R = ?
Ammeter 24 volts o amperes FIGURE 8-48. A circuit with unknown resistance.
resented in schematic form by the symbol for a battery. The negative sign is located near one side to indicate the negative terminal of the source, or battery. The opposite side is marked positive with a + symbol. Arrows are sometimes used to indicate the direction of current flow from the negative terminal, through the conducting wires and other circuit devices, to the positive terminal of the source.
Figure 8-48 shows that the values of voltage and current are known. To find the quantity of resistance in the circuit, Ohm's law can be transposed to solve for R.
Transposing the basic formula I = to R = and substituting the known circuit values in the equation, R = 8 amperes 24 volts - = 3 ohms, or 3 g.
Ohm's law can also be transposed to determine the voltage applied to a circuit when current flow and resistance are known, as shown in figure 8-49.
In this circuit the unknown circuit quantity, the voltage, is represented by the symbol E. The value of resistance is 3 ohms, and the current flow is 8 amperes. (The word amperes is often shortened to "amps.") Transposing Onm's law from its basic formula, the equation to solve for E becomes E = I X R.
R = 30 Ammeter E = ?
8 amperes FIGURE 8-49. Circuit with unknown voltage.
Amperes (I)
6.0
5.5
5.0
4.5
4.0
3.5
3.0
2.5 R - 20 Ohms
2.0 (Constant)
1.5
1.0
0.5 10 20 30 40 50 60 70 80 90 100 110120 Volts (E) FIGURE 8-50. Voltage vs. current in a constant-resistance circuit.
Substituting the known values in the equation, E = 8X3 È = 24 volts or 24 V.
The relationship between the various circuit quantities can be further demonstrated if the resistance in a circuit is held constant. In such a case, the current will increase or decrease in direct proportion to the increase or decrease of voltage appled to the circuit. For example, if the voltage applied to a circuit is 120 volts and the resistance of the circuit is 20 ohms, the current flow will be 120 20' or 6 amperes. If this resistance remains constant at 20 ohms, a graph of voltage-current relationship, as shown in figure 8-50, can be ploted.
The relationship between voltage and current in this example shows voltage plotted horizontally along the X axis in values from 0 to 120 volts, and the corresponding values of current are plotted vertically in values from 0 to 6.0 amperes along the Y axis. A straight line drawn through all the points where the voltage and current lines meet represents the equation I = 20 E and is called a linear relationship. The constant, 283
Electromotive force Current = Resistance Volts I= = Amperes = R Ohms Electromotive force Resistance = Current R=I E Ohms = - Volts Amperes Electromotive force = current X resistance E = IR Volts = amperes × ohms FIGURE 8-51. Ohm's law.
20, represents the resistance, which is assumed not to change in this example. This graph represents an important characteristic of the basic law, that the current varies directly with the applied voltage if the resistance remains constant.
The basic equations derived from Ohm's law are summarized, together with the units of measurements of circuit quantities, in figure 8-51.
The various equations which may be derived by transposing the basic law can be easily obtained by using the triangles in figure 8-52.
The triangles containing E, I, and R are divided into two parts, with E above the line and I X R below it. To determine an unknown circuit quantity when the other two are known, cover the unknown quantity with a thumb.
The location of the remaining uncovered letters in the triangle will indicate the mathematical operation to be performed. For example, to find I, refer to (a) of figure 8-52, and cover I with the thumb. The uncovered letters indicate that E is to be divided by R, or 1 = R To find R, refer to
(b) of figure 8-52, and cover R with the thumb. The result indicates that E is to be divided by Lor R = - To find E, refer to (e) of figure 8-52, and cover E with the thumb. The result indicates I is to be multiplied by R, or E = I X R.
This chart is useful when learning to use Ohm's law. It should be used to supplement the beginner's knowledge of the algebraic method.
Power In addition to the volt, ampere, and ohm, there is one other unit frequently used in electrical circuit calculations. This is the unit of power.
