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RELIABILITY LEVEL PER 1,000 HRS.
BLACK 0 1 PERCENT 1.0% BROWN 1 10 1 .1% RED 2 100 2 .01% ORANGE 3 1,000 . 001% YELLOW 4 10,000 — — GREEN 5 100,000 — — BLUE 6 1,000,000 — — VIOLET 7 10,000,000 — — GRAY 8 100,000,000 — — WHITE 9 1,000,000,000 — — GOLD — .1 5 — SILVER — .01 10 — NO COLOR — — 20 —
Table 1-1.—Standard Color Code for Resistors
THE WATTAGE RATING OF A RESISTOR is related to the resistor’s physical size, that is, the greater the surface area exposed to the air, the larger the rating.
THE STANDARD COLOR CODE for resistors is used to determine the following:
1. Ohmic value 2. Tolerance 3. Reliability level (on some resistors)
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ANSWERS TO QUESTIONS Q1. THROUGH Q65.
A1. Anything that occupies space and has weight. Solids, liquids, gases.
A2. A substance which cannot be reduced to a simpler substance by chemical means.
A3. A substance consisting of two or more elements. A4. A compound is a chemical combination of elements that cannot be separated by physical means. A
mixture is a physical combination of elements and compounds that are not chemically combined. A5. A chemical combination of two or more atoms. A6. Electrons-negative, protons-positive, and neutrons-neutral.
A7. Kinetic energy.
A8. Invisible light photons (ultraviolet) bombard the phosphor atom in the light tube. The phosphor atoms emit visible light photons.
A9. The number of electrons in the outer shell. A10. An atom with more or less than its normal number of electrons. A11. The number of valence electrons. A12. Through the accumulation of excess electrons. A13. By friction. A14. Negative.
A15. Like charges repel, and unlike charges attract with a force directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
A16. The space between and around charged bodies.
A17. Leaving positive, entering negative.
A18. Motors, generators, speakers, computers, televisions, tape recorders, and many others.
A19. Those materials that are attracted by magnets and have the ability to become magnetized.
A20. The relative ease with which they are magnetized.
A21. A material that exhibits low reluctance and high permeability, such as iron or soft steel. A22. The ability of a material to retain magnetism. A23. They are very similar; like charges repel, unlike charges attract, like poles repel —unlike poles
attract.
A24. To the magnetic north pole.
A25. South pole at the right, north pole at the left.
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A26. The domain theory is based upon the electron spin principle; Weber’s theory uses the concept of tiny molecular magnets.
A27. To enable you to "see" the magnetic field. A28. No specific pattern, sawdust is a nonmagnetic material. A29. An imaginary line used to illustrate magnetic effects. A30. Electrostatic lines of force do not form closed loops. A31. By shielding or surrounding the instrument with a soft iron case, called a magnetic shield or
screen.
A32. In pairs, with opposite poles together to provide a complete path for magnetic flux. A33. The ability to do work.
A34. Kinetic energy.
A35. Potential energy.
A36. Difference of potential. A37. 2100 volts.
A38. (a) 250 kV, (b) 25 V, (c) 1 9� A39. A voltage source.
A40. Friction, pressure, heat, light, chemical action, and magnetism.
A41. Pressure.
A42. Heat.
A43. Chemical.
A44. Magnetic.
A45. Electron theory assumes that electron flow is from negative to positive. A46. The speed of light (186,000 miles per second, 300,000,000 meters per second). A47. Current increases as voltage increases.
A48. 0.35 amperes.
A49.
A50. When the need for conductivity is great enough to justify the additional expense. A51. Wire B.
A52. Positive.
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A53. Zero temperature coefficient. A54. The mho (v), siemens. A55. They are reciprocals of each other. A56.
A57. Its ability to dissipate heat.
A58. 1. Change value with age. 2. Limited power capacity.
A59. The wirewound resistor.
A60. The rheostat may have two connections, one fixed and one moveable; the potentiometer always has three connections, one moveable and two fixed.
A61. The rheostat.
A62. The bands are green, black, black, and gold.
A63. 470,000 ohms (470 kilohms). A64. The resistor’s chance of failure is 0.001 percent for 1000 hours of operation. A65. 6,800,000 ohms (6.8 megohms), with 5% tolerance, and a 1% reliability level.
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Upon completing this chapter, you will be able to:
1. State the purpose of a cell.
2. State the purpose of the three parts of a cell.
3. State the difference between the two types of cells.
4. Explain the chemical process that takes place in the primary and secondary cells.
5. Recognize and define the terms electrochemical action, anode, cathode, and electrolyte.
6. State the causes of polarization and local action and describe methods of preventing these effects.
7. Identify the parts of a dry cell.
8. Identify the various dry cells in use today and some of their capabilities and limitations.
9. Identify the four basic secondary cells, their construction, capabilities, and limitations.
10. Define a battery, and identify the three ways of combining cells to form a battery.
11. Describe general maintenance procedures for batteries including the use of the hydrometer, battery capacity, and rating and battery charging.
12. Identify the five types of battery charges.
13. Observe the safety precautions for working with and around batteries.
The purpose of this chapter is to introduce and explain the basic theory and characteristics of batteries. The batteries which are discussed and illustrated have been selected as representative of many models and types which are used in the Navy today. No attempt has been made to cover every type of battery in use, however, after completing this chapter you will have a good working knowledge of the batteries which are in general use.
First, you will learn about the building block of all batteries, the CELL. The explanation will explore the physical makeup of the cell and the methods used to combine cells to provide useful voltage, current, and power. The chemistry of the cell and how chemical action is used to convert chemical energy to electrical energy are also discussed.
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In addition, the care, maintenance, and operation of batteries, as well as some of the safety precautions that should be followed while working with and around batteries are discussed.
Batteries are widely used as sources of direct-current electrical energy in automobiles, boats, aircraft, ships, portable electric/electronic equipment, and lighting equipment. In some instances, they are used as the only source of power; while in others, they are used as a secondary or standby power source.
A battery consists of a number of cells assembled in a common container and connected together to function as a source of electrical power.
A cell is a device that transforms chemical energy into electrical energy. The simplest cell, known as either a galvanic or voltaic cell, is shown in figure 2-1. It consists of a piece of carbon (C) and a piece of zinc (Zn) suspended in a jar that contains a solution of water (H20) and sulfuric acid (H2S0 4) called the electrolyte.
The cell is the fundamental unit of the battery. A simple cell consists of two electrodes placed in a container that holds the electrolyte.
In some cells the container acts as one of the electrodes and, in this case, is acted upon by the electrolyte. This will be covered in more detail later.
The electrodes are the conductors by which the current leaves or returns to the electrolyte. In the simple cell, they are carbon and zinc strips that are placed in the electrolyte; while in the dry cell (fig. 2-2), they are the carbon rod in the center and zinc container in which the cell is assembled.
2-3
The electrolyte is the solution that acts upon the electrodes. The electrolyte, which provides a path for electron flow, may be a salt, an acid, or an alkaline solution. In the simple galvanic cell, the electrolyte is in a liquid form. In the dry cell, the electrolyte is a paste.
The container which may be constructed of one of many different materials provides a means of holding (containing) the electrolyte. The container is also used to mount the electrodes. In the voltaic cell the container must be constructed of a material that will not be acted upon by the electrolyte.
Q1. What is the purpose of a cell?
Q2. What are the three parts of a cell?
Q3. What is the purpose of each of the three parts of a cell?
A primary cell is one in which the chemical action eats away one of the electrodes, usually the negative electrode. When this happens, the electrode must be replaced or the cell must be discarded. In the galvanic-type cell, the zinc electrode and the liquid electrolyte are usually replaced when this happens. In the case of the dry cell, it is usually cheaper to buy a new cell.
A secondary cell is one in which the electrodes and the electrolyte are altered by the chemical action that takes place when the cell delivers current. These cells may be restored to their original condition by forcing an electric current through them in the direction opposite to that of discharge. The automobile storage battery is a common example of the secondary cell.
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Q4. What are the two types of cells?
Q5. What is the main difference between the two types of cells?
If a load (a device that consumes electrical power) is connected externally to the electrodes of a cell, electrons will flow under the influence of a difference in potential across the electrodes from the CATHODE (negative electrode), through the external conductor to the ANODE (positive electrode).
A cell is a device in which chemical energy is converted to electrical energy. This process is called ELECTROCHEMICAL action.
The voltage across the electrodes depends upon the materials from which the electrodes are made and the composition of the electrolyte. The current that a cell delivers depends upon the resistance of the entire circuit, including that of the cell itself. The internal resistance of the cell depends upon the size of the electrodes, the distance between them in the electrolyte, and the resistance of the electrolyte. The larger the electrodes and the closer together they are in the electrolyte (without touching), the lower the internal resistance of the cell and the more current the cell is capable of supplying to the load.
Q6. What is electrochemical action?
Q7. What is another name for the (a) positive electrode, and the (b) negative electrode?
When a current flows through a primary cell having carbon and zinc electrodes and a diluted solution of sulfuric acid and water (combined to form the electrolyte), the following chemical reaction takes place.
The current flow through the load is the movement of electrons from the negative electrode of the cell (zinc) and to the positive electrode (carbon). This causes fewer electrons in the zinc and an excess of electrons in the carbon. Figure 2-1 shows the hydrogen ions (H2) from the sulfuric acid being attracted to the carbon electrode. Since the hydrogen ions are positively charged, they are attracted to the negative charge on the carbon electrode. This negative charge is caused by the excess of electrons. The zinc electrode has a positive charge because it has lost electrons to the carbon electrode. This positive charge attracts the negative ions (S04) from the sulfuric acid. The negative ions combine with the zinc to form zinc sulfate. This action causes the zinc electrode to be eaten away. Zinc sulfate is a grayish-white substance that is sometimes seen on the battery post of an automobile battery.
The process of the zinc being eaten away and the sulfuric acid changing to hydrogen and zinc sulfate is the cause of the cell discharging. When the zinc is used up, the voltage of the cell is reduced to zero.
In figure 2-1 you will notice that the zinc electrode is labeled negative and the carbon electrode is labeled positive. This represents the current flow outside the cell from negative to positive.
The zinc combines with the sulfuric acid to form zinc sulfate and hydrogen. The zinc sulfate dissolves in the electrolyte (sulfuric acid and water) and the hydrogen appears as gas bubbles around the carbon electrode. As current continues to flow, the zinc gradually dissolves and the solution changes to zinc sulfate and water. The carbon electrode does not enter into the chemical changes taking place, but simply provides a return path for the current.
2-5
Q8. In the primary cell, why are negative ions attracted to the negative terminal of the cell?
Q9. How do electrons get from the negative electrode to the positive electrode?
Q10. What causes the negative electrode to be eaten away?
As stated before, the differences between primary and secondary cells are, the secondary cell can be recharged and the electrodes are made of different materials. The secondary cell shown in figure 2-3 uses sponge lead as the cathode and lead peroxide as the anode. This is the lead-acid type cell and will be used to explain the general chemistry of the secondary cell. Later in the chapter when other types of secondary cells are discussed, you will see that the materials which make up the parts of a cell are different, but that the chemical action is essentially the same.
2-6
Figure 2-3 view A shows a lead-acid secondary cell that is fully charged. The cathode is pure sponge lead, the anode is pure lead peroxide, and the electrolyte is a mixture of sulfuric acid and water.
Figure 2-3 view B shows the secondary cell discharging. A load is connected between the cathode and anode; current flows negative to positive as shown. This current flow creates the same process as was explained for the primary cell with the following exceptions.
In the primary cell the zinc cathode was eaten away by the sulfuric acid. In the secondary cell the sponge-like construction of the cathode retains the lead sulfate formed by the chemical action of the sulfuric acid and the lead. In the primary cell the carbon anode was not chemically acted upon by the sulfuric acid. In the secondary cell the lead peroxide anode is chemically changed to lead sulfate by the sulfuric acid.
When the cell is fully discharged it will be as shown in figure 2-3 view C. The anode and cathode retain some lead peroxide and sponge lead but the amounts of lead sulfate in each is maximum. The electrolyte has a minimum amount of sulfuric acid. With this condition no further chemical action can take place within the cell.
As you know, the secondary cell can be recharged. Recharging is the process of reversing the chemical action that occurs as the cell discharges. To recharge the cell, a voltage source, such as a generator, is connected as shown in figure 2-3 view D. The negative terminal of the voltage source is connected to the cathode of the cell and the positive terminal of the voltage source is connected to the anode of the cell. With this arrangement the lead sulfate is chemically changed back to sponge lead in the cathode, lead peroxide in the anode, and sulfuric acid in the electrolyte. After all the lead sulfate is chemically changed, the cell is fully charged as shown in figure 2-3 view A. Once the cell has been charged, the discharge-charge cycle may be repeated.
Q11. Refer to figure 2-3(B). Why is the sulfuric acid decreasing?
Q12. Refer to figure 2-3(D). How is it possible for the sulfuric acid to be increasing?
