This is a PowerPoint presentation whose contents are obtained from Earl D. Gates Introduction to Basic Electricity and Electronics. No Copyright Rules Intended to Be Violated. Only the manner of presentation is deemed original. Details about Ohm and Kirchhoff were obtained from Wikipedia.
2. After completing this chapter, the student will be able to:
• Identify the three basic parts of a circuit.
• Identify three types of circuit configurations.
• Describe how current flow can be varied in a circuit.
• State Ohm’s law with reference to current, voltage, and resistance.
• Solve problems using Ohm’s law for current, resistance, or voltage in series, parallel, and
series-parallel circuits.
Objectives
Guibelondo | Ohm’s Law | 2
3. After completing this chapter, the student will be able to:
• Describe how the total current flow differs between series and parallel circuits.
• Describe how the total voltage drop differs between series and parallel circuits.
• Describe how the total resistance differs between series and parallel circuits.
• State and apply Kirchhoff’s current and voltage laws.
• Verify answers using Ohm’s law with Kirchhoff’s laws.
Objectives
Guibelondo | Ohm’s Law | 3
4. • Ohm’s law defines the relationship among three fundamental quantities: current, voltage, and
resistance.
• It states that current is directly proportional to voltage and inversely proportional to resistance.
• This chapter examines Ohm’s law and how it is applied to a circuit. Some of the concepts were
introduced in previous chapters.
Introduction
Guibelondo | Ohm’s Law | 4
5. • As stated earlier, current flows from a point with an excess of electrons to a point with a
deficiency of electrons.
• The path that the current follows is called an electric circuit. All electric circuits consist of a
voltage source, a load, and a conductor.
• The voltage source establishes a difference of potential that forces the current to flow. The
source can be a battery, a generator, or another of the devices described in Chapter 12:
Voltage.
Electric
Circuits
Guibelondo | Ohm’s Law | 5
6. • The load consists of some type of resistance to current flow. The resistance may be high or
low, depending on the purpose of the circuit.
• The current in the circuit flows through a conductor from the source to the load. The
conductor must give up electrons easily. Copper is used for most conductors.
Electric
Circuits
Guibelondo | Ohm’s Law | 6
7. • The path the electric current takes to the load may be through any of three types of circuits: a
series circuit, a parallel circuit, or a series-parallel circuit.
Electric
Circuits
Guibelondo | Ohm’s Law | 7
8. • A series circuit (Figure 14-1) offers a single continuous path for the current flow, going from
the source to the load.
Electric
Circuits
Guibelondo | Ohm’s Law | 8
9. • A parallel circuit (Figure 14-2) offers more than one path for current flow. It allows the source
to apply voltage to more than one load.
Electric
Circuits
Guibelondo | Ohm’s Law | 9
10. • A series-parallel circuit (Figure 14-3) is a combination of the series and parallel circuits.
Electric
Circuits
Guibelondo | Ohm’s Law | 10
11. • Current in an electric circuit flows from the negative side of the voltage source through the
load to the positive side of the voltage source (Figure 14-4).
Electric
Circuits
Guibelondo | Ohm’s Law | 11
12. • As long as the path is not broken, it is a closed circuit and current flows (Figure 14-5).
Electric
Circuits
Guibelondo | Ohm’s Law | 12
13. • However, if the path is broken, it is an open circuit and no current can flow (Figure 14-6).
Electric
Circuits
Guibelondo | Ohm’s Law | 13
14. • Changing either the voltage applied to the circuit or the resistance in the circuit can vary the
current flow in an electric circuit.
• The current changes in exact proportion to the change in the voltage or resistance.
Electric
Circuits
Guibelondo | Ohm’s Law | 14
15. • If the voltage is increased, the current also increases. If the voltage is decreased, the
current also decreases (Figure 14-7).
Electric
Circuits
Guibelondo | Ohm’s Law | 15
16. • On the other hand, if the resistance is increased, the current decreases (Figure 14-8).
