
PRINCIPLES OF ELECTRICITYINTRODUCTIONThis chapter describes the basic direct current (DC) circuit and the basic schematic diagram of that circuit. The schematic diagram is used when working in electricity and electronics. This chapter also describes the series DC circuit and the parallel DC circuit. It explains how to determine the total resistance, current, voltage, and power in a series, parallel, or seriesparallel network through the use of Ohm's and Kirchhoff's Laws. BASIC ELECTRIC CIRCUITThe 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) that changes the electrical energy into a more useful form of energy (light), and a switch to control the energy delivered to the load. A load is any device through which an electrical current flows and which changes this electrical energy into a more useful form. The following are common examples of loads:
A source is the device that furnishes the electrical energy used by the load. It may be a simple dry cell (as in a flashlight), a storage battery (as in an automobile), or a power supply (such as a battery charger). A switch permits control of the electrical device by interrupting the current delivered to the load. SCHEMATIC REPRESENTATIONThe engineer's main aid in troubleshooting a circuit in a piece of equipment is the schematic diagram. This is a picture of the circuit that uses symbols to represent the various circuit components. A relatively small diagram can show large or complex circuits. Before studying the basic schematic, review the appendix, which shows the symbols used in the schematic diagram. These symbols and others like them are used throughout the study of electricity and electronics. The schematic in Figure 31 represents a flashlight. In the deenergized state, the switch (S1) is open. There is not a complete path for current (I) through the circuit, so the bulb (DS1) does not light. In the energized state, the switch (S1) is closed. Current flows 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. OHM'S LAWIn 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, called Ohm's Law, 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. Where: I = current in amperes E = voltage in volts R = resistance in ohms As stated in Ohm's Law, current is inversely proportional to resistance. As the resistance in a circuit increases, the current decreases proportionately. In the equation I = E/R, if any two quantities are known, you can determine the third one. Refer to Figure 31 view 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 you can determine the current in the circuit by using these values in the equation: If the flashlight were a twocell flashlight, twice the voltage or 3.0 volts would be applied to the circuit. You can determine the current in the circuit using this voltage in the equation: As the applied voltage is doubled, the current flowing through the circuit doubles. This demonstrates that the current is directly proportional to the applied voltage.If the value of resistance of the lamp is doubled, you can determine the current in the circuit: The current has been reduced to onehalf 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 onehalf of its former value. Figures 32 and 33 are diagrams for determining resistance and voltage in a basic circuit, respectively. Using Ohms's Law, the resistance of a circuit can be determined 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, you can determine resistance (R), the only parameter not known: The Ohm's Law equation and its various forms may be obtained readily using Figure 34. The circle containing E, I, and R is divided into two parts, with E above the line and 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  POWERPower, 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 freed objects, does not constitute work. Voltage is an electrical force that 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. 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 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. The basic unit of power is the watt. Power in watts equals 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. The basic power formula is  P = I x E Where: I = current in the circuit E = voltage The amount of power changes when either voltage or current or both are changed. 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. Four of the most important electrical quantities are voltage (E), current (I), resistance (R), and power (P). The relationships among these quantities are used throughout the study of electricity. Previously, P was expressed in terms of alternate pairs of the other three basic quantities (E, I, and R). In practice, any one of these quantities can be expressed in terms of any two of the others. Figure 35 is a summary of 12 basic formulas. The four quantities E, I, R, and P are at the center of the figure. Next to each quantity are three segments. In each segment, the basic quantity is expressed in terms of two other basic quantities and no two segments are alike. For example, you can use the formula wheel in Figure 35 to find the formula to solve this problem. A circuit has a source voltage of 24 volts and a measured current of 10 amperes. What would the power rate be? Find Pin the center of the wheel. IE or current multiplied by voltage fits the supplied information.