Q13. Refer to figure 2-3(D). When all the lead sulfate has been converted, what is the condition of the cell?
The chemical action that occurs in the cell while the current is flowing causes hydrogen bubbles to form on the surface of the anode. This action is called POLARIZATION. Some hydrogen bubbles rise to the surface of the electrolyte and escape into the air, some remain on the surface of the anode. If enough bubbles remain around the anode, the bubbles form a barrier that increases internal resistance. When the internal resistance of the cell increases, the output current is decreased and the voltage of the cell also decreases.
A cell that is heavily polarized has no useful output. There are several methods to prevent polarization or to depolarize the cell.
One method uses a vent on the cell to permit the hydrogen to escape into the air. A disadvantage of this method is that hydrogen is not available to reform into the electrolyte during recharging. This problem is solved by adding water to the electrolyte, such as in an automobile battery. A second method is to use material that is rich in oxygen, such as manganese dioxide, which supplies free oxygen to combine with the hydrogen and form water.
2-7
A third method is to use a material that will absorb the hydrogen, such as calcium. The calcium releases hydrogen during the charging process. All three methods remove enough hydrogen so that the cell is practically free from polarization.
When the external circuit is removed, the current ceases to flow, and, theoretically, all chemical action within the cell stops. However, commercial zinc contains many impurities, such as iron, carbon, lead, and arsenic. These impurities form many small electrical cells within the zinc electrode in which current flows between the zinc and its impurities. Thus, the chemical action continues even though the cell itself is not connected to a load.
Local action may be prevented by using pure zinc (which is not practical), by coating the zinc with mercury, or by adding a small percentage of mercury to the zinc during the manufacturing process. The treatment of the zinc with mercury is called amalgamating (mixing) the zinc. Since mercury is many times heavier than an equal volume of water, small particles of impurities weighing less than mercury will float to the surface of the mercury. The removal of these impurities from the zinc prevents local action. The mercury is not readily acted upon by the acid. When the cell is delivering current to a load, the mercury continues to act on the impurities in the zinc. This causes the impurities to leave the surface of the zinc electrode and float to the surface of the mercury. This process greatly increases the storage life of the cell.
Q14. Describe three ways to prevent polarization.
Q15. Describe local action
.
The development of new and different types of cells in the past decade has been so rapid that it is virtually impossible to have a complete knowledge of all the various types. A few recent developments are the silver-zinc, nickel-zinc, nickel-cadmium, silver-cadmium, organic and inorganic lithium, and mercury cells.
The dry cell is the most popular type of primary cell. It is ideal for simple applications where an inexpensive and noncritical source of electricity is all that is needed.
The dry cell is not actually dry. The electrolyte is not in a liquid state, but is a moist paste. If it should become totally dry, it would no longer be able to transform chemical energy to electrical energy.
The construction of a common type of dry cell is shown in figure 2-4. These dry cells are also referred to as Leclanche' cells. The internal parts of the cell are located in a cylindrical zinc container. This zinc container serves as the negative electrode (cathode) of the cell. The container is lined with a nonconducting material, such as blotting paper, to separate the zinc from the paste. A carbon electrode is located in the center, and it serves as the positive terminal (anode) of the cell. The paste is a mixture of several substances such as ammonium chloride, powdered coke, ground carbon, manganese dioxide, zinc chloride, graphite, and water.
2-8
This paste, which is packed in the space between the anode and the blotting paper, also serves to hold the anode rigid in the center of the cell. When the paste is packed in the cell, a small space is left at the top for expansion of the electrolytic paste caused by the depolarization action. The cell is then sealed with a cardboard or plastic seal.
Since the zinc container is the cathode, it must be protected with some insulating material to be electrically isolated. Therefore, it is common practice for the manufacturer to enclose the cells in cardboard and metal containers.
The dry cell (fig. 2-4) is basically the same as the simple voltaic cell (wet cell) described earlier, as far as its internal chemical action is concerned. The action of the water and the ammonium chloride in the paste, together with the zinc and carbon electrodes, produces the voltage of the cell. Manganese dioxide is added to reduce polarization when current flows and zinc chloride reduces local action when the cell is not being used.
A cell that is not being used (sitting on the shelf) will gradually deteriorate because of slow internal chemical changes (local action). This deterioration is usually very slow if cells are properly stored. If unused cells are stored in a cool place, their shelf life will be greatly preserved. Therefore, to minimize deterioration, they should be stored in refrigerated spaces.
The blotting paper (paste-coated pulpboard separator) serves two purposes—(1) it keeps the paste from making actual contact with the zinc container and (2) it permits the electrolyte from the paste to filter through to the zinc slowly. The cell is sealed at the top to keep air from entering and drying the electrolyte. Care should be taken to prevent breaking this seal.
Q16. What serves as the cathode of a dry cell?
Q17. Why is a dry cell called a DRY cell?
Q18. What does the term "shelf life" mean?
2-9
The mercuric-oxide zinc cell (mercury cell) is a primary cell that was developed during World War II. Two important assets of the mercury cell are its ability to produce current for a long period of time and a long shelf life when compared to the dry cell shown in figure 2-4. The mercury cell also has a very stable output voltage and is a power source that can be made in a small physical size.
With the birth of the space program and the development of small transceivers and miniaturized equipment, a power source of small size was needed. Such equipment requires a small cell which is capable of delivering maximum electrical energy at a constant discharge voltage. The mercury cell, which is one of the smallest cells, meets these requirements.
Present mercury cells are manufactured in three basic types as shown in figure 2-5. The wound- anode type, shown in figure 2-5 view A, has an anode composed of a corrugated zinc strip with a paper absorbent. The zinc is mixed with mercury, and the paper is soaked in the electrolyte which causes it to swell and press against the zinc and make positive contact. This process ensures that the electrolyte makes contact with the anode.
In the pressed-powder cells, shown in figure 2-5 views B and C, the zinc powder for the anode is mixed prior to being pressed into shape. The absorbent shown in the figure is paper soaked in the electrolyte. The space between the inner and outer containers provides passage for any gas generated by an improper chemical balance or impurities present within the cell.
If the anode and cathode of a cell are connected together without a load, a SHORT CIRCUIT condition exists. Short circuits (shorts) can be very dangerous. They cause excessive heat, pressure, and current flow which may cause serious damage to the cell or be a safety hazard to personnel.
2-10
Do not short the mercury cell. Shorted mercury cells have exploded with considerable force.
There are many different types of primary cells. Because of such factors as cost, size, ease of replacement, and voltage or current needs, many types of primary cells have been developed. The following is a brief description of some of the primary cells in use today.
The Manganese Dioxide-Alkaline-Zinc Cell is similar to the zinc-carbon cell except for the electrolyte used. This type of cell offers better voltage stability and longer life than the zinc-carbon type. It also has a longer shelf life and can operate over a wide temperature range. The manganese dioxide- alkaline-zinc cell has a voltage of 1.5 volts and is available in a wide range of sizes. This cell is commonly referred to as the alkaline cell.
The Magnesium-Manganese Dioxide Cell uses magnesium as the anode material. This allows a higher output capacity over an extended period of time compared to the zinc-carbon cell. This cell produces a voltage of approximately 2 volts. The disadvantage of this type of cell is the production of hydrogen during its operation.
The Lithium-Organic Cell and the Lithium-Inorganic Cell are recent developments of a new line of high-energy cells. The main advantages of these types of cells are very high power, operation over a wide temperature range, they are lighter than most cells, and have a remarkably long shelf life of up to 20 years.
Lithium cells contain toxic materials under pressure. Do not puncture, recharge, short-circuit, expose to excessively high temperatures, or incinerate. Use these batteries/cells only in approved equipment. Do not throw in trash.
Q19. Why should a mercury cell NOT be shorted?
Q20. What factors should be considered when selecting a primary cell for a power source?
Secondary cells are sometimes known as wet cells. There are four basic type of wet cells, the lead- acid, nickel-cadmium, silver-zinc, and silver-cadmium.
The lead-acid cell is the most widely used secondary cell. The previous explanation of the secondary cell describes exactly the manner in which the lead-acid cell provides electrical power. The discharging and charging action presented in electrochemical action describes the lead-acid cell.
You should recall that the lead-acid cell has an anode of lead peroxide, a cathode of sponge lead, and the electrolyte is sulfuric acid and water.
2-11
The nickel-cadmium cell (NICAD) is far superior to the lead-acid cell. In comparison to lead-acid cells, these cells generally require less maintenance throughout their service life in regard to the adding of electrolyte or water. The major difference between the nickel-cadmium cell and the lead-acid cell is the material used in the cathode, anode, and electrolyte. In the nickel-cadmium cell the cathode is cadmium hydroxide, the anode is nickel hydroxide, and the electrolyte is potassium hydroxide and water.
The nickel-cadmium and lead-acid cells have capacities that are comparable at normal discharge rates, but at high discharge rates the nickel-cadmium cell can deliver a larger amount of power. In addition the nickel-cadmium cell can:
1. Be charged in a shorter time,
2. Stay idle longer in any state of charge and keep a full charge when stored for a longer period of time, and
3. Be charged and discharged any number of times without any appreciable damage.
Due to their superior capabilities, nickel-cadmium cells are being used extensively in many military applications that require a cell with a high discharge rate. A good example is in the aircraft storage battery.
The silver-zinc cell is used extensively to power emergency equipment. This type of cell is relatively expensive and can be charged and discharged fewer times than other types of cells. When compared to the lead-acid or nickel-cadmium cells, these disadvantages are overweighed by the light weight, small size, and good electrical capacity of the silver-zinc cell.
The silver-zinc cell uses the same electrolyte as the nickel-cadmium cell (potassium hydroxide and water), but the anode and cathode differ from the nickel-cadmium cell. The anode is composed of silver oxide and the cathode is made of zinc.
The silver-cadmium cell is a fairly recent development for use in storage batteries. The silver- cadmium cell combines some of the better features of the nickel-cadmium and silver-zinc cells. It has more than twice the shelf life of the silver-zinc cell and can be recharged many more times. The disadvantages of the silver-cadmium cell are high cost and low voltage production.
The electrolyte of the silver-cadmium cell is potassium hydroxide and water as in the nickel- cadmium and silver-zinc cells. The anode is silver oxide as in the silver-zinc cell and the cathode is cadmium hydroxide as in the nicad cell. You may notice that different combinations of materials are used to form the electrolyte, cathode, and anode of different cells. These combinations provide the cells with different qualities for many varied applications.
Q21. What are the four basic types of secondary (wet) cells?
Q22. What are the advantages of a nicad cell over a lead-acid cell?
Q23. What type of cell is most commonly used for emergency systems?
Q24. What three cells use the same electrolyte?
2-12
A battery is a voltage source that uses chemical action to produce a voltage. In many cases the term battery is applied to a single cell, such as the flashlight battery. In the case of a flashlight that uses a battery of 1.5 volts, the battery is a single cell. The flashlight that is operated by 6 volts uses four cells in a single case and this is a battery composed of more than one cell. There are three ways to combine cells to form a battery.
In many cases, a battery-powered device may require more electrical energy than one cell can provide. The device may require either a higher voltage or more current, and in some cases both. Under such conditions it is necessary to combine, or interconnect, a sufficient number of cells to meet the higher requirements. Cells connected in SERIES provide a higher voltage, while cells connected in PARALLEL provide a higher current capacity. To provide adequate power when both voltage and current requirements are greater than the capacity of one cell, a combination SERIES-PARALLEL network of cells must be used.
Assume that a load requires a power supply of 6 volts and a current capacity of 1/8 ampere. Since a single cell normally supplies a voltage of only 1.5 volts, more than one cell is needed. To obtain the higher voltage, the cells are connected in series as shown in figure 2-6.
Figure 2-6 view B is a schematic representation of the circuit shown in figure 2-6 view A. The load is shown by the resistance symbol and the battery is indicated by one long and one short line per cell.
In a series hookup, the negative electrode (cathode) of the first cell is connected to the positive electrode (anode) of the second cell, the negative electrode of the second to the positive of the third, etc.
2-13
The positive electrode of the first cell and negative electrode of the last cell then serve as the terminals of the battery. In this way, the voltage is 1.5 volts for each cell in the series line. There are four cells, so the output terminal voltage is 1.5 × 4, or 6 volts. When connected to the load, 1/8 ampere flows through the load and each cell of the battery. This is within the capacity of each cell. Therefore, only four series-connected cells are needed to supply this particular load.
When connecting cells in series, connect alternate terminals together (−−−− to +, −−−− to +, etc.) Always have two remaining terminals that are used for connection to the load only. Do not connect the two remaining terminals together as this is a short across the battery and would not only quickly discharge the cells but could cause some types of cells to explode.
