1. Unit Seven Series-Parallel Circuits
ET115 DC Electronics
Unit Seven Series-Parallel Circuits
John Elberfeld

2. Schedule Unit Topic Chpt Labs Quantities, Units, Safety 1 2 (13)
Voltage, Current, Resistance
Ohm’s Law (35)
Energy and Power (41)
Series Circuits Exam I 4 7 (49)
Parallel Circuits (65)
Series-Parallel Circuits (75)
Thevenin’s, Power Exam (133)
Superposition Theorem (81)
Magnetism & Magnetic Devices 7 Lab Final
Course Review and Final Exam

3. Unit 7 Objectives - I
Identify series-parallel relationships in a combinational resistive circuit.
Determine total resistance, current, and power in a series-parallel resistivecircuit.
Draw the schematic for the Wheatstone bridge.
Explain how to determine if the Wheatstone bridge is balanced.
Determine the effect of one or more resistive loads on a voltage divider

4. Unit 7 Objectives – II
Determine the bleeder current in a resistive voltage divider.
Determine the loading effect of a voltmeter on a circuit given the internal resistance of the voltmeter.
Define bipolar voltage divider
Use proper prototype board wiring and test procedures for DC resistive circuit components including using the digital multimeter..

5. Unit 7 Objectives – III Construct basic DC circuits on a protoboard.
Use a digital multimeter (DMM) to measure a predetermined low voltage on a power supply.
Measure resistances and voltages in a DC circuit using a DMM.
Apply Ohm’s Law, Thevenin’s theorem, KVL and KCL to practical circuits.

6. Reading Assignment Read and study
Chapter 6: Series-Parallel Circuits: Pages (First half of chapter)

7. Lab Assignment
Experiment 10, “Series-Parallel Combination Circuits,” beginning on page 75 of DC Electronics: Lab Manual and MultiSim Guide.
Complete all measurements, graphs, and questions and turn in your lab before leaving the room

8. Written Assignments Complete the Unit 7 Homework sheet
Show all your work!
Be prepared for a your second MAJOR EXAM on questions similar to those on the homework.

9. Ohms Law MEMORIZE: V = I R Ohms Law
If you increase the voltage, you increase the current proportionally
3 times the voltage gives you three times the current
Resistance (ohms) is the proportionality constant and depends on the atomic structure of the material conducting the current

10. Power Formula Power = Work / time P = V I
Voltage is the work done per coulomb, and current is the number of coulombs per second passing by a point.
The product of voltage and current gives the work done per second, or power.

11. Series-Parallel Combination of Resistances
This type of circuit consists of a combination of resistances in series and parallel.
Series-parallel combination of resistances can be classified as:
Simple series-parallel circuits
Complex series-parallel circuits

12. Parallel Resistors
For resistors to be in parallel, both ends of each resistor must be in direct electrical contact with each other – with no resistance between the ends
Parallel Not Parallel (blue) R1 R2
R1||R2

13. Series-Parallel Combination of Resistances
To differentiate between a simple series-parallel circuit and a complex circuit:
observe simple series and parallel combinations of resistances in the given circuit.
replace the simple series and simple parallel combinations by their single equivalent resistor.
repeat the simplification process to verify whether the circuit is reduced to a single resistance circuit.
if a single resistor can replace ALL the resistors, it is a simple series-parallel circuit

14. Recognizing Series Components
RT= R1+R2+R3||R4||(R5+R6)

15. Analysis
Analysis of the circuit also requires one to recognize the various paths for current flow.
The ability to recognize the points where current branches out and where current converges (sums) is vital.

16. EXAMPLE
Electron Flow →

17. Series-Parallel Circuits
Pure series-parallel circuits can be reduced to a single equivalent resistor:
Example R1 R1
R2||R3 R2 R3
RT= R1+R2||R3

18. Example Another simple series-parallel circuit R1 R1 R3 R2 R2||R3 R4
RT= R1+R2||R3+R4||R5 R5

19. RT=R2||[R1+R3||(R4+R5)] R1 R4 R1 R3 R3 R4+R5 R5 R2 R2 RT R2 R1

20. Complex Circuit Unable to simplify to a single resistor Complex Simple
Where the battery connects to the circuit is important!!!!
Complex Simple

