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ELEC 103 UNIT 2 Unit two is essentially made up of three individual areas that all utilize the information from the prior unit. The three major areas of this unit, along with the subdivisions, are as follows: 1. Scientific Notation 2. Engineering Notation (Shorthand Notation) 3. Ohms Law & Power Calculations in: a. Series Circuits b. Parallel Circuits c. SeriesParallel Circuits Scientific notation is a method of representing numbers in a form that makes it easier to enter the number into a calculator and also easier to utilize the number in normal calculations. In the electronics industry, numbers are used that are very large, as in the case of resistors, or very small, as in the case of the values for capacitors. Since most calculators have only a seven or eight number display, it is very difficult to enter a number like 0.000000000012. It simply does not fit. However, this is the value of a capacitor that would normally be referred to as a 12 pico Farad capacitor in normal conversation, but would be entered into a calculator as 1.2 x10-11. The number 1.2 x10-11 is exactly equal to the number 0.000000000012 but is simply presented in another form. The symbols after the number 1.2, are written in scientific notation. You must learn how to use scientific notation if you are going to use the calculator for scientific calculations. You must be able to perform scientific calculations in this field of endeavor. Learning SCIENTIFIC NOTATION means learning how to write a number in a form that can be used for calculator entries and also converted into ENGINEERING NOTATION (shorthand notation) for conversational and writing usage. An example of converting a number into scientific notation and then converting the scientific notation into engineering notation follows: The number 0.000000000012 Farads can be written as 1.2 x10-11 Farads and can also be written as 12 pico Farads. They all mean exactly the same thing. The regular number format is difficult to verbalize or use in normal conversation and almost impossible to use on a calculator. However, the scientific notation format can be used on a calculator very nicely. Of Course, the "shorthand" notation can be verbalized and put into print much more readily than the other two formats. There is a need for all three formats in our field and therefore, you must learn all three formats. NUMBER FORMAT TO SCIENTIFIC NOTATION CONVERSION The rules for converting a number into scientific notation are simple if you follow the few steps necessary to perform the conversion. 1. Move the decimal point from where it is presently located to a position located to the right of the first integer. In the number used in the previous example, the decimal would be moved from where it is located to a position between the one and the two. You would write the number as 1.2, which places the decimal to the right of the first integer. 2. When the number is written it should be followed by x10. The x10 is going to be referred to as the carrier. The carrier is only in place to carry the important information. Therefore, the number should presently be written as 1.2 x10. 3. Count the number of places the decimal point was moved and write that number as a superscript of the carrier. The converted number should now be written as 1.2 x1011 since the decimal point was moved eleven places from where it was located to the present position after the first integer. Page 1 ELEC 103 UNIT 2 4. The last part of the conversion procedure requires you to decide which way you would need to move the decimal to get it back to where it was located before you moved it to the present location. If you need to move the decimal point to the left to return it to the original location, you must place a minus sign in front of the superscript number. That is the situation with the number being presently used in this example. Therefore, in this example, the number 0.000000000012 Farads when converted into scientific notation is 1.2 x10-11 Farads. 5. If you need to move the decimal point to the right to return it to the original location, you must place a plus sign in front of the superscript number. When this condition exists, the plus sign can be placed in position or not placed in position. The sign is understood to be PLUS when no sign is present and that is the preferred method. To put this procedure into practice will require you to go through a few examples and then try to do a few problems on your own. There is a resistor that has a value of two million four hundred thousand Ohms that needs to be converted into scientific notation. The number is 2,400,000 Ohms. Move the decimal to the right of the first integer. Alternatively, if you want to say it differently, say, move the decimal between the first and second numbers. You would write 2.4 followed by the carrier. This would make the number 2.4 x10 with the superscript number and sign still to be added. Count the number of places the decimal point was moved and use that number in the superscript position. The conversion presently is 2.4 x106 with the sign in front of the superscript still to be added. It should be obvious that the decimal being located between the two and the four, would have to be moved to the right to make the number the same as it was originally. Therefore, a plus sign should be placed in front of the superscript number. The conversion would make 2,400,000 Ohms equal 2.4 x10+6 Ohms, or 2.4 x106 Ohms, but the exponent without the sign is preferred. Since events happen very quickly in electronics, we often discuss events as happening in small parts of a second. An event happened in 0.00008 seconds and we need to convert that time into scientific notation to make some calculations. The first step is to move the decimal point from the present location to a new location to the right of the first integer. This would require moving the decimal point to a position to the right of the number eight. Therefore, you would write the number followed by the carrier, which is 8 x10. The second step is to count the number of places the decimal has been moved and use that number as the superscript number for the carrier. Therefore, you would write 8 x105 since the decimal was moved five places. The third step is to determine in which direction you must move the decimal to return the number to its original format. If you must move the decimal to the left to return the number to its original format, a minus sign must be placed in front of the superscript number. On the other hand, if you must move the decimal to the right to return the number to its original format a plus sign must be placed in front of the superscript number. In this example, you should have determined the direction needed to return the number to its original format was to the left and therefore, the completed conversion is 8 x10-5 seconds. As you probably know, our ears can hear frequencies between 20 Hertz (cycles per second) and 20,000 Hertz (Hz). We need to use 19,500 Hz in scientific notation format, so therefore, must perform a conversion. The first step in the conversion is to move the decimal point from its present location after the last zero to a position between the first and second integer. That means the number, converted thus far, would be 1.95 x10. The second step in the conversion is to count the number of places the decimal point was moved. You should count a movement of four places. This means the number four can be added as the superscript of the carrier. That means the number, converted thus far, would be 1.95 x 104. The third and last step is to determine which direction the decimal point must be moved to return the number to Page 2 ELEC 103 UNIT 2 its original position. It should be obvious that the decimal must be moved to the right to get it back to its original position. The final conversion of 19,500 Hz to scientific notation is such that the number is 1.95 x 10+4 or, as preferred, it should be written without the plus sign as 1.95 x 104 Hz. 1. 0.005 Amperes In the figure on the right you see a listing of several numbers. Before going 2. 0.0000021 Sec any further, you should make an effort to convert each one of the numbers in 3. 27000 Ohms Figure 21 into scientific notation. The answers, in scientific notation 4. 