The unit used to measure power in d.c. electrical circuits is the watt. Power is defined as the rate of doing work and is equal to the product of the voltage and current in a d.c. circuit. When the current in amperes (I) is multiplied by e.m.f. in volts (E), the result is power measured in watts (P). This indicates that the electrical power delivered to a circuit varies directly with the applied voltage and current flowing in the circuit.
Expressed as an equation, this becomes P = IE.
This equation may be transposed to determine any one of the three circuit quantities as long as the other two are known. Thus, if the power is read directly from a wattmeter and the voltage is measured with a voltmeter, the intensity of the current (I) flowing in the circuit can be determined P by transposing the basic equation to I = Similarly, the voltage (E) can be found by transposing the basic power formula to E = T P Since some of the values used to determine the power delivered to a circuit are the same as those used in Ohm's law, it is possible to substitute Ohm's law values for equivalents in the power formula.
In Cher's law; I = If this vales I" is substituted for I in the power formula, it becomes P= IXE;P = EX ROP = R This equation, P = R' illustrates that the power in watts delivered to a circuit varies directly with the square of the applied voltage and inversely with the circuit resistance.
R To find I (amperes) place thumb over I and To find R (ohms) place thumb over R and divide E by R as indicated. divide as indicated.
(a) (b) R To find E (volts) place thumb over X and multiply as indicated.
(c) E = volts 1 = amperes R = ohms FIGURE 8-52. Ohm's law chart.
The watt is named for James Watt, the inventor of the steam engine. Watt devised an experiment to measure the power of a horse in order to find a means of measuring the mechanical power of his steam engine. One horsepower is required to move 33,000 pounds 1 foot in 1 minute. Since power is the rate of doing work, it is equivalent to the work divided by time. Stated as a formula, this is 33,000 ft.-lb.
Power = 60 sec. (1 min.) ' Or P = 550 ft.-lb./sec.
Electrical power can be rated in a similar manner. For example, an electric motor rated as a 1-horsepower motor requires 746 watts of electrical energy. But the watt is a relatively small unit of power. Much more common is the kilowatt or 1,000 watts. (The prefix kilo means 1,000.) In measuring amounts of electrical energy consumed, the kilowatt hour is used. For example, if a 100- watt bulb consumes electrical energy for 20 hours, it has used 2,000 watt hours or 2 kilowatt hours of electrical energy.
Electrical power that is lost in the form of heat when current flows through an electrical device is often referred to as power loss. This heat is usually dissipated into the surrounding air and serves no useful purpose, except when used for heating. Since all conductors possess some resistance, circuits are designed to reduce these losses. Referring again to the basic power formula, P = I X E, it is possible to substitute the Ohm's law values for E in the power formula to obtain a power formula that directly reflects the power losses in a resistance.
P= IXE; E = IXR.
Substituting the Ohm's law value for E (I X R) in the power formula, P = IXIXR.
Collecting terms, this gives, P = PR.
From this equation, it can be seen that the power in watts in a circuit varies as the square of the circuit current in amperes and varies directly with the circuit resistance in ohms.
Finally, the power delivered to a circuit can be expressed as a function of current and resistance by transposing the power equation P = I R.
Transposing to solve for current gives 12 = R and by extracting the square root of both sides of the equation, 1 = 1 Thus, the current through a 500-watt, 100-ohm load (resistance) is as follows:
1 = VR - 100 500 = 2.24 amperes.
The electrical equations derived from Ohm's law and the basic power formula do not reveal all about the behavior of circuits. They do indicate the numerical relation between the volt, ampere, ohm, and watt. Figure 8-53 provides a summary of all the possible transpositions of these formulas in a 12-segment circle.
SERIES D. C. CIRCUITS The series circuit is the most basic of electrical circuits. All other types of circuits are elaborations or combinations of series circuits. Figure 8-54 is an example of a simple series circuit. It is a circuit because it provides a complete path for current to flow from the negative to the positive terminal of the battery. It is a series circuit because there is only one possible path in which current can flow, as indicated by the arrows showing the direction of electron movement. It is also called a series circuit because current must pass through the circuit components, the battery and the resistor, one after the other, or "in series."