In this case, assume an electrical load requires only 1.5 volts, but will require 1/2 ampere of current. (Assume that a single cell will supply only 1/8 ampere.) To meet this requirement, the cells are connected in parallel, as shown in figure 2-7 view A and schematically represented in 2-7 view B. In a parallel connection, all positive cell electrodes are connected to one line, and all negative electrodes are connected to the other. No more than one cell is connected between the lines at any one point; so the voltage between the lines is the same as that of one cell, or 1.5 volts. However, each cell may contribute its maximum allowable current of 1/8 ampere to the line. There are four cells, so the total line current is 1/8 × 4, or 1/2 ampere. In this case four cells in parallel have enough capacity to supply a load requiring 1/2 ampere at 1.5 volts.
2-14
Figure 2-8 depicts a battery network supplying power to a load requiring both a voltage and a current greater than one cell can provide. To provide the required 4.5 volts, groups of three 1.5-volt cells are connected in series. To provide the required 1/2 ampere of current, four series groups are connected in parallel, each supplying 1/8 ampere of current.
The connections shown have been used to illustrate the various methods of combining cells to form a battery. Series, parallel, and series-parallel circuits will be covered in detail in the next chapter, "Direct Current."
Some batteries are made from primary cells. When a primary-cell battery is completely discharged, the entire battery must be replaced. Because there is nothing else that can be done to primary cell batteries, the rest of the discussion on batteries will be concerned with batteries made of secondary cells.
Q25. What does the term battery normally refer to?
Q26. What are the three ways of combining cells, and what is each used for?
Secondary cell batteries are constructed using the various secondary cells already described. The lead-acid battery is one of the most common batteries in use today and will be used to explain battery construction. The nickel-cadmium battery is being used with increasing frequency and will also be discussed.
Figure 2-9 shows the makeup of a lead-acid battery. The container houses the separate cells. Most containers are hard rubber, plastic, or some other material that is resistant to the electrolyte and mechanical shock and will withstand extreme temperatures. The container (battery case) is vented through vent plugs to allow the gases that form within the cells to escape. The plates in the battery are the cathodes and anodes that were discussed earlier. In figure 2-10 the negative plate group is the cathode of the individual cells and the positive plate group is the anode. As shown in the figure, the plates are interlaced with a terminal attached to each plate group. The terminals of the individual cells are connected together by link connectors as shown in figure 2-9. The cells are connected in series in the battery and the
2-15
positive terminal of one end cell becomes the positive terminal of the battery. The negative terminal of the opposite end cell becomes the negative terminal of the battery.
The terminals of a lead-acid battery are usually identified from one another by their size and markings. The positive terminal, marked (+) is sometimes colored red and is physically larger than the negative terminal, marked (−).
2-16
The individual cells of the lead-acid battery are not replaceable, so in the event one cell fails the battery must be replaced.
The nickel-cadmium battery is similar in construction to the lead-acid battery with the exception that it has individual cells which can be replaced. The cell of the nicad battery is shown in figure 2-11.
The construction of secondary cell batteries is so similar, that it is difficult to distinguish the type of battery by simply looking at it. The type of battery must be known to properly check or recharge the battery. Each battery should have a nameplate that gives a description of its type and electrical characteristics.
Q27. Other than the type of cell used, what is the major difference between the construction of the lead- acid and nicad battery?
Q28. How is the type of battery most easily determined?
The following information concerns the maintenance of secondary-cell batteries and is of a general nature. You must check the appropriate technical manuals for the specific type of battery prior to performing maintenance on any battery.
For a battery to work properly, its electrolyte (water plus active ingredient) must contain a certain amount of active ingredient. Since the active ingredient is dissolved in the water, the amount of active ingredient cannot be measured directly. An indirect way to determine whether or not the electrolyte contains the proper amount of active ingredient is to measure the electrolyte's specific gravity. Specific gravity is the ratio of the weight of a certain amount of a given substance compared to the weight of the same amount of pure water. The specific gravity of pure water is 1.0. Any substance that floats has a specific gravity less than 1.0. Any substance that sinks has a specific gravity greater than 1.0.
2-17
The active ingredient in electrolyte (sulfuric acid, potassium hydroxide, etc.) is heavier than water. Therefore, the electrolyte has a specific gravity greater than 1.0. The acceptable range of specific gravity for a given battery is provided by the battery's manufacturer. To measure a battery's specific gravity, use an instrument called a HYDROMETER.
A hydrometer, shown in figure 2-12, is a glass syringe with a float inside it. The float is a hollow glass tube sealed at both ends and weighted at the bottom end, with a scale calibrated in specific gravity marked on its side. To test an electrolyte, draw it into the hydrometer using the suction bulb. Draw enough electrolyte into the hydrometer to make the float rise. Do not draw in so much electrolyte that the float rises into the suction bulb. The float will rise to a point determined by the specific gravity of the electrolyte. If the electrolyte contains a large amount of active ingredient, its specific gravity will be relatively high. The float will rise higher than it would if the electrolyte contained only a small amount of active ingredient.
To read the hydrometer, hold it in a vertical position and read the scale at the point that surface of the electrolyte touches the float. Refer to the manufacturer's technical manual to determine whether or not the battery's specific gravity is within specifications.
Note: Hydrometers should be flushed with fresh water after each use to prevent inaccurate readings. Storage battery hydrometers must not be used for any other purpose.
Q29. What is the purpose of the hydrometer?
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Q30. Which electrolyte has more active ingredient? Electrolyte A, specific gravity 1.015? Electrolyte B, specific gravity 1.125?
The routine maintenance of a battery is very simple. Terminals should be checked periodically for cleanliness and good electrical connection. The battery case should be inspected for cleanliness and evidence of damage. The level of electrolyte should be checked and if the electrolyte is low, distilled water should be added to bring the electrolyte to the proper level. Maintenance procedures for batteries are normally determined by higher authority and each command will have detailed procedures for battery care and maintenance.
All types of batteries should be handled with care:
1. NEVER SHORT THE TERMINALS OF A BATTERY.
2. CARRYING STRAPS SHOULD BE USED WHEN TRANSPORTING BATTERIES.
3.
4. NO SMOKING, ELECTRIC SPARKS, OR OPEN FLAMES SHOULD BE PERMITTED NEAR CHARGING BATTERIES.
5. CARE SHOULD BE TAKEN TO PREVENT SPILLING OF THE ELECTROLYTE.
In the event electrolyte is splashed or spilled on a surface, such as the floor or table, it should be diluted with large quantities of water and cleaned up immediately.
If the electrolyte is spilled or splashed on the skin or eyes, IMMEDIATELY flush the skin or eyes with large quantities of fresh water for a minimum of 15 minutes. If the electrolyte is in the eyes, be sure the upper and lower eyelids are pulled out sufficiently to allow the fresh water to flush under the eyelids. The medical department should be notified as soon as possible and informed of the type of electrolyte and the location of the accident.
The CAPACITY of a battery is measured in ampere-hours. The ampere-hour capacity is equal to the product of the current in amperes and the time in hours during which the battery will supply this current. The ampere-hour capacity varies inversely with the discharge current. For example, a 400 ampere-hour battery will deliver 400 amperes for 1 hour or 100 amperes for 4 hours.
Storage batteries are RATED according to their rate of discharge and ampere-hour capacity. Most batteries are rated according to a 20-hour rate of discharge. That is, if a fully charged battery is completely discharged during a 20-hour period, it is discharged at the 20-hour rate. Thus, if a battery can deliver 20 amperes continuously for 20 hours, the battery has a rating of 20 amperes × 20 hours, or 400 ampere-hours. Therefore, the 20-hour rating is equal to the average current that a battery is capable of supplying without interruption for an interval of 20 hours. (Note: Aircraft batteries are rated according to a 1-hour rate of discharge.)
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All standard batteries deliver 100 percent of their available capacity if discharged in 20 hours or more, but they will deliver less than their available capacity if discharged at a faster rate. The faster they discharge, the less ampere-hour capacity they have.
The low-voltage limit, as specified by the manufacturer, is the limit beyond which very little useful energy can be obtained from a battery. This low-voltage limit is normally a test used in battery shops to determine the condition of a battery.
Q31. When should safety precautions pertaining to batteries be observed?
Q32. How long should a 200 ampere-hour battery be able to deliver 5 amperes?
It should be remembered that adding the active ingredient to the electrolyte of a discharged battery does not recharge the battery. Adding the active ingredient only increases the specific gravity of the electrolyte and does not convert the plates back to active material, and so does not bring the battery back to a charged condition. A charging current must be passed through the battery to recharge it.
Batteries are usually charged in battery shops. Each shop will have specific charging procedures for the types of batteries to be charged. The following discussion will introduce you to the types of battery charges.
The following types of charges may be given to a storage battery, depending upon the condition of the battery:
1. Initial charge
2. Normal charge
3. Equalizing charge
4. Floating charge
5. Fast charge
When a new battery is shipped dry, the plates are in an uncharged condition. After the electrolyte has been added, it is necessary to charge the battery. This is accomplished by giving the battery a long low- rate initial charge. The charge is given in accordance with the manufacturer's instructions, which are shipped with each battery. If the manufacturer's instructions are not available, reference should be made to the detailed instructions for charging batteries found in current Navy directives.
A normal charge is a routine charge that is given in accordance with the nameplate data during the ordinary cycle of operation to restore the battery to its charged condition.
An equalizing charge is a special extended normal charge that is given periodically to batteries as part of a maintenance routine. It ensures that all the sulfate is driven from the plates and that all the cells
2-20
are restored to a maximum specific gravity. The equalizing charge is continued until the specific gravity of all cells, corrected for temperature, shows no change for a 4-hour period.
In a floating charge, the charging rate is determined by the battery voltage rather than by a definite current value. The floating charge is used to keep a battery at full charge while the battery is idle or in light duty. It is sometimes referred to as a trickle charge and is accomplished with low current.
A fast charge is used when a battery must be recharged in the shortest possible time. The charge starts at a much higher rate than is normally used for charging. It should be used only in an emergency, as this type charge may be harmful to the battery.
Normally, the charging rate of Navy storage batteries is given on the battery nameplate. If the available charging equipment does not have the desired charging rates, the nearest available rates should be used. However, the rate should never be so high that violent gassing (explained later in this text) occurs.
The charge must be continued until the battery is fully charged. Frequent readings of specific gravity should be taken during the charge and compared with the reading taken before the battery was placed on charge.
When a battery is being charged, a portion of the energy breaks down the water in the electrolyte. Hydrogen is released at the negative plates and oxygen at the positive plates. These gases bubble up through the electrolyte and collect in the air space at the top of the cell. If violent gassing occurs when the battery is first placed on charge, the charging rate is too high. If the rate is not too high, steady gassing develops as the charging proceeds, indicating that the battery is nearing a fully charged condition.
A mixture of hydrogen and air can be dangerously explosive. No smoking, electric sparks, or open flames should be permitted near charging batteries.
Q33. Can a battery be recharged by adding more electrolyte?
Q34. If violent gassing occurs during a battery charge, what action should be taken?
In this chapter you learned that batteries are widely used as sources of direct-current. You were introduced to electrochemical action and the way it works in a cell, the cell itself, the type and parts of a cell, and how cells are connected together to form batteries. You learned the construction and maintenance of batteries and some of the safety precautions in handling and working with batteries.
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Several new terms were introduced in this chapter. The following is a summary of the chapter on batteries.
A CELL is a device that transforms chemical energy into electrical energy. The cell has three parts; the electrodes, the electrolyte, and the container. There are two basic cells: primary and secondary.
THE ELECTRODES are the current conductors of the cell.
THE ELECTROLYTE is the solution that acts upon the electrodes.
THE CONTAINER holds the electrolyte and provides a means of mounting the electrodes.
THE PRIMARY CELL is a cell in which the chemical action finally destroys one of the electrodes, usually the negative. The primary cell cannot be recharged.
THE SECONDARY CELL is a cell in which the chemical action alters the electrodes and electrolyte. The electrodes and electrolyte can be restored to their original condition by recharging the cell.
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ELECTROCHEMICAL ACTION is the process of converting chemical energy into electrical energy.
THE ANODE is the positive electrode of a cell.
THE CATHODE is the negative electrode of a cell.
PRIMARY CELL CHEMISTRY is the process in which electrons leaving the cathode to the load cause a positive charge which attracts negative ions from the electrolyte. The negative ions combine with the material of the cathode and form a substance such as lead-sulfate. Electrons from the load to the anode create a negative charge which attracts positive ions (hydrogen) from the electrolyte.
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SECONDARY CELL CHEMISTRY is the process in which the electrolyte acts upon and chemically changes both electrodes. This process also depletes the amount of active material in the electrolyte. A charging current applied to the cell reverses the process and restores the cell to its original condition.
POLARIZATION is the effect of hydrogen surrounding the anode of a cell which increases the internal resistance of the cell. Polarization can be prevented by venting the cell, adding a material rich in oxygen, or adding a material that will absorb hydrogen.
LOCAL ACTION is the continuation of current flow within the cell when there is no external load. It is caused by impurities in the electrode and can be prevented by the use of mercury amalgamated with the material of the electrode.