21. Complex Circuit
Unable to simplify to a single resistor

22. Practice
Classify this circuit as series-parallel or complex

23. Practice
Classify this circuit as simple series-parallel or complex

24. Practice Classify this circuit as series-parallel or complex
It is the electrical connections, not the drawing, that determines the type of circuit

25. Practice
Classify this circuit as series-parallel or complex

26. Voltage Drop in Series-Parallel Resistive Circuit
To find the equivalent resistance of a series-parallel resistive circuit:
Identify the resistance combinations as either series or parallel.
Simplify, combine and recombine series and parallel combinations
Find the value of the total resistance of the circuit.
Once you know the equivalent resistance, you can calculate voltage or current

27. Total Resistance
One of the common approaches is called Outside Toward the Source approach.
To implement this method, one begins farthest from the source and works back toward the source.

28. Equivalent Resistance
The analysis of the circuit uses equivalent resistance as circuit reductions are performed.
For instance, if a 6-k and a 3-k resistor are in parallel, their equivalent series resistance is 2 k.

29. Current In a Series-Parallel Circuit
The analysis of current in this circuit is a fundamental step.
Kirchhoff’s Current Law must be followed to see how current divides and sums together.

30. Circuit Analysis Tools
The circuit tools used to determine circuit parameters have already been covered. They include:
-Ohm’s law Power equations
-KCL and KVL -Voltage divider rule

31. Voltage In a Series-parallel Circuit
Voltage distribution throughout the circuit follows the laws appropriate to series and parallel connections.

32. Circuit Analysis
Combine the two parallel resistors to one equivalent resistor
1/R23 = 1/20Ω + 1/30Ω = 3/60Ω + 2/60Ω
1/R23 = 5/60Ω
R23= 12 Ω
R1 = 10Ω
R1 = 10Ω
V = 25V
R3 = 30Ω
R23 = 12Ω
V = 25V
R2 = 20Ω

33. Circuit Analysis
Combine the two parallel resistors to one equivalent resistor using POS
R23 = 20Ωx30Ω / (20Ω+30Ω) = 12 Ω
Same results!
R1 = 10Ω
R1 = 10Ω
V = 25V
R3 = 30Ω
R23 = 12Ω
V = 25V
R2 = 20Ω

34. Analysis Combine the series resistors R123 = 10 Ω + 12 Ω = 22 Ω
V = 25V
V = 25V
R23 = 12Ω

35. Analysis
V = 25 V, RT = 22Ω, IT = 1.14 A
VR1 = 11.4V, VR2 = VR3 = 25V-11.4V=13.6V
I R2= VR2/20 Ω = .68A
I R3= VR3/30 Ω = .45A
Check: IT = IR1 = IR2 + IR3 = 1.14 A
VT = VR1+VR2 = 25V
R1 = 10Ω
R3 = 30Ω
V = 25V
R2 = 20Ω

36. Current in Parallel Resistive Circuit
In the circuit shown, the current flowing through R1, R2, R3, and R4 has two separate paths called branches.
Applying KCL, the current, I, is equal to the sum of the branch currents, I1 and I2.
←Conventional Flow

37. What is the equivalent resistance of the following circuit?
Class Activity
What is the equivalent resistance of the following circuit?
R1 = 5.6kΩ
V = 30V
R2 = 10kΩ
R5 = 3.3kΩ
R3 = 4.7kΩ
R4 = 2.2kΩ

38. What is the equivalent resistance of the following circuit?
Description
What is the equivalent resistance of the following circuit?
R1 = 5.6kΩ
V = 30V
R2 = 10kΩ
R5 = 3.3kΩ
R3 = 4.7kΩ
R4 = 2.2kΩ

39. Class Activity
R2 and R3 are….
In series with each other AND the series combination is in parallel with R5
R1 and R4 are in series with the combination of R5 in parallel with R2 and R3
RT= R1+R4+R5||(R2+R3)
V = 30V
R2 = 10kΩ
R1 = 5.6kΩ
R3 = 4.7kΩ
R5 = 3.3kΩ
R4 = 2.2kΩ

40. Predict Total Resistance
The total resistance must be between:
5.6k Ω kΩ = 7.8kΩ Minimum
5.6k Ω kΩ kΩ = 11.1 kΩ Maximum
WHY???