I = 10 amps E = 24 volts
P = IE P = 10 x 24 P = 240 watts Power Rating 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 150watt lamp to a 100watt lamp. The higher wattage rating of the 150watt lamp indicates it can convert 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 150watt 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 a resistor uses is considerably less than the power rating of the resistor because a 50 percent 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 selected. Resistors of the same resistance value are available indifferent wattage values. Carbon resistors, for example, are commonly made in wattage ratings of 1/8, 1/4, l/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, with special types being 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 megawatt (1,000,000 watts), the kilowatt (1,000 watts), and the milliwatt (1/1,000 of a watt). Power Conversion and Efficiency 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 used by electrical devices 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 electrical devices use is measured in watthours. This practical unit of electrical energy equals 1 watt of power used continuously for 1 hour. The term "kilowatt hour" (kWh), used more often on a daily basis, equals 1,000 watthours. The efficiency (EFF) 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 that cannot be used. To calculate the amount of power converted by an electrical device is simple. The length of time (t) the device is operated and the input power in horsepower (HP) rating are needed (1 horsepower equals 746 watts). Horsepower, a unit of work, is often found as a rating on electrical motors. Example: A 3/4HP motor operates 8 hours a day. How much power is converted by the motor per month? How many kWh does this represent?
t = 8 hours x 30 days P = 3/4 HP
P = HP x 746 watts P = 3/4 x 746 watts P = 559 watts Use the following to convert watts to watthours: P = work x time P = 559 watts x 8 hours x 30 days P = 134,000 watthours per month
NOTE: These figures are approximate. Use the following to convert to kWh: If the motor actually uses 137 kWh per month, what is the efficiency of the motor?
Power converted = 134 kWh per month Power used = 137 kWh per month
SERIES DC CIRCUITSWhen two unequal charges are connected by a conductor, a complete pathway for current exists. An electric circuit is a complete conducting pathway. It consists of the conductor and the path through the voltage source. Inside the voltage source, current flows from the positive terminal, through the source, and emerges at the negative terminal. Characteristics A series circuit is a circuit that contains only one path for current flow. Figure 36 shows the basic circuit and a more complex series circuit. The basic circuit has only one lamp, and the series circuit has three lamps connected in series. Resistance in a Series Circuit. The current in a series circuit must flow through each lamp to complete the electrical path in the circuit (Figure 36). Each additional lamp offers added resistance. In a series circuit, the total circuit resistance (Rt) equals the sum of the individual resistances (Rt = R1 + R2 + R3 + Rn).
NOTE: The subscript n denotes any number of additional resistances that might be in the equation. Example: Figure 37 shows a series circuit consisting of three resistors (10 ohms, 1.5 ohms, and 30 ohms). What is the total resistance?
R1 = 10 ohms R2 = 15 ohms R3 = 30 ohms
Rt = Rl + R2 + R3 Rt = 10 ohms + 15 ohms + 30 ohms Rt = 55 ohms 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 (Figure 38). Rt  Rl  R2 = R3 40 ohms  10 ohms  10 ohms = 20 ohms Current in a Series Circuit. 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 (Figure 39). If this were done, each meter would be found to indicate the same value of current. Voltage in a Series Circuit, The loads in a circuit consume voltage (energy). This is called a voltage drop. Voltage drop across the resistor in a circuit, consisting of a single resistor and a voltage source, is the total voltage across the circuit and equals 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. An examination of the circuit in Figure 310 proves this. In this circuit, a source potential (Et) of 20 volts is consumed by a series circuit consisting of two 5ohm resistors. The total resistance of the circuit (Rt) equals the sum of the two individual resistance or 10 ohms. Using Ohm's Law, calculate the circuit current (I) as follows:
Et = 20 volts Rt = 10 ohms
The value of the resistors is 5 ohms each, and the current through the resistors is 2 amperes. With these values known, you can calculate the voltage drops across the resistors. Calculate the voltage (El) across R1 as follows:
I1 = 2 amps R1 = 5 ohms
E1 = I1 x R1 El = 2 amps x 5 ohms El = 10 volts R2 is the same ohmic value as R1 and carries the same current. Therefore, the voltage drop across R2 is also equal to 10 volts. Adding these two 10volt drops together gives a total drop of 20 volts, equal to the applied voltage. For series circuit, then  Et = El + E2 + 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 30ohm resistor is 2 amperes. To solve the problem, first draw and label a circuit diagram (Figure 311). R1 = 20 ohms R2 = 30 ohms R3 = 50 ohms I = 2 amps
Et = El + E2 + E3 El = R1 x I1 (I1 = the current through resistor R1) E2 = R2 x I2 E3 = R3 x I3
Et = (Rl x I1) + (R2 x I2) + (R3 x I3) Et = (20 ohms x 2 amps) + (30 ohms x 2 amps) + (50 ohms x 2 amps) Et = 40 volts + 60 volts + 100 volts Et = 200 volts
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 applied voltage determines the value of the voltage dropped by a resistor. It 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. Power in a Series Circuit. Each of the loads in a series circuit consumes power that is dissipated in the form of heat. Since this power must come from the source, the total power supplied must be equal to the power consumed by the circuit's loads. In a series circuit, the total power equals the sum of the power dissipated by the individual loads. Total power (Pt) equals  Pt = Pl + 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 (Figure 312).