DRY CELL is the type commonly referred to as the "flashlight battery." Since the electrolyte is not in liquid form, but is a paste, the term dry cell is used. In most dry cells the case is the cathode.
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SHELF LIFE is the period the cell may be stored and still be usable.
MERCURY CELLS should never be shorted because of the danger of explosion.
DRY CELLS are of many types, each having advantages and disadvantages. The type selected for use depends on such factors as cost, size, ease of replacement, and voltage or current needs.
THE LEAD-ACID CELL is the most widely used secondary cell. The lead-acid cell produces electricity by electrochemical action. The anode is lead peroxide, the cathode is sponge lead, and the electrolyte is sulfuric acid and water.
THE NICKEL-CADMIUM CELL, commonly called the NICAD, has the following advantages over the lead-acid cell; charges in a shorter period of time, delivers a larger amount of power, stays idle longer, and can be charged and discharged many times. The anode is nickel hydroxide, the cathode is cadmium hydroxide, and the electrolyte is potassium hydroxide and water.
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THE SILVER-ZINC CELL is used mostly for emergency equipment. It is light, small, and has a large power capacity for its size. The anode is silver oxide, the cathode is zinc, and the electrolyte is potassium hydroxide and water.
THE SILVER-CADMIUM CELL combines the better features of the nickel-cadmium and silver- zinc cells. The anode is silver-oxide, the cathode is cadmium hydroxide, and the electrolyte is potassium hydroxide.
A BATTERY is a voltage source in a single container made from one or more cells. The cells can be combined in series, parallel, or series-parallel.
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SERIES CONNECTED CELLS provide a higher voltage than a single cell, with no increase in current.
PARALLEL CONNECTED CELLS provide a higher current than a single cell, with no increase in voltage.
SERIES-PARALLEL CONNECTED CELLS provide a higher voltage and a higher current than a single cell.
TYPES OF BATTERIES can be determined from nameplate data.
HYDROMETER provides the means to check the specific gravity of the electrolyte.
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SAFETY PRECAUTIONS should always be followed when working with or around batteries.
CAPACITY is an indication of the current-supplying capability of the battery for a specific period of time; e.g., 400 ampere-hour.
RATING is the capacity of the battery for a specific rate of discharge. In most batteries the rating is given for a 20 hour discharge cycle; e.g., 20 amperes for 20 hours.
BATTERY CHARGE is the process of reversing the current flow through the battery to restore the battery to its original condition. The addition of active ingredient to the electrolyte will not recharge the battery. There are five types of charges:
1. Initial charge
2. Normal charge
3. Equalizing charge
4. Floating charge
5. Fast charge
GASSING is the production of hydrogen gas caused by a portion of the charge current breaking down the water in the electrolyte. Steady gassing is normal during the charging process. Violent gassing indicates that the charge rate is too high.
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ANSWERS TO QUESTIONS Q1. THROUGH Q34.
A1.A cell is a device that converts chemical energy to electrical energy.
A2.The electrodes, the electrolyte, and the container.
A3.The electrodes are the current conductors of the cell. The electrolyte is the solution that acts upon the electrodes. The container holds the electrolyte and provides a means of mounting the electrodes.
A4.Primary and secondary.
A5.The secondary cell can be restored to its original condition by an electric current. The primary cell cannot.
A6.The process of converting chemical energy into electrical energy.
A7.(a) The anode, (b) the cathode.
A8.The positive charge caused by electrons leaving the negative electrode attracts the negative ions.
A9.By current flow through the load.
A10.The chemical action between the negative electrode and the electrolyte.
A11.The sulfuric acid is chemically acting upon the anode and cathode which creates a current flow through the load.
A12.The charging currents causes the lead sulfate in the anode and cathode to be changed back to lead peroxide, sponge lead, and sulfuric acid.
A13.Fully charged.
A14.Vent the cell, add a material rich in oxygen, and use a material that will absorb hydrogen.
A15.Current flow in a cell with no external load.
A16.The zinc container.
A17.The electrolyte is not a liquid but is in the form of a paste.
A18.The period that a cell can be stored and still be useable.
A19.The danger of explosion.
A20.Cost, size, ease of replacement, and voltage or current needs.
A21.Lead-acid, nickel-cadmium (NICAD), silver-zinc, and silver-cadmium.
A22.Can be charged in a shorter time, can deliver a larger amount of power, and stays idle longer.
A23.Silver-zinc cell.
A24.Silver-cadmium, silver-zinc, and nickel-cadmium.
A25.A voltage source in a single container made from one or more cells.
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A26.Series, to increase voltage but not current. Parallel, to increase current but not voltage. Series- Parallel, to increase both current and voltage.
A27.The cells in the nicad battery can be replaced.
A28.By looking at the nameplate data.
A29.To measure the amount of active ingredient in the electrolyte.
A30.Electrolyte B. It is heavier per unit volume.
A31.At all times.
A32.Forty hours.
A33.No, a current must be passed through the battery.
A34.Reduce the charging rate.
3-1
Upon completing this chapter, you will be able to:
1. Identify the term schematic diagram and identify the components in a circuit from a simple schematic diagram.
2. State the equation for Ohm’s law and describe the effects on current caused by changes in a circuit.
3. Given simple graphs of current versus power and voltage versus power, determine the value of circuit power for a given current and voltage.
4. Identify the term power, and state three formulas for computing power.
5. Compute circuit and component power in series, parallel, and combination circuits.
6. Compute the efficiency of an electrical device.
7. Solve for unknown quantities of resistance, current, and voltage in a series circuit.
8. Describe how voltage polarities are assigned to the voltage drops across resistors when Kirchhoff’s voltage law is used.
9. State the voltage at the reference point in a circuit.
10. Define open and short circuits and describe their effects on a circuit.
11. State the meaning of the term source resistance and describe its effect on a circuit.
12. Describe in terms of circuit values the circuit condition needed for maximum power transfer.
13. Compute efficiency of power transfer in a circuit.
14. Solve for unknown quantities of resistance, current, and voltage in a parallel circuit.
15. State the significance of the polarity assigned to a current when using Kirchhoff’s current law.
16. State the meaning of the term equivalent resistance.
17. Compute resistance, current, voltage, and power in voltage dividers.
18. Describe the method by which a single voltage divider can provide both positive and negative voltages.
19. Recognize the safety precautions associated with the hazard of electrical shock.
20. Identify the first aid procedures for a victim of electrical shock.
3-2
The material covered in this chapter contains many new terms that are explained as you progress through the material. The basic dc circuit is the easiest to understand, so the chapter begins with the basic circuit and from there works into the basic schematic diagram of that circuit. The schematic diagram is used in all your future work in electricity and electronics. It is very important that you become familiar with the symbols that are used.
This chapter also explains how to determine the total resistance, current, voltage, and power in a series, parallel, or combination circuit through the use of Ohm’s and Kirchhoff’s laws. The voltage divider network, series, parallel, and series-parallel practice problem circuits will be used for practical examples of what you have learned.
The flashlight is an example of a basic electric circuit. It contains a source of electrical energy (the dry cells in the flashlight), a load (the bulb) which changes the electrical energy into a more useful form of energy (light), and a switch to control the energy delivered to the load.
Before you study a schematic representation of the flashlight, it is necessary to define certain terms. The LOAD is any device through which an electrical current flows and which changes this electrical energy into a more useful form. Some common examples of loads are a lightbulb, which changes electrical energy to light energy; an electric motor, which changes electrical energy into mechanical energy; and the speaker in a radio, which changes electrical energy into sound. The SOURCE is the device which furnishes the electrical energy used by the load. It may consist of a simple dry cell (as in a flashlight), a storage battery (as in an automobile), or a power supply (such as a battery charger). The SWITCH, which permits control of the electrical device, interrupts the current delivered to the load.
The technician’s main aid in troubleshooting a circuit in a piece of equipment is the SCHEMATIC DIAGRAM. The schematic diagram is a "picture" of the circuit that uses symbols to represent the various circuit components; physically large or complex circuits can be shown on a relatively small diagram. Before studying the basic schematic, look at figure 3-1. This figure shows the symbols that are used in this chapter. These, and others like them, are referred to and used throughout the study of electricity and electronics.
3-3
The schematic in figure 3-2 represents a flashlight. View A of the figure shows the flashlight in the off or deenergized state. The switch (S1) is open. There is no complete path for current (I) through the circuit, and the bulb (DS1) does not light. In figure 3-2 view B, switch S1 is closed. Current flows in the direction of the arrows from the negative terminal of the battery (BAT), through the switch (S1), through the lamp (DS1), and back to the positive terminal of the battery. With the switch closed the path for current is complete. Current will continue to flow until the switch (S1) is moved to the open position or the battery is completely discharged.
3-4
Q1. In figure 3-2, what part of the circuit is the (a) load and (b) source? Q2. What happens to the path for current when S1 is open as shown in figure 3-2(A)? Q3. What is the name given to the "picture" of a circuit such as the one shown in figure 3-2?
In the early part of the 19th century, George Simon Ohm proved by experiment that a precise relationship exists between current, voltage, and resistance. This relationship is called Ohm’s law and is stated as follows:
The current in a circuit is DIRECTLY proportional to the applied voltage and INVERSELY proportional to the circuit resistance. Ohm’s law may be expressed as an equation:
As stated in Ohm’s law, current is inversely proportional to resistance. This means, as the resistance in a circuit increases, the current decreases proportionately.
3-5
if any two quantities are known, the third one can be determined. Refer to figure 3-2(B), the schematic of the flashlight. If the battery (BAT) supplies a voltage of 1.5 volts and the lamp (DS1) has a resistance of 5 ohms, then the current in the circuit can be determined. Using this equation and substituting values:
If the flashlight were a two-cell flashlight, we would have twice the voltage, or 3.0 volts, applied to the circuit. Using this voltage in the equation:
You can see that the current has doubled as the voltage has doubled. This demonstrates that the current is directly proportional to the applied voltage.
If the value of resistance of the lamp is doubled, the equation will be:
The current has been reduced to one half of the value of the previous equation, or .3 ampere. This demonstrates that the current is inversely proportional to the resistance. Doubling the value of the resistance of the load reduces circuit current value to one half of its former value.
By using Ohm’s law, you are able to find the resistance of a circuit, knowing only the voltage and the current in the circuit.
In any equation, if all the variables (parameters) are known except one, that unknown can be found. For example, using Ohm’s law, if current (I) and voltage (E) are known, resistance (R) the only parameter not known, can be determined:
1. Basic formula:
2. Remove the divisor by multiplying both sides by R:
3-6
3. Result of step 2: R x I = E
4. To get R alone (on one side of the equation) divide both sides by I:
5. The basic formula, transposed for R, is:
Refer to figure 3-3 where E equals 10 volts and I equals 1 ampere. Solve for R, using the equation just explained.
Solution:
3-8
The Ohm’s law equation and its various forms may be obtained readily with the aid of figure 3-5. The circle containing E, I, and R is divided into two parts, with E above the line and with I and R below the line. To determine the unknown quantity, first cover that quantity with a finger. The position of the uncovered letters in the circle will indicate the mathematical operation to be performed. For example, to find I, cover I with a finger. The uncovered letters indicate that E is to be divided by R, or
To find the formula for E, cover E with your finger. The result indicates that I is to be multiplied by R, or E = IR. To find the formula for R, cover R. The result indicates that E is to be divided by I, or
You are cautioned not to rely wholly on the use of this diagram when you transpose the Ohm’s law formulas. The diagram should be used to supplement your knowledge of the algebraic method. Algebra is a basic tool in the solution of electrical problems.
Q4. According to Ohm’s law, what happens to circuit current if the applied voltage (a) increases, (b) decreases?
Q5. According to Ohm’s law, what happens to circuit current if circuit resistance (a) increases, (b) decreases?
Q6. What is the equation used to find circuit resistance if voltage and current values are known? GRAPHICAL ANALYSIS OF THE BASIC CIRCUIT
One of the most valuable methods of analyzing a circuit is by constructing a graph. No other method provides a more convenient or more rapid way to observe the characteristics of an electrical device.
3-9
The first step in constructing a graph is to obtain a table of data. The information in the table can be obtained by taking measurements on the circuit under examination, or can be obtained theoretically through a series of Ohm’s law computations. The latter method is used here.
Since there are three variables (E, I, and R) to be analyzed, there are three distinct graphs that may be constructed.
To construct any graph of electrical quantities, it is standard practice to vary one quantity in a specified way and note the changes which occur in a second quantity. The quantity which is intentionally varied is called the independent variable and is plotted on the horizontal axis. The horizontal axis is known as the X-AXIS. The second quantity, which varies as a result of changes in the first quantity, is called the dependent variable and is plotted on the vertical, or Y-AXIS. Any other quantities involved are held constant.
For example, in the circuit shown in figure 3-6, if the resistance was held at 10 ohms and the voltage was varied, the resulting changes in current could then be graphed. The resistance is the constant, the voltage is the independent variable, and the current is the dependent variable.