5400000 Hz format, for each one of the four numbers in the figure will be available in FIGURE 21 Figure 22 on the next page. Keep in mind the three step procedure involved in the conversion and follow it very carefully for smooth conversions. SCIENTIFIC NOTATION TO NUMBER FORMAT CONVERSION Next on the agenda is the conversion from scientific notation back into the original number format. This conversion is the opposite of the conversion routine you have been learning using the numbers in Figure 21 and Figure 22. Most importantly, you should learn how to convert regular numbers into scientific notation. The conversion from scientific notation into number format is easy if you know how the number got into scientific notation in the first place. Suppose the number 8 x103 Amperes is the number, in scientific notation, that is going to be converted back into its original number format. The first step in the conversion is to look at the superscript attached to the carrier. The superscript number tells you how many places to move the decimal, and the sign indicates which way to move the decimal point. Therefore, in the present example, 8 x10-3 Amperes, the decimal should be moved three places. The second step is to determine which way the decimal is to be moved. Since the sign is minus, the decimal should be moved three places to the left from its present position. The third step is to move the decimal to the new position and attach the correct units to the number. Therefore, 8 x10-3 Amperes converts to 0.008 Amperes. It is obvious that once you know how to convert into scientific notation it is quite easy to convert back to number format. The first step in the conversion is to determine the number of places to move the decimal point to return the decimal to its original position. You can determine that number by reading the superscript number attached to the carrier. In this example, the number of places to move the decimal is twelve places. The second step is to determine the direction to move the decimal. If the sign of the superscript number is minus, the decimal will be moved to the left. If the sign of the superscript number is plus, or there is no sign, the decimal will be moved to the right. In this example, the decimal will be moved to the left. The third step involves moving the decimal the appropriate number of places in the proper direction and attaching the unit designator to the number. Of course, the carrier and the carrier superscript are discarded. The number for this example is 9.1 x10-12 Farads, which converts to 0.0000000000091 Farads. In Figure 23, you will find five numbers in scientific Scientific Notation Number Format notation. Perform a scientific notation format to number 1. 3.9 x104 Ohms format conversion on each one of the numbers in the figure. -9 The answers are on the next page in Figure 24. WAIT ! 2. 5 x10 Farads Do not look until you have worked out the answers, then 3. 9.050 x107 Hz look. 4. 5.4 x10-6 Amps 5. 1.5 x102 Volts FIGURE 23 Page 3 ELEC 103 UNIT 2 The answers to the two previous questions. Number Format Scientific Notation Scientific Notation Number Format 1. 3.9 x104 Ohms 39,000 Ohms 2. 5 x10-9 Farads 0.000000005 Farads 3. 9.050 x107 Hz 90,500,000 Hz 0.005Amps = 5 x 10-3 Amps 0.0000021 Sec = 2.1 x10-6 Sec 27000 Ohms = 2.7 x104 Ohms 4. 5.4 x10-6 Amps 0.0000054 Amps 54000000 Hz = 5.4 x107 Hz 5. 1.5 x102 Volts 150 Volts FIGURE 24 FIGURE 22 ENGINEERING NOTATION Engineering notation requires a letter or a word be substituted in place of the scientific notation carrier. In other words, if the number under the scientific notation column, in problem four of Figure 24 was going to be used in a written report it would not be written as you see it, 5.4 x10 -6 Amps. The carrier and its superscript would be thrown away and replaced with a letter or a word that is known to be equal to that particular carrier. The written version would be 5.4µA or could also be written as 5.4 microAmps. There are standard symbols and words that are the equivalent of specific scientific notation format. These symbols and words make up the shorthand notation that can be used to replace scientific notation and number format for the written or spoken word. These symbols, shorthand words and their equivalent value in scientific notation are called engineering notation. One of the first things to look at in Figure 25 is the PREFIX SYMBOL POWER OF 10 scientific notation equivalent column. Note that all the Tera T x1012 exponents are in three unit increments. Specifically running from plus 12 to minus 12. This introduces a Giga G x109 problem when you want to convert a number from Mega M x106 scientific notation format into engineering format. Kilo k x103 Suppose you had a number similar to number one in Figure 24 and you wanted to convert it into Units x100 engineering notation. The exponent of the carrier is plus milli m x10-3 four and there is only a plus three or plus six listed for engineering notation conversion. The solution is to put micro x10-6 4 the number 3.9 x10 into engineering notation format nano n x10-9 with a carrier similar to one of those in Figure 25 under power of 10. pico p x10-12 You should use one of those listed that will allow you to FIGURE 25 have a whole number followed by one of engineering notation numbers that appear in the table. The first step is to convert the scientific notation format back to number format. To do that, the decimal point should be moved four places to the right and the number becomes 39,000. The second step is to move the decimal point either three places, six places, or nine places etc., so the number and the carrier exponent agree with one of the listings in the engineering format table. If the decimal is moved three places, the number becomes 39 x103. If the decimal is moved six places, the number becomes 0.039 x106. The one that is acceptable is 39 x103 since it is a whole number and Page 4 ELEC 103 UNIT 2 not a decimal number. Now that the number has been converted into engineering notation format with a ± 3, ± 6, ± 9, or ± 12 superscript value, the shorthand substitution can be made. The number 39 x103 can be made into 39k by substituting k for x103. Look at the second number in Figure 24. Convert that number into engineering notation (shorthand format) using two possible formats, and in addition, show two versions of each format. To put the second number in Figure 24 into engineering format is easy, since x10-9 is listed in the engineering format table. The number 5 x10-9 Farads could be written as 5 nano Farads and could also be written as 5 nF. The second conversion will be more difficult. Just remember that the number, when converted, should be a whole number and not a decimal number. To convert the scientific notation format into engineering format, move the decimal in increments of three until you have a whole number. Since, in the previous paragraph, the decimal was moved until the number of decimal places moved was nine, the decimal should be moved an additional three places in the same direction. This makes the number 5000, and therefore, the decimal would have been moved a total of twelve places. Therefore, the number should be written, in engineering notation, as 5000 x1012 Farads. Of course, this is exactly the same as 5 x10-9 Farads but in another form. The number, 5000 x10-12 Farads, can be converted to 5000 pF and can also be converted to 5000 pico Farads. All three numbers in the last sentence are equal. In fact, 5 nF, 5 nano Farads, 5000 pF, and 5000 pico Farads are all equal but just appear in different forms. When a number appears with the decimal place after the first number and is followed by the carrier and exponent, it is said to be in scientific notation. However, if one wants to use engineering notation, the decimal does not need to be located after the first number. The primary goal is to have the carrier exponent be made into ± 3, ± 6, ± 9, or ± 12. When this occurs, the word substitution can be made for the carrier and its exponent. The requirement that the decimal must be placed to the right of the first integer in scientific notation, is not a requirement in engineering notation. Engineering notation requires that the main number be a whole number and the carrier exponent be ± 3, ± 6, ± 9, or ± 12 followed by the appropriate unit designator. Unlike scientific notation, the main number is not required to be a single integer. The main number can be any number of integers. However, it must be followed by the appropriate carrier exponent so the equivalent shorthand notation can be substituted for the carrier and its exponent. For Figure 22, convert the numbers in the scientific notation column into engineering notation. The answers will be found in Figure 26 on the next page. This unit, so far, has covered the conversion of ordinary numbers into scientific notation. The conversion of scientific notation into ordinary numbers. The conversion of ordinary numbers into engineering notation. In addition, of course, if you can convert from scientific notation into ordinary numbers, you can convert from engineering notation into ordinary numbers. The answers to the conversions from scientific notation into engineering notation appear in word format and in symbol format. OHMS LAW Ohms Law was discovered in 1827 by Georg Simon Ohm. Ohms Law applies to an entire circuit or to any part of a circuit. Thus, the potential difference or voltage drop across any part of the circuit or conductor equals the current (I) flowing in the conductor multiplied by the resistance (R) of that conductor, or E = IxR. Page 5 ELEC 103 UNIT 2 Scientific Notation Engineering Notation -3 1. 5 x10 Amps = 5 milliAmps or 5 mA 2. 2.1 x10-6 Sec = 2.1 microSec or 2.1µs 3. 2.7 x104 Ohms = 27 kiloOhms or 27 k 4. 5.4 x107 Hz = 54 MegaHertz or 54 MHz On the left is the answer to the question from the previous page. FIGURE 26 Ohms Law applies to all linear conductors. Ohms Law does not apply to nonlinear conductors and devices such as ionized gases, semiconductors, and other devices when operated in the nonlinear part of their characteristics. Figure 27 shows Ohms Law in three forms. The main form, I = E=IxR E R, is the form used to describe Ohms Law. The Law states that the current (I) is directly proportional to the voltage (E) I=ER and inversely proportional to the resistance (R). The current is directly proportional to the voltage means that if the voltage is R = E I increased, the current in the circuit will increase in the same FIGURE 27 proportion. Figure 28 shows a mathematical analysis of the Directly Current Inversely statements in the last paragraph to help you better Current To understand Ohms Law. The current is directly Proportional To The Proportional Voltage The Resistance proportional to the voltage means that if the voltage is I=ER I = E R doubled the current will double. This is shown in Figure 28 in the boxes numbered one and two. If you compare 1. 