R1 = 5 ohms R2 = 10 ohms R3 = 15 ohms Et = 120 volts
Rt = Rl + R2 + R3 Rt = 5 ohms + 10 ohms + 15 ohms Rt = 30 ohms Calculate the circuit current by using the total resistance and the applied voltage: I = Et I = 120 volts I = 4 amps Calculate the power for each resistor using the power formulas:
P1 = I^{2} x R1 P1 = (4 amps)^{2} x 5 ohms P1 = 80 watts
P2 = I^{2} x R1 P2 = (4 amps)^{2} x 10 ohms P2 = 160 watts
P3 = I^{2} x R3 P3 = (4 amps)^{2} x 15 ohms P3 = 240 watts
Pt = Pl + P2 + P3 Pt = 80 watts + 160 watts + 240 watts Pt = 480 watts To check the answer, calculate the total power delivered by the source: P source = I source x E source P source = 4 amps x 120 volts P source = 480 watts The total power equals the sum of the power used by the individual resistors. Rules for Series DC Circuits Listed below are the important factors governing the operation of a series circuit. For ease of study, they are set up as a group of rules. They must be completely understood before studying more advanced circuit theory. 1. The same current flows through each part of a series circuit.
2. The total resistance of a series circuit equals the sum of the individual resistances.
3. The total voltage across a series circuit equals 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 equals the sum of the individual powers used by each circuit component.
Series Circuit Analysis The following sample problems show the procedure for solving series circuits: Example: Three resistors of 5 ohms, 10 ohms, and 15 ohms are connected in series with a power source of 90 volts (Figure 313). a. What is the total resistance? b. What is the circuit current? c. What is the voltage drop across each resistor? d. What is the power of each resistor? e. What is the total power of the circuit? In solving the circuit, find the total resistance first. Next, calculate the circuit current. Once the current is known, calculate the voltage drops and power dissipations.
R1 = 5 ohms R2 = 10 ohms R3 = 15 ohms Et = 90 volts
Rt = R1 + R2 + R3 Rt = 5 ohms + 10 ohms + 15 ohms Rt = 30 ohms
I = 3 amps
El = (I)(R1) El = 3 amps x 5 ohms E1 = 15 volts E2 = (I)(R2) E2 = 3 amps x 10 ohms E2 = 30 volts E3 = (I)(R3) E3 = 3 amps x 15 ohms E3 = 45 volts
P1 = (I)(E1) P1 = 3 amps x 15 volts P1 = 45 watts P2 = (I)( E2) P2 = 3 amps x 30 volts P2 = 90 watts P3 = (I)(E3) P3 = 3 amps x 45 volts P3 = 135 watts
Pt = (Et)(I) Pt = 90 volts x 3 amps Pt = 270 watts or Pt = P1 + P2 + P3 Pt = 45 watts + 90 watts + 135 watts Pt = 270 watts Example: Four resistors (Rl = 10 ohms, R2 = 10 ohms, R3 = 50 ohms, and R4 = 30 ohms) are connected in series with a power source (Figure 314). 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?