Figure 3-7 shows the graph and a table of values. This table shows R held constant at 10 ohms as E is varied from 0 to 20 volts in 5-volt steps. Through the use of Ohm’s law, you can calculate the value of current for each value of voltage shown in the table. When the table is complete, the information it contains can be used to construct the graph shown in figure 3-7. For example, when the voltage applied to the 10-ohm resistor is 10 volts, the current is 1 ampere. These values of current and voltage determine a point on the graph. When all five points have been plotted, a smooth curve is drawn through the points.
3-10
Through the use of this curve, the value of current through the resistor can be quickly determined for any value of voltage between 0 and 20 volts.
Since the curve is a straight line, it shows that equal changes of voltage across the resistor produce equal changes in current through the resistor. This fact illustrates an important characteristic of the basic law—the current varies directly with the applied voltage when the resistance is held constant.
When the voltage across a load is held constant, the current depends solely upon the resistance of the load. For example, figure 3-8 shows a graph with the voltage held constant at 12 volts. The independent variable is the resistance which is varied from 2 ohms to 12 ohms. The current is the dependent variable. Values for current can be calculated as:
3-11
This process can be continued for any value of resistance. You can see that as the resistance is halved, the current is doubled; when the resistance is doubled, the current is halved.
This illustrates another important characteristic of Ohm's law—current varies inversely with resistance when the applied voltage is held constant.
Q7. Using the graph in figure 3-7, what is the approximate value of current when the voltage is 12.5 volts?
Q8. Using the graph in figure 3-8, what is the approximate value of current when the resistance is 3 ohms?
Power, whether electrical or mechanical, pertains to the rate at which work is being done. Work is done whenever a force causes motion. When a mechanical force is used to lift or move a weight, work is done. However, force exerted WITHOUT causing motion, such as the force of a compressed spring acting between two fixed objects, does not constitute work.
Previously, it was shown that voltage is an electrical force, and that voltage forces current to flow in a closed circuit. However, when voltage exists but current does not flow because the circuit is open, no work is done. This is similar to the spring under tension that produced no motion. When voltage causes electrons to move, work is done. The instantaneous RATE at which this work is done is called the electric power rate, and is measured in WATTS.
A total amount of work may be done in different lengths of time. For example, a given number of electrons may be moved from one point to another in 1 second or in 1 hour, depending on the RATE at which they are moved. In both cases, total work done is the same. However, when the work is done in a
3-12
short time, the wattage, or INSTANTANEOUS POWER RATE, is greater than when the same amount of work is done over a longer period of time.
As stated, the basic unit of power is the watt. Power in watts is equal to the voltage across a circuit multiplied by current through the circuit. This represents the rate at any given instant at which work is being done. The symbol P indicates electrical power. Thus, the basic power formula is P = E x I, where E is voltage and I is current in the circuit. The amount of power changes when either voltage or current, or both voltage and current, are caused to change.
In practice, the ONLY factors that can be changed are voltage and resistance. In explaining the different forms that formulas may take, current is sometimes presented as a quantity that is changed. Remember, if current is changed, it is because either voltage or resistance has been changed.
Figure 3-9 shows a basic circuit using a source of power that can be varied from 0 to 8 volts and a graph that indicates the relationship between voltage and power.
The resistance of this circuit is 2 ohms; this value does not change. Voltage (E) is increased (by increasing the voltage source), in steps of 1 volt, from 0 volts to 8 volts. By applying Ohm’s law, the current (I) is determined for each step of voltage. For instance, when E is 1 volt, the current is:
3-13
Power (P), in watts, is determined by applying the basic power formula:
You should notice that when the voltage was increased to 2 volts, the power increased from .5 watts to 2 watts or 4 times. When the voltage increased to 3 volts, the power increased to 4.5 watts or 9 times. This shows that if the resistance in a circuit is held constant, the power varies directly with the SQUARE OF THE VOLTAGE.
Another way of proving that power varies as the square of the voltage when resistance is held constant is:
3-14
Another important relationship may be seen by studying figure 3-10. Thus far, power has been calculated with voltage and current (P = E x I), and with voltage and resistance
Referring to figure 3-10, note that power also varies as the square of current just as it does with voltage. Thus, another formula for power, with current and resistance as its factors, is P = I 2R. This can be proved by:
3-15
Up to this point, four of the most important electrical quantities have been discussed. These are voltage (E), current (I), resistance (R), and power (P). You must understand the relationships which exist among these quantities because they are used throughout your study of electricity. In the preceding paragraphs, P was expressed in terms of alternate pairs of the other three basic quantities E, I, and R. In practice, you should be able to express any one of these quantities in terms of any two of the others.
Figure 3-11 is a summary of 12 basic formulas you should know. The four quantities E, I, R, and P are at the center of the figure. Adjacent to each quantity are three segments. Note that in each segment, the basic quantity is expressed in terms of two other basic quantities, and no two segments are alike.
3-16
For example, the formula wheel in figure 3-11 could be used to find the formula to solve the following problem:
A circuit has a voltage source that delivers 6 volts and the circuit uses 3 watts of power. What is the resistance of the load?
Since R is the quantity you have been asked to find, look in the section of the wheel that has R in the center. The segment
contains the quantities you have been given. The formula you would use is
The problem can now be solved.
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Q9. What is the term applied to the rate at which a mechanical or electrical force causes motion? Q10. How can the amount of current be changed in a circuit? Q11. What are the three formulas for electrical power?
Electrical components are often given a power rating. The power rating, in watts, indicates the rate at which the device converts electrical energy into another form of energy, such as light, heat, or motion. An example of such a rating is noted when comparing a 150-watt lamp to a 100-watt lamp. The higher wattage rating of the 150-watt lamp indicates it is capable of converting more electrical energy into light energy than the lamp of the lower rating. Other common examples of devices with power ratings are soldering irons and small electric motors.
In some electrical devices the wattage rating indicates the maximum power the device is designed to use rather than the normal operating power. A 150-watt lamp, for example, uses 150 watts when operated at the specified voltage printed on the bulb. In contrast, a device such as a resistor is not normally given a voltage or a current rating. A resistor is given a power rating in watts and can be operated at any combination of voltage and current as long as the power rating is not exceeded. In most circuits, the actual power used by a resistor is considerably less than the power rating of the resistor because a 50% safety factor is used. For example, if a resistor normally used 2 watts of power, a resistor with a power rating of 3 watts would be used.
Resistors of the same resistance value are available in different wattage values. Carbon resistors, for example, are commonly made in wattage ratings of 1/8, 1/4, 1/2, 1, and 2 watts. The larger the physical size of a carbon resistor the higher the wattage rating. This is true because a larger surface area of material radiates a greater amount of heat more easily.
When resistors with wattage ratings greater than 5 watts are needed, wirewound resistors are used. Wirewound resistors are made in values between 5 and 200 watts. Special types of wirewound resistors are used for power in excess of 200 watts.
As with other electrical quantities, prefixes may be attached to the word watt when expressing very large or very small amounts of power. Some of the more common of these are the kilowatt (1,000 watts), the megawatt (1,000,000 watts), and the milliwatt (1/1,000 of a watt).
Q12. What is the current in a circuit with 5 ohms of resistance that uses 180 watts of power? (refer to figure 3-12)
Q13. What type of resistor should be used in the circuit described in question 12? Q14. What is the power used in a circuit that has 10 amperes of current through a 10-ohm resistor?
3-18
The term power consumption is common in the electrical field. It is applied to the use of power in the same sense that gasoline consumption is applied to the use of fuel in an automobile.
Another common term is power conversion. Power is used by electrical devices and is converted from one form of energy to another. An electrical motor converts electrical energy to mechanical energy. An electric light bulb converts electrical energy into light energy and an electric range converts electrical energy into heat energy. Power used by electrical devices is measured in energy. This practical unit of electrical energy is equal to 1 watt of power used continuously for 1 hour. The term kilowatt hour (kWh) is used more extensively on a daily basis and is equal to 1,000 watt-hours.
The EFFICIENCY of an electrical device is the ratio of power converted to useful energy divided by the power consumed by the device. This number will always be less than one (1.00) because of the losses in any electrical device. If a device has an efficiency rating of .95, it effectively transforms 95 watts into useful energy for every 100 watts of input power. The other 5 watts are lost to heat, or other losses which cannot be used.
Calculating the amount of power converted by an electrical device is a simple matter. You need to know the length of time the device is operated and the input power or horsepower rating. Horsepower, a unit of work, is often found as a rating on electrical motors. One horsepower is equal to 746 watts. Example: A 3/4-hp motor operates 8 hours a day. How much power is converted by the motor per month? How many kWh does this represent?
3-19
(NOTE: These figures are rounded to the nearest 1000.) To convert to kWh
If the motor actually uses 137 kWh per month, what is the efficiency of the motor?
Solution:
Q15. How much power is converted by a 1-horsepower motor in 12 hours? Q16. What is the efficiency of the motor if it actually uses 9.5 kWh in 12 hours?
When two unequal charges are connected by a conductor, a complete pathway for current exists. An electric circuit is a complete conducting pathway. It consists not only of the conductor, but also includes the path through the voltage source. Inside the voltage source current flows from the positive terminal, through the source, emerging at the negative terminal.
3-20
A SERIES CIRCUIT is defined as a circuit that contains only ONE PATH for current flow. To compare the basic circuit that has been discussed and a more complex series circuit, figure 3-13 shows two circuits. The basic circuit has only one lamp and the series circuit has three lamps connected in series.
Referring to figure 3-13, the current in a series circuit must flow through each lamp to complete the electrical path in the circuit. Each additional lamp offers added resistance. In a series circuit, THE TOTAL CIRCUIT RESISTANCE (RT) IS EQUAL TO THE SUM OF THE INDIVIDUAL RESISTANCES.
As an equation: RT = R1 + R2 + R3 + . . . R n
NOTE: The subscript n denotes any number of additional resistances that might be in the equation.
Example: In figure 3-14 a series circuit consisting of three resistors: one of 10 ohms, one of 15 ohms, and one of 30 ohms, is shown. A voltage source provides 110 volts. What is the total resistance?
3-21
In some circuit applications, the total resistance is known and the value of one of the circuit resistors has to be determined. The equation RT = R1 + R2 + R3 can be transposed to solve for the value of the unknown resistance.
Example: In figure 3-15 the total resistance of a circuit containing three resistors is 40 ohms. Two of the circuit resistors are 10 ohms each. Calculate the value of the third resistor (R 3).
3-22
Since there is only one path for current in a series circuit, the same current must flow through each component of the circuit. To determine the current in a series circuit, only the current through one of the components need be known.
The fact that the same current flows through each component of a series circuit can be verified by inserting meters into the circuit at various points, as shown in figure 3-16. If this were done, each meter would be found to indicate the same value of current.
3-23
The voltage dropped across the resistor in a circuit consisting of a single resistor and a voltage source is the total voltage across the circuit and is equal to the applied voltage. The total voltage across a series circuit that consists of more than one resistor is also equal to the applied voltage, but consists of the sum of the individual resistor voltage drops. In any series circuit, the SUM of the resistor voltage drops must equal the source voltage. This statement can be proven by an examination of the circuit shown in figure 3-17. In this circuit a source potential (ET) of 20 volts is dropped across a series circuit consisting of two 5-ohm resistors. The total resistance of the circuit (RT) is equal to the sum of the two individual resistances, or 10 ohms. Using Ohm’s law the circuit current may be calculated as follows:
3-24
Since the value of the resistors is known to be 5 ohms each, and the current through the resistors is known to be 2 amperes, the voltage drops across the resistors can be calculated. The voltage (E1) across R1 is therefore:
By inspecting the circuit, you can see that R2 is the same ohmic value as R1 and carries the same current. The voltage drop across R2 is therefore also equal to 10 volts. Adding these two 10-volts drops together gives a total drop of 20 volts, exactly equal to the applied voltage. For a series circuit then:
ET = E1 = E 2 + E3 = . . . En
Example: A series circuit consists of three resistors having values of 20 ohms, 30 ohms, and 50 ohms, respectively. Find the applied voltage if the current through the 30 ohm resistor is 2 amps. (The abbreviation amp is commonly used for ampere.)
To solve the problem, a circuit diagram is first drawn and labeled (fig 3-18).
3-26
NOTE: When you use Ohm’s law, the quantities for the equation MUST be taken from the SAME part of the circuit. In the above example the voltage across R2 was computed using the current through R2 and the resistance of R2.
The value of the voltage dropped by a resistor is determined by the applied voltage and is in proportion to the circuit resistances. The voltage drops that occur in a series circuit are in direct proportion to the resistances. This is the result of having the same current flow through each resistor—the larger the ohmic value of the resistor, the larger the voltage drop across it.
Q17. A series circuit consisting of three resistors has a current of 3 amps. If R1 = 20 ohms, R2= 60 ohms, and R 3 = 80 ohms, what is the (a) total resistance and (b) source voltage of the circuit?
Q18. What is the voltage dropped by each resistor of the circuit described in question 17? Q19. If the current was increased to 4 amps, what would be the voltage drop across each resistor in the
circuit described in question 17?
