLESSON DESCRIPTION:Upon completion of this lesson, you will be able to solve for unknownvalues of voltage, current and resistance in series-parallel circuits.Lesson 2 replaces SA0704.TERMINAL LEARNING OBJECTIVEACTION:Solve for unknown values of voltage, current andresistance in series-parallel circuits.CONDITION:Given the information provided in this lesson.STANDARD:To demonstrate competency of this task, you must achievea minimum of 70 percent on the subcourse examination.2-1IT0334
PART ASERIES CIRCUITS CONNECTED IN PARALLELA series-parallel circuit may consist of series circuits connected inparallel or it may consist of parallel circuits connected in series.You have had the necessary laws and formulas needed to solve seriesparallel circuit problems. The important thing to remember is thatwhen working with resistances connected in series, use series-circuitlaws; and when working with resistances connected in parallel, useparallel-circuit laws.The following formulas are based on the laws of series circuits.Total resistance equals the sum of the individual resistances.Rt R1 R2 R3 .Current flow is the same acrosseach resistance.It I1 I2 I3 .Total voltage equals the sum of thevoltage drops across each resistance.Et E1 E2 E3 .The following formulas are based on the laws of parallelcircuits.Total resistance is determined by the reciprocal method.1Rt 1R1 1R2 1 .R3Total current is equal to thesum of the current of eachbranch.It I1, I2 I3 .Total voltage is equal to thevoltage of each branch.Et - E1 E2 E3 .IT03342-2
The circuit in figure A consists of two series circuits connected inparallel across a battery.Figure B shows resistancesR1 and R2 connected inseries with each other toform one path for currentfrom the battery.Figure C shows resistancesR3 and R4 connected inseries to form a secondpath for current from thebattery.Refer to figure D.a.b.c.How many series circuits are in this circuit?How many paths are there for current?How are these series circuits connected?2-3IT0334
a.b.c.THREETHREEPARALLELDraw connecting lines between the followingcomponents to form two series circuitsconnected in parallel.No new formulas are needed to find the valuesof series-parallel circuits. Instead, asimplified circuit is made of the originalseries-parallel circuit, and the necessarycircuit laws are used for the simplifiedcircuit.The circuit in figure A below is electricallythe same as the circuit in figure B.In figure A, the circuit has two paths forcurrent; each path (branch) has a total of 10ohms resistance. R1 R2 gives the totalresistance of the first branch. R3 R4 givesthe total resistance of the second branch.Figure B is the simplified circuit of thecircuit in figure A. To solve for Rt in figureB, use - circuit laws.(series/parallel)Rt ohmsIT03342-4
PARALLELSimplified circuits are very useful whensolving series-parallel circuit problems.5 OHMSDraw and use simplified circuits when solvingthe problems in this program.Figure B is the simplified drawing of thecircuit in figure A. To solve for the totalresistance of the circuit below, first, solvefor the combined resistance of the seriesnetwork R1, R2, and R3. This combined value isrepresented in the simplified circuit as Ra.Second, solve for the combined value of R4 andR5. This combined value is represented in thesimplified circuit as Rb.a.The circuit in figure A consists ofseries circuits connected in parallel.b.Ra. in figure B represents whichresistances in figure A?c.Rb in figure B represents whichresistances in figure A?d.What is the value of Ra?e.What is the value of Rb?f.Rt ohms.2-5IT0334
a.b.TWOR1, R2, R3c.R4, R5d.e.f.100 OHMS100 OHMS50 OHMSThe total current (It) in series circuitsconnected in parallel is equal to totalvoltage (E
Lesson 1: DC Series Circuits 1-1 Practice Exercise 1-29 Answer Key and Feedback 1-34 Lesson 2: Series-Parallel Circuits 2-1 Part A: Series Circuits Connected in Parallel 2-2 Part B: Parallel Circuits Connected in Series 2-15 IT0334 ii
Contemporary Electric Circuits, 2nd ed., Prentice-Hall, 2008 Class Notes Ch. 9 Page 1 Strangeway, Petersen, Gassert, and Lokken CHAPTER 9 Series–Parallel Analysis of AC Circuits Chapter Outline 9.1 AC Series Circuits 9.2 AC Parallel Circuits 9.3 AC Series–Parallel Circuits 9.4 Analysis of Multiple-Source AC Circuits Using Superposition 9.1 AC SERIES CIRCUITS
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circuits, all the current flows through one path. In parallel circuits, current can flow through two or more paths. Investigations for Chapter 9 In this Investigation, you will compare how two kinds of circuits work by building and observing series and parallel circuits. You will explore an application of these circuits by wiring two switches .
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Excellence Model and its suitability for measuring the levels of business excellence is being demonstrated. The second part is dedicated to the exploration and presentation of alternative ways for attaining the highest possible levels of excellence, while, in the third part a comprehensive comparison among them is being conducted. Finally, in the fourth part, the basic findings are summarized .
