DC Circuits Part 3 W - BPA.gov

2y ago
37 Views
2 Downloads
473.36 KB
25 Pages
Last View : 23d ago
Last Download : 2m ago
Upload by : Lee Brooke
Transcription

DC CircuitsPart 3Combination CircuitsWe have learned about the series circuit and the parallel circuit. Both circuits can be examinedby the use of Ohm’s Law. Let’s now combine the two circuits to build a more complex circuit.This type of circuit will be called a series/parallel circuit or a combination circuit. Ohm’s Lawcan also be used to examine this type of circuit.A series/parallel circuit is a circuit in which the electrical components areconnected in series and parallel groups.In the electrical industry you hardly ever run across a circuit that is only a series or a parallelcircuit. Most all circuits are combination circuits that contain both series elements and parallelelements.The illustration above is a simple representation of a combination circuit. Many such circuitsfound in the industry are quite complex and contain many amounts of both series and parallelelements. Even with all of the complexity of these circuits the basic laws still apply; i.e.Kirchhoff’s Laws and Ohm’s Law. You will need a sharp eye to decipher the elements containedin complex circuits.71

Notice that in the circuit below, the total current of the circuit will pass through the seriessection of the circuit and the current will branch out in the parallel portion of the circuitaccording to the resistances within the parallel section.The voltage drop across the components in a combination circuit act according across each groupof components as if they were in an individual series or parallel circuits. That is the voltages dropacross each resistor in the series portion of a combination circuit and the voltage to the commonpoints of the parallel portion of a combination circuit is the same at each common point. Let’slook at a simple combination circuit to visualize this concept.72

In the previous lessons on series and parallel we solved for total resistance of the circuit or anequivalent resistance. When dealing with combination circuits, in order to solve for unknownvalues, an equivalent circuit can be derived. The Equivalent circuit is a circuit that has beensimplified from many components down to only one equivalent component and the source. In theseries circuit we simply added the values of the resistances which then equated the sum to onesingle large valued resistor as an equivalent circuit. In the parallel circuit we had several methodsto equate the equivalent resistance depending upon the values of the resistor branches. In thecombination circuit to, find the equivalent circuit; you will be using the practices taught for bothseries and parallel circuits.Again we will use the same simple combination circuit so we may solve for the following values;(1) total resistance, (2) total current, (3) current through the parallel branches, (4) voltage dropacross each series component and (5) total power including power dissipated across eachcomponent.There are different methods in solving for the unknown values, but with practice you will be ableto choose the most logical methods. We have all of the resistances labeled therefore we canproceed to simplify the circuit to achieve an equivalent or total resistance. We will tackle theparallel section first. We learned several methods to solve for parallel resistors in parallel, but themost logical method for two unequal resistors is the product over the sum.R 3R 4 R3 R450 25 50 251250 16.667Ω7573

The circuit can now be simplified by replacing the parallel components with an equivalentresistance of 16.667Ω. For ease of understanding, a new schematic will be used.As the circuit suggests, the parallel section is replaced with an equivalent resistance of 16.667Ω.The circuit is now simplified to just a series circuit in which the three resistances need only to besummed.R 1 R 2 R equivalent R tTotal resistance.100Ω 75Ω 16.667Ω 191.667ΩThe circuit can now be illustrated with the following diagram.74

We can now solve for the total current within this circuit by using the known total values andOhm’s law.I ER50V191.667ΩI 0.2608A or 260.8mAI Total current.The current through the series portion is equal to the total current of the circuit or 260.8mA.Once the current enters the parallel section of the circuit, the current will branch to the twobranches according to the resistances within the branches.The total power of this circuit can also be easily ascertained by one of the PIER circle equations.P IEPt 0.2608A 50VTotal Power.Pt 13.04WThe next logical step is to derive the voltage drops across the series resistors and the equivalentresistor.75

The voltage drop across the series components can easily be found by using Ohm’s law. Thevoltage drop across the parallel components will equal the voltage drop across the equivalentresistor. Since the equivalent resistor is also in series, the same method of deriving the voltagedrop can be made.V1 I t R 1V1 0.2608A 100ΩVoltage across R1V1 26.08VV2 I t R 2V2 0.2608A 75ΩVoltage across R2V2 19.56VThe remaining voltage should be the voltage across the parallel components. We can derive thedrop by adding the series voltage drops and subtracting the sum from the total voltage(Kirchhoff’s Law) or we can again use Ohms Law.Veq I t R eqVeq 0.2608A 16.667ΩVoltage across the parallelVeq 4.35VOrVt V1 V2 Veq 0( 50V ) ( -26.08V ) ( -19.56V )Veq ( -50V ) 26.08V 19.56V Veq 0Using Voltage Kirchhoff’s LawVeq -4.36V or 4.36V76