2A= 4V 2 4. 2A= 4V 2 the boxes that are numbered one and three, you see that 2. 4A= 8V 2 5. 1A= 4V 4 when the voltage is halved the current halves. This means 3. 1A= 2V 2 6. 4A= 4V 1 the current does just the same as the voltage does, and FIGURE 28 does it in the same proportion. So when the voltage doubles the current doubles and when the voltage halves the current halves. The boxes numbered four, five and six, in Figure 28, show an inverse proportionality exists between the current and the resistance. If the resistance is doubled, the current will halve and when the resistance is halved, the current will double. This is a mathematical analysis of the verbal expression of Ohms Law, which states the current is directly proportional to the voltage and inversely proportional to the resistance. Figure 27 shows a circle with Ohms Law in mathematical format inside the divided circle. It is a way to help you remember the three different versions of Ohms Law that are shown to the right of the circle. In a problem, when you want to calculate the voltage, you place your finger over the E in the divided circle, which represents the voltage. In addition, as you can see, the the R and I are next to each other indicating multiplication as the uppermost formula also indicates. If you place a finger over the I within the circle, you will see the E over the R as the middle formula indicates. The circle is a method for you to use to remember the three versions of Ohms Law so you can calculate the voltage, the current, and the resistance. However, after using Ohms Law for some problem solving, you will rely less on the circle method to remember Ohms Law. Page 6 ELEC 103 UNIT 2 OHMS LAW AND SERIES CIRCUITS You have seen series circuits previously, in unit one, when you performed voltage and current measurements. In this section of the present unit, you will be learning how to solve series circuit problems for voltage, current, resistance and power. The definition of a series circuit is, the series circuit provides only one path for the current to flow in the circuit, from the negative terminal of the source to the positive terminal of the source through a resistive path. There are three rules that apply to the series circuit. You must know the rules, in addition to Ohms Law, to solve series circuit problems. The rules are related to voltage, current and resistance and can be written in word format and in mathematical format. 1. The rule for voltage states that the applied voltage or source voltage is equal to the sum of the individual resistor voltage drops no matter how many resistors there are in the circuit. The applied voltage (VA), the source voltage (VS), and the total voltage (VT) are all used interchangeably. The mathematical expression would be written as VT = VR1 + VR2 + ••• +VRN 2. The rule for current states that the current flowing in the circuit is the same everywhere in the circuit. The total circuit current is the same as the current flowing in each resistor in the circuit. Since there is only one path for current to flow it cannot be different in different parts of the circuit. The mathematical expression is IT = IR1 = IR2 = ••• =IRN 3. The rule for resistance states that the total resistance of the circuit is equal to the sum of the individual resistances, no matter how many resistors there are in the circuit. The mathematical expression for total circuit resistance is RT = R1 + R2 + ••• +RN Figure 29 contains a series circuit that will be used to introduce you to the techniques used to solve series circuit problems for voltage drops, circuit current, and resistances. The method that is being presented to solve series circuit problems is not the only method available. If you wish to use another method, feel free to do so. In the circuit of Figure 29, there are three GIVEN values. The voltage drop across R1 is 6 Volts. The resistance of R2 is 3.9k and the total current flowing in the circuit is 2mA.You are requested to FIND the total or source voltage and the resistance of R1. To help remember the values that have been GIVEN and the values that are to be calculated, Figure 29 a tabulation format will be utilized. Figure 210 illustrates the tabular format that will be E I R used for assistance in solving series circuit problems. As R1 6V * you look at the table, notice there is a row for each R2 3.9k resistor in the circuit. There is also a row for the circuit Total Ckt * 2mA totals. For each resistor there is a box for voltage, current and resistance as there also is for total voltage, total FIGURE 2-10 current and total resistance. The GIVEN quantities are placed in their respective positions. The values that are to be calculated are marked with an asterisk in the appropriate box. Most importantly, Ohms Law can be used with values from one row only. You cannot pick a voltage from one row, a resistance from another row and calculate a current. You must use only values on the same row or horizontal line associated with the component in which you are interested. For instance, the total circuit current must be used with the total circuit resistance to calculate the total circuit voltage. Page 7 ELEC 103 UNIT 2 Additionally, the three rules for series circuits that were introduced earlier can be used with the tabular format shown in Figure 210. For example, the voltage drop across R1 when added to the voltage drop across R2 will equal the total circuit voltage. In addition, the resistance of R1 added to the resistance of R2 will equal the total circuit resistance. In addition, since the current is the same everywhere in the circuit, if you know one current you know all the currents. The current flowing through R1 is the same as the current flowing through R2, which is the total circuit current. The columns or vertical boxes are associated with the rules for voltage, for current, and for resistance. The rows or horizontal boxes are associated with the Ohms Law formula. To start the problem solving procedure, the table should be drawn as shown in Figure 210. Draw one row for each resistor in the circuit then draw the last row for the total circuit values. There should be a column for voltage, a column for current, and a column for resistance. Next, fill in the appropriate boxes with the GIVEN information and finally, place an asterisk in the boxes where values are to be calculated. Once all the preliminary work has been performed, the first step is to apply the three rules stated earlier, to the table. The first rule, the one for voltage, will not benefit us. Since the total voltage is equal to the sum of the individual resistor voltage drops, the formula requires only one unknown value. In this example, only one voltage is known and two voltages are unknown. Therefore, the voltage rule cannot help this time. The second rule, the rule for current can provide some help. The second rule states that the current is the same everywhere in the circuit. Therefore, since the total circuit current is known, the current through R2 and the current through R1 will be the same. Look at Figure 211, which shows the appropriate boxes, filled in with the newly acquired information. The third rule, the one for resistance, will not benefit us. E I R Since the total resistance is equal to the sum of the R1 6V 2mA * individual resistances in the circuit, the formula requires R2 2mA 3.9k there be only one unknown value. In this example this is not true, there are two unknowns. Total Ckt * 2mA Now that the first step has been completed, the second FIGURE 2-11 step can be started. The second step consists of looking across the rows for a row with two known values and only one unknown value across the row. There are two rows that have that qualification, the first row and the second row. You might say to yourself, which row is the place to start. The answer to that question is that you should start where ever you see a calculation possibility. If you can calculate the value for each box in the table, it makes sense that you know everything about the circuit. Therefore, start where you see a calculation possibility and calculate everything. When finished, you will have the answer to any question that might be asked about the circuit. Looking at the first row in Figure 211, the value for resistance (R) is unknown but the values for voltage (E) and current (I) are known. Therefore, using the circle method shown in Figure 27, place your finger over R, the letter used for resistance. You should see that you need to divide the voltage across R1 (ER1), by the current flowing through R1 (IR1), in order to calculate the resistance of R1. The equation should look like this: (R1 = ER1 IR1). Substituting the proper values into the equation, you should be dividing 6 Volts, the voltage across the resistor R1, by 2 mA, the current flowing through R1. Therefore, you should see 6V divided by 0.002A, which equals 3000 Ohms. The calculated resistance should be converted into engineering notation. When the calculated value has been converted into engineering notation, the calculated value should be placed in the resistance box of the R1 row of tabulated results for this problem. Page 8 ELEC 103 UNIT 2 Figure 212 shows the results up to the present time. E I R REMEMBER, before starting the problem, and after each R1 6V 2mA 3.0k calculation you should apply the rules to each of the columns. This time, the