The circuit shown in figure 8-55 contains the basic components required for any circuit: a source of power (battery), a load or current-limiting resistance (resistor), and a conductor (wire). Most practical circuits contain at least two other items:
a control device (switch) and a safety device (fuse). With all five components in the circuit it would appear as shown in figure 8-55, which is a d.c. series circuit.
IR FIGURE 8-53. Summary of basic equations using the volt, ampere, ohm, and watt.
In the d.c. or direct-current circuit, current flows in one direction from the negative terminal of the battery through the switch (which must be closed) through the load resistance and the fuse to the positive terminal of the battery.
To discuss the behavior of electric current in a
d.c. series circuit, figure 8-56 is redrawn in figure 8-57 to include three ammeters and two resistors. Since an ammeter measures the intensity of current flow, three have been located in the circuit to measure the current flowing at various points in the circuit.
With the switch closed to complete the circuit, all three ammeters will indicate the same amount of current. This is an important characteristic of all series circuits: No matter how many components are included in a series circuit, the current is the same intensity throughout the circuit.
While it is true that an increase in the number of circuit components will increase the resistance to current flow in the circuit, whatever the value of current flowing in the circuit, it will be the same value at all points in the circuit.
In figure 8-56, the current through resistor R1 is labeled 11 and the current through resistor R2 is labeled I2. If the total current in the circuit is Ir, the formula describing the current flow is, 17 = 11 = 12.
If the number of resistors is increased to five, the formula will be IT = 11 = 12 = 13 = 14 = 15.
FIGURE 8-54. A series circuit.
Without indicating how much current is flowing, it will always be true that the current through any resistor in a series circuit will be the same as that through any other resistor.
Figure 8-57 is a series circuit containing two resistances. In order to determine the amount of current flow in this circuit, it is necessary to know how much resistance or opposition the current flow will encounter. Thus, the second characteristic of series circuits is: Total resistance in a series circuit is the sum of the separate resistances in the circuit. Stated as a formula, this becomes Rr = R1 + Rz.
In figure 8-57, this is Rr = R1 (5 Q) + Rz (10 2) or Rr = 5 + 10 = 15 Q.
The total resistance of the circuit in figure 8-57 is 15 ohms. It is important to remeber that, if the circuit were altered to include 10, 20, or even 100 resistors, the total resistance would still be the sum of all the separate resistances. It is also true that there is a certain negligible resistance in the battery, as well as in the fuse and the switch. These small values of resistance will not be considered in determining the value of current flow in this circuit.
The Ohm's law formula for finding current is 1 = E/R. Since the battery voltage is 30 volts and the total circuit resistance is 15 ohms, the equation becomes 1 = 158 30 V = 2 amperes.
Switch o FIGURE 8-56. Current flow in a series circuit.
The current flow is 2 amperes (sometimes the word anperes is shortened to amps), and this value of current is everywhere in the circuit.
To consider what effect a change in resistance will have on current flow when the voltage remains constant, the total resistance is doubled to 30 ohms. Using Ohm's law E 30 V = 1 ampere.
It can be seen that current will be reduced to half its former value when resistance is doubled.
On the other hand, if voltage remains constant, and resistance is reduced to half its former value the current will double its original value.
I = R = 1 = 1.59 - 4 amperes- Thus, if voltage remains constant and resistance increases, current must decrease. Conversely, if resistance decreases, current must increase.
However, if resistance is held constant and voltage is doubled, the current flow will double its original value. If the voltage applied to the circuit in figure 8-58 is doubled to 60 volts and the original value of resistance is maintained at Load • 30V R, = 102< _Battery Fuse R, = 58 FIGURE 8-55. A d.c. series circuit. FIGURE 8-57. A series circuit with two resistors.
E, = 30 volts A, = 1082 E = 150 volts Rz + Ę =? 100 Rz E, = ?