R1 = 10 ohms R2 = 10 ohms R3 = 50 ohms R4 = 30 ohms I = 0.5 amp
Et = (I)(Rt) Rt = Rl + R2 + R3 + R4 Rt = 10 ohms + 10 ohms + 50 ohms + 30 ohms Rt = 100 ohms Et = 0.5 amp x 100 ohms Et = 50 volts
El = (I)(R1) El = 0.5 amp x 10 ohms El = 5 volts E2 = (I)(R2) E2 = 0.5 amp x 10 ohms E2 = 5 volts E3 = (I)(R3) E3 = 0.5 amp x 50 ohms E3 = 25 volts E4 = (I)(R4) E4 = 0.5 amp x 30 ohms E4 = 15 volts
P1 = (I)(E1) P1 = 0.5 amp x 5 volts P1 = 2.5 watts P2 = (I)(E2) P2 = 0.5 amp x 5 volts P2 = 2.5 watts P3 = (I)(E3) P3 = 0.5 amp x 25 volts P3 = 12.5 watts P4 = (I)(E4) P4 = 0.5 amp x 15 volts P4 = 7.5 watts
Pt = Pl + P2 + P3 + P4 Pt = 2.5 watts + 2.5 watts + 12.5 watts + 7.5 watts Pt = 25 watts or When applying Ohm's Law to a series circuit, 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.
Et = It x Rt To find total current 
NOTE: In a series circuit, It equals I. However, the distinction between It and I in the formula should be noted because future circuits may have several currents. Then it would be necessary to differentiate between It and other currents. To compute any quantity (E, I, R, or P) associated with a single given resistor, obtain the values used in the formula from that particular resistor. For example, to find the value of an unknown resistance, use the voltage across and the current through that particular resistor.
KIRCHHOFF'S VOLTAGE LAWIn 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 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." To state Kirchhoff's Law another way, the voltTo find total 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) Et = It x Rt 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 later chapters). 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. Example: Three resistors are connected across a 50volt source. What is the voltage across the third resistor if the voltage drops across the first two resistors are 25 volts and 15 volts?
Et = El + E2 + E3 Since the voltages of El and E2 as well as the voltage supply Et are given, the equation can be rewritten with the known values: 50 volts = 25 volts + 15 volts + Ex (the unknown factor)
Ex = 50 volts  25 volts  15 volts Ex = 10 volts Using this same idea, many electrical problems can be solved, not by knowing all the mysterious properties of electricity, but by understanding the basic principles of math. This algebraic expression can be used for all equations, not just for voltage, current, and resistance. CIRCUIT TERMS AND CHARACTERISTICSThe following terms and characteristics used in electrical circuits are used throughout the study of electricity and electronics. Open Circuit A circuit is open when a break interrupts a complete conducting pathway. Although an open circuit normally occurs when a switch is used to deenergize a circuit, one may also develop accidentally. To restore a circuit to proper operation, the opening must be located, its cause determined, and repairs made. Sometimes an open circuit can be located visually by 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 circuit has on circuit conditions enables a technician to use test equipment to locate the open component. In Figure 315, the series circuit consists of two resistors and a fuse. Notice the effects on circuit conditions when the fuse opens. Current ceases to flow. Therefore, there is no longer a voltage drop across the resistors. Each end of the open circuit conducting path becomes an extension of the battery terminals and the voltage felt across the open circuit equals the applied voltage (Et). An open circuit has infinite resistance. Infinity represents a quantity so large it cannot be measured. (The symbol for infinity is . In an open circuit, Rt = .) Short Circuit 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 0 ohms. A short often occurs as a result of improper wiring or broken insulation. In Figure 316, a short is caused by improper wiring. Note the effect on current flow. Since the resistor (R1) has in effect been replaced with apiece of wire, practically all the current flows through the short and very little current flows through the resistor (R1). Electrons flow through the short (a path of almost zero resistance) and the remainder of the circuit by passing through the 10ohm resistor (R2) 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 10ohm resistor (R2) becomes heated. As it trys to dissipate this heat, the resistor will probably be destroyed. Figure 317 shows a pictorial wiring diagram, rather than a schematic diagram, to indicate how broken insulation might cause a short circuit. Source Resistance A meter connected across the terminals of a good 1.5volt 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. Figure 318 illustrates this principle, showing the internal resistance of a battery as Ri. In the schematic, the internal resistance is indicated by an additional resist or in series with the battery. 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 R1. 