voltage rule is of no benefit and the current rule R2 2mA 3.9k is of no benefit. However, the resistance rule will allow us to Total Ckt * 2mA calculate the total circuit resistance. The sum of the FIGURE 2-12 individual resistances is equal to the total circuit resistance. Therefore, the total resistance is equal to the sum of the 3k and the 3.9k resistors, which is 6.9k. The lower right hand box, which represents the total circuit resistance, can be filled with 6.9k. Figure 213 shows the results up to the present time. E I R Since the rules have just been applied, it is time to do R1 6V 2mA 3.0k another calculation. The second row and the last row R2 2mA 3.9k both qualify for a calculation since both rows have only one unknown item. One of the FIND items was the total Total Ckt * 2mA 6.9k voltage, so let us make the calculation using the data in FIGURE 2-13 the last row. Refer to Figure 27. Place your finger over the E, which represents voltage, which is the unknown in this calculation. You should see that the current (I) should be multiplied by the resistance (R) in order to calculate the voltage (E). Therefore, the equation should look like this: (ET = IT x RT). Substituting the numbers into the equation, you will multiply 2mA by 6.9k, which means you will see 0.002A times 6900, which is equal to 13.8 Volts. The calculated value should be placed in the total circuit voltage box of tabulated results for this problem. This can be seen in Figure 214, which shows the results tabulated up to the present time, including the last calculation. The rules should be applied after each calculation. E I R Therefore, if you look at Figure 214, you should see R1 6V 2mA 3.0k that only one rule can be applied in this situation. The voltage rule is the rule that can be applied since we know R2 2mA 3.9k two of the three voltages. The voltage rule states that the Total Ckt 13.8V 2mA 6.9k total circuit voltage equals the sum of the individual resistor voltage drops. Since we know the total circuit FIGURE 2-14 voltage and one of the resistor voltage drops, the equation would look like this: (ER2 = ET ER1). Subtracting 6 Volts from 13.8 Volts will give you the voltage drop across R2. This calculated value should be placed in the appropriate box of tabulated results for the problem. All the boxes are filled and therefore, the problem has been totally solved. You can pick out the values you were requested to FIND, with no difficulty, since everything concerning the circuit is known and is in one of the boxes in the tabulated format. The final tabulated display for the original problem, E I R stated in Figure 29 and just concluded, is presented in R1 6V 2mA 3.0k Figure 215. Look at it and review the features of using R2 7.8V 2mA 3.9k the tabulated format. The three rules are associated with the columns. Look down the voltage column and note Total Ckt 13.8V 2mA 6.9k that 6 Volts plus 7.8 Volts equals 13.8 Volts. The current FIGURE 2-15 column shows that the current is the same everywhere in the circuit. Finally, the resistance column shows that the total resistance is equal to the sum of the individual resistances. Also notice that the numbers on any row agree with any Ohms Law calculation Page 9 ELEC 103 UNIT 2 you might perform using the data on that particular row. The tabulated format helps you remember the rules, stores given and calculated values in a useful manner, and helps you use correct values for calculations. SERIES CIRCUITS AND POWER CALCULATIONS When a current flows through a resistance there is work being done, and heat is being generated. An example of this is the light bulb in you home. A 15W light bulb is not as bright as a 100W light bulb. However, a 100W light bulb is considerably hotter than a 15W light bulb. Both bulbs have current flowing through the resistance of the filament to create the light that we utilize. However, since work is being done, heat is being generated and is being dissipated into the atmosphere. Power (P) can be calculated by multiplying the voltage (E) by the current (I) as shown in Figure 216. The unit of power is the WATT. Every resistive component that has a Figure 216 current flowing through it, will have heat generated by the work that is being performed. The component must get rid of the generated heat, dissipate the heat, so it does not overheat and malfunction. To be able to calculate the power a component must dissipate, or a source must supply to a circuit, you can use the circle format shown in Figure 216. This circle format works exactly the same as the one you used for Ohms Law. The three basic forms of the power formula are also displayed so you can verify the power circle method. Remember, the unit of power is the WATT and power is calculated by multiplying voltage times current. Refer to Figure 215 and visualize another column next to the resistance column. This column would have a P for power in the uppermost box. The rule for the power is the total power being supplied to the circuit is equal to the sum of the power being dissipated by the individual E I R P components in the circuit. The additional 6V 2mA 3.0k 12.0mW column, the power column, has been attached to R1 7.8V 2mA 3.9k 15.6mW the tabulated results of the last problem and is R2 shown in Figure 217. Total Ckt. 13.8V 2mA 6.9k 27.8mW If you multiply the voltage by the current on a FIGURE 217 specific row, the product will be the power. Add the power being dissipated by each resistor and the sum will equal the total power being supplied to the circuit by the source. That power, for this circuit, is 27.8 milliWatts. There are other formulas that are available to calculate the power. For the present, however, the formula introduced in Figure 216 will be sufficient for all power calculations you will be required to perform. Here is another series circuit problem that requires the use of Ohms Law and the Power formula. You must remember to use the rules that are applied to the vertical columns, before you attempt a calculation. In addition, after each calculation you should apply the rules again. The circuit is shown in Figure 218. The problem, as you can see, is a single source series circuit with three resistors. The values that are GIVEN are as follows: The voltage drop across resistor R1 is 6.0 Volts and the power being dissipated by R1 is 18 milliWatts. The total circuit power is 60.0 milliWatts and resistor R2 has a Figure 218 Page 10 ELEC 103 UNIT 2 value of 1.5k. The values that you are to E I R P calculate are as follows: The voltage drop across R1 6.0V 18.0mW R3, (ER3), and the total circuit resistance (RT). 1.5k The table for this example is shown in Figure R2 219 with the GIVEN values in their proper R3 * places, and the values you are to FIND marked Total Ckt. * 60.0mW with an asterisk. FIGURE 219 What is the first thing you are going to do to start solving the problem? That is right! Apply the rules to the voltage column, the current column, the resistance column and the power column. However, for this problem, the rules just do not work the first time as they did in the previous problem. So, proceed to the next step, which means you are to look for a row where two pieces of data are available to you. For this example, the first row is the only row where two pieces of data are available. The next problem you have is to find the proper formula. However, of the two available, you have only one that has the voltage (E) and the power (P) in the same formula. Since both the formula we are using have three values, the third value can be calculated when the other two values are known. The power formula in Figure 216, should be utilized for this calculation. Since voltage and power are the known values, the current is the unknown value. The calculation IR1 = PR1 ER1 will involve dividing 18 milliWatts by 6 Volts to calculate the current. The calculated current, IR1, is 3 mA. Therefore, that value can be placed in the current box on the R1 row. Before and after each calculation, you must use the rules in association with the columns. The rule for current applies nicely at this time, and allows you to fill in all the current boxes for the problem with the 3mA calculated current. The current law states that the current is the same everywhere, in a series circuit. Therefore, the current through R2, the current through R3 and the total circuit current, IT, are the same as the current flowing through R1, which was just calculated. Let us review what has occurred so far. The problem was setup using the tabulation method and the rules were applied to the columns with no success. The first row had a calculation possibility using the power formula. The current flowing through R1 was calculated as 3mA. Again, the rules were applied to the table after the calculation was completed. The rule for current was able to be applied to the current column in the table. If one current in a series circuit is known, all currents in the series circuit are known. Therefore, 3mA could be placed in all the current boxes of Figure 220. A E I R P calculation was performed and the calculation 6.0V 3mA 18.0mW was followed by the application of the rules to R1 3mA 1.5k the appropriate column. The next step will be to R2 look for a calculation possibility. That means you R3 * 3mA should look for a row where two pieces of data Total Ckt. 3mA * 60.0mW are available. Obviously, there are three rows FIGURE 220 where that condition exists. The first row, the second row and the fourth row each have two known values. Which one should be calculated first? It