Q20. What would have to be done to the circuit described in question 17 to increase the current to 4 amps?
Each of the resistors in a series circuit consumes power which is dissipated in the form of heat. Since this power must come from the source, the total power must be equal to the power consumed by the circuit resistances. In a series circuit the total power is equal to the SUM of the power dissipated by the individual resistors. Total power (PT) is equal to:
PT = P1 + P2 + P3 . . .Pn
Example: A series circuit consists of three resistors having values of 5 ohms, 10 ohms, and 15 ohms. Find the total power when 120 volts is applied to the circuit. (See fig. 3-19.)
3-27
Given:
Solution: The total resistance is found first.
By using the total resistance and the applied voltage, the circuit current is calculated.
By means of the power formulas, the power can be calculated for each resistor:
To check the answer, the total power delivered by the source can be calculated:
3-28
The total power is equal to the sum of the power used by the individual resistors.
The important factors governing the operation of a series circuit are listed below. These factors have been set up as a group of rules so that they may be easily studied. These rules must be completely understood before the study of more advanced circuit theory is undertaken.
1. The same current flows through each part of a series circuit.
2. The total resistance of a series circuit is equal to the sum of the individual resistances.
3. The total voltage across a series circuit is equal to the sum of the individual voltage drops.
4. The voltage drop across a resistor in a series circuit is proportional to the ohmic value of the resistor.
5. The total power in a series circuit is equal to the sum of the individual powers used by each circuit component.
To establish a procedure for solving series circuits, the following sample problems will be solved.
Example: Three resistors of 5 ohms, 10 ohms, and 15 ohms are connected in series with a power source of 90 volts as shown in figure 3-20. Find the total resistance, circuit current, voltage drop of each resistor, power of each resistor, and total power of the circuit.
3-29
In solving the circuit the total resistance will be found first. Next, the circuit current will be calculated. Once the current is known, the voltage drops and power dissipations can be calculated.
3-30
Example: Four resistors, R1 = 10 ohms, R 2 = 10 ohms, R 3 = 50 ohms, and R4 = 30 ohms, are connected in series with a power source as shown in figure 3-21. The current through the circuit is 1/2 ampere.
a. What is the battery voltage?
b. What is the voltage across each resistor?
c. What is the power expended in each resistor?
d. What is the total power?
3-32
Solution (d):
An important fact to keep in mind when applying Ohm’s law to a series circuit is to consider whether the values used are component values or total values. When the information available enables the use of Ohm’s law to find total resistance, total voltage, and total current, total values must be inserted into the formula. To find total resistance:
3-33
To find total voltage:
To find total current:
NOTE: IT is equal to I in a series circuit. However, the distinction between IT and I in the formula should be noted. The reason for this is that future circuits may have several currents, and it will be necessary to differentiate between IT and other currents.
To compute any quantity (E, I, R, or P) associated with a single given resistor, the values used in the formula must be obtained from that particular resistor. For example, to find the value of an unknown resistance, the voltage across and the current through that particular resistor must be used.
To find the value of a resistor:
To find the voltage drop across a resistor:
To find current through a resistor:
Q21. A series circuit consists of two resistors in series. R1 = 25 ohms and R2 = 30 ohms. The circuit current is 6 amps. What is the (a) source voltage, (b) voltage dropped by each resistor, (c) total power, and (d) power used by each resistor?
In 1847, G. R. Kirchhoff extended the use of Ohm’s law by developing a simple concept concerning the voltages contained in a series circuit loop. Kirchhoff’s voltage law states:
"The algebraic sum of the voltage drops in any closed path in a circuit and the electromotive forces in that path is equal to zero."
3-34
To state Kirchhoff’s law another way, the voltage drops and voltage sources in a circuit are equal at any given moment in time. If the voltage sources are assumed to have one sign (positive or negative) at that instant and the voltage drops are assumed to have the opposite sign, the result of adding the voltage sources and voltage drops will be zero.
NOTE: The terms electromotive force and emf are used when explaining Kirchhoff’s law because Kirchhoff’s law is used in alternating current circuits (covered in Module 2). In applying Kirchhoff’s law to direct current circuits, the terms electromotive force and emf apply to voltage sources such as batteries or power supplies.
Through the use of Kirchhoff’s law, circuit problems can be solved which would be difficult, and often impossible, with knowledge of Ohm’s law alone. When Kirchhoff’s law is properly applied, an equation can be set up for a closed loop and the unknown circuit values can be calculated.
To apply Kirchhoff’s voltage law, the meaning of voltage polarity must be understood.
In the circuit shown in figure 3-22, the current is shown flowing in a counterclockwise direction. Notice that the end of resistor R1, into which the current flows, is marked NEGATIVE (í���7KH�HQG�RI�51 at which the current leaves is marked POSITIVE (+). These polarity markings are used to show that the end of R1 into which the current flows is at a higher negative potential than the end of the resistor at which the current leaves. Point A is more negative than point B.
Point C, which is at the same potential as point B, is labeled negative. This is to indicate that point C is more negative than point D. To say a point is positive (or negative) without stating what the polarity is based upon has no meaning. In working with Kirchhoff’s law, positive and negative polarities are assigned in the direction of current flow.
3-35
Kirchhoff’s voltage law can be written as an equation, as shown below:
Ea + Eb + Ec + . . . En = 0
where Ea, Eb, etc., are the voltage drops or emf’s around any closed circuit loop. To set up the equation for an actual circuit, the following procedure is used.
1. Assume a direction of current through the circuit. (The correct direction is desirable but not necessary.)
2. Using the assumed direction of current, assign polarities to all resistors through which the current flows.
3. Place the correct polarities on any sources included in the circuit.
4. Starting at any point in the circuit, trace around the circuit, writing down the amount and polarity of the voltage across each component in succession. The polarity used is the sign AFTER the assumed current has passed through the component. Stop when the point at which the trace was started is reached.
5. Place these voltages, with their polarities, into the equation and solve for the desired quantity.
Example: Three resistors are connected across a 50-volt source. What is the voltage across the third resistor if the voltage drops across the first two resistors are 25 volts and 15 volts?
Solution: First, a diagram, such as the one shown in figure 3-23, is drawn. Next, a direction of current is assumed (as shown). Using this current, the polarity markings are placed at each end of each resistor and also on the terminals of the source. Starting at point A, trace around the circuit in the direction of current flow, recording the voltage and polarity of each component. Starting at point A and using the components from the circuit:
Substituting values from the circuit:
3-36
Using the same idea as above, you can solve a problem in which the current is the unknown quantity.
Example: A circuit having a source voltage of 60 volts contains three resistors of 5 ohms, 10 ohms, and 15 ohms. Find the circuit current.
Solution: Draw and label the circuit (fig. 3-24). Establish a direction of current flow and assign polarities. Next, starting at any point—point A will be used in this example—write out the loop equation.
3-37
Since the current obtained in the above calculations is a positive 2 amps, the assumed direction of current was correct. To show what happens if the incorrect direction of current is assumed, the problem will be solved as before, but with the opposite direction of current. The circuit is redrawn showing the new direction of current and new polarities in figure 3-25. Starting at point A the loop equation is:
3-38
Notice that the AMOUNT of current is the same as before. The polarity, however, is NEGATIVE. The negative polarity simply indicates the wrong direction of current was assumed. Should it be necessary to use this current in further calculations on the circuit using Kirchhoff’s law, the negative polarity should be retained in the calculations.
In many practical applications a circuit may contain more than one source of emf. Sources of emf that cause current to flow in the same direction are considered to be SERIES AIDING and the voltages are added. Sources of emf that would tend to force current in opposite directions are said to be SERIES OPPOSING, and the effective source voltage is the difference between the opposing voltages. When two opposing sources are inserted into a circuit current flow would be in a direction determined by the larger source. Examples of series aiding and opposing sources are shown in figure 3-26.
3-39
A simple solution may be obtained for a multiple-source circuit through the use of Kirchhoff’s voltage law. In applying this method, the same procedure is used for the multiple-source circuit as was used above for the single-source circuit. This is demonstrated by the following example.
Example: Using Kirchhoff’s voltage equation, find the amount of current in the circuit shown in fig 3-27.
Figure 3-27.-Solving for circuit current using Kirchhoff's voltage equation.
3-40
Solution: As before, a direction of current flow is assumed and polarity signs are placed on the drawing. The loop equation will be started at point A.
Q22. When using Kirchhoff’s voltage law, how are voltage polarities assigned to the voltage drops across resistors?
Q23. Refer to figure 3-27, if R1 was changed to a 40-ohm resistor, what would be the value of circuit current (I
T)? Q24. Refer to figure 3-27. What is the effective source voltage of the circuit using the 40-ohm resistor?
Before you learn about the types of circuits other than the series circuit, you should become familiar with some of the terms and characteristics used in electrical circuits. These terms and characteristics will be used throughout your study of electricity and electronics.
A reference point is an arbitrarily chosen point to which all other points in the circuit are compared. In series circuits, any point can be chosen as a reference and the electrical potential at all other points can be determined in reference to that point. In figure 3-28 point A shall be considered the reference point. Each series resistor in the illustrated circuit is of equal value. The applied voltage is equally distributed across each resistor. The potential at point B is 25 volts more positive than at point A. Points C and D are 50 volts and 75 volts more positive than point A respectively.
3-41
When point B is used as the reference, as in figure 3-29, point D would be positive 50 volts in respect to the new reference point. The former reference point, A, is 25 volts negative in respect to point B.
3-42
As in the previous circuit illustration, the reference point of a circuit is always considered to be at zero potential. Since the earth (ground) is said to be at a zero potential, the term GROUND is used to denote a common electrical point of zero potential. In figure 3-30, point A is the zero reference, or ground, and the symbol for ground is shown connected to point A. Point C is 75 volts positive in respect to ground.
In most electrical equipment, the metal chassis is the common ground for the many electrical circuits. When each electrical circuit is completed, common points of a circuit at zero potential are connected directly to the metal chassis, thereby eliminating a large amount of connecting wire. The electrons pass through the metal chassis (a conductor) to reach other points of the circuit. An example of a chassis grounded circuit is illustrated in figure 3-31.
3-43
Most voltage measurements used to check proper circuit operation in electrical equipment are taken in respect to ground. One meter lead is attached to a grounded point and the other meter lead is moved to various test points. Circuit measurement is explained in more detail in NEETS Module 3.
A circuit is said to be OPEN when a break exists in a complete conducting pathway. Although an open occurs when a switch is used to deenergize a circuit, an open may also develop accidentally. To restore a circuit to proper operation, the open must be located, its cause determined, and repairs made.
Sometimes an open can be located visually by a close inspection of the circuit components. Defective components, such as burned out resistors, can usually be discovered by this method. Others, such as a break in wire covered by insulation or the melted element of an enclosed fuse, are not visible to the eye. Under such conditions, the understanding of the effect an open has on circuit conditions enables a technician to make use of test equipment to locate the open component.
In figure 3-32, the series circuit consists of two resistors and a fuse. Notice the effects on circuit conditions when the fuse opens.
3-44
Current ceases to flow; therefore, there is no longer a voltage drop across the resistors. Each end of the open conducting path becomes an extension of the battery terminals and the voltage felt across the open is equal to the applied voltage (EA).
An open circuit has INFINITE resistance. INFINITY represents a quantity so large it cannot be measured. The symbol for infinity is ��,Q�DQ�RSHQ�FLUFXLW��5T = �
A short circuit is an accidental path of low resistance which passes an abnormally high amount of current. A short circuit exists whenever the resistance of a circuit or the resistance of a part of a circuit drops in value to almost zero ohms. A short often occurs as a result of improper wiring or broken insulation.
In figure 3-33, a short is caused by improper wiring. Note the effect on current flow. Since the resistor has in effect been replaced with a piece of wire, practically all the current flows through the short and very little current flows through the resistor. Electrons flow through the short (a path of almost zero resistance) and the remainder of the circuit by passing through the 10-ohm resistor and the battery. The amount of current flow increases greatly because its resistive path has decreased from 10,010 ohms to 10 ohms. Due to the excessive current flow the 10-ohm resistor becomes heated. As it attempts to dissipate this heat, the resistor will probably be destroyed. Figure 3-34 shows a pictorial wiring diagram, rather than a schematic diagram, to indicate how broken insulation might cause a short circuit.
3-46
A meter connected across the terminals of a good 1.5-volt battery reads about 1.5 volts. When the same battery is inserted into a complete circuit, the meter reading decreases to something less than 1.5 volts. This difference in terminal voltage is caused by the INTERNAL RESISTANCE of the battery (the opposition to current offered by the electrolyte in the battery). All sources of electromotive force have some form of internal resistance which causes a drop in terminal voltage as current flows through the source.