41. Current Analysis Combine resistors in series
R23 = 10kΩ + 4.7kΩ = 14.7kΩ
R1 = 5.6kΩ
V = 30V
R5 = 3.3kΩ
R23 = 14.7kΩ
R4 = 2.2kΩ

42. Current Analysis Combine parallel resistors (POS)
R235 = 14.7 kΩ x 3.3kΩ / (14.7kΩ + 3.3kΩ)
R235 = 2.70 kΩ
R1 = 5.6kΩ
V = 30V
R235 = 2.7 kΩ
R4 = 2.2kΩ

43. Current Analysis Combine series resistors
R12345 = 5.6 kΩ + 2.7kΩ +2.2 kΩ =10.5 kΩ
R12345 = 10.5kΩ
V = 30V

44. More Analysis IT = VT/RT = 30V / 10.5 KΩ = 2.85 mA
VR1 = 16V, VR235 = 7.7V, VR4 = 6.3V
R1 = 5.6kΩ
V = 30V
R235 = 2.7 kΩ
R4 = 2.2kΩ

45. Current Analysis VR235 = 7.7V IR23 = 523μA, IR5 = 2.33mA R1 = 5.6kΩ
V = 30V
R5 = 3.3kΩ
R23 = 14.7kΩ
R4 = 2.2kΩ

46. Current VR235 = 7.7V, IR23 = 523μA VR2 = 5.23V, VR3 = 2.46V,
R1 = 5.6kΩ
V = 30V
R2 = 10kΩ
R5 = 3.3kΩ
R3 = 4.7kΩ
R4 = 2.2kΩ

47. Check You Results 30V =VR1+VR2+VR3+VR4 = VR1+VR5+VR4 IT = IR23 + IR5
R1 = 5.6kΩ
V = 30V
R2 = 10kΩ
R5 = 3.3kΩ
R3 = 4.7kΩ
R4 = 2.2kΩ

48. More Analysis Which resistor has the largest voltage drop?
Is it necessarily the biggest resistor?
Definition: I1 is the current through R1, I2 the current through R2, etc.
What must I1 be equal to?
What must I1 be great than?
Which currents equal IT – the total current output of the power source?

49. Power In A Series-parallel Circuit
Analysis of power distribution in the circuit follows the same rules from pure series and pure parallel circuits.
Total power is a sum of all individual power levels in the circuit components.

50. Power in Series-Parallel Resistive Circuit
Total power generated by the source of a series parallel circuit is dissipated in various branches of the circuit. where Pd = power dissipated in the circuit
Pc = power consumed by the circuit,
P1, P2 …..P= are the powers dissipated in the components of the circuit

51. Effects Of Opens
As noted earlier, the location of the open has a great influence on how the electrical parameters change with the open.
Total current decreases as a result.

52. Troubleshooting An Open
As the open becomes more dramatic, the technician will note that current and power levels greatly decrease.

53. Current Measurements
Measuring current in the parallel portion will indicate if an open exists.
Measuring voltage in the series portion will indicate opens via the absence of voltage drops.

54. Effects Of Shorts
The location of the short determines how dramatic the impact will be to the circuit.
A general rule is that the closer the short is to the source, the greater the impact will be to circuit operation.

55. What IF What is the current IF R2 is OPEN? RT = R1+R5+R4 = 11.1kΩ
IT= 2.70mA (was 2.85 mA - DEcrease)
R1 = 5.6kΩ
V = 30V
R2 = 10kΩ
R5 = 3.3kΩ
R3 = 4.7kΩ
R4 = 2.2kΩ

56. What IF What is the current IF R1 is OPEN? RT = infinite
IT = 0 A (was 2.85 mA - DEcrease)
R1 = 5.6kΩ
V = 30V
R2 = 10kΩ
R5 = 3.3kΩ
R3 = 4.7kΩ
R4 = 2.2kΩ

57. What IF What is the current IF R5 is SHORTED? RT = R1+R4 = 7.8kΩ
IT = 3.9mA (was 2.85 mA - INcrease)
R1 = 5.6kΩ
V = 30V
R2 = 10kΩ
R5 = 0
R3 = 4.7kΩ
R4 = 2.2kΩ

58. Application of Kirchhoff’s Current Law (KCL)
The following circuit consists of four resistances and one voltage source, along with the total and branch currents.
Applying KCL at node B:

59. A Completed Example

60. Class Activity
Find the current I4 flowing through the resistance R4 in the following circuit.

61. Plan of Attack
How would you tackle this problem?

62. Plan of Attack
How would you tackle this problem?

63. Plan of Attack
How would you tackle this problem?

64. Loaded Voltage Dividers
The Voltage Divider is a common series-parallel type circuit.
As loads are placed on the circuit, the analysis becomes a bit more difficult.

