Module 1—Introduction to Matter, Energy, and Direct Current
i - ix
, 1-1 to 1-10
1-11 to 1-20
, 1-21 to 1-30
1-41 to 1-50
,1-51 to 1-60
1-61 to 1-65
, 2-1 to 2-10
2-11 to 2-20
, 2-21 to 2-29
3-1 to 3-10
, 3-11 to 3-20
3-21 to 3-30
, 3-31 to 3-40
3-41 to 3-50
,3-51 to 3-60,
3-61 to 3-70
, 3-71 to 3-80
3-81 to 3-90
, 3-91 to 3-100
3-101 to 110
, 3-111 to 3-120
3-121 to 3-126
An important fact to keep in mind when applying Ohm’s law to a series circuit is to consider whether the
values used are component values or total values. When the information available enables the use Ohm’s law to find
total resistance, total voltage, and total current, total values must be inserted into the formula. To find total
To find total voltage:
To find total current:
NOTE: IT is equal to I in a series circuit. However, the distinction between IT and I in the formula
should be noted. The reason for this is that future circuits may have several currents, and it will be necessary
to differentiate between IT and other currents.
To compute any quantity (E, I, R, or P) associated with a
single given resistor, the values used in the formula must be obtained from that particular resistor. For example,
to find the value an unknown resistance, the voltage across and the current through that particular resistor must
To find the value a resistor:
To find the voltage drop across a resistor:
To find current through a resistor:
Q21. A series circuit consists two resistors in series. R1 = 25 ohms and R2 = 30 ohms. The circuit
current is 6 amps. What is the (a) source voltage, (b) voltage dropped by each resistor, (c) total power, and (d)
power used by each resistor?
Kirchhoff's VOLTAGE LAW
In 1847, G. R. Kirchhoff extended the use Ohm’s law by developing a simple concept concerning the voltages
contained in a series circuit loop. Kirchhoff's voltage law states:
“The algebraic sum 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 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) at that
instant and the voltage drops are assumed to have the opposite sign, the result 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 Module 2). 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 Kirchhoff's law, circuit problems can be solved which would
be difficult, and ten impossible, with knowledge 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.
To apply Kirchhoff's voltage law, the meaning voltage polarity must be understood.
In the circuit
shown in figure 3-22, the current is shown flowing in a counterclockwise direction. Notice that the end resistor
R1, into which the current flows, is marked NEGATIVE (—). The end R1 at which the current leaves is marked
POSITIVE (+). These polarity markings are used to show that the end Ri into which the current flows is at a
higher negative potential than the end the resistor at which the current leaves. Point A is more negative than
Figure 3-22.—Voltage polarities.
Point C, which is at the same potential as point B, is labeled negative. This is to indicate that point C
is more negative than point D. To say a point is positive (or negative) without stating what the polarity is based
upon has no meaning. In working with Kirchhoff's law, positive and negative polarities are
assigned in the
direction current flow.
APPLICATION of Kirchhoff's VOLTAGE LAW
Kirchhoff's voltage law can be written
as an equation, as shown below: Ea + Eb + Ec + . . . En = 0 where Ea, Eb, etc., are the voltage drops or emf’s
around any closed circuit loop. To set up the equation for an actual circuit, the following procedure is used.
1. Assume a direction current through the circuit. (The correct direction is desirable but not necessary.)
2. Using the assumed direction current, assign polarities to all resistors through which the current flows.
3. Place the correct polarities on any sources included in the circuit.
4. Starting at any point in
the circuit, trace around the circuit, writing down the amount and polarity the voltage across each component in
succession. The polarity used is the sign AFTER the assumed current has passed through the component. Stop when
the point at which the trace was started is reached.
5. Place these voltages, with their polarities, into
the equation and solve for the desired quantity. Example: Three resistors are connected across a 5O-volt source.
What is the voltage across the third resistor if the voltage drops across the first two resistors are 25 volts and
Solution: First, a diagram, such as the one shown in figure 3-23, is drawn. Next, a direction
current is assumed (as shown). Using this current, the polarity markings are placed at each end each resistor and
also on the terminals the source. Starting at point A, trace around the circuit in the direction current flow,
recording the voltage and polarity each component. Starting at point A and using the components from the circuit:
Substituting values from the circuit:
Figure 3-23.—Determining unknown voltage in a series circuit.