Now that the series voltage drops are known, the power dissipated across the series resistors cannow easily be determinedWe will again refer to the PIER circle to obtain the proper equation to solve for the powerdissipation across the series resistors. (The PIER circle is only a tool. Try to memorize as manyof the formulas in the circle and remember how to derive the remaining equations from the onesyou have stored in your memory. You may not have the luxury of having a copy of the circlewhen you are working in the field.)P1 I t E1P2 I t E 2P1 0.2608A 26.08VP2 0.2608A 19.56VP1 6.8WP2 5.1W77

Power dissipated across R1 & R2 in WattsWe can now look at the parallel components with the data that we have derived from itsequivalent circuit. We basically now have a voltage that is common to both branches within theparallel circuit.By using this Voltage the currents and power can now be derived for these branches. We will useOhm’s law to find the current in Branch R3.I3 E eqR34.35V50ΩI 3 0.087A or 87mAI3 Current of Branch R378

To get you thinking, let’s explore another way to derive the current in Branch R4. We wouldnormally choose the same method as we did for Branch R3 but this is just a lesson to learn from.Let’s look at the current to resistance ratios. The relationships are inversely proportional.R3I 4R4I350ΩI4 25Ω .087A50 .087I4 25I 4 0.174A or 174mACurrent of Branch R4We must now prove that the currents are correct. Any ideas on how to do this?Let’s use Kirchhoff’s current law which states:The sum of the currents entering and leaving a point (node) in a circuit equals zero.The mathematical equation for this law looks like this;I1 I 2 I 3 . I n 0Using the currents derived from our equations the results should be:Now to solve for our last piece of the puzzle is to find the power dissipated across the remainingtwo resistors that are connected parallel in this circuit. We will use an equation derived fromOhm’s Law.P3 I 3E eqP3 0.087A 4.35VPower of branch R3P3 0.378W or 378mW79

It is time again for another experiment. Normally we would solve for the R4 branch using thesame method as we did for branch R3. Let’s refer to the PIER circle and grab another equation tosolve for power.P4 2E eqR44.35V 2Power of branch R425Ω18.923P4 25P4 0.757W or 757mWP4 It’s time again for the proof. The sum of the power dissipated across the components equals thetotal power dissipated in the circuit. Earlier we determined that the power dissipated in the circuitwas 13.1 Watts. This was derived from the source voltage times the total current of the circuit.Now all we have to do is to add up the powers found dissipated across each component that wehave previously solved for.P1 P2 P3 P4 Pt6.8W 5.1W 0.378W 0.757W Pt13.035W PtDue to rounding of numbers, the two totals are not exact yet are basically the same number.Please complete the exercises contained in Practice 13 – Combination Circuit Analysis –80

So far we have worked on a simple combination circuit. More complex circuits may consist ofseveral sections of series and parallel components. In fact there may be series components withinone of the branches of a parallel circuit or two or more parallel branches in series. Let’s look atsome different circuits and locate the series or parallel sections within the circuits.As one can see, there are infinite arrangements that a circuit may have. You must be able toidentify the different series and parallel sections within a circuit so you can then logicallydetermine the variables of that circuit.Let’s look at the first circuit and define itssections.As you can see, resistors 1 and 6 can beeasily spotted since they are the onlyconnected resistors to the source; that isthey are connected in series with thesource.Behind resistor 1 the conductor is acommon connection to three moreresistors. The conductor is highlighted inorange. The common connection definesa parallel section of componentsconnected to the voltage source.Resistor 5 is located in one of the parallelbranches of the circuit yet it is connected in series with another resistor of that branch. You must81

treat the relationship of these two resistors as a series circuit within the larger circuit. The tworesistances must be summed to obtain an equivalent series resistance which then becomes thevalue of one of the parallel resistances.The next circuit is the most complicated. A proper sequence of steps must be met to ensure thatthe values of the circuit are properly determined.When you first observe this circuit the large parallel section jumps out at you. For those whocan’t see it, the blue circle encompasses it. Within the parallel section of this circuit one branchhas three series resistors which can be summed to an equivalent resistance and the other branchhas a second parallel section within its branch which must be determined then summed withanother resistor (purple) in series within that branch. Once the equivalent resistances of the twobranches of the large parallel section are determined (blue circle), the whole section can besummed to an equivalent resistance.There was another parallel section of the circuit located in the lower left corner of the circuit.This section will be independently determined to obtain an equivalent resistance. Once this isdone, the circuit is nothing more than a series circuit, two resistors in red and two equivalentresistances, one defining the region within the blue circle and one defined by the red circle in thelower left corner of the circuit. This series circuit can then be easily determined.82