300 60,2 FIGURE 8-58. Voltage drops in a circuit. FIGURE 8-59. Applying Ohm's law.
15 ohms, resistors are added (10 V + 20 V), a value equal to the applied voltage, 30 volts, is obtained. This = - 159 60 V - = 4 amperes, confirms the basic formula for series circuits:
Er = E1 + Er.
and if voltage is reduced to half its original value, with resistance constant, current will decrease to half its original value.
In any d.c. series circuit, a missing quantity such as voltage, resistance, or current can be calculated by using Ohm's law if any two of the 1 = R E = 15 g = 1 ampere. 15 V Thus, if resistance remains constant, and voltage increases, current must also increase. If voltage decreases, current decreases also.
It is important to distinguish between the terms "voltage" and "voltage drop" in discussing quantities are known.
Figure 8-59 is a series circuit containing three known values of resistance and an applied voltage of 150 volts. Using these values, the unknown circuit quantities can be determined by applying Ohm's law as follows:
R1 = 30 g series circuits. Voltage drop refers to the loss in R2 = 60 9 electrical pressure caused by forcing electrons through a resistance. In figure 8-58, the applied R3 = 10 Q voltage (the battery) is 30 volts and is labeled Er. Rr = Since there are two resistances in the circuit, there will be two separate voltage drops. These Ir = two voltage drops will be the loss in electrical ERI = pressure used to force electrons through the resistances. The amount of electrical pressure ER, = required to force a given number of electrons ER: = through a resistance is proportional to the size of the resistance. Thus, the voltage drop across Ri Total resistances:
will be twice that across Ra since R, has two times RT = R1 + Rz + R3 the resistance value of Rz. The drop across R, is labeled E1, and that across R, is E2. The current, = 30 + 60 + 10 I, is the same throughout the circuit. = 100 R.
Using: Total Current:
E = IR E1 = IR1 E2 = 1R2 Ez = 2a X 5 Ix = RI Er E1 = 2a × 10 Ez = 10 V. 150 V E1 = 20 V = 100 If the voltage drops (used) across the two = 1.5 amperes.
Voltage Drops:
E = IR ER, = IT X R1 = 1.5 amps X 30 = 45 V.
ERr = IT X Rz = 1.5 amps X 60 = 90 V.
ERS = IT X R3 = 1.5 amps X 10 = 15 V.
Do these values for voltage drops equal the applied voltage?
ET = ER: + ERZ + ERS Er = 150 volts 150 V = 45 V + 90 V + 15 V.
The sum of the voltage drops equals the applied voltage.
Kirchhoff Laws In 1847, a German physicist, G. R. Kirchhoff, elaborated on Ohm's law and developed two statements that are known as Kirchhoff's laws for current and voltage. An understanding of these laws enables the aircraft technician to gain a better understanding of the behavior of electriity. Using Kirchhoff's laws, it is possible to find:
(1) The current in each branch of a network circuit when both the resistance and the electromotive force in each branch are known, or (2) the electromotive force in each branch when both the resistance of, and the current in, each branch are known. These laws are stated as follows:
Current Law-The algebraie sum of the.
currents at any junction of conductors in a circuit is zero. This means that the amount of current flowing away from a point in a circuit is equal to the amount flowing to that point.
Voltage Law- The algebraic sum of the applied voltage and the voltage drop around any closed circuit is zero, which means that the voltage drop around any closed circuit is equal to the applied voltage.
When applying Kirchhoff's laws, use the follow- 50 102 + 5 v 10 v 30 v 15 v < 15 g FIGURE 8-60. Polarity of voltage drops.
ing procedures to simplify the work:
I. When the direction of current is not apparent, assume a direction of flow.
If the assumption is wrong, the answer will be numerically correct but preceded by a negative sign.
2. Place polarity markings (plus and minus signs) on all resistors and batteries in the circuit being solved. The assumed direction of current flow will not affect the polarities of the batteries, but it will affect the polarity of the voltage drop on resistors.