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. Power Transfer and Efficiency Maximum power is transferred from the source to the load when the resistance of the load equals the internal resistance of the source. The table and the graph in Figure 319 illustrate this theory. 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's 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 in Figure 319. 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. When the amount of power involved is 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. When 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 the source is capable of supplying. PARALLEL DC CIRCUITSThe series circuit has only one path for current. Another basic type of circuit is the parallel circuit. While 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. Characteristics A parallel circuit has more than one current path connected to a common voltage source. Parallel circuits, therefore, must contain two or more resistances that are not connected in series. Figure 320 shows an example of a basic parallel circuit. Start at the voltage source (Et) and trace counterclockwise around the circuit in Figure 320. 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. Voltage in a Parallel Circuit. 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 (section of a circuit that has a complete path for current). In Figure 320, this voltage equals the applied voltage (Et). This can be expressed in equation form: Et = E1 = E2 = En Voltage measurements taken across the resistors of a parallel circuit verify this equation (Figure 321). Each meter indicates the same amount of voltage. Notice that the voltage across each resistor is the same as applied voltage. Example: The current through a resistor of a parallel circuit is 12 amperes and the value of the resistor is 10 ohms. Determine the source voltage. Figure 322 shows the circuit.
R2 = 10 ohms I2 = 12 amps
E2 = (I2)(R2) E2 = 12 amps x 10 ohms E2 = 120 volts Et = E2 Et = 120 volts Current in a Parallel Circuit. Ohm's Law states that the current in a circuit is inversely proportional to the circuit resistance. This 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 following illustrations show the behavior of current in parallel circuits using example circuits with different values of resistance for a given value of applied voltage. Figure 323 view A shows a basic series circuit. Here, the total current must pass through the single resistor. The amount of current can be determined as follows:
Et = 50 volts R1 = 10 ohms
View B 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 view A.
Et = 50 volts R1 = 10 ohms R2 = 10 ohms
If 5 amperes of current flow 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 leaves the negative terminal of the battery and flows to point a (view B). Point a, called anode, is a connecting point for the two resistors. At node a, the total current divides into two currents of 5 amperes each. These two currents flow through their respective resistors and rejoin at node b. The total current then flows from node 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 onehalf of the total current. Each individual current path in the circuit of view B is a branch. Each branch carries a current that is a portion of the total current. Two or more branches form a network. The characteristics of current in a parallel circuit can be expressed in terms of the following general equation: It = I1 + I2 + ... In Compare Figure 324 view A with the circuit in Figure 323 view B. Notice that doubling the value of the second branch resistor (R2) has no effect on the current in the first branch (I1). However, it does reduce the second branch current (I2) to onehalf 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:
Et = 50 volts R1 = 10 ohms R2 = 20 ohms Solution: I2 = 2.5 amps It = I1 + I2 It = 5 amps + 2.5 amps It = 7.5 amps The amount of current flow in the branch circuits and the total current in the circuit in Figure 324 view B are determined by the following computations:
R1 = 10 ohms R2 = 10 ohms R3 = 10 ohms Es = E1 = E2 = E3 13 = 5 amps It = I1 + I2 + I3 It = 5 amps + 5 amps + 5 amps It = 15 amps Notice that the sum of the ohmic values of the resistors in both circuits in Figure 324 is equal (30 ohms) and that the applied voltage is the same value (50 volts). However, the total current in Figure 324 view B (15 amperes) is twice the amount in Figure 324 view A (7.5 amperes). 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, "The algebraic sum of the currents entering and leaving any node of conductors is equal to zero." This law stated mathematically is  Ia + Ib + ... In = 0 Where: Ia, Ib, . . . In = the current entering and leaving the node. Currents entering the node are considered positive, and currents leaving the node are 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 13 in Figure 325.