Electric
Circuits
Guibelondo | Ohm’s Law | 16
17. Georg Simon Ohm
Ohm’s Law
Introduction
Guibelondo | Ohm’s Law | 17
Born 16 March 1789
Erlangen, Brandenburg-Bayreuthin the Holy Roman
Empire
(present-day Germany)
Died 6 July 1854 (aged 65)
Munich, Kingdom of Bavaria in the German Confederation
(present-day Germany)
Residence Brandenburg-Bayreuth, Bavaria
Nationality German
Alma mater University of Erlangen
Known for Ohm's law
Ohm's phase law
Ohm's acoustic law
Awards Copley Medal (1841)
Scientific career
Fields Physics (studies of electricity)
Institutions University of Munich
Doctoral advisor Karl Christian von Langsdorf
18. • In 1827, George Ohm first observed Ohm’s law, or the relationship among current, voltage,
and resistance.
• Ohm’s law states that the current in an electric circuit is directly proportional to the voltage
and inversely proportional to the resistance in a circuit.
Ohm’s Law
Introduction
Guibelondo | Ohm’s Law | 18
19. • This may be expressed as follows:
• Where: I = current in amperes
E = voltage in volts
R = resistance in ohms
• Whenever two of the three quantities are known, the third quantity can always be determined.
Ohm’s Law
Introduction
Guibelondo | Ohm’s Law | 19
𝐶𝑢𝑟𝑟𝑒𝑛𝑡 =
𝑉𝑜𝑙𝑡𝑎𝑔𝑒
𝑅𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝐼 =
𝐸
𝑅
or
20. • Example: How much current flows in the circuit shown in Figure 14-9?
• Answer: IT = 0.012 amp, or 12 milliamps
Ohm’s Law
Introduction
Guibelondo | Ohm’s Law | 20
21. • Example: In the circuit shown in Figure 14-10, how much voltage is required to produce 20
milliamps of current flow?
• Answer: ET = 24 volts
Ohm’s Law
Introduction
Guibelondo | Ohm’s Law | 21
22. • Example: What resistance value is needed for the circuit shown in Figure 14-11 to draw 2
amperes of current?
• Answer: RT = 60 ohms
Ohm’s Law
Introduction
Guibelondo | Ohm’s Law | 22
23. SERIES CIRCUIT
• In a series circuit (Figure 14-12), the same current flows throughout the circuit.
Ohm’s Law
Applications
Guibelondo | Ohm’s Law | 23
𝐼 𝑇 = 𝐼 𝑅1
= 𝐼 𝑅2
= 𝐼 𝑅3
… = 𝐼 𝑅 𝑛
24. SERIES CIRCUIT
• The total voltage in a series circuit is equal to the voltage drop across the individual loads
(resistance) in the circuit.
• The total resistance in a series circuit is equal to the sum of the individual resistances in the
circuit.
Ohm’s Law
Applications
Guibelondo | Ohm’s Law | 24
𝐸 𝑇 = 𝐸 𝑅1
+ 𝐸 𝑅2
+ 𝐸 𝑅3
+ … + 𝐸 𝑅 𝑛
𝑅 𝑇 = 𝑅1 + 𝑅2 + 𝑅3 + … + 𝑅 𝑛
25. PARALLEL CIRCUIT
• In a parallel circuit (Figure 14-13), the same voltage is applied to each branch in the circuit.
Ohm’s Law
Applications
Guibelondo | Ohm’s Law | 25
𝐸 𝑇 = 𝐸 𝑅1
= 𝐸 𝑅2
= 𝐸 𝑅3
… = 𝐸 𝑅 𝑛
26. PARALLEL CIRCUIT
• The total current in a parallel circuit is equal to the sum of the individual branch currents in the
circuit.
• The reciprocal of the total resistance is equal to the sum of the reciprocals of the individual branch
resistances.
• The total resistance in a parallel circuit will always be smaller than the smallest branch resistance.
Ohm’s Law
Applications
Guibelondo | Ohm’s Law | 26
𝐼 𝑇 = 𝐼 𝑅1
+ 𝐼 𝑅2
+ 𝐼 𝑅3
+ … + 𝐼 𝑅 𝑛
1
𝑅 𝑇
=
1
𝑅1
+
1
𝑅2
+
1
𝑅3
+ … +
1
𝑅 𝑛
27. • Ohm’s law states that the current in a circuit (series, parallel, or series-parallel) is directly
proportional to the voltage and inversely proportional to the resistance.
