143Analyzing Series Circuits143
NORMAL ACTUAL * NORMAL ACTUAL *
TP1 –25 V [95] 94 kΩ [205]
TP2 –17.8 V [110] 67 kΩ [107]
TP3 –15.1 V [120] 57 kΩ [8]
TP4 –10.4 V [76] 39 kΩ [149]
R1 7.18 V [220] 27 kΩ [45]
R2 2.66 V [83] 10 kΩ [15]
R3 4.79 V [141] 18 kΩ [67]
R4 10.4 V [102] 39 kΩ [86]
VA 25 V [127] — —
* Numbers in brackets refer to value numbers in Appendix A.
Table 4-5. PShooting Data for the Circuit in Figure 4-87
5. Troubleshoot the circuit shown in Figure 4-87 using the PShooter data provided in Table 4-6.
Chapter Summary
• A series circuit is characterized by having one and only one path for current flow. The current is the same through every component in every part of the circuit. Each resistor in a series circuit drops a portion of the applied voltage. The portion of voltage dropped is in direct proportion to the resistance of the resistor as compared to the total resistance in the circuit. Total resistance in a series circuit is simply the sum of the individual resis-tances. Kirchhoff ’s Voltage Law states that the algebraic sum of the voltages (drops and sources) in any closed loop is equal to zero. This means that the sum of the voltage drops across the circuit resistance must always be equal to the value of applied voltage.
• Each resistor in a series circuit dissipates a certain amount of power, which is supplied by the applied voltage source. Total power consumption from the voltage source is equal to the sum of the individual resistor power dissipations.
• Fully specified series circuits can generally be analyzed by direct substitution of component values into an applicable equation. Analysis of partially specified circuits requires creative application of Ohm’s Law, Kirchhoff ’s Voltage Law, power equations, and basic series circuit theory. In order to compute a circuit quantity, you must select an equation with only one unknown value. It is essential that properly subscripted values be used in all calculations to avoid accidental mixing of quantity types.
• Ground is a point in the circuit from which voltage measurements are referenced.
Ground generally takes the physical form of a metal chassis, a wide PCB trace, or an entire plane in a multilayer PCB.
TESTPOINT
VOLTAGE RESISTANCE
NORMAL ACTUAL* NORMAL ACTUAL*
TP1 –25 V [38] 94 kΩ [113]
TP2 –17.8 V [41] 67 kΩ [131]
TP3 –15.1 V [88] 57 kΩ [93]
TP4 –10.4 V [151] 39 kΩ [73]
R1 7.18 V [51] 27 kΩ [58]
R2 2.66 V [27] 10 kΩ [24]
R3 4.79 V [144] 18 kΩ [135]
R4 10.4 V [188] 39 kΩ [89]
VA 25 V [70] — —
* Numbers in brackets refer to value numbers in Appendix A.
Table 4-6. PShooting Data for the Circuit in Figure 4-87
• A series circuit may have more than one voltage source. If the multiple voltage sources all have the same polarity, the sources are said to be series-aiding. The effective voltage of series-aiding sources is the sum of the individual sources. When two voltage sources are connected with opposite polarities, they are said to be series-opposing. The net voltage of two series-opposing voltage sources is the difference between the individual source voltages. The polarity of the resulting voltage is the same as the larger of the two individual sources. If a series circuit has both series-aiding and series-opposing voltage sources, then the circuit has a complex voltage source. The effective voltage of a complex voltage source is equal to the algebraic sum of the individual source voltages.
• An unloaded voltage divider is an application of a series circuit. Voltage dividers are used to provide one or more lower voltages from a single higher source voltage.
Unloaded voltage dividers cannot supply current to other circuits or devices. Because they are series circuits, unloaded voltage dividers can be analyzed and designed using the same equations and techniques used with other series circuits.
• Logical, systematic troubleshooting of series circuits is an important technician skill since it provides the foundation for troubleshooting of more complex circuits. The possible defects in a series circuit can be loosely categorized into two classes: opened (or increased resistance) and shorted (decreased resistance). When a resistor increases in value in a series circuit, all voltages in the circuit will measure nearer to the value of the supply terminal on the same side of the open. When a resistor decreases in value, all voltages in the circuit will measure nearer to the value of the supply terminal on the opposite side of the defect.
Review Questions Review Questions
Section 4.1: Current Through Each Component
1. Current is the same in all parts of a series circuit. (True or False)
2. If a certain series circuit has 1.2 A leaving the voltage source, what percentage of this current will flow through the first resistor in the series chain? The last resistor in the series chain?
Section 4.2: Voltage Across Each Resistor
3. The voltage across a resistor is (directly, indirectly) proportional to its resistance value.
4. If one resistor in a series circuit is increased, what happens to the amount of voltage dropped across it?
5. If one resistor in a series circuit is increased, what happens to the amount of voltage dropped across the remaining resistors?
