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U.S. ARMY MEDICAL DEPARTMENT CENTER AND SCHOOLFORT SAM HOUSTON, TEXAS 78234-6100BASIC ELECTRICALCIRCUITSSUBCOURSE MD0903EDITION 200

DEVELOPMENTThis subcourse is approved for resident and correspondence course instruction. Itreflects the current thought of the Academy of Health Sciences and conforms to printedDepartment of the Army doctrine as closely as currently possible. Development andprogress render such doctrine continuously subject to change.ADMINISTRATIONFor comments or questions regarding enrollment, student records, or shipments,contact the Nonresident Instruction Section at DSN 471-5877, commercial (210) 2215877, toll-free 1-800-344-2380; fax: 210-221-4012 or DSN 471-4012, e-mailaccp@amedd.army.mil, or write to:COMMANDERAMEDDC&SATTN MCCS HSN2105 11TH STREET SUITE 4192FORT SAM HOUSTON TX 78234-5064Approved students whose enrollments remain in good standing may apply to theNonresident Instruction Section for subsequent courses by telephone, letter, or e-mail.Be sure your social security number is on all correspondence sent to the Academy ofHealth Sciences.CLARIFICATION OF TRAINING LITERATURE TERMINOLOGYWhen used in this publication, words such as "he," "him," "his," and "men" are intendedto include both the masculine and feminine genders, unless specifically stated otherwiseor when obvious in context.

TABLE OF CONTENTSLessonPAGEINTRODUCTION.1BASIC ELECTRICAL CIRCUITS.MD0903i1

CORRESPONDENCE COURSE OFTHE US ARMY MEDICAL DEPARTMENT CENTER AND SCHOOLSUBCOURSE MDO903BASIC ELECTRICAL CIRCUITSINTRODUCTIONThis subcourse is designed to give you a basic knowledge of simple circuitsthat carry electricity from a power source to some kind of electrical equipment. With aknowledge of these fundamentals, you will be able to make better use of electricalequipment and to better understand future textual materials that mention electricalfactors in the function of equipment.Subcourse Components:This subcourse consists of programmed text.Lesson 1. Basic Electrical CircuitsStudy Suggestions:Here are some suggestions that may be helpful to you in completing thissubcourse:--Read and study each lesson carefully.--Complete the subcourse lesson.Credit Awarded:To receive credit hours, you must be officially enrolled and complete anexamination furnished by the Nonresident Instruction Section at Fort Sam Houston,Texas. Upon successful completion of the examination for this subcourse, you will beawarded 3 credit hours.You can enroll by going to the web site http://atrrs.army.mil and enrolling under"Self Development" (School Code 555).A listing of correspondence courses and subcourses available through theNonresident Instruction Section is found in Chapter 4 of DA Pamphlet 350-59, ArmyCorrespondence Course Program Catalog. The DA PAM is available at the followingwebsite: 903ii

SUBCOURSE MD0903LESSON 1Basic Electrical Circuits.ASSIGNMENTFrames 1 through 100.OBJECTIVEAfter completing the programmed text, you should beable to choose correct answers to questions aboutbasic electrical circuits, current, resistance, amperes,volts, and equivalent.INSTRUCTIONSThis text is set up differently from most subcoursesIt is a workbook that utilizes programmed instruction.The numbered "frames" present information and/ora question about presented information. You shouldwork through the frames in the order presented.Answer each question that is presented. To checkyour answers, go to the shaded box of the NEXTframe. For example. the solution to the questionpresented in Frame 2 is found in the shaded box ofFrame 3.SUGGESTIONSRead Subcourse MD0902, Basic Electricity, beforetaking this subourse.After going through the programmed text at a relativelyslow pace, go back through it several times as rapidlyas you can. This will not take long and will help youfeel more knowledgeable as you study. The purpose ofthe programmed text is memorization as well asunderstanding.MD09031-1

FRAME 1The diagram below will help you to recall that current is aflow of through a conductor.FRAME 2Solution to Frame 1Below are several series circuits. Study the carefully.electronsIn a series circuit, there (is only one/are more than one) pathfor the current to flow.Solution to Frame 2FRAME 3is only oneThe above is a series circuit because.FRAME 4Solution to Frame 3Label each circuit as either “Series” or “Not Series.”a.b.c.d.it has only one path forcurrent to flowMD09031-2