• In determining unknown quantities in a circuit, follow these steps:
• Draw a schematic of the circuit and label all known quantities.
• Solve for equivalent circuits and redraw the circuit.
• Solve for the unknown quantities.
Ohm’s Law
Applications
Guibelondo | Ohm’s Law | 27
𝐼 ∝
𝐸
𝑅
28. • Ohm’s Law is true for any point in a circuit and can be applied at any time.
• The same current flows throughout a series circuit, and the same voltage is present at any
branch of a parallel circuit.
Ohm’s Law
Applications
Guibelondo | Ohm’s Law | 28
29. • Example: What is the total current flow in the circuit shown in Figure 14-14?
• Answer: IT = 0.0054 amp, or 5.4 milliamp
Ohm’s Law
Applications
Guibelondo | Ohm’s Law | 29
30. • Example: How much voltage is dropped across resistor R2 in the circuit shown in Figure 14-
16?
• Answer: ER2 = 17.50 volts
Ohm’s Law
Applications
Guibelondo | Ohm’s Law | 30
31. • Example: What is the value of R2 in the circuit shown in Figure 14-18?
• Answer: R2 = 2,033.9 ohms
Ohm’s Law
Applications
Guibelondo | Ohm’s Law | 31
32. • Example: What is the current through R3 in the circuit shown in Figure 14-19?
• Answer: IR3 = 13.9 milliamps
Ohm’s Law
Applications
Guibelondo | Ohm’s Law | 32
33. • Practice Problem: What is the total circuit current in Figure 14-21?
• IT = ?
• ET = 12 volts
• R1 = 500 ohms
• R2 = 1200 ohms
• R3 = 2200 ohms
• Answer: IT = ?
Ohm’s Law
Applications
Guibelondo | Ohm’s Law | 33
34. Gustav Robert Kirchhoff
Kirchhoff’s
Current Law and Voltage Law
Guibelondo | Ohm’s Law | 34
Born 12 March 1824
Königsberg, Province of East Prussia, Kingdom of Prussia
(present-day Kaliningrad, Russia)
Died 17 October 1887 (aged 63)
Berlin, Province of Brandenburg, Kingdom of Prussia
(present-day Germany)
Residence Prussia / German Empire
Nationality Prussian (1824-1871)
German (1871-1887)
Alma mater University of Königsberg
Known for Kirchhoff's circuit laws
Kirchhoff's law of thermal radiation
Kirchhoff's laws of spectroscopy
Kirchhoff's law of thermochemistry
Awards Rumford medal (1862)
Davy Medal (1877)
Matteucci Medal (1877)
Janssen Medal (1887)
35. Gustav Robert Kirchhoff
Kirchhoff’s
Current Law and Voltage Law
Guibelondo | Ohm’s Law | 35
Scientific career
Fields Physics
Chemistry
Institutions University of Berlin
University of Breslau
University of Heidelberg
Doctoral advisor Franz Ernst Neumann
Notable students Loránd Eötvös
Edward Nichols
Gabriel Lippmann
Dmitri Ivanovich Mendeleev
Max Planck
Jules Piccard
Max Noether
Heike Kamerlingh Onnes
Ernst Schröder
36. Kirchhoff’s
Current Law and Voltage Law
Guibelondo | Ohm’s Law | 36
• In 1847, G. R. Kirchhoff extended Ohm’s law with two important statements that are referred to as
Kirchhoff’s laws.
37. Kirchhoff’s
Current Law
Guibelondo | Ohm’s Law | 37
• The first law, known as Kirchhoff’s current law, states the following:
➢The algebraic sum of all the currents entering and leaving a junction is equal to 0.
• Here is another way of stating Kirchhoff ’s current law:
➢The total current flowing into a junction is equal to the sum of the current flowing out of that
junction.
• A junction is defined as any point of a circuit at which two or more current paths meet. In a
parallel circuit, the junction is where the parallel branches of the circuit connect.