The third example seems to be quite busy butupon further inspection you will find a rathersimple circuit. There is a pattern of threeresistors connected in parallel connected toanother section of three resistors in parallel.This pattern of connections repeats four moretimes for a total of six parallel groups ofresistors connected in series. The six parallelsections are individually determined and thesix equivalent quantities are then handled as aseries circuit containing six resistors.The last circuit is a circuit thatcontains two large strings (series)of resistors connected in parallel.That is, there are only twoparallel branches connected to avoltage source. You only have tosum the series connectedresistors to obtain an equivalentbranch resistance for eachbranch. The circuit can then beeasily solved as a simple parallelcircuit.Once you get the hang of it, youwill be able to identify the different sections within a circuit and then determine the logicalapproach to solving for the values that need to be determined within the circuit. Yes, this meanspractice.83

Now that we can identify the relations components have within a circuit, let’s dive in and beginto solve for specific quantities from given values within a more complicated circuit.I won’t be able to run through all of the combinations possible with this circuit but we will lookat a few switch options to see the affects to the circuit.With all switches open, the circuit consists of only R1 in series with the 130 volt source. With allof the switches closed, a series parallel portion of the circuit is connected in parallel to theresistor R1. Switch 1 can be considered the master switch because no matter what condition theother switches are, switch 1 will ensure that the series parallel portion of the circuit can bedisconnected from the source voltage. Switch 2 has a special purpose. If the switch is closed, R2is bypassed (or shorted) so it is essentially out of the circuit. This switching method removescomponents from series circuits without turning the circuit completely off. Switch 3 or switch 4can individually remove either R3 or R4 from the circuit.We will now find the (1) total current, (2) the total resistance, (3) the total power, (4) the currentthru the branches, (5) the voltage drops across the resistors, (6) the power dissipated across eachresistor for the following switch conditions.Switch 1 openWith switch 1 open the only resistor in play is R1. The circuit is a simple series resistor across a130 volt source. The analysis is as follows:E source 130 voltsR total 2500Ωgivengiven84

Current of circuit is;ER130VTotal CurrentIt 2500ΩI t 0.052A or 52mAI The total power is also the power across the resistor which is:P IEP 0.052 130P 6.76WTotal PowerWhat’s interesting will be the sudden change of these totals as we manipulate the switches.Switches 1, 3, 4, closed; 2 openStep 1: let’s simplify the parallel section of the circuit. This contains R3 and R4 in parallel. Theparallel contains only two branches so we can use the product over the sum method.R eq R 3R 4R3 R425 12525 1253125 150 20.833ΩR eq R eqR eqStep 2: Sum the series resistances of R2 and Req. This equals the total equivalent resistance of thebranch.R branch R eq R 2R branch 20.833Ω 150ΩR branch 170.833ΩStep 3: Sum R1 and Rbranch as two parallelbranches toobtain total resistance of circuit. Since we have used the product over the sum method forparallel resistance, let’s do the reciprocal of the sum of the reciprocals to obtain the parallelequivalent resistance which in this case is the total resistance of the circuit.85

Rt 111 R1R branch1Rt 11 2500Ω 170.833Ω1Rt 0.0004 0.005853661Rt 0.00625366Total ResistanceR t 159.91ΩStep 4: The total circuit current can now be determined by Ohm’s Law.It EsourceRt130V159.91ΩI t 0.8129A or 812.9mAIt Total CurrentStep 5: Derive the total power of the circuit.P t IEPt 0.8129A 130VTotal powerPt 105.677WLet’s stop at this point to analyze these two totals with the first totals when all of the switcheswere open.All switches openSwitch 1,3,4 closed; 2 openTotal current52mA812.9mATotal power6.76W105.68WTotal resistance2500Ω159.9ΩIf you look at both current and power, they both increased by a factor of 15.63. On the otherhand the resistance has decreased by a factor of 15.63. So is it true that resistance is inverselyproportional to both current and power?86