FRAME 5Solution to Frame 4What would be the reading on the ammeter in the seriescircuit below?I E Ra. Seriesb. Not Seriesc. Not Seriesd. SeriesFRAME 6Solution to Frame 5No matter where you measure the current in the seriescircuit below, the current readings would all be the.12v 6 amp2ΩFRAME 7Solution to Frame 6In any part of a series circuit, the current is theas long as the circuit is not changed.same (6 amp)FRAME 8Solution to Frame 7Write in the current reading of each ammeter connected inthe series circuit below.samea.MD0903b.1-3

FRAME 9Solution to Frame 8In any series circuit, the total resistance (Rt) is the sum (ortotal) of all the single resistances. In the series circuit below,Rt is the of R1 and R2.a. 1 ampFRAME 10Solution to Frame 9a. The total resistance (Rt) in the series circuit below is10 Ω 40 Ω Ω.sum (or total)b. 1.ampb. Total resistance (Rt) is ohms.FRAME 11Solution to Frame 10In the series circuit belowa. 5b. 55R1 R2 R3 R4 The total resistance (Rt) is .MD09031-4

FRAME 12Solution to Frame 11So far, you have learned that in series circuits:R1 3 ΩR2 2 ΩR3 5 ΩR4 10 Ω(Rt) 20 Ωa. There is (only one/more than one) path for the current toflow.b. Current has (the same value/different values) everywherein the circuit.c. To get Rt (total resistance), we (add/subtract/multiply) allthe individual resistances.FRAME 13Solution to Frame 12To find It (current in any series circuit), you must use Rt inthe formula It EtRta. only oneb. the same valuec. addTo find It in the circuit above, you must use in theformula .FRAME 14Solution to Frame 13To find I in the series circuit below, you must use(10/20/30/200) Ω in the formula It EtRtRtMD09031-5It EtRt

FRAME 15Solution to Frame 14In a series circuit with only one resistor, R1 and Rt must bethe same. In the series circuit below, there is only oneresistor. This means the R1 and Rt (are/are not) the same.They are both equal to .30FRAME 16Solution to Frame 15To find the current in the circuit below, you would substitutethe number (4/6/10/24) for Rt in the formula It Et .RtareFRAME 17Solution to Frame 16In the circuit below, Et . Rt .10Find the current It .MD09031-62Ω

FRAME 18Solution to Frame 17If you calculated current (I) in the circuit below and used thisformula:90va.It Et, your answer would be (right/wrong).R1b.It Et, your answer would be (right/wrong).R2c.It Et, your answer would be (right/wrong).Rt30Ω3 ampIt Et 90v 3 ampRt 30ΩFRAME 19Solution to Frame 18In the circuit below, find I.a. wrongExample:b. wrong:c. rightIt Et 10 vRt (2 3) Ω 10v 2 amps5ΩYou do this one: MD09031-7

FRAME 20Solution to Frame 19In the circuit below, It .It Et 12 v 2 ampsRt6ΩFRAME 21Solution to Frame 20So far, you have learned that in series circuits:3 ampa. There is/are (only one/more than one) path for thecurrent to flow.It EtRt 90 v30 Ωb. Current has the (same/different) value everywhere in thecircuit. 3 ampc. To get Rt, we (sum/subtract) all individual resistances.d. To find It, you must use (R1/Rt) in the formulaIt Et .Rt.FRAME 22Solution to Frame 21The voltage applied by a battery is called the applied voltagea. only one(Ea).b. same.This battery will apply a voltage called the.c. sumFRAME 23Solution to Frame 22In the circuit below, the Ea (applied voltage) is volts.applied voltage (Ea)MD09031-8d. Rt

FRAME 24Solution to Frame 23To move a wagon uphill, you must apply a force. To moveelectrons through a resistor, a battery must also apply.24FRAME 25Solution to Frame 24When you move a wagon uphill, force is used up. Whenelectrons are pushed through a resistance, electromotiveforce (EMF) is also .a force (or a voltage)FRAME 26Solution to Frame 25When EMF is used up, the voltage drops. Across anyresistance, EMF is used up and the voltage .used upFRAME 27Solution to Frame 26The drop in voltage is called voltage drop.dropsAcross the resistor above, we have a 10v .FRAME 28Solution to Frame 27In the diagram below, the voltage drop across R1 isand across R2, it is .voltage dropMD09031-9