This principle is illustrated in figure 3-35, where the internal resistance of a battery is shown as Ri. In the schematic, the internal resistance is indicated by an additional resistor in series with the battery. The battery, with its internal resistance, is enclosed within the dotted lines of the schematic diagram. With the switch open, the voltage across the battery terminals reads 15 volts. When the switch is closed, current flow causes voltage drops around the circuit. The circuit current of 2 amperes causes a voltage drop of 2 volts across Ri. The 1-ohm internal battery resistance thereby drops the battery terminal voltage to 13 volts. Internal resistance cannot be measured directly with a meter. An attempt to do this would damage the meter.
The effect of the source resistance on the power output of a dc source may be shown by an analysis of the circuit in figure 3-36. When the variable load resistor (RL) is set at the zero-ohm position (equivalent to a short circuit), current (I) is calculated using the following formula:
3-47
This is the maximum current that may be drawn from the source. The terminal voltage across the short circuit is zero volts and all the voltage is across the resistance within the source.
3-48
If the load resistance (RL) were increased (the internal resistance remaining the same), the current drawn from the source would decrease. Consequently, the voltage drop across the internal resistance would decrease. At the same time, the terminal voltage applied across the load would increase and approach a maximum as the current approaches zero amps.
Maximum power is transferred from the source to the load when the resistance of the load is equal to the internal resistance of the source. This theory is illustrated in the table and the graph of figure 3-36. When the load resistance is 5 ohms, matching the source resistance, the maximum power of 500 watts is developed in the load.
The efficiency of power transfer (ratio of output power to input power) from the source to the load increases as the load resistance is increased. The efficiency approaches 100 percent as the load resistance approaches a relatively large value compared with that of the source, since less power is lost in the source. The efficiency of power transfer is only 50 percent at the maximum power transfer point (when the load resistance equals the internal resistance of the source). The efficiency of power transfer approaches zero efficiency when the load resistance is relatively small compared with the internal resistance of the source. This is also shown on the chart of figure 3-36.
The problem of a desire for both high efficiency and maximum power transfer is resolved by a compromise between maximum power transfer and high efficiency. Where the amounts of power involved are large and the efficiency is important, the load resistance is made large relative to the source resistance so that the losses are kept small. In this case, the efficiency is high. Where the problem of matching a source to a load is important, as in communications circuits, a strong signal may be more important than a high percentage of efficiency. In such cases, the efficiency of power transfer should be only about 50 percent; however, the power transfer would be the maximum which the source is capable of supplying.
You should now understand the basic concepts of series circuits. The principles which have been presented are of lasting importance. Once equipped with a firm understanding of series circuits, you hold the key to an understanding of the parallel circuits to be presented next.
Q25. A circuit has a source voltage of 100 volts and two 50-ohm resistors connected in series. If the reference point for this circuit is placed between the two resistors, what would be the voltage at the reference point?
Q26. If the reference point in question 25 were connected to ground, what would be the voltage level of the reference point?
Q27. What is an open circuit? Q28. What is a short circuit? Q29. Why will a meter indicate more voltage at the battery terminal when the battery is out of a circuit
than when the battery is in a circuit?
Q30. What condition gives maximum power transfer from the source to the load? Q31. What is the efficiency of power transfer in question 30? Q32. A circuit has a source voltage of 25 volts. The source resistance is 1 ohm and the load resistance
is 49 ohms. What is the efficiency of power transfer?
3-49
The discussion of electrical circuits presented up to this point has been concerned with series circuits in which there is only one path for current. There is another basic type of circuit known as the PARALLEL CIRCUIT with which you must become familiar. Where the series circuit has only one path for current, the parallel circuit has more than one path for current.
Ohm’s law and Kirchhoff’s law apply to all electrical circuits, but the characteristics of a parallel dc circuit are different than those of a series dc circuit.
A PARALLEL CIRCUIT is defined as one having more than one current path connected to a common voltage source. Parallel circuits, therefore, must contain two or more resistances which are not connected in series. An example of a basic parallel circuit is shown in figure 3-37.
Start at the voltage source (Es) and trace counterclockwise around the circuit. Two complete and separate paths can be identified in which current can flow. One path is traced from the source, through resistance R1, and back to the source. The other path is from the source, through resistance R2, and back to the source.
You have seen that the source voltage in a series circuit divides proportionately across each resistor in the circuit. IN A PARALLEL CIRCUIT, THE SAME VOLTAGE IS PRESENT IN EACH BRANCH. (A branch is a section of a circuit that has a complete path for current.) In figure 3-37 this voltage is equal to the applied voltage (Es). This can be expressed in equation form as:
Voltage measurements taken across the resistors of a parallel circuit, as illustrated by figure 3-38 verify this equation. Each meter indicates the same amount of voltage. Notice that the voltage across each resistor is the same as the applied voltage.
3-50
Example: Assume that the current through a resistor of a parallel circuit is known to be 4.5 milliamperes (4.5 mA) and the value of the resistor is 30,000 ohms (30 N ���'HWHUPLQH�WKH�VRXUFH voltage. The circuit is shown in figure 3-39.
Given:
Solution:
3-51
Since the source voltage is equal to the voltage of a branch:
To simplify the math operation, the values can be expressed in powers of ten as follows:
If you are not familiar with the use of the powers of 10 or would like to brush up on it, Mathematics, Vol. 1, NAVEDTRA 10069-C, will be of great help to you.
Q33. What would the source voltage (ES) in figure 3-39 be if the current through R2 were 2 milliamps? Current in a Parallel Circuit
Ohm’s law states that the current in a circuit is inversely proportional to the circuit resistance. This fact is true in both series and parallel circuits.
There is a single path for current in a series circuit. The amount of current is determined by the total resistance of the circuit and the applied voltage. In a parallel circuit the source current divides among the available paths.
The behavior of current in parallel circuits will be shown by a series of illustrations using example circuits with different values of resistance for a given value of applied voltage.
Part (A) of figure 3-40 shows a basic series circuit. Here, the total current must pass through the single resistor. The amount of current can be determined.
3-52
Given:
Solution:
Part (B) of figure 3-40 shows the same resistor (R1) with a second resistor (R2) of equal value connected in parallel across the voltage source. When Ohm’s law is applied, the current flow through each resistor is found to be the same as the current through the single resistor in part (A).
3-53
Given:
Solution:
It is apparent that if there is 5 amperes of current through each of the two resistors, there must be a TOTAL CURRENT of 10 amperes drawn from the source.
The total current of 10 amperes, as illustrated in figure 3-40(B), leaves the negative terminal of the battery and flows to point a. Since point a is a connecting point for the two resistors, it is called a JUNCTION. At junction a, the total current divides into two currents of 5 amperes each. These two currents flow through their respective resistors and rejoin at junction b. The total current then flows from junction b back to the positive terminal of the source. The source supplies a total current of 10 amperes and each of the two equal resistors carries one-half the total current.
Each individual current path in the circuit of figure 3-40(B) is referred to as a BRANCH. Each branch carries a current that is a portion of the total current. Two or more branches form a NETWORK.
From the previous explanation, the characteristics of current in a parallel circuit can be expressed in terms of the following general equation:
IT = I1 + I 2 + . . . In
3-54
Compare part (A) of figure 3-41 with part (B) of the circuit in figure 3-40. Notice that doubling the value of the second branch resistor (R2) has no effect on the current in the first branch (IR1), but does reduce the second branch current (IR2) to one-half its original value. The total circuit current drops to a value equal to the sum of the branch currents. These facts are verified by the following equations.
Given:
Solution:
3-55
The amount of current flow in the branch circuits and the total current in the circuit shown in figure 3-41(B) are determined by the following computations.
Given:
3-56
Solution:
Notice that the sum of the ohmic values in each circuit shown in figure 3-41 is equal (30 ohms), and that the applied voltage is the same (50 volts). However, the total current in 3-41(B) (15 amps) is twice the amount in 3-41(A) (7.5 amps). It is apparent, therefore, that the manner in which resistors are connected in a circuit, as well as their actual ohmic values, affect the total current.
The division of current in a parallel network follows a definite pattern. This pattern is described by KIRCHHOFF’S CURRENT LAW which states:
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"The algebraic sum of the currents entering and leaving any junction of conductors is equal to zero." This law can be stated mathematically as:
Ia + lb + . . . I n + 0
where: Ia, Ib, etc., are the currents entering and leaving the junction. Currents ENTERING the junction are considered to be POSITIVE and currents LEAVING the junction are considered to be NEGATIVE. When solving a problem using Kirchhoff’s current law, the currents must be placed into the equation WITH THE PROPER POLARITY SIGNS ATTACHED.
Example: Solve for the value of I3 in figure 3-42.
Given:
Solution:
Ia + lb + . . . I a + 0
The currents are placed into the equation with the proper signs.
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I3 has a value of 2 amperes, and the negative sign shows it to be a current LEAVING the junction. Example. Using figure 3-43, solve for the magnitude and direction of I3.
Given:
Solution:
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I3 is 2 amperes and its positive sign shows it to be a current entering the junction. Q34. There is a relationship between total current and current through the individual components in a
circuit. What is this relationship in a series circuit and a parallel circuit?
Q35. In applying Kirchhoff’s current law, what does the polarity of the current indicate? Resistance in a Parallel Circuit
In the example diagram, figure 3-44, there are two resistors connected in parallel across a 5-volt battery. Each has a resistance value of 10 ohms. A complete circuit consisting of two parallel paths is formed and current flows as shown.
Computing the individual currents shows that there is one-half of an ampere of current through each resistance. The total current flowing from the battery to the junction of the resistors, and returning from the resistors to the battery, is equal to 1 ampere.
The total resistance of the circuit can be calculated by using the values of total voltage (ET) and total current (IT).
NOTE: From this point on the abbreviations and symbology for electrical quantities will be used in example problems.
Given:
Solution:
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This computation shows the total resistance to be 5 ohms; one-half the value of either of the two resistors.
Since the total resistance of a parallel circuit is smaller than any of the individual resistors, total resistance of a parallel circuit is not the sum of the individual resistor values as was the case in a series circuit. The total resistance of resistors in parallel is also referred to as EQUIVALENT RESISTANCE (Req). The terms total resistance and equivalent resistance are used interchangeably.
There are several methods used to determine the equivalent resistance of parallel circuits. The best method for a given circuit depends on the number and value of the resistors. For the circuit described above, where all resistors have the same value, the following simple equation is used:
This equation is valid for any number of parallel resistors of EQUAL VALUE. Example: Four 40-ohm resistors are connected in parallel. What is their equivalent resistance?
Given:
Solution:
Figure 3-45 shows two resistors of unequal value in parallel. Since the total current is shown, the equivalent resistance can be calculated.
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Given:
Solution:
1, R2). An important point to remember is that the equivalent resistance of a parallel circuit is always less than the resistance of any branch.
Equivalent resistance can be found if you know the individual resistance values and the source voltage. By calculating each branch current, adding the branch currents to calculate total current, and dividing the source voltage by the total current, the total can be found. This method, while effective, is somewhat lengthy. A quicker method of finding equivalent resistance is to use the general formula for resistors in parallel:
If you apply the general formula to the circuit shown in figure 3-45 you will get the same value for HTXLYDOHQW�UHVLVWDQFH��� ��DV�ZDV�REWDLQHG�LQ�WKH�SUHYLRXV�FDOFXODWLRQ�WKDW�XVHG�VRXUFH�YROWDJH�DQG�WRWDO current.
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Given:
Solution:
Convert the fractions to a common denominator.
Since both sides are reciprocals (divided into one), disregard the reciprocal function.
There are other simplifications of the general formula for resistors in parallel which can be used to calculate the total or equivalent resistance in a parallel circuit.
RECIPROCAL METHOD.—This method is based upon taking the reciprocal of each side of the equation. This presents the general formula for resistors in parallel as:
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This formula is used to solve for the equivalent resistance of a number of unequal parallel resistors. You must find the lowest common denominator in solving these problems. If you are a little hazy on finding the lowest common denominator, brush up on it in Mathematics Volume 1, NAVEDTRA 10069 (Series).
Example: Three resistors are connected in parallel as shown in figure 3-46. The resistor values are: R1 = 20 ohms, R2 = 30 ohms, R 3 = 40 ohms. What is the equivalent resistance? (Use the reciprocal method.)
Given:
Solution:
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PRODUCT OVER THE SUM METHOD.—A convenient method for finding the equivalent, or total, resistance of two parallel resistors is by using the following formula.
This equation, called the product over the sum formula, is used so frequently it should be committed to memory.
Example: What is the equivalent resistance of a 20-ohm and a 30-ohm resistor connected in parallel, as in figure 3-47?
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Given:
Solution:
Q36. Four equal resistors are connected in parallel, each resistor has an ohmic value of 100 ohms, what is the equivalent resistance?
Q37. Three resistors connected in parallel have values of 12 N �����N ��DQG����N ��:KDW�LV�WKH equivalent resistance?
Q38. Two resistors connected in parallel have values of 10 N �DQG����N ��:KDW�LV�WKH�HTXLYDOHQW resistance?