Therefore, the voltage drop should be marked so that the end of a resistor into which the current is assumed to flow is negative, and the end from which it leaves is positive.
In the statements of Kirchhoff's laws, the term algebraic sum was used. An algebraic sum differs from an arithmetic sum in that both the magnitude and the sign of each number must be considered. In electrical circuits a voltage drop occurs when current flows through a resistor.
The magnitude of the voltage is determined by the size of the resistor and the amount of current flow. The polarity (sign) of the voltage drop is determined by the direction of the current flow.
For example, observe the polarities of the applied electromotive force (e.m.f.) and the voltage drop as shown in figure 8-60. The applied e.m.f. causes electrons to flow through the opposition offered by the resistances. The voltage drop across each resistance is therefore opposite in polarity to that of the applied e.m.f. Note that the side of each resistor where the current enters is labeled the negative side.
2 = P = 1 amp d 150 E, = 30 volts B FIGURE: 8-61. Circuit demonstrating Kirchhoff's laws, (A) current law and (B) voltage law.
A portion of a circuit which illustrates Kirchhoff's current law is shown in figure 8-61.
The current flowing through resistor Ri has an intensity of four amperes. The current flowing through resistor R, has a magnitude of one ampere and is flowing into the same junction as the current through Ri. Using Kirchhoff's current law, it is possible to determine how much current is flowing through R2 and whether it is flowing toward or away from the common junction. This is expressed in equation form as:
11 + 12 + Is = 0.
Substituting the current values in the equation gives 4 + 12+ (-1) =0 12=1+4 12 = 5
- 4 + (-1) +5 = 0.
Kirchhoff's current law finds a wider application in more complex parallel or series-parallel circuits.
Figure 8-61 (B) is a series d.c. circuit which is used to demonstrate Kirchhoff's voltage law.
The total resistance is the sum of R1, R2, and Rs which is 30 ohms. Since the applied voltage is 30 volts, the current flowing in the circuit is 1 ampere. Therefore, the voltage drops across R1, R2, and R, are 5 volts, 10 volts, and 15 volts, respectively. The sum of the voltage drops is equal to the applied voltage of 30 volts.
This circuit may also be solved by using the polarities of the voltages and showing that the algebraic sum of the voltages is zero. Consider voltages positive if the (+) sign is met first and negative if the (-) sign is met first when tracing the current flow. By starting at the battery and going in the direction of current flow (as indicated by the arrows) the following equation may be set up:
Total voltage (El) = + 30 - 5 - 10 - 15 E. = O.
The point to start around the circuit and the polarity to use are arbitrary and are a matter of choice for each circuit.
PARALLEL D. C. CIRCUITS A circuit in which two or more electrical resistances, or loads, are connected across the same voltage source is a parallel circuit. The parallel circuit differs from the series circuit in that more than one path is provided for current flowthe more paths added in parallel, the less opposition to flow of electrons from the source. In a series circuit the addition of resistance increases the opposition to current flow. The minimum requirements for a parallel circuit are the following:
(1) a power source.
(2) conductors.
(3) a resistance or load for each current path.
(4) two or more paths for current flow.
Figure 8-62 shows a parallel circuit with three paths for current flow. Points A, B, C, and D are connected to the same conductor and are at the same electrical potential. In a similar manner, points E, F, G, and H are at the same potential.
Since the applied voltage appears between points A and E, the same voltage is applied between points B and F, points C and G, and between points D and H. Thus, when resistors are connected in parallel across a voltage source, each resistor has the same applied voltage, although the currents through the resistors may differ de- 290
The total current, Ir, is Ir = I1 + I2 + 13 Ir = 4 amps + .24 amps + .5 amps Ir = 1.14 amps.
In a parallel circuit, Ir = I1 + I2 + Is. By Ohm's law the following relationships can be obtained:
FIGURE 8-62. A parallel circuit.
pending on the values of resistance. The voltage in a parallel circuit may be expressed as follows:
Er = E1 = E2 = E3.