38. Kirchhoff’s
Current Law
Guibelondo | Ohm’s Law | 38
• In Figure 14-22, point A is one junction and point B is the second junction. Following the current
in the circuit, IT flows from the voltage source into the junction at point A. There the current splits
among the three branches as shown. Each of the three branch currents (I1, I2, and I3) flows out of
junction A.
39. Kirchhoff’s
Current Law
Guibelondo | Ohm’s Law | 39
• According to Kirchhoff ’s current law, which states that the total current into a junction is equal to
the total current out of the junction, the current can be stated as
• Following the current through each of the three branches finds them coming back together at
point B. Currents I1, I2, and I3 flows out into junction B, and IT flows out. Kirchhoff ’s current law
formula at this junction is the same as at junction A:
𝐼 𝑇 = 𝐼1 + 𝐼2 + 𝐼3
𝐼1 + 𝐼2 + 𝐼3 = 𝐼 𝑇
40. • Practice Problem: Refer to Figure 14-23. What are the values of I2 and I3?
• Answer: I2 = ? I3 = ?
Kirchhoff’s
Current Law
Guibelondo | Ohm’s Law | 40
41. Guibelondo | Ohm’s Law | 41
• Kirchhoff ’s second law is referred to as Kirchhoff’s voltage law, and it states the following:
➢The algebraic sum of all the voltages around a closed circuit equals 0.
• Here is another way of stating Kirchhoff’s voltage law:
➢The sum of all the voltage drops in a closed circuit equals the voltage source.
Kirchhoff’s
Voltage Law
42. Guibelondo | Ohm’s Law | 42
• In Figure 14-24 there are three voltage drops and one voltage source (voltage rise) in the circuit. If
the voltages are summed around the circuit as shown, they equal 0. Notice that the voltage
source (ET) has a sign opposite that of the voltage drops. Therefore the algebraic sum equals 0.
Kirchhoff’s
Voltage Law
𝐸 𝑇 − 𝐸1 − 𝐸2 − 𝐸3 = 0
43. Guibelondo | Ohm’s Law | 43
• Looking at this another way, the sum of all the voltage drops equals the voltage source.
• Both of the formulas shown are stating the same thing and are equivalent ways of expressing
Kirchhoff’s voltage law.
• The key to remember is that the polarity of the voltage source in the circuit is opposite to that
of the voltage drops.
Kirchhoff’s
Voltage Law
𝐸 𝑇 = 𝐸1 + 𝐸2 + 𝐸3
44. • Practice Problem: Refer to Figure 14-25. What is the total voltage applied to the circuit?
• Answer: ET = ?
Kirchhoff’s
Voltage Law
Guibelondo | Ohm’s Law | 44
45. Summary
Guibelondo | Ohm’s Law | 45
• An electric circuit consists of a voltage source, a load, and a conductor.
• The current path in an electric circuit can be series, parallel, or series-parallel.
• A series circuit offers only one path for current to flow.
• A parallel circuit offers several paths for the flow of current.
• A series-parallel circuit provides a combination of series and parallel paths for the flow of current.
• Current flows from the negative side of the voltage source through the load to the positive side of
the voltage source.
• Changing either the voltage or the resistance can vary current flow in an electric circuit.
46. Summary
Guibelondo | Ohm’s Law | 46
• Ohm’s law gives the relationship of current, voltage, and resistance.
• Ohm’s law states that the current in an electric circuit is directly proportional to the voltage
applied and inversely proportional to the resistance in the circuit.
• Ohm’s law applies to all series, parallel, and series-parallel circuits.
• To determine unknown quantities in a circuit:
• Solve for equivalent circuits.
• Draw a schematic of the circuit and label all quantities and redraw the circuit.
• Solve for all unknown quantities.
𝐼 =
𝐸
𝑅
47. Summary
Guibelondo | Ohm’s Law | 47
• Kirchhoff ’s current law: The algebraic sum of all the currents entering and leaving a junction is
equal to 0; it may be restated as the total current flowing into a junction is equal to the sum of the
current flowing out of that junction.
• Kirchhoff ’s voltage law: The algebraic sum of all the voltages around a closed circuit equals 0; it
may be restated as follows: The sum of all the voltage drops in a closed circuit equals the voltage
source.