really makes no difference, since eventually, you will probably fill all the boxes. The second row has known values for the current (I) and the resistance (R). Therefore, you should pick a formula that has both the current and the resistance in the formula. Of course, ER2 = IR2 x RR2 is the formula that should be used for this calculation. The voltage across the resistor R2, (ER2), will be the calculated value. The voltage drop ER2, is calculated by multiplying 3mA, (IR2), times 1.5k, (R2), for a value of 4.5 Volts. That value can be placed in the box representing the voltage drop across resistor R2. The rules should be applied after a calculation. However, nothing can be learned from applying the rules, at this time, so Page 11 ELEC 103 UNIT 2 the procedure continues with another calculation. A calculation could be performed using data on the first row, the second row or the circuit totals row. Since the fourth row, the circuit totals row, has an asterisk, it will be used for the calculation. The known values are total circuit power (PT) and total circuit current (IT). The power formula in Figure 216 will show that voltage equals power divided by current. The equation should be written with subscripts and should look like this ET = PT IT. Substituting into the equation will show 60mW divided by 3mA for a total voltage of 20 Volts. The calculated voltage should be placed in the E I R P appropriate box and Figure 221 will show all 6.0V 3mA 18.0mW the circuit information known, up to and R1 4.5V 3mA 1.5k including the last calculation. If you look at the R2 voltage column of the figure, you will notice all R3 * 3mA boxes are filled except one. The voltage rule Total Ckt. 20.0V 3mA * 60.0mW states, the total voltage equals the sum of the FIGURE 221 individual voltage drops. Since ET = ER1 + ER2 + ER3, then transposing the equation will make ER3 = ET ER1 ER2. Substituting into the equation, produces a value for ER3 equal to 9.5 Volts. The calculated voltage should be placed in the table in the proper box and the rules should be applied to the columns. Again, there will be no success when the rules are applied, so you should look for another calculation to perform. There is one asterisk remaining in the table and it is in the circuit totals row. There is sufficient information to calculate the value necessary to fill that box. If that box is filled, the problem is over even though not all the boxes will be filled. What formula would you use to calculate the total circuit resistance, (RT)? The only formula with resistance in it is the Ohms Law formula and it should take the form of R T = ET IT. Substituting into the equation should show 20V divided by 3mA for a total circuit resistance, R T, equal to 6.7k. That finishes the calculations necessary to answer the FIND questions set forward in the original problem statement. If you want to calculate the information E I R necessary to fill the boxes in the table 6.0V 3mA 2.0k completely, you should do it first before looking R1 4.5V 3mA 1.5k at Figure 222 for the answers. This is a good R2 time to review the rules by checking the column's R3 9.5V 3mA 3.2k in the figure to the left. Also, perform some Total Ckt. 20.0V 3mA 6.7k Ohms Law calculations on various rows and FIGURE 222 make certain that the numbers in the boxes turn out correctly when they are multiplied or divided as the case may demand. P 18.0mW 13.5mW 28.5mW 60.0mW OHMS LAW AND PARALLEL CIRCUITS When you studied series circuits, you learned there were three rules associated with series circuits that you needed to know. If you did not know the rules, in addition to Ohms Law, you could not solve series circuit problems. The same is true with parallel circuits as was true with series circuits. There are three rules that apply to parallel circuits and it is necessary for you to learn the three rules before attempting to solve parallel circuit problems. The rules, as before, are related with voltage, current, and resistance and can be written in word format and in mathematical format. 1. The rule for voltage states the voltage is the same across all branches of the parallel circuit and is equal to the applied or source voltage. The applied voltage (VA), the source voltage Page 12 ELEC 103 UNIT 2 2. 3. (VS), and the total voltage (VT) are all used interchangeably. The mathematical expression would be written as: VT = VB1 = VB2 = ••• = VBN The rule for current states the total current flowing in the circuit is equal to the sum of the individual branch currents. Since there is more than one path for current to flow, it is different in different parts of the circuit. The mathematical expression would be written as: IT = IB1 + IB2 + ••• + IBN The rule for resistance states the total resistance 1 of the circuit is smaller than the smallest RT = 1 1 1 + + ... + branch resistance. The mathematical expression RB1 RB2 RBN for total circuit resistance is: The total resistance is equal to the reciprocal of the sum of the reciprocals of the branch resistances, no matter how many branches there are in the circuit. The above equation can also be manipulated, using two branches, to produce another equation that is referred to as the product over the sum formula. It can be used to calculate the total resistance when the circuit consists of only two branches and a source. Here is the formula: RB1 x RB2 RT = RB1 + RB2 Before starting a parallel circuit problem, you should spend some time going over the three rules for parallel circuits. To help you apply the parallel circuit rules to an actual circuit, refer to Figure 223 which is a parallel circuit with a source and two branches. Each branch has a single resistor, although there could be more than one resistor in a branch. In this circuit, branch one will be defined as the vertical space between the points marked "A" and "B" which includes the 6k resistor. Branch two of the circuit will be defined as the space between the points marked "C" and "D" which includes the 3k resistor and the milliammeter labeled, Ir. The voltage rule states the voltage across each branch of a parallel circuit is the same and is equal to the source voltage. Therefore, if a voltmeter was placed across branch one, with the negative lead placed at point "A" and the positive lead placed at point "B", the meter would indicate twelve volts. If a voltmeter was placed across branch two, with the negative lead placed at point "C" and the positive lead placed at point "D", the meter would indicate twelve volts. In addition, if the voltmeter was placed across the source it would also measure twelve volts. Of course, you would know where to place the voltmeter leads for that measurement! The second rule states the total current flowing in the circuit is equal to the sum of the individual branch currents. If you read the milliammeter labeled It, in Figure 223, it would read the total circuit current. The reading would be greater than the milliammeter labeled Ir, since it is measuring the current in branch two. If there was another milliammeter placed in series with the 6k resistor, and the current it measured was added to the current measured by Ir, it would be equal to the current measured by the milliammeter labeled, It. The current Figure 223 flowing through the 6k resistor is the branch one current. The current flowing through the 3k resistor is the branch two current. The total circuit current can be measured where the milliammeter, It, is now located or it could be measured between the positive end of the source and point "B". Page 13 ELEC 103 UNIT 2 To help you put together what has been discussed thus E I R far, you should refer to the table shown in Figure 224. This tabulation of the parallel circuit information will B1 12V 6k work the same as the tables used with series circuits. B2 12V 3k Since the source voltage was given, the rule for voltage 12V could be applied to the voltage column. That enables Total Ckt you to fill in all voltage boxes with the same value as FIGURE 224 the source voltage. In addition, the resistor values can be obtained from the schematic and placed in the appropriate boxes. If you study the table, you should see three calculation possibilities. The current flowing in branch one, the current flowing in branch two, and the total circuit resistance. The total circuit resistance is equal to the reciprocal of the sum of the reciprocals. However, since there are only two branches, the product over the sum formula can be used. We will use both formulas to calculate the total circuit resistance. The first formula will be the product over the sum: RB1 x RB2 6x103 x 3x103 18x106 2 RT = = = 2x103 = 2k 3 3 3 RB1 + RB2 6x10 + 3x10 9x10 The next formula is the reciprocal of the sum of the reciprocals formula. You will need a scientific calculator for this calculation and if you do not have one available, you should locate one before attempting the calculation. On the calculator keyboard, locate the reciprocal key, which is the key marked 1/X. The input sequence will be as follows: 6 1/X + 3 1/X = 1/X The number or symbol inside the box is the key that should be pressed on the calculator keyboard. Since the resistors are in the same k units, the k units were factored out and will be placed after the answer. If you enter the sequence, as shown, the answer will appear on the calculator display after the last key has been pressed. Since 2k was calculated earlier as the total circuit resistance, it should be no different using the reciprocal of the sum of the reciprocals formula. The formula is: RT 1 1 1 + 6k 3k