Of course.Now let’s continue and find the variables of the components in this circuit. Remember that wehad a parallel circuit containing R1 and Rbranch. This is where we will begin and we willdetermine current across these branches.Step 6: Find current of R1 by using Ohm’s Law.I1 ER1130V2500ΩI1 0.052A or 52mAI1 Resistance of branch R1Do notice that this has not changed from the first switch condition when all switches were open!Step 7: Find current of Rbranch by using Ohm’s Law.I branch ER branch130VCurrent of branch Rbranch170.833Ω 0.7609A or 760.9mAI branch I branchTo check our work we can now add the two current and they should total the calculated totalcurrent as found above.0.052 0.7609 0.8129 or 812.9mA ”Ain’t” we good!!Step 8: Find power dissipated across R2. Since R2 is in series in the Rbranch circuit, the totalcurrent of the branch will flow through R2. The power dissipated across R2 can be determined asfollows:P2 I 22 R 2P2 0.76092 150Power dissipated across R2P2 86.84WStep 9: Determine voltage drops of R2 and Req. R2 is in series with the parallel components of R3and R4. An equivalent resistance has been determined to replace the parallel section forcomputations. That equivalent resistance is 20.833Ω87

Proof:V2 IR 2Veq IR eqV2 0.7609A 150ΩVeq 0.7609A 20.833ΩV2 114.135VVeq 15.85VV2 Veq 130V114.135 15.85 129.985Vbasically the identical answer.Since the voltage of Veq is 15.85V, this also represents the common voltage of the parallel circuitcomposing of R3 and R4.Step 10: Determine the current in the branches of R3 and R4. Ohm’s law can be used for bothvariables.I3 EdR3I4 15.85V25ΩI 3 0.634A or 634mAEdR415.85V125ΩI 4 0.1268A or 126.8mAI3 I4 Let’s again check top see if we are on the right path. The sum of all of the current paths shouldequal the total current that was calculated above. That was 812.9mA. We do rememberKirchhoff’s current law?88

Step 11: For the final step, we only need to find the power dissipated across the resistors R3 andR4. The wattage will be determined as follows;P3 I 3E dP3 0.634A 15.85VPower dissipated across R3P3 10.05WP4 I 4 E dP4 0.1268A 15.85VPower dissipated across R4P4 2WDid we calculate the correct power dissipation across all of the four resistors? To find proof wemust sum all of the power across all of the resistors and this should equal the original total powercalculated which is 105.68 watts.P1 P2 P3 P4 Pt6.76W 86.84 W 10.05W 2W 105.65WBasically the same.Now it’s your turn. I want you to calculate for all of the values (1) total current, (2) the totalresistance, (3) the total power, (4) the current thru the branches, (5) the voltage drops across theresistors, (6) the power dissipated across each resistor for the following switch condition.Switch 1, 2, 3 and 4 all closed89

One more thought before we go on to the next subject. I want you to compare this simple closeschematic with the circuit that we have been working on.You have just analyzed a Power Circuit Breaker close scheme.90

Circuit Re-drawsSo far we have analyzed circuit schemes that are offered in a format that can be easily defined sothat the variables can be determined. Sometimes we are forced to actually trace out a circuit fromthe actual wire board to obtain a schematic to work from. These tracings are not quite asorganized as a manufacturer’s schematic. Let’s look at some typical circuits that don’t meet ourstandard layout.I will present the drawing in rough form then we will begin our procedure to simplify thedrawing to a workable layout. The eventual task is to determine the equivalent series circuit.Drawing 1Our first step is to define all of thecommon connection points withinthe circuit. We will begin at thenegative post of the battery. There isa node that three resistors areconnected to. Being an electrician isbeing in a world of color, so we willcolor code this node yellow and allof the conductors connected to thisnode.The four resistors, 10Ω, 15Ω, 25Ω and 20Ω are all connected together to one common point. Wewill highlight these connections green. There is a third common connection point located on thethree resistors, 5Ω, 25Ω, 20Ω and the positive post of the battery. This common point with allconductors will be color highlighted in pink.91

Our next task is to re-draw the circuit and use the color highlighted connection points as aconnector bus. This circuit has three busses.The schematic now seems to have a far better order of appearance to it. It can now be simplifiedto the series equivalent circuit remembering that was our original intent for this circuit. There aretwo parallel circuit sections each containing a pair of resistors. (15Ω, 10Ω), (25Ω, 20Ω)These two circuitsbe simplified to twoequivalent resistorsconnected in series.canThe calculations toobtain the twoequivalentresistances wereperformed using theproduct over themethod forsimplifying parallel circuits.sum15Ω 10Ω15Ω 10Ω150 25 6Ω25Ω 20Ω25Ω 20Ω500 45 11.11ΩR eq1 R eq2 R eq1R eq2R eq1R eq292