65. Voltage Divider Review
VCC
Under ideal conditions, I1 = I2 = I
V=IR (Ohm’s Law!!!)
VCC = I (R1 + R2)
I = VCC / (R1 + R2)
VR2 = R2 I
VR2 = R2 VCC / (R1 + R2) I1 R1
VR2 R2 I2

66. Voltage Divider Review
VCC
Under ideal conditions, I1 = I2 = I3 = I
VCC = I (R1 + R2 + R3)
I = VCC / (R1 + R2 + R3)
VR3 = R3 I
VR3 = R3 VCC / (R1 + R2 + R3) R1 I1 I2 R2
VRr I3 R3

67. Loaded Voltage Dividers
An unloaded voltage divider circuit has no other components connected across it.
The circuit shows a loaded voltage divider circuit, where resistor R3 is connected across the points, A and ground.
The resistors R2 and R3 are in parallel. Therefore, their voltages will be same and currents through them will be inversely proportional to their resistances.

68. Analysis
VCC
If the applied voltage is 30V, and R1= 4.7kΩ and R2 = 5.6kΩ, find VR2. I1 R1
VR2 R2 I2

69. Calculations VR2 = 30V 5.6kΩ / ( 4.7kΩ + 5.6kΩ) VR2 = 16.3 V
IR2 = 16.3 V/ 5.6kΩ = 2.91 mA
Check:
IT = 30V / ( 4.7kΩ + 5.6kΩ)
IT = 2.91 mA
VCC I1 R1
VR2 R2 I2

70. Analysis
VCC
If the applied voltage is 30V, and R1= 4.7kΩ, R2 = 5.6kΩ, and R3 = 8.7kΩ, find all currents and find VR2 I1 R1
VR2 R3 R2 I3 I2

71. Calculations R2 is parallel to R3 (POS!)
R23 = 5.6k Ω 8.7kΩ / (5.6kΩ + 8.7kΩ)
R23 = 3.4kΩ
V2 = 30V3.4k Ω /(3.4KΩ + 4.7kΩ)
V2 = 12.6 V
I2 = 12.6V/5.6k Ω = 2.25mA
I3 = 12.6V/8.7k = 1.45 mA
IT = 30V / (3.4kΩ + 4.7kΩ)
IT = 3.70 mA
CHECK: IT = I2 + I3
VCC I1 R1
VR2 R3 R2 I3 I2

72. Voltmeters
You measure voltage by placing a voltmeter across a resistor in a active circuit.
A perfect voltmeter has infinite resistance and draws no current, but in reality, a voltmeter might have a resistance of 10 MΩ
Any measuring device affects the value being measured

73. Analysis
VCC
If the applied voltage is 30V, and R1= 200 kΩ, R2 = 100 kΩ, and R3 = 10 MΩ (a voltmeter), find all currents and find VR2
Ideally,
VR2 = 30V(100 kΩ)/(300 kΩ)
VR2 = 10 V I1 R1
VR2 R3 R2 I3 I2

74. Calculations R2 is parallel to R3 (POS!)
R23 = 100kΩ 10 MΩ / (100kΩ + 10 MΩ )
R23 = kΩ
V2 = 30V kΩ =9.93 V (99.01kΩ+200kΩ)
V2 = 9.93 V
The meter causes less than 1% drop in measured voltage, but this increases as the size of the measured resistors increase
VCC I1 R1
VR2 R3 R2 I3 I2

75. Special Conditions
If two equal resistors are in parallel, the current AND voltage is the same for both resistors
If two equal resistors are in parallel, the equivalent resistance is equal to just ½ of one of the resistors

76. Example IR1= 25V/4.7kΩ = 5.32 mA IR2= 25V/4.7kΩ = 5.32 mA
IT = IR1 + IR2 = 5.32 mA mA = 10.6mA
RT = VT/IT = 25V / 10.6mA = 2.36 k Ω
Check: RT = 4.7kΩ 4.7kΩ / (4.7kΩ 4.7kΩ)
RT = 2.35 kΩ
Currents same, RT = R1/2
V = 25V
R1 = 4.7kΩ
R2 = 4.7kΩ