Therefore, you are starting the entry routine in the lower left of the formula and are proceeding just as the formula directs you to proceed. That is, take the reciprocal of 6k and add that to the reciprocal of 3k and since that is all the branches being used, hit equals to get a sum of the reciprocals. Take the reciprocal of the sum to get the total resistance. If you have more than two branches, just extend the entry routine to include the resistance of the other branches. REMEMBER, if you put all the resistors into the same units, you can enter just the number into the calculator and attach the units that were factored out from the resistances, to the answer. The calculated value for total circuit resistance can be placed in its box. It should now be obvious that the total circuit resistance is smaller than the smallest branch resistance. There are three current calculations that can be performed. You will only need to perform two of the calculations though, since the current rule for parallel circuits will allow you to calculate the third current. Page 14 ELEC 103 UNIT 2 The calculations for current should all be made using Ohms E I R Law, in the classic configuration, which is I = E R. Of B1 12V 6k course, you would use the appropriate subscripts with the 12V 3k formula to insure the correct number values are substituted into B2 the formula. To calculate the current in branch one, the formula Total Ckt 12V 2k would read IB1 = EB1 RB1, which would mean dividing 12V FIGURE 225 by 6k, for a branch one current of 2mA. The branch two current could be calculated in the same manner. Divide 12V, the voltage across branch two, by 3k, the resistance of branch two, for a branch two current of 4mA. The current rule for parallel circuits states the total circuit current equals the sum of the branch currents. In this circuit, the sum of the branch currents is 6mA. Therefore, the total circuit current will be equal to 6mA. This can be proven quite easily by dividing the source voltage, 12V, by the total circuit resistance, 2k. This calculation proves the total circuit current is 6mA and is equal to the sum of the branch currents. As in series circuits, the power can be calculated, using the power formula, and the total power is equal to the sum of the power being dissipated by the individual circuit components. The power calculation results and the other circuit E I R P calculation results are shown in Figure 226. Use the 12V 2mA 24mW 6k columns of the tabulated results to check the rules for B1 parallel circuits, and do a few calculations across the B2 12V 4mA 48mW 3k rows using Ohms Law and the power formula. Total Ckt 12V 6mA 72mW 2k The voltage is the same across each branch of a FIGURE 226 parallel circuit and is equal to the voltage applied to the circuit. Since all the voltages in the voltage column are the same, the voltage law seems to work. The total circuit current is equal to the sum of the individual branch currents. The two branch currents do indeed add up to be equal to the total circuit current, so the current rule checks. The total resistance will be smaller than the smallest branch resistance. The total circuit resistance has a value of 2k and was calculated two ways for confirmation. The total power being dissipated by the circuit is equal to the sum of the power being dissipated by each branch. Those are the rules, and if you check the columns you will see the values in the column boxes obey the parallel circuit rules. The most difficult rule to understand is the resistance rule. However, you should remember that the current for each branch is being supplied by the source. If the source must supply current for many branches, the resistance it sees must be smaller than any branch resistance since as more current flows the resistance seen by the source must be getting smaller. So each time a branch is added to the parallel circuit and more current must be supplied by the source, the source sees the total circuit resistance getting smaller. Another parallel circuit is shown in Figure 227. Here is a list of the data that will be utilized to solve the problem. GIVEN: FIND: IB1 = 5mA RT = PR1 = 50mW PR4 = RB2 = 25k IB2 = Figure 227 R2 = 3k IT = R3 = 10k Page 15 ELEC 103 UNIT 2 A tabular format has been used for each circuit that has been solved so far, and that will continue in this problem. However, instead of one table you might want to use additional tables to help you keep aware of what you know and what you can solve with the rules and the formula. A multiple table format will be introduced with this problem to demonstrate how useful it can be. There will be the main circuit table as has been done in the past. However, this time a table will be used for branch one and a table will be used for branch two. When you look at the circuit in Figure 227, put your thumb over the resistors R3 and R4. You should be looking at a two resistor series circuit. Therefore, a table will be created for the series E I R P Main circuit that is really branch one. In addition, a 5mA table will also be created for the series circuit B1 made up of the source and resistors R3 and R4. B2 * 25k The table for the main circuit is the one on top, * * Figure 228. The table for the series circuit made Total Main up of the source, R1 and R2, is the middle table, FIGURE 228 Figure 229. The table for the series circuit made E I R P up of the source, R3 and R4, is the lower table, B1 Figure 230. The GIVEN values have been R1 50mW placed in their boxes and the FIND boxes have R2 3k been marked. 5mA The entire TOTAL B1 row of the middle table is Total B1 exactly the same as the B1 row in the main circuit FIGURE 229 table. Moreover, the TOTAL B2 row of the lower table is exactly the same as the B2 row in the B2 E I R P main circuit table. An example would be the total 20k branch one current in Figure 229, of 5mA, can R3 be used as the B1 row current in the main circuit R4 table of Figure 228. In addition, the total branch Total B2 * 25k two resistances in Figure 230, of 25k, can be FIGURE 230 used as the B2 row resistance in the main circuit table of Figure 228. REMEMBER, use the series circuit rules with the lower two tables and use the parallel circuit rules with the uppermost table. The first step is to apply the proper rules to each of the three tables to discover any new information. The main circuit table provides no new information. However, the middle table, the branch one table, gives up some information. Since the total branch current is 5mA, the current through R1 and the current through R2 is the same current. This means all current boxes in the middle table can be filled with 5mA. The lower table, the branch two tables, also gives up some information. The total resistance is equal to the sum of the individual resistances so, therefore, the resistance of R4 can be calculated. Calculate the value of R4. The second step is to look for calculation possibilities. Two possibilities exist and both are in the branch one table. You will see them when you insert the information, learned by applying the rules, in the boxes of the lower two tables. The first calculation will use the power formula and will take the form; ER1 = PR1 IR1 which means you will be dividing 50mW by 5mA. Write the ER1 voltage in the appropriate box. The second calculation will use the Ohms Law formula and will take the form; ER2 = Page 16 ELEC 103 UNIT 2 IR2 x RR2 which means you will be multiplying 5mA times 3k. Write the ER2 voltage in the appropriate box. The sequence of events that has taken place, while solving this problem, will be reviewed at this time. You should look at the tables on this page while the review is being presented. The tables were drawn and the GIVEN values were placed in their appropriate boxes, while the FIND boxes were marked with an asterisk. The proper rules were applied to tables, parallel E I R P Main circuit rules to the upper table and series circuit 5mA 25V rules to the middle and lower tables. The B1 information gained from the application of the B2 * 25V 25k rules is indicated by bold type. The two currents, * 25V in the middle table, could be entered into their Total Main boxes since the current through R1 and R2 is the FIGURE 231 same as the branch one current, which was given. E I R P The resistance of R4 could be calculated using the B1 50mW resistance rule for series circuits. Two R1 10V 5mA calculations were performed and they produced R2 15V 5mA 3k the two values that are not bold print. The rules 5mA 25V were applied to all tables and the middle table Total B1 gave up some information. The total voltage is FIGURE 232 equal to the sum of the individual voltage drops. E I R P Therefore, the value that is bold print was placed B2 in its box. Of course, if the voltage across a R3 20k branch is known, the voltage across all branches 5k is known. The applied voltage is also known R4 since they are all equal. Therefore, the value of Total B2 * 25V 25k 25V could be placed in all the main circuit FIGURE 233 voltage boxes and in the box representing the total voltage across branch two. The main circuit table has a calculation possibility that, if performed, will fill one of the important FIND boxes. There are other calculation possibilities, but this one in the main circuit table is one of the most important since the answer brings us one step closer to a solution. The calculation will use the Ohms Law formula and will take the form; I = E R which means you will be dividing 25V by 25mA. Write the branch two current, I , in the appropriate