I1 = 10 amps I2 = 3 amps I4 = 5 amps
Ia + Ib + ... In = 0 The currents are placed into the equation with the proper signs: I1+ I2+ I3+ I4= 0 10 amps + (3 amps) + 13 + (5 amps) = 0 I2 + 2amps = 0 I2 = 2 amps I2 has a value of 2 amperes. The negative sign shows it to be a current leaving the node. Resistance in a Parallel Circuit. The example diagram (Figure 326) has two resistors connected in parallel across a 5volt 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 1/2 ampere of current through each resistance. The total current flowing from the battery to the node of the resistors and returning from the resistors to the battery equals 1 ampere. The total resistance of the circuit is calculated using the values of total voltage (Et) and total current (It):
Et = 5 volts It = 1 amp
This computation shows the total resistance to be 5 ohms, onehalf the value of either of the two resistors. The total resistance of a parallel circuit is smaller than any of the individual resistors. Thus, the 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). Several methods are 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: Where: Rt = total parallel resistance R = ohmic value of one resistor N = number of resistors This equation is valid for any number of parallel resistors of equal value. Example: Four 40ohm resistors are connected in parallel. What is their equivalent resistance?
R1 = R2 = R3 = R4 R1 = 40 ohms
Figure 327 shows two resistors of unequal value in parallel. Since the total current is shown, the equivalent resistance can be calculated.
Et = 30 volts It = 15 amps
Reciprocal Method. This method is based on taking the reciprocal of each side of the equation. This presents the general formula for resistors in parallel as  This formula is generally used to solve for the equivalent resistance of any number of unequal parallel resistors. Unlike the equal value or the productoverthesum method, the reciprocal method is the only formula that can be used to determine the equivalent resistance in any combination of parallel resistances. You must find the lowest common denominator in solving these problems. Example: Three resistors are connected in parallel as shown in Figure 328. The resistor values are R1 = 20 ohms, R2 = 30 ohms, and R3 = 40 ohms. What is the equivalent resistance? Use the reciprocal method.
R1 = 20 ohms R2 = 30 ohms R3 = 40 ohms
ProductOvertheSum Method. A convenient method for finding the equivalent, or total, resistance of two parallel resistors is by using the productover thesum formula: Example: What is the equivalent resistance of a 20ohm and a 30ohm resistor connected in parallel, as in Figure 329?
R1 = 20 R2 =30
The productoverthesum method can only be used with two resistance values at a time. If three or more resistors are to be calculated, combine any two ohmic values into an equivalent resistance using the formula. Repeat the formula again, and this time, combine the remaining ohmic value with the recently derived equivalent resistance. Combining additional resistance values with equivalent resistance may be continued throughout the parallel circuit. Power in a Parallel Circuit. Power computations in a parallel circuit are basically 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 equals the sum of the power dissipated by the individual resistors. Like the series circuit, the total power consumed by the parallel circuit is  Pt = Pl + P2 + ... Pn Example Find the total power consumed by the circuit in Figure 330.
R1 = 10 ohms I1 = 5 amps R2 = 25 ohms I2= 2 amps R3 = 50 ohms I3 = 1 amp
P= I^{2}R P1 = (I1)^{2} x R1 P1 = (5 amps)^{2} x 10 ohms P1 = 250 watts P2 = (I2)^{2} x R2 P2 = (2 amps)^{2} x 25 ohms P2 = 100 watts P3 = (I3)^{2} x R3 P3 = (1 amp)^{2} x 50 ohms P3 = 50 watts Pt = Pl + P2 + P3 Pt = 250 watts + 100 watts + 50 watts Pt = 400 watts Since the total current and source voltage are known, the total power can also be computed:
Et = 50 volts It = 8 amps
Pt = Et x It Pt = 50 volts x 8 amps Pt = 400 watts Equivalent Circuits 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 331 shows a parallel circuit with three resistors of equal value and the redrawn equivalent circuit. The parallel circuit in view A shows the original circuit. To create the equivalent circuit, first calculate the equivalent resistance:
R1 = 45 ohms R2 = 45 ohms R3 = 45 ohms
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 (Figure 331 view B). The reduction of the electrical circuit from a complex parallel circuit to the simple single resistor series circuit may appear to drastically distort the original circuit and apply only to the mathematical electrical rules. However, this is the basic electrical schematic that a power source sees. The generator or battery only sees one single series electrical load. The load determines the total resistance (Rt) that the generator must deal with. Based on this, the generator supplies the current (It), pushed through the circuits by the voltage (Et). The electrical wiring system of series and parallel combinations and various electrical loads will require the current to be divided up effectively, as seen with Kirchhoff's Current Law. Rules for Parallel DC Circuits The following are rules for parallel DC circuits:
Et = El = E2 = E3 = En 2. The total current of a parallel circuit equals the sum of the individual branch currents of the circuit. It = I1 + I2 + I3 + In 3. The total resistance of a parallel circuit is found by the general formula (1/Rt = 1/Rl + 1/R2 + 1/R3 + 1/Rn) or one of the formulas derived from this general formula. 4. The total power consumed in a parallel circuit equals the sum of the power consumptions of the individual resistance. Pt = P1 + P2 + P3 + Pn Parallel Circuit Problems 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:
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 on the known quantities. 4. Substitute the known values in the selected equation 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. Calculate the value for total resistance, total power, total current, source voltage, the power used by each resistor, and the current through resistors R2 R3, R4, and R5.