77. Example All resistors are 20Ω. The current in R3 is 10mA.
Find the voltage drop across and the current through all the other resistors. R1
RL1 R2
RL2 R3
RL3

78. Analysis VR3 = 10mA 20 Ω = .2 V VL3 = VR3 so IL3 = .2 V/ 20 Ω = 10 mA
IR2 = IR3 + IL3 = 10 mA + 10 mA = 20 mA
VR2 = 20mA 20 Ω = .4 V
VL2 = VR2 + VR3 so IL2 = .6 V/ 20 Ω = 30 mA
IR1 = IR2 + IL2 = 20 mA + 30 mA = 50 mA
VR1 = 50mA 20 Ω = 1 V
VL1 = VR1+ VR2 + VR3 so
IL1 = 1.6 V/ 20 Ω = 80 mA
IT = IR1 + IL1 = 50 mA + 80 mA = 130 mA

79. Checks VT = VR1 + VR2 + VR3 = .2V+.4V+1.0V = 1.6V
RT=VT/IT = 1.6V / 130mA = 12.3 Ω

80. Check your work – Find RT
20Ω
20Ω
20Ω
20Ω
??V
18V
20Ω
20Ω
20Ω
20Ω
20Ω
20Ω
10Ω

81. Simplify
20k
20k
20Ω
20Ω
??V
20Ω
??V
20Ω
20Ω
30 Ω
10 Ω

82. Simplify
20Ω
20Ω
20Ω
20Ω
??V
??V
20Ω
12Ω
30 Ω

83. Simplify
20Ω
20Ω
20Ω
18V
12Ω
18V
32Ω

84. Checks!!!!
= 12.3 Ω
20Ω
??V
??V
32Ω

85. Bridge Circuits
In a balanced bridge circuit, the current through and the voltage drop across R5 = 0:
If any one of the outer resistances is increased, the voltage drop across that resistance will also increase. Therefore, it is an unbalanced bridge circuit. B A

86. The Wheatstone Bridge
This circuit is a series-parallel circuit that is very popular in controls and industrial applications.
There are two states for the bridge:
Balanced Unbalanced
VA = VB VA ≠ VB

87. Why??
In a balanced bridge, the voltage drop across R3 = voltage drop across R4
There is NO current across the resistor connecting R3 to R4
VR3 = VInR3/(R3+R1)
VR4 = VInR4/(R4+R2)
VR3 = VR4 B A

88. Calculations VInR3/(R3+R1) = VInR4/(R4+R2)
Multiply both sides by (R3+R1)(R4+R2) and divide both by Vin
R3 (R4+R2) =R4 (R3+R1)
R3 R4+ R3 R2 =R4 R3+ R4 R1
R3 R2 =R4 R1
R1 / R3 =R2 / R4
In most problems, given 3, you find the 4th value

89. Class Activity
For the following balanced bridge circuit, calculate the value of the resistance, R.

90. Lab Experiment
With a wire, resistance is proportional to the length of the wire
The longer the wire, the more resistance it has
A wire exactly 1 meter long is connected to a battery
A slider that moves up and down the wire makes an electrical connection to it

91. Lab Setup
73 cm
XXX Ω
0 mA
35 V A R L
27 cm
1kΩ

92. Suppose you forgot? VL = 35V 27 cm / (27cm + 73 cm) = 9.45 V = VR
I1k = 9.45 V / 1 kΩ = 9.45 mA
VXXX = 35 V – 9.45 V = V
RXXX = V/I = V / 9.45 ma = 2.70 kΩ
R1 / R3 =R2 / R4
73/27 = X/1k Ω
X = 1k 73/ 27 = 2.70 kΩ

93. Practice
33 cm
XXX Ω
0 mA
20 V A R L
67 cm
5kΩ

94. Unit 7 Summary
1. Identifying series/parallel relationships in combinational circuits 2. Calculating resistance, current or voltage in combinational resistive circuits 3. Applying KVL and KCL to combinational resistive circuits 4. Wheatstone bridge 5. Loaded voltage-dividers 6. Meter loading