6. The largest resistor in a series circuit will always drop the (least, most) voltage.
7. The (smallest, largest) resistor in a series circuit will drop the least voltage.
8. Equal-valued resistors in a series circuit will have equal voltage drops. Explain why this is true.
Section 4.3: Kirchhoff’s Voltage Law
9. Electron current exits the (positive, negative) end of a resistor.
10. Electron current returns to the (positive, negative) side of a voltage source.
11. The algebraic sum of the voltage drops and voltage sources in a closed loop is always equal to __________________.
12. What is the relative magnitude of the source voltage in a series circuit as compared to the sum of the resistor voltage drops?
13. Write the closed-loop equation to describe the circuit shown in Figure 4-88.
14. Use Kirchhoff ’s Voltage Law to determine whether the voltages shown on the circuit in Figure 4-89 are correct.
15. Label the polarities of the voltage drops in Figure 4-90.
VA R2
R3 18 V
3 V R1
6 V
9 V
+ –
+ – + –
Figure 4-88. Circuit for Problem 13.
VA
R3
R4
50 V 7 V
R1 5 V
10 V
– +
R2 30 V
– +
– – + +
Figure 4-89. Circuit for Problem 14.
VA
2 V
10 V
2 V
3.5 V R1
R4
R2
2.5 V R3
Figure 4-90. Circuit for Problem 15.
16. Use Kirchhoff ’s Voltage Law to determine whether the voltages shown on the circuit in Figure 4-91 are correct.
Section 4.4: Computing Voltage, Current, Resistance, and Power
17. A certain series circuit has the following resistors: 10 kΩ, 4.7 kΩ, 5.6 kΩ, and 8.2 kΩ.
Calculate the value of total circuit resistance.
18. If a series circuit is composed of three 27-kΩ resistors, what is the total circuit resistance?
19. How much current flows in the circuit shown in Figure 4-92?
20. A certain series circuit has three 10-kΩ resistors. Each resistor has 325 mA of current flowing through it. What is the value of total current through the volt-age source?
21. A circuit has a 5-kΩ resistor and a 10-kΩ resistor in series. If the 5-kΩ resistor has 100 µA of current flow, how much current will flow through the 10-kΩ resistor?
22. The resistors in a particular series circuit dissipate 1.2 W, 3.5 W, and 2.0 W each.
How much power must be supplied by the voltage source?
23. Compute the total power in the circuit shown in Figure 4-93.
VA
Figure 4-91. Circuit for Problem 16.
VA R2
Figure 4-92. Circuit for Problem 19.
VA R2
Figure 4-93. Circuit for Problem 23.
24. Refer to Figure 4-94 and complete the solution matrix shown in Table 4-7.
25. Refer to Figure 4-95 and complete the solution matrix shown in Table 4-8.
VA R2
R3
20 V 1.8 kΩ
5 V R1
I1 = 2 mA
+ –
Figure 4-94. Circuit for Problem 24.
COMPONENT RESISTANCE VOLTAGE CURRENT POWER
R1 2 mA
R2 1.8 kΩ
R3 5 V
Total 20 V
Table 4-7. Solution Matrix for the Circuit in Figure 4-94
VA R2
R3 PT = 100 W
100 Ω R1 25 V
50 V +
+ –
–
Figure 4-95. Circuit for Problem 25.
COMPONENT RESISTANCE VOLTAGE CURRENT POWER
R1 25 V
R2
R3 100 Ω 50 V
Total 100 W
Table 4-8. Solution Matrix for the Circuit in Figure 4-95
Section 4.5: Ground and Other Reference Points
26. Ground is the reference point from which other voltages are measured. (True or False) 27. If VA = +2 V and VB = –4 V, what is the value of VAB?
28. If VED = 10 V and VE = 50 V, what is the value of VD? 29. If VBC = –6 V and VC = +20 V, what is the value of VB? Refer to Figure 4-96 for Problems
30 through 35.
30. What is the value of VB? 31. What is the value of VF?
32. Determine the voltage at point B with reference to point C.
33. Determine the value of VBE. 34. What is the value of VFA? 35. Determine the value of VEB.
Section 4.6: Multiple Voltage Sources
36. If a 10-V battery and a 25-V battery are connected in a series-opposing configuration, what is the effective source voltage?
37. A typical flashlight uses two 1.5-V cells (D cells) connected in a series-aided configu-ration. How much voltage does this arrangement provide for the lamp?
38. If you accidentally reversed one of the cells in a two-cell flashlight, what would be the effective voltage?
39. A certain electronic game requires four D cells (1.5 V each) to be connected as series-aiding voltage sources. If one of the cells is acci-dentally reversed, what will be the effective voltage available to the game circuitry?
40. Suppose you have three 28-V bat-teries and you need to obtain the highest possible voltage. How would you connect the batteries?
41. Determine the effective voltage for the circuit shown in Figure 4-97.
VT
Figure 4-96. Circuit for Problems 30 through 35.
Figure 4-97. What is the effective voltage in this circuit?
42. Determine the effective voltage for the circuit shown in Figure 4-98.
Section 4.7: Applied Technology: Voltage Dividers