R1 = 20 ohms R2 = 30 ohms R3 = 18 ohms R4 = 18 ohms R5 = 18 ohms Il = 9 amps
Rt, Et, It, Pt, I2, I3, I4, I5, P1, P2, P3, P4, P5 This may seem to be a large amount of mathematical manipulation. However, the stepbystep approach simplifies the calculation. The first step in solving this problem is to draw the circuit and indicate the known values (Figure 332). There are several ways to approach this problem. With the given values, you could first solve for Rt, the power used by Rl, or the voltage across R1 which 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 R1 is known, this value will help in calculating other unknowns. Therefore, the logical unknown to solve for is the source voltage (the voltage across R1).
R1 = 20 ohms I1 = 9 amps El = Et
Et = Rl x Il Et = 9 amps x 20 ohms Et = 180 volts Now that source voltage is known, you can solve for current in each branch:
Et = 180 volts R2 = 30 ohms R3 = 18 ohms R4 = 18 ohms R5 = 18 ohms
I4 = 10 amps I5 = 10 amps Solve for total resistance:
R1 = 20 ohms R2 = 30 ohms R3 = 18 ohms R4 = 18 ohms R5 = 18 ohms
An alternate method for solving Rt can be used. By observation, you can see that R3, R4, and R5 are equal ohmic value. Therefore, an equivalent resistor can be substituted for these three resistors in solving for total resistance.
R3 = R4 = R5 = 18 ohms
The circuit is now redrawn again using a resistor labeled Req 1 in place of R3, R4, and R5 (Figure 333). An equivalent resistor can be calculated and substituted for R1 and R2 by use of the productoverthesum formula:
R1 = 20 ohms R2 = 30 ohms Solution: The circuit is now redrawn using a resistor labeled Req 2 in place of R1 and R2 (Figure 334). Two resistors are now left in parallel. The productoverthesum method can now be used to solve for total resistance:
R1 = 6 ohms R2 = 12 ohms
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 335, and the total current can be calculated.
Et = 180 volts Rt = 4 ohms
This solution can be checked by using the values already calculated for the branch currents:
I1 = 9 amps I2 = 6 amps I3 = 10 amps I4 = 10 amps I5 = 10 amps
It = I1 + I2 + ... In It = 9 amps + 6 amps + 10 amps + 10 amps + 10 amps It = 45 amps Now that total current is known, the next logical step is to find total power.