box. Of course, it can be placed in two boxes. The main circuit table, branch two row current box and the branch two circuit table, Total B row, current box. Remember, the Total B row in the bottom table and the B2 row in the main circuit table are exactly the same. After the calculation, apply the rules to the table columns. The parallel circuit current rule will help us this time. The total circuit current is equal to the sum of the individual branch currents. This rule will allow you to add the 5mA branch one current to the 1mA branch two current and obtain the total circuit current. Fill in the appropriate boxes. Another calculation can be performed using the main circuit table data. The total resistance can be calculated using Ohms Law and will take the form; RT = E I which means you will be dividing 25V by 6mA for a total circuit resistance equal to 4.17k. There is only one item that remains to be calculated, the power being dissipated by resistor R , P . There is more than one way to go about calculating that value, so the following procedure is just one way. Page 17 ELEC 103 UNIT 2 Calculating P involves two calculations following this method. The first calculation will be to calculate the voltage across R using the current flowing through and the resistance of R. E = I x R is the formula, which means you will be multiplying 1mA times 5k for a voltage across R equal to 5 Volts. The second calculation uses the power formula and takes the form; PR4 = ER4 x IR4 which means you will be multiplying 5V by 1mA for a power being dissipated by R4 equal to 5mW. That completes the solution of the problem even though not all the boxes have been filled. It is recommended that you completely fill the boxes in Figures 234, 235, and 236 so you can gain some additional experience. The values for each box appear in the three figures on the next page, but do the work before checking the next page for answers. In the meantime, the total circuit resistance will be calculated using both formulas. The first formula used will be the product over the sum using the branch resistances of: RB1 = 5k and RB2 = 25k E I B1 25V 5mA B2 Total Main 25V 25V Main R P 1mA 25k 6mA 4.17k FIGURE 234 E I R B1 R1 10V 5mA R2 15V 5mA Total B1 25V 5mA 50mW 3k B2 FIGURE 235 E I R R3 1mA 20k R4 1mA 5k 1mA 25k Total B2 25V P P FIGURE 236 RB1 x RB2 5x103 x 25x103 125x106 2 RT = = = 4.17x103 = 4.17k 3 3 3 RB1 + RB2 5x10 + 25x10 30x10 The next formula is the reciprocal of the sum of the reciprocals formula. The input sequence will be as follows: 5 1/X + 25 1/X = 1/X The number or symbol inside the box is the key, or keys that should be pressed on the calculator keyboard. Since the resistors are in the same k units, the k units were factored out and will be placed after the answer. If you enter the sequence, as shown, the answer will appear on the calculator display after the last key has been pressed. Just add k after the number that appears on the display and you should have the same answer that was calculated twice previously. E I R P Main 25V 5mA 125mW The total circuit resistance was calculated using B1 5k all three methods just to demonstrate the B2 25V 1mA 25k 25mW methods. It is not necessary to do so in normal 25V 6mA 4.17k 150mW Total Main parallel circuit problem solving. The correct FIGURE 237 values have been placed in all the boxes in the three figures to the left. The solution to the problem should be presented as follows: Page 18 ELEC 103 UNIT 2 GIVEN: FIND: E I R P B1 IB1 = 5mA RT = 4.17k R1 10V 5mA 50mW 2k PR1 = 50mW PR4 = 5mW R2 15V 5mA 3k 75mW RB2 = 25k IB2 = 1mA 25V 5mA Total B1 5k 125mW R2 = 3k IT = 6mA FIGURE 238 R3 = 20k As stated earlier, the table method, demonstrated E I R P in the last few problems, is a way of keeping B2 track of information you already know and also R3 20V 1mA 20k 20mW the data that is available for possible calculations. R4 5V 1mA 5mW 5k You might not want to use the table method, but 25V 1mA 25k 25mW Total B2 it is one of the technique's that can be used to help you learn how to solve series circuit and FIGURE 239 parallel circuit problems. It makes it easier to remember the rules and easier to apply the rules to circuit problems. This completes the section on Ohms Law and Parallel Circuits. OHMS LAW AND SERIESPARALLEL CIRCUITS SeriesParallel circuits are, as you most likely have already guessed, a combination of series circuits and parallel circuits. When working with seriesparallel circuits you must apply the rules for series circuits to the appropriate series part of the circuit and the rules for parallel circuits to their part of the seriesparallel circuit. Therefore, instead of working with only three rules you will be working with six rules. The Ohms Law formula and the power formula are still used as they have been used in the past. The figure to the right is one of the most simple seriesparallel circuits that can be drawn. It is being shown to make it easy for you to identify the series portion of the circuit and the parallel portion of the circuit. When the total circuit current is flowing through a component that is the series circuit portion of the seriesparallel circuit. When only part of the total circuit current is flowing through a component that is the parallel circuit portion of the seriesparallel circuit. Referring again to Figure 240, the resistor R1 would be the series circuit portion of the overall seriesparallel circuit schematic shown in the figure. All of the circuit current leaving the negative terminal of the voltage source must flow through resistor R1 to complete the path back to the positive terminal of the voltage source. Therefore, R1 is identified as the series portion of the seriesparallel circuit. When looking at resistors R2 and R3, it should be obvious they are in parallel with each other and will each share some of the current entering the parallel combination. Neither R2 nor R3 will have the total circuit current flowing through them, but will have just a portion of the total circuit current flowing through them, as is the usual case in a parallel circuit. Resistors R2 and R3 can be identified as the parallel circuit portion of the seriesparallel circuit schematic illustrated in Figure 240. Figure 240 It is much easier to work with the circuit schematic shown in Figure 241 than it is to work with the circuit schematic shown in the previous figure. The circuit is exactly the same in both figures. Page 19 ELEC 103 UNIT 2 However, redrawing the circuit so the schematic presents all the resistors in a vertical configuration makes it much easier to identify the series and the parallel parts of the circuit. In addition, redrawing the circuit will make it much easier to solve seriesparallel circuit problems as the circuits become more complex. When working on seriesparallel circuit problems, one must learn how to redraw the circuit so the circuit can be sufficiently simplified so Ohms Law can be applied to the simplified circuit. What is meant by simplifying the drawing? A procedure usually requires you to take the following steps. 1. Redraw the circuit so all components are oriented in a vertical configuration, similar to Figure 241. 2. Look as far away from the voltage source as possible and identify one of two situations. If there is only one resistor located as far away from the voltage source as possible, it is probably in parallel with another resistor. If there is more than one resistor located as far away from the voltage source as possible, it is probably in series with other resistors. Figure 241 3. If there is only one resistor, put it in parallel with the other resistor and calculate the equivalent resistance of the two resistors. Redraw the circuit, replacing the two resistors with the equivalent resistance in the redrawn circuit. 4. If there is more than one resistor, add its value to the value of the other resistors in series with it. Redraw the circuit, replacing those resistors you added together with a resistor of equivalent value. 5. Repeat steps 2, 3, and 4 until there is only one resistor and the voltage source remaining in the circuit. At that point, you will have the total circuit resistance represented by the single resistance. The total circuit current can then be calculated using the voltage divided by the resistance of the lone resistor. An example of the procedure just stated in the five steps can be demonstrated using the circuit of Figure 240. The circuit is redrawn in Figure 241 so all components are oriented in a vertical position. Looking as far away from the voltage source as possible, using Figure 241, you should see resistor R3 out there all alone. If you look carefully, you should see resistor R2 is in parallel with resistor R3. If you use R2 and R3 in the parallel formula for total resistance, you will calculate a value equivalent to R2 and R3 in parallel. You should redraw the Figure 242 circuit using all other resistors in the circuit, but replace resistors R2 and R3 with resistor Rx. The redrawn circuit is in Figure 242. The next step is to apply the rules again to the circuit just redrawn. The result of applying the rules to the circuit in Figure 242 is the circuit in Figure 243. Of course, the schematic shown in Figure 243 is a series circuit and series circuits can be solved very easily, at this point in your experience, if component values are made available. Well, let us assign values to the components in Figure 240 and work through a typical seriesparallel circuit problem. Figure 243 Using Figure 240 as the basic seriesparallel circuit