Where Er is the applied voltage, E, is the voltage across R1, Ez is the voltage across R2, and Es is the voltage across R3 (figure 8-62).
The current in a parallel circuit divides among the various branches in a manner depending on the resistance of each branch (see figure 8-63).
A branch containing a small value of resistance will have a greater current flow than a branch containing a high resistance. Kirchhoff's current law states that the current flowing toward a point is equal to the current flowing away from that point. Thus, the current flow in a circuit may be expressed mathematically as follows:
IT = 11 + 12 + 13 where IT is the total current and I1, I2, and Is are the currents through R1, R2, and R3, respectively.
Kirchhoff's and Ohm's law can be applied to find the total current flow in the circuit shown in figure 8-63.
The current flow through the branch containing resistance R1 is I1 = Ri E = 15 6 = 4 amps.
The current through Re is Iz = Rz E = 25 6 = .24 amps.
The current through R3 is Із = = 12 -_ = .5 amps.
Rz İT = Substituting these values in the equation for total current, In a parallel circuit Er = E1 = E2 = E}.
Therefore, Dividing through by E gives, This equation is the reciprocal formula for finding the total or equivalent resistance of a parallel circuit. Another form of the equation may be derived by solving for Rr.
RT = An analysis of the equation for total resistance in a parallel circuit shows that Rr is always less R 6V {15 & 25 gl | 1212 FIGURE 8-63. Current flow in a parallel circuit.
A, = 200 R, = 100 { R, = 100 FICURE 8-65. A series-parallel circuit.
FIGURE 8-64. A series-parallel circuit.
than the smallest resistance in a parallel circuit.
Thus a 10-ohm, a 20-ohm, and a 40-ohm resistor connected in parallel have a total resistance of less than 10 ohms.
If there are only two resistors in a parallel circuit, the reciprocal formula is Simplified, this becomes:
Ry Rz Rr = Re + R2 This simplified, shorter formula can be used when two resistances are in parallel. Another method can be used for any number of resistors in parallel if they are of equal resistance. The resistance value of one resistor is divided by the number of resistors in parallel to determine the total resistance. Expressed mathematically this becomes:
Rr = Where Rr is the total resistance, R is the resistance of one resistor, and N is the number of resistors.
SERIES-PARALLEL D. C. CIRCUITS Most circuits in electrical equipment are not series or parallel circuits. They are usually seriesparallel circuits, which are combinations of series and parallel circuits. A series-parallel circuit consists of groups of parallel resistors connected in series with other resistors. An example of a series-parallel circuit is shown in figure 8-64.
The requirements for a series-parallel circuit are as follows:
(1) power source (battery).
(2) conductors (wires).
(3) load (resistances).
(4) more than one path for current flow.
(5) a control (switch).
(6) safety device (fuse).
While series-parallel circuits may appear extremely complex, the same rule used for series and parallel circuits can be applied to simplify and solve them.
The easiest method of handling series-parallel circuits is to break them apart and redraw them as equivalent circuits. The circuit in figure 8-65 is an example of a simple series-parallel circuit that can be redrawn to illustrate this procedure.
In this circuit the same voltage is applied to Re and Re; thus they are in parallel. The equivalent resistance of these two resistors is equal to the value of one resistor divided by the number of resistors in parallel. This is true only when the parallel resistors have the same ohmic value.
If this rule is applied, the circuit can be redrawn as shown in figure 8-66.
This has converted the original series-parallel circuit into a simple series circuit containing two resistances. To further simplify the circuit, the two series resistances can be added, and the circuit can be redrawn as shown in figure 8-67.
Although the last redrawing of the circuit could have been omitted and the calculations done mentally, this circuit illustrates clearly that one 25-ohm resistor is the resistive equiv- R, = 202 52 2 R. and R equivalent FIGURE 8-66. A redrawn series-parallel circuit.
R, 602 ww R, 25g equivalent 100g 30 2 R; 602 m Re 602 FIGURE 8-67. An equivalent series-parallel circuit alent of the three resistors of the original circuit.