Power computations in a parallel circuit are essentially the same as those used for the series circuit. Since power dissipation in resistors consists of a heat loss, power dissipations are additive regardless of how the resistors are connected in the circuit. The total power is equal to the sum of the power dissipated by the individual resistors. Like the series circuit, the total power consumed by the parallel circuit is:
Example: Find the total power consumed by the circuit in figure 3-48.
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Since the total current and source voltage are known, the total power can also be computed by:
Given:
Solution:
In the study of electricity, it is often necessary to reduce a complex circuit into a simpler form. Any complex circuit consisting of resistances can be redrawn (reduced) to a basic equivalent circuit containing the voltage source and a single resistor representing total resistance. This process is called reduction to an EQUIVALENT CIRCUIT.
Figure 3-49 shows a parallel circuit with three resistors of equal value and the redrawn equivalent circuit. The parallel circuit shown in part A shows the original circuit. To create the equivalent circuit, you must first calculate the equivalent resistance.
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Given:
Solution:
Once the equivalent resistance is known, a new circuit is drawn consisting of a single resistor (to represent the equivalent resistance) and the voltage source, as shown in part B. Rules for Parallel DC Circuits
1. The same voltage exists across each branch of a parallel circuit and is equal to the source voltage.
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2. The current through a branch of a parallel network is inversely proportional to the amount of resistance of the branch.
3. The total current of a parallel circuit is equal to the sum of the individual branch currents of the circuit.
4. The total resistance of a parallel circuit is found by the general formula:
or one of the formulas derived from this general formula.
5. The total power consumed in a parallel circuit is equal to the sum of the power consumptions of the individual resistances.
Problems involving the determination of resistance, voltage, current, and power in a parallel circuit are solved as simply as in a series circuit. The procedure is the same — (1) draw the circuit diagram, (2) state the values given and the values to be found, (3) select the equations to be used in solving for the unknown quantities based upon the known quantities, and (4) substitute the known values in the equation you have selected and solve for the unknown value.
Example: A parallel circuit consists of five resistors. The value of each resistor is known and the current through R1 is known. You are asked to calculate the value for total resistance, total power, total current, source voltage, the power used by each resistor, and the current through resistors R2, R 3, R4, and R5.
Given:
Find:
This may appear to be a large amount of mathematical manipulation. However, if you use the step- by-step approach, the circuit will fall apart quite easily.
The first step in solving this problem is for you to draw the circuit and indicate the known values as shown in figure 3-50.
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There are several ways to approach this problem. With the values you have been given, you could first solve for RT, the power used by R1, or the voltage across R1, which you know is equal to the source voltage and the voltage across each of the other resistors. Solving for RT or the power used by R1 will not help in solving for the other unknown values.
Once the voltage across Rl is known, this value will help you calculate other unknowns. Therefore the logical unknown to solve for is the source voltage (the voltage across R1).
Given:
Solution:
Now that source voltage is known, you can solve for current in each branch.
Given:
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Solution:
Since R 3 = R4 = R5 and the voltage across each branch is the same:
Solving for total resistance.
Given:
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Solution:
An alternate method for solving for RT can be used. By observation, you can see that R 3, R 4, and R5 are of equal ohmic value. Therefore an equivalent resistor can be substituted for these three resistors in solving for total resistance.
Given:
Solution:
The circuit can now be redrawn using a resistor labeled Req1 in place of R 3, R4, and R5 as shown in figure 3-51.
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An equivalent resistor can be calculated and substituted for Rl and R2 by use of the product over the sum formula.
Given:
Solution:
The circuit is now redrawn again using a resistor labeled Req2 in place of R1 and R2 as shown in figure 3-52.
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You are now left with two resistors in parallel. The product over the sum method can now be used to solve for total resistance.
Given:
Solution:
This agrees with the solution found by using the general formula for solving for resistors in parallel.
The circuit can now be redrawn as shown in figure 3-53 and total current can be calculated.
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Given:
Solution:
This solution can be checked by using the values already calculated for the branch currents.
Given:
Solution:
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Now that total current is known, the next logical step is to find total power.
Given:
Solution:
Solving for the power in each branch.
Given:
Solution:
Since IR3 = IR4 = IR5 then, PR3 = PR4 = PR5 = 1800 W. The previous calculation for total power can now be checked.
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Given:
Solution:
Q39. What term identifies a single resistor that represents total resistance of a complex circuit? Q40. The total power in both series and parallel circuits is computed with the formula: PT = P1 + P2 +
P3 +...Pn. Why can this formula be used for both series and parallel circuits? Q41. A circuit consists of three resistors connected in parallel across a voltage source. Rl� ��� ��52 = �� ��5�3 � ��� ��DQG�3R3 = 360 watts. Solve for RT, ES and IR2. (Hint: Draw and label the circuit first.)
In the preceding discussions, series and parallel dc circuits have been considered separately. The technician will encounter circuits consisting of both series and parallel elements. A circuit of this type is referred to as a COMBINATION CIRCUIT. Solving for the quantities and elements in a combination circuit is simply a matter of applying the laws and rules discussed up to this point.
The basic technique used for solving dc combination-circuit problems is the use of equivalent circuits. To simplify a complex circuit to a simple circuit containing only one load, equivalent circuits are substituted (on paper) for the complex circuit they represent. To demonstrate the method used to solve combination circuit problems, the network shown in figure 3-54(A) will be used to calculate various circuit quantities, such as resistance, current, voltage, and power.
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Examination of the circuit shows that the only quantity that can be computed with the given information is the equivalent resistance of R2 and R 3.
Given:
Solution:
Now that the equivalent resistance for R2 and R 3 has been calculated, the circuit can be redrawn as a series circuit as shown in figure 3-54(B).
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The equivalent resistance of this circuit (total resistance) can now be calculated. Given:
Solution:
The original circuit can be redrawn with a single resistor that represents the equivalent resistance of the entire circuit as shown in figure 3-54(C).
To find total current in the circuit:
Given:
Solution:
To find total power in the circuit:
Given:
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Solution:
To find the voltage dropped across Rl, R2, and R3, refer to figure 3-54(B). Req1 represents the parallel network of R2 and R 3. Since the voltage across each branch of a parallel circuit is equal, the voltage across Req1 (Eeq1) will be equal to the voltage across R2 (ER2) and also equal to the voltage across R3 (ER3).
Given:
Solution:
To find power used by Rl:
Given:
Solution:
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To find the current through R2 and R 3, refer to the original circuit, figure 3-54(A). You know ER2 and ER3 from previous calculation.
Given:
Solution:
To find power used by R2 and R 3, using values from previous calculations:
Given:
Solution:
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Now that you have solved for the unknown quantities in this circuit, you can apply what you have learned to any series, parallel, or combination circuit. It is important to remember to first look at the circuit and from observation make your determination of the type of circuit, what is known, and what you are looking for. A minute spent in this manner may save you many unnecessary calculations.
Having computed all the currents and voltages of figure 3-54, a complete description of the operation of the circuit can be made. The total current of 3 amps leaves the negative terminal of the battery and flows through the 8-ohm resistor (R1). In so doing, a voltage drop of 24 volts occurs across resistor R1. At point A, this 3-ampere current divides into two currents. Of the total current, 1.8 amps flows through the 20-ohm resistor. The remaining current of 1.2 amps flows from point A, down through the 30-ohm resistor to point B. This current produces a voltage drop of 36 volts across the 30-ohm resistor. (Notice that the voltage drops across the 20- and 30-ohm resistors are the same.) The two branch currents of 1.8 and 1.2 amps combine at junction B and the total current of 3 amps flows back to the source. The action of the circuit has been completely described with the exception of power consumed, which could be described using the values previously computed.
It should be pointed out that the combination circuit is not difficult to solve. The key to its solution lies in knowing the order in which the steps of the solution must be accomplished.
Figure 3-55 is a typical combination circuit. To make sure you understand the techniques of solving for the unknown quantities, solve for ER1.
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It is not necessary to solve for all the values in the circuit to compute the voltage drop across resistor R1 (E R1). First look at the circuit and determine that the values given do not provide enough information to solve for ER1 directly.
If the current through R1 (IR1) is known, then ER1 can be computed by applying the formula:
The following steps will be used to solve the problem.
1. The total resistance (RT) is calculated by the use of equivalent resistance. Given:
Solution:
Redraw the circuit as shown in figure 3-55(B). Given:
Solution:
Solution:
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Redraw the circuit as shown in figure 3-55(C). Given:
Solution:
2. The total current (IT) is now computed. Given:
Solution:
3. Solve for the voltage dropped across Req2. This represents the voltage dropped across the network R1, R2, and R 3 in the original circuit.
Given:
Solution:
4. Solve for the current through Req1. (Req1 represents the network R1 and R2 in the original circuit.) Since the voltage across each branch of a parallel circuit is equal to the voltage across the equivalent resistor representing the circuit:
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Given:
Solution:
5. Solve for the voltage dropped across R1 (the quantity you were asked to find). Since Req1 represents the series network of R1 and R2 and total current flows through each resistor in a series circuit, IR1 must equal IReq1.
Given:
Solution:
Q42. Refer to figure 3-55(A). If the following resistors were replaced with the values indicated: R1 = ��� ��5�3 = lk ��ZKDW�LV�WKH�WRWDO�SRZHU�LQ�WKH�FLUFXLW"�:KDW�LV�(R2?
You will notice that the schematic diagrams you have been working with have shown parallel circuits drawn as neat square figures, with each branch easily identified.
In actual practice the wired circuits and more complex schematics are rarely laid out in this simple form. For this reason, it is important for you to recognize that circuits can be drawn in a variety of ways, and to learn some of the techniques for redrawing them into their simplified form. When a circuit is redrawn for clarity or to its simplest form, the following steps are used.
1. Trace the current paths in the circuit.
2. Label the junctions in the circuit. 3. Recognize points which are at the same potential.
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4. Visualize a rearrangement, "stretching" or "shrinking," of connecting wires.
5. Redraw the circuit into simpler form (through stages if necessary). To redraw any circuit, start at the source, and trace the path of current flow through the circuit. At
points where the current divides, called JUNCTIONS, parallel branches begin. These junctions are key points of reference in any circuit and should be labeled as you find them. The wires in circuit schematics are assumed to have NO RESISTANCE and there is NO VOLTAGE drop along any wire. This means that any unbroken wire is at the same voltage all along its length, until it is interrupted by a resistor, battery, or some other circuit component. In redrawing a circuit, a wire can be "stretched" or "shrunk" as much as you like without changing any electrical characteristic of the circuit.
Figure 3-56(A) is a schematic of a circuit that is not drawn in the box-like fashion used in previous illustrations. To redraw this circuit, start at the voltage source and trace the path for current to the junction marked (a). At this junction the current divides into three paths. If you were to stretch the wire to show the three current paths, the circuit would appear as shown in figure 3-56(B).
While these circuits may appear to be different, the two drawings actually represent the same circuit. The drawing in figure 3-56(B) is the familiar box-like structure and may be easier to work with. Figure 3-57(A) is a schematic of a circuit shown in a box-like structure, but may be misleading. This circuit in reality is a series-parallel circuit that may be redrawn as shown in figure 3-57(B). The drawing in part (B) of the figure is a simpler representation of the original circuit and could be reduced to just two resistors in parallel.
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Figure 3-58(A) shows a complex circuit that may be redrawn for clarification in the following steps.
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NOTE: As you redraw the circuit, draw it in simple box-like form. Each time you reach a junction, a new branch is created by stretching or shrinking the wires.
Start at the negative terminal of the voltage source. Current flows through R1 to a junction and divides into three paths; label this junction (a). Follow one of the paths of current through R2 and R3 to a junction where the current divides into two more paths. This junction is labeled (b).
The current through one branch of this junction goes through R5 and back to the source. (The most direct path.) Now that you have completed a path for current to the source, return to the last junction, (b). Follow current through the other branch from this junction. Current flows from junction (b) through R4 to the source. All the paths from junction (b) have been traced. Only one path from junction (a) has been completed. You must now return to junction (a) to complete the other two paths. From junction (a) the current flows through R7 back to the source. (There are no additional branches on this path.) Return to junction (a) to trace the third path from this junction. Current flows through R6 and R8 and comes to a junction. Label this junction (c). From junction (c) one path for current is through R9 to the source. The other path for current from junction (c) is through R10 to the source. All the junctions in this circuit have
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now been labeled. The circuit and the junction can be redrawn as shown in figure 3-58(C). It is much easier to recognize the series and parallel paths in the redrawn circuit.
Q43. What is the total resistance of the circuit shown in figure 3-59? (Hint: Redraw the circuit to simplify and then use equivalent resistances to compute for RT.)
Q44. What is the total resistance of the circuit shown in figure 3-60?
Q45. What effect does the internal resistance have on the rest of the circuit shown in figure 3-60? EFFECTS OF OPEN AND SHORT CIRCUITS
Earlier in this chapter the terms open and short circuits were discussed. The following discussion deals with the effects on a circuit when an open or a short occurs.
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