schematic, let the Page 20 ELEC 103 UNIT 2 source be VS = 30 Volts. The resistors in the circuit will be assigned the following values: GIVEN values: R1 = 1.8k, R2 = 2K, and R3 = 3k. You should FIND the current through R3, (IR3), and the voltage dropped across R1, (VR1). A table can be used to help you solve seriesparallel circuit problems, but, with all seriesparallel circuit problems, the only use the table has is to keep a record of component values and to provide the proper data for use with Ohms Law. The rules can no longer be applied to the columns as they were with series circuits and with parallel circuits. The column's are a mixture of values and will be unusable for the application of the rules. On the next page you will see the GIVEN and FIND values, the table that will be used to record the information that is given, and calculated. In addition, you will see the three circuit schematics that will be necessary to solve this seriesparallel circuit problem. In the meantime, an explanation of how the table is setup and how it will be used might be beneficial. The table is somewhat similar to the previous tables you have seen and used. This time, in addition to the rows for the circuit resistances, a row is added for each resistor that is created by you as you reduce the circuit down to just the source and one resistor. The last circuit with one resistor is, in effect, equal to the total circuit resistance. The first step to take in solving the seriesparallel circuit E I R problem is to look at the original schematic, which is R1 * 1.8k shown in Figure 240. Next, setup the table as you see to 2.0k the left. If you do not know how many extra rows to add, R2 * 3.0k leave room for one or two with simple circuits and more R3 with complex circuits. The second step is to redraw the RX original circuit so all the resistors in the circuit are TOTAL 30V orientated in a vertical configuration. The original circuit FIGURE 244 has been redrawn in Figure 245. Redrawing the original circuit does not seem to make sense for such a simple circuit. However, redrawing the circuit will definitely make circuit problems easier to solve, as the circuits become more complex. After you redraw the circuit, you should look as far away from the voltage source as possible. FIGURE 245 FIGURE 246 FIGURE 247 This circuit, when it has been redrawn has only one resistor in the position as far away from the source as possible. That resistor is obviously, R3. Therefore, if you look for what is in parallel with resistor R3, you should see resistor R2 is in parallel with R3. The next step is to use the reciprocal of the sum of the reciprocals formula, or the product over the sum formula to calculate the equivalent resistance of the two resistors. Using the values R2 = 3k and R3 = 2k, one will calculate an equivalent resistance of 1.2k when using either formula. This equivalent resistance should be given a new name, different from any other resistor in the circuit. This resistor is labeled RX, but could have been labeled anything you wish. This resistor, RX, gets a row in the table and will be treated as a real resistor in the circuit of Page 21 ELEC 103 UNIT 2 Figure 246. The value for RX, that was just calculated, should be entered into the table. The next step is to look as far away from the voltage source as possible, and check for one of the two possible situations discussed earlier. The second situation exists, so the values of the resistors, when added, will be represented by a single resistor. Since this will be the only resistor in the circuit, it will be given the name RT, thereby representing the total circuit resistance. The following is a summation of the calculations required for the circuit in Figure 247. The total circuit current, It, for the circuit in Figure 247 E I R can be calculated using ET = 30V and RT = 3.0k. The * 1.8k calculated total circuit current is entered in the table to the R1 left. Since there is only one resistor in the circuit of Figure R2 2.0k 247, nothing else can be calculated concerning the circuit R3 * 3.0k (Power has not been requested). Since everything about the 1.2k circuit is known, you should move to the previous RX 30V 10mA 3.0k schematic. In this problem, move to Figure 246 where TOTAL you will determine everything you can about the circuit FIGURE 248 shown there. Since the two circuits, the circuit of Figure 246 and the circuit of Figure 247, are equivalent circuits, what you know about one circuit you know about both circuits. Since the current flowing in the circuit of Figure 247 is R1 * 10mA 1.8k 10mA, the current flowing in the circuit of Figure 246 2.0k must also be 10mA. If that is true, then the current flowing R2 * 3.0k through R1 and the current flowing through RX must both R3 be 10mA, since the circuit is a series circuit. Put that value RX 10mA 1.2k in the table to the left for both resistors as is shown. The TOTAL 30V 10mA 3.0k usual procedure is to determine everything about a circuit FIGURE 249 that is possible, before moving on to the previous circuit. If the current through both the resistors is known and the resistance of the resistors is known, the voltage drops across the resistors can be calculated. Look at the table in Figure 249. The rows for R1 and RX have unknown values for voltage and known values for current and resistance. If the voltage drops for resistor R1, (ER1), and for resistor RX, (ERX), are calculated, you then will know everything about the circuit shown in Figure 246. You can then move on to the previous circuit, the circuit shown in Figure 245. Enter the voltage drops you calculated for ER1 and for ERx into the appropriate boxes of the table in Figure 249. Since the resistors R1 and RX are in a series circuit, their voltage drops should add up to be equal to the applied voltage of 30 Volts. Does that happen? When you move to the circuit in Figure 245, you should be aware that the resistor RX in Figure 246 was representing the parallel combination of R2 and R3 in this circuit. If RX was representing the parallel combination of R2 and R3, then the voltage across RX must be the voltage across the parallel combination of R2 and R3. Additionally, the sum of the currents flowing through R2 and R3 must be equal to the current flowing through RX. Enter the voltage drops into the appropriate boxes in the table of Figure 249. After entering the voltages in the table, it should be obvious the only calculations remaining are the two current calculations. It is easy to make those calculations and add them together to check if they equal the current flowing through RX. The boxes should all be filled and the results can be seen in Figure 250. Page 22 ELEC 103 UNIT 2 The procedure you have followed to solve this problem R1 18V 10mA 1.8k will be used to solve more complex problems in other R2 12V 6mA 2.0k courses. Therefore, it might be beneficial if you spend 12V 4mA 3.0k some time reviewing the procedure used in solving the R3 problem. Also, review the calculations that were RX 12V 10mA 1.2k performed along the way, and make certain you TOTAL 30V 10mA 3.0k understand how the rules for series circuits and the rules FIGURE 250 for parallel circuits were applied in this problem. Homework problems for this unit are listed on the next page. Please submit the homework on 8½ x 11 paper with no ragged edges. Draw all circuit schematics and show all work. YOUR HOMEWORK WILL NOT BE ACCEPTED UNLESS THESE CONDITIONS ARE MET. HOMEWORK SHOULD BE SUBMITTED TO YOUR PROFESSOR. Page 23 ELEC 103 UNIT 2 HOMEWORK PROBLEMS FOR UNIT TWO 1. Draw the circuit and calculate the current that flows through four series connected resistors when R1 = 150, R2 = 250, R3 = 125, R4 = 75 and the source voltage is 12 Volts. 2. Draw the circuit for three series connected resistors supplied by a 25 Volt source. Calculate the voltage drop across each resistor and the power dissipated by each resistor. The resistor values are; R1 = 6k, R2 = 13k, and R3 = 11k. 3. A small transistor radio normally draws 15mA from a 9 Volt battery. Calculate the value of the resistor that must be connected in series with the radio if it is to be supplied from a 12 Volt battery. Also, calculate the power rating for the series resistor. HINT: Draw the circuit schematic. 4. Draw the circuit of four parallel connected resistors being supplied by a 25 Volt source. Three of the resistors have values of R1 = 1k, R2 = 12k, and R3 = 8.2k. If the total circuit current is measured as 36.5mA, determine the value of the fourth resistor. 5. Three parallel connected resistors; R1 = 1.2k, R2 = 3.9k, and R3 = 4.7k are supplied by a voltage source with an unknown value. What is the total circuit resistance? 6. Two resistors, connected in parallel, draw a total of 75mA from the voltage source. R1 = 620 and R2 = 880. Calculate IR1, IR2 and the source voltage. Also, determine the power dissipated by each resistor. 7. Analyze the circuit shown in Figure 251 to determine all resistor currents and voltages. Also, determine the total circuit current and the power dissipated by resistors R1 and R3. VS = 9 Volts, R1 = 3.3k, R2 = 12k, R3 = 6.8k, and R4 = 18k. 8. If resistor R4 is open circuited in the circuit of Figure 251, what is the value of the total circuit current that is flowing under that condition? 9. If resistor R4 is short circuited in the circuit of Figure 251, FIGURE 251 what is the value of the total circuit current that is flowing under that condition? 10. Draw the circuits for problems number 8 and number 9. Label all components in each circuit. Page 24