FRAME 29Solution to Frame 28You have learned that the symbol for voltage is E.(R1) 5vFor voltage drop across R1, you will use the symbol E1.(R2) 20vFor voltage drop across R2, you will use the symbol E2.For voltage drop across R3, you will use the symbol .FRAME 30Solution to Frame 29Total voltage drop (Et) is the sum of all individual voltagedrops. In the circuit below, Et is the ofE1 and E2. Et .E3FRAME 31Solution to Frame 30Et (total voltage drop) in this circuit is volts.total (or sum)15vMD09031-10

FRAME 32Solution to Frame 31Et in the circuit below is .30FRAME 33Solution to Frame 32Et (total voltage drop) in the circuit below is .15vEt (applied voltage) is also .FRAME 34Solution to Frame 33In the series circuit below, the Ea (applied voltage) is ,and the Et (total voltage) is . Et and Ea are(the same/different) in any series circuit.10vMD09031-1110v

FRAME 35Solution to Frame 34Et in the circuit below is ; Ea is .18v18vthe sameFRAME 36Solution to Frame 356vIf Et 24v, then Ea 6vIf Ea 6v then Et 6vIf any series circuit, Et and Ea are .FRAME 37Solution to Frame 36One way to find I (current) in a series circuit is to use Et inthe formula It EtRt24v6vthe same (equal)To find It in the circuit above, use in the formula.MD09031-12

FRAME 38Solution to Frame 37Find I in the circuit below.EtExample:You work this problem.It Et RtIt 20v10ΩIt 2 ampsIt EtRt It FRAME 39Solution to Frame 38Find I in the circuit below.EtRtExample:You do this:24v12Ω2 ampIt Et RtIt 27v9Ω 3 ampsMD0903It 1-13

FRAME 40Solution to Frame 39I (current) in the series circuit below is .It EtRt 5v5Ω 1 ampFRAME 41Solution to Frame 40You just learned that to find I in a series circuit, you use theformula: It Et.Rt3 ampSince Ea Et, you (may/may not)also use the formula9v3Ω 3 ampIt EaRtSolution to Frame 41FRAME 42In the circuit below I . mayMD09031-14

FRAME 43Solution to Frame 42Find I in the circuit below:3 ampI Ea 27v 3 ampRt 9Ωdrop at side resistor 12 v.drop at bottom resistor 9vFRAME 44Solution to Frame 43So far you have learned that in a series circuit:8 ampa. There is/are (only one/more than one) path for the currentto flow.It Et 80v 8 ampRt 10Ωb. I (current) has (the same/a different) value(s) everywherein the circuitc. To get Rt, you (sum/subtract) the individual voltageresistances.d. To get Et, you (sum/subtract) the individual voltage drops.e. Et and Ea (are/are not) the same.f. To find I, you (must/must not) use Rt.g. To find I, you (must/must not) use either Et or Ea.FRAME 45Solution to Frame 44The statement “The greater the resistance, the greater thevoltage drop” does not tell you how to calculate the exactvoltage drop. However, Ohm’s law does allow you to.a. only oneb. the samec. sumd. sume. aref. mustg. mustMD09031-15

FRAME 46Solution to Frame 45To calculate E (voltage drop across R1) exactly, you use theformula E1 I1 x R1:calculate the exactvoltage dropTo calculate E1 in the circuit above, you use the formula.FRAME 47Solution to Frame 46Find E1 in the circuits below:E1 I1 x R1Example:You do this oneE1 I x R1E1 3 amp x 2Ω 6v MD09031-16

Solution to Frame 47FRAME 48Example:E2 I x R2You do this one:E2 3 amp x 4Ω 12v E1 I1 x R1 4 amp x 6Ω 24vFRAME 49Solution to Frame 48In the circuit below I 2 amp. Find E2 (voltage drop acrossR2).E2 I x R2E2 . 3 amp x 5Ω 15vFRAME 50Solution to Frame 49Compute E2 in the circuit below: E2 .24vE2 I x R2 2 amp x 12Ω 24vMD09031-17