The two equivalent parallel resistors are now in series and can the simply be added to obtain theseries resistance of the circuit section. The value of that resistance is 17.11Ω.The last step to obtain the equivalent series circuit is to sum the two remaining parallel branches.One branch consists of the equivalent resistance of 17.11Ω and the other branch is the remaining5Ω resistor. We can again apply the product over the sum method to solve this parallel circuit.17.11Ω 5Ω17.11Ω 5Ω85.55Rt 22.11R t 3.87ΩRt Let’s try another circuit but this time we will only simplify it to the common busses.Drawing 2As you can see, some of these rough circuit drawings can be quite busy. We will now define thiscircuit by highlighting the common connections associated with the conductors.93

The pink highlight represents a common connection between the battery positive post andresistors 35Ω, 25Ω and 10Ω. The green highlight represents common connections for seriescomponents. The yellow highlight represents the common connection point for resistors 20Ω,15Ω, 30Ω, and 5Ω. The following drawing can now be redrawn to a simpler form using thecommon busses.The circuit is now re-drawn to a simplerformat so the calculations concerning anyvariable can now be made.Please complete the exercises contained in Practice 14 – Larger Complex Circuits –94

Review1.A combination circuit contains elements of a series circuit and elements of a parallel circuit.2.The six parameters of a series circuit apply to the series portion of the combination circuit.3.The six parameters of a parallel circuit apply to the parallel portion of a combination circuit.4.To find values across components of a combination circuit, it may be necessary to simplify the circuit to aseries equivalent circuit to obtain the circuit totals.5.Sections of the combination circuit must be identified as series sections or parallel sections so the rulesinvolving these types of circuits can be applied.6.Kirchhoff’s voltage and current laws still apply.7.Abstract circuit will have to be organized so a proper analysis of the circuit may be made.You have completed Trade Theory – DC Circuits. You must complete the Final Exam with ascore of 75% or better to receive credit for this course. Use the link below to take you to theFinal Exam.DCCircuitsQuizFinalExam v0.09.tbk95

DC Circuits . Part 3. Combination Circuits. We have learned about the series circuit and the parallel circuit. Both circuits can be examined by the use of Ohm’s Law. Let’s now combine the two circuits to build a more complex circuit. This type of circuit will be called a series/parallel circuit or a combination circuit. Ohm’s Law

Related Documents:

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

1 Introduction to RL and RC Circuits Objective In this exercise, the DC steady state response of simple RL and RC circuits is examined. The transient behavior of RC circuits is also tested. Theory Overview The DC steady state response of RL and RC circuits are essential opposite of each other: that is, once steady state is reached, capacitors behave as open circuits while inductors behave as .

/ Voltage Divider Circuits Voltage Divider Circuits AC Electric Circuits Question 1 Don’t just sit there! Build something!! Learning to mathematically analyze circuits requires much study and practice. Typically, students practice by working through lots of samp

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 .

Lab Experiment 7 Series-Parallel Circuits and In-circuit resistance measurement Series-Parallel Circuits Most practical circuits in electronics are made up combinations of both series and parallel circuits. These circuits are made up of all sorts of components such as resistors, capacitors, inductors, diodes, transistors and integrated circuits.

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

Part No : MS-HTB-4 Part No : MS-HTB-6M Part No : MS-HTB-6T Part No : MS-HTB-8 Part No : MS-TBE-2-7-E-FKIT Part No : MS-TC-308 Part No : PGI-63B-PG5000-LAO2 Part No : RTM4-F4-1 Part No : SS 316 Part No : SS 316L Part No : SS- 43 ZF2 Part No : SS-10M0-1-8 Part No : SS-10M0-6 Part No : SS-12?0-2-8 Part No : SS-12?0-7-8 Part No : SS-1210-3 Part No .

EXTERIOR WALLS Weathertightness AAMA 501-15 TF & F Air leakage ASTM E 283-04(2012) TF & F Water penetration ASTM E 331-00(2016) TF & F Structural performance ASTM E 330/330M-14 TF & F CURTAIN WALLING Impact resistance of opaque wall components - hard body impact tests BS 8200:1985 TF Impact resistance of opaque wall components - soft body impact tests BS 8200:1985 TF Impact resistance BS EN .