Figure 8-68 contains a more complex seriesparallel circuit.
The first step in simplifying this circuit is to reduce each group of parallel resistors to a single equivalent resistor. The first group is the parallel combination of Re and R3. Since these resistors have unequal values of resistance, the formula for two parallel resistances is used:
Rz Rz - 120·10 - 4800 Ra = Rr + R3 = 30 Q.
160 Then the parallel combination of R, and R3 can be replaced with a single 30 I resistor, as shown in figure 8-69.
Next, the equivalent resistance of the parallel combination of R4, R5, and Re can be determined using the formula Ri = R/N: where, R& is the equivalent resistance of R4, R5, and R6, R is the value of one of the resistors, and N is the number of resistors in parallel, A =Ñ = 60 - = 20 Q.
The parallel combination of R4, R5, and Re can 100 22 120g R, 60 g m R, 60 g
80.2 m 100 g FIGURE 8-69. Series-parallel circuit with one equivalent resistance.
now be redrawn as a single 20 I resistor, as shown in figure 8-70.
The original series-parallel circuit has now been replaced with its equivalent series circuit. This circuit could be redrawn again to replace the five resistors in series with one 330-ohm resistor.
This can be proved by using the total resistance formula for series circuits:
Rr = R: + Ra + Ro + Rr + Res = 100 + 30 + 20 + 80 + 100 = 330 ohms.
The first series-parallel circuit used is redrawn to discuss the behavior of current flow (figure 8-71).
Unlike the parallel circuit, the branch currents, I1 and 12, cannot be established using the applied voltage. Since R, is in series with the parallel combination of R2 and R3, a portion of the applied voltage is dropped across Ri. In order to find the branch currents, total resistance and total current must be found first. Since R, and R3 are 4052 Re 600 1000 3022 2022 R 802 - 100g Re 10052 FIGURE 8-68. A more complex series-parallel circuit. FIGURE 8-70. Series-parallel equivalent circuit.
The voltage drops across parallel resistors are 210 always equal. It should also be remembered that, when the voltage is held constant and the resistance of any resistor in a series-parallel circuit 28V 31480 is increased, the total current will decrease. This should not be confused as adding another parallel 140 resistor to a parallel combination, which could reduce total resistance and increase total current FICURE 8-71. Current flow in a series-parallel VOLTAGE DIVIDERS circuit.
Voltage dividers are devices which make it equal resistance, possible to obtain more than one voltage from a single power source.
Roger - N= 2 = 70. A voltage divider usually consists of a resistor, or resistors connected in series, with fixed or Total resistance is movable contacts and two fixed terminal contacts. As current flows through the resistor, Rr = Ri + Requiv. different voltages can be obtained between the = 210+70 contacts. A typical voltage divider is shown in figure 8-72.
= 28 0. A load is any device which draws current. A Using Ohm's law, total current is Is = Rr Er = 28 h = 1 ampere. 28 V The total current, 1 ampere, flows through Ri and divides at point A, with part of the current flowing through R2, and the other part through Re. Since R, and R3 are of equal size, it is obvious that half of the total current, or .5 amps, will flow through each branch.
The voltage drops in the circuit are determined by Ohm's law:
E = IR.
ER, = ITRI heavy load means a heavy current drain. In addition to the current drawn by the various loads, there is a certain amount drawn by the voltage divider itself. This is known as bleeder To understand how a voltage divider works, examine the illustration in figure 8-73 carefully and observe the following:
Each load draws a given amount of current: I1, I2, I3. In addition to the load currents, some bleeder current (Is) flows. The current I, is drawn from the power source and is equal to the sum of all currents.
The voltage at each point is measured with respect to a common point. Note that the common = 1 × 21 = 21 Volts.
E = IR.
ER2 = 11R2 = .5 × 14 = 7 Volts.
E = IR.
ERs = 12 R$ = .5 × 14 = 7 Volts. FIGURE 8-72. A voltage divider circuit.