Et = 180 volts It = 45 amps
P = EI Pt = Et x It Pt = 180 volts x 45 amps Pt = 8,100 watts = 8.1 KW Solve for the power in each branch:
Et = 180 volts I1 = 9 amps I2 = 6 amps I3 = 10 amps I4 = 10 amps I5 = 10 amps
P = EI Pl = Et x Il P1 = 189 volts x 9 amps P1 = 1,620 watts P2 = Et x I2 P2 = 180 volts x 6 amps P2 = 1,080 watts P3 = Et x I3 P3 = 180 volts x 10 amps P3 = 1,800 watts Since I3 = I4 = I5, then P3 = P4 = P5 = 1,800 watts. The previous calculation for total power can now be checked:
P1 = 1,620 watts P2 = 1,080 watts P3 = 1,800 watts P4 = 1,800 watts P5 = 1,800 watts
Pt = Pl + P2 + P3 + P4 + P5 Pt = 1,620 watts + 1,080 watts + 1,800 watts + 1,800 watts + 1,800 watts Pt = 8,100 watts Pt = 8.1 KW SERIESPARALLEL DC CIRCUITSEngineers encounter circuits consisting of both series and parallel elements. This type of circuit is called a seriesparallel network. Solving for the quantities and elements in a seriesparallel network is simply a matter of applying the laws and rules discussed up to this point. COMBINATIONCIRCUIT PROBLEMSThe basic technique used for solving DC combinationcircuit 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 seriesparallel networks problems, the network in Figure 336 view A will be used to calculate various circuit quantities, such as resistance, current, voltage, and power. Examination of the circuit shows that the only quantity that can be computed with the given information is the equivalent resistance of R2 and R3.
R2 = 20 ohms R3 = 30 ohms
Now that the equivalent resistance for R2 and R3 has been calculated, the circuit can be redrawn as a series circuit (view B). The equivalent resistance of this circuit (total resistance) can now be calculated:
R1 = 8 ohms (resistors in series) R2,3 = 12 ohms
Rt = Rl + R2,3 Rt = 8 + 12 Rt = 20 ohms The original circuit can be redrawn with a single resistor that represents the equivalent resistance of the entire circuit (view C). To find total current in the circuit 
Et = 60 volts Rt = 20 ohms
To find total power in the circuit 
Et = 60 volts It = 3 amps
Pt = 60 volts x 3 amps Pt = 180 watts To find the voltage dropped across Rl, R2, and R3, refer to Figure 336 view B. R2,3 represents the parallel network of R2 and R3. Since the voltage across each branch of a parallel circuit is equal, the voltage across R2,3 will be the same across R2 and R3.
It = 3 amps (current through each part of a series circuit equals total current.) Rl = 8 ohms R2,3 = 12 ohms
E1 = I1 x R1 El = 3 amps x 8 ohms El = 24 volts E2 = E3 = E2,3 E2,3 = It x R2,3 E2,3 = 3 amps x 12 ohms E2,3 = 36 volts E2 = 36 volts E3 = 36 volts To find power used by R1 
El = 24 volts It = 3 amps
Pl = E1 x Rt P1 = 24 volts x 3 amps PI = 72 watts To find the current through R2 and R3, refer to the original circuit (Figure 335 view A). E2 and E3 are known from previous calculation.
E2 = 36 volts E3 = 36 volts R2 = 20 ohms R3 = 30 ohms
To find power used by R2 and R3, using values from previous calculations 
E2 = 36 volts E3 = 36 volts I2 = 1.8 amps I3 = 1.2 amps
P2 = E2 x I2 P2 = 36 volts x 1.8 amps P2 = 64.8 watts P3 = E3 x I3 P3 = 36 volts x 1.2 amps P3 = 43.2 watts After computing all the currents and voltages of Figure 336, a complete description of the operation of the circuit can be made. The total current of 3 amperes leaves the negative terminal of the battery and flows through the 8ohm resistor (R1). In so doing, a voltage drop of 24 volts occurs across resistor R1. At point A, this 3ampere current divides into two currents. Of the total current, 1.8 amperes flows through the 20ohm resistor. The remaining current of 1.2 amperes flows from point A, down through the 30ohm resistor to point B. This current produces a voltage drop of 36 volts across the 30ohm resistor. Notice that the voltage drops across the 20 and 30ohm resistors are the same.) The two branch currents of 1.8 and 1.2 amperes combine at node B, and the total current of 3 amperes 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. The seriesparallel network is not difficult to solve. The key to its solution lies in knowing the order to apply the steps of the solution. First look at the circuit. From this observation, determine the type of circuit, what is known, and what must be determined. 