Using the same idea as above, you can solve a problem in which the current is the unknown quantity.
Example: A circuit having a source voltage 60 volts contains three resistors 5 ohms, 10 ohms, and 15 ohms. Find
the circuit current.
Solution: Draw and label the circuit (fig. 3-24). Establish a direction current flow
and assign polarities. Next, starting at any point—point A will be used in this example—write out the loop
Figure 3-24.—Correct direction assumed current.
Since the current obtained in the above calculations is a positive 2 amps, the assumed direction current
was correct. To show what happens if the incorrect direction current is assumed, the problem will be solved as
before, but with the opposite direction current. The circuit is redrawn showing the new direction current and new
polarities in figure 3-25. Starting at point A the loop equation is:
Figure 3-25.—Incorrect direction assumed current.
Notice that the AMOUNT current is the same as before. The polarity, however, is NEGATIVE. The negative
polarity simply indicates the wrong direction current was assumed. Should it be necessary to use this current in
further calculations on the circuit using Kirchhoff's law, the negative polarity should be retained in the
Series Aiding and Opposing Sources
In many practical applications a circuit may
contain more than one source emf. Sources emf that cause current to flow in the same direction are considered to
be SERIES AIDING and the voltages are added. Sources emf that would tend to force current in opposite directions
are said to be SERIES OPPOSING, and the effective source voltage is the difference between the opposing voltages.
When two opposing sources are inserted into a circuit current flow would be in a direction determined by the
larger source. Examples series aiding and opposing sources are shown in figure 3-26.
Figure 3-26.—Aiding and opposing sources.
A simple solution may be obtained for a multiple-source circuit through the use Kirchhoff's voltage
law. In applying this method, the same procedure is used for the multiple-source circuit as was used above for the
single-source circuit. This is demonstrated by the following example.
Example: Using Kirchhoff's voltage
equation, find the amount current in the circuit shown in fig 3-27.
Figure 3-27.-Solving for circuit current using Kirchhoff's voltage equation.
Solution: As before, a direction current flow is assumed and polarity signs are placed on the
drawing. The loop equation will be started at point A.
E2 + ER1 + E 1 + E3 + ER2 = 0
Q22. When using Kirchhoff's voltage law, how are voltage polarities assigned to the voltage drops across
Q23. Refer to figure 3-27, if R1 was changed to a 40-ohm resistor, what would be the value
circuit current (IT)?
Q24. Refer to figure 3-27. What is the effective source voltage the circuit using
the 40-ohm resistor?
CIRCUIT TERMS and CHARACTERISTICS
Before you learn about the types circuits other than the series circuit, you should become familiar with
some the terms and characteristics used in electrical circuits. These terms and characteristics will be used
throughout your study electricity and electronics.
point is an arbitrarily chosen point to which all other points in the circuit are compared. In series circuits,
any point can be chosen as a reference and the electrical potential at all other points can be determined in
reference to that point. In figure 3-28 point A shall be considered the reference point. Each series resistor in
the illustrated circuit is equal value. The applied voltage is equally distributed across each resistor. The
potential at point B is 25 volts more positive than at point A. Points C and D are 50 volts and 75 volts more
positive than point A respectively.
Introduction to Matter, Energy, and Direct Current,
to Alternating Current and Transformers, Introduction to Circuit Protection,
Control, and Measurement
, Introduction to Electrical Conductors, Wiring Techniques,
and Schematic Reading
, Introduction to Generators and Motors
Introduction to Electronic Emission, Tubes, and Power Supplies,
Introduction to Solid-State Devices and Power Supplies
Introduction to Amplifiers, Introduction to
Wave-Generation and Wave-Shaping Circuits
, Introduction to Wave Propagation, Transmission
Lines, and Antennas
, Microwave Principles,
, Introduction to Number Systems and Logic Circuits, Introduction
to Microelectronics, Principles of Synchros, Servos, and Gyros
Introduction to Test Equipment
, Radar Principles,
The Technician's Handbook,
Master Glossary, Test Methods and Practices,
Introduction to Digital Computers,
Magnetic